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With triple majors in Chemistry, Mathematics, and Classics, Professor Raffi Hovasapian returns to teach his favorite subject, AP Chemistry. His discussions on the topics are very detailed to insure students understand what is happening at the atomic level. Topics include everything on the AP Chemistry exam including Stoichiometry, Gases, Equilibrium, Acids & Bases, Thermodynamics, and Electrochemistry. With his 10+ years teaching and tutoring experience, Raffi explains difficult chemistry concepts through essential theory followed by plenty of worked out examples. Professor Hovasapian also works out an entire AP exam including all multiple choice and free response questions.

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I. Review
  Naming Compounds 41:24
   Intro 0:00 
   Periodic Table of Elements 0:15 
   Naming Compounds 3:13 
    Definition and Examples of Ions 3:14 
    Ionic (Symbol to Name): NaCl 5:23 
    Ionic (Name to Symbol): Calcium Oxide 7:58 
    Ionic - Polyatoms Anions: Examples 12:45 
    Ionic - Polyatoms Anions (Symbol to Name): KClO 14:50 
    Ionic - Polyatoms Anions (Name to Symbol): Potassium Phosphate 15:49 
    Ionic Compounds Involving Transition Metals (Symbol to Name): Co₂(CO₃)₃ 20:48 
    Ionic Compounds Involving Transition Metals (Name to Symbol): Palladium 2 Acetate 22:44 
    Naming Covalent Compounds (Symbol to Name): CO 26:21 
    Naming Covalent Compounds (Name to Symbol): Nitrogen Trifluoride 27:34 
    Naming Covalent Compounds (Name to Symbol): Dichlorine Monoxide 27:57 
    Naming Acids Introduction 28:11 
    Naming Acids (Name to Symbol): Chlorous Acid 35:08 
    % Composition by Mass Example 37:38 
  Stoichiometry 37:19
   Intro 0:00 
   Stoichiometry 0:25 
    Introduction to Stoichiometry 0:26 
    Example 1 5:03 
    Example 2 10:17 
    Example 3 15:09 
    Example 4 24:02 
    Example 5: Questions 28:11 
    Example 5: Part A - Limiting Reactant 30:30 
    Example 5: Part B 32:27 
    Example 5: Part C 35:00 
II. Aqueous Reactions & Stoichiometry
  Precipitation Reactions 31:14
   Intro 0:00 
   Precipitation Reactions 0:53 
    Dissociation of ionic Compounds 0:54 
    Solubility Guidelines for ionic Compounds: Soluble Ionic Compounds 8:15 
    Solubility Guidelines for ionic Compounds: Insoluble ionic Compounds 12:56 
    Precipitation Reactions 14:08 
    Example 1: Mixing a Solution of BaCl₂ & K₂SO₄ 21:21 
    Example 2: Mixing a Solution of Mg(NO₃)₂ & KI 26:10 
  Acid-Base Reactions 43:21
   Intro 0:00 
   Acid-Base Reactions 1:00 
    Introduction to Acid: Monoprotic Acid and Polyprotic Acid 1:01 
    Introduction to Base 8:28 
    Neutralization 11:45 
    Example 1 16:17 
    Example 2 21:55 
    Molarity 24:50 
    Example 3 26:50 
    Example 4 30:01 
    Example 4: Limiting Reactant 37:51 
    Example 4: Reaction Part 40:01 
  Oxidation Reduction Reactions 47:58
   Intro 0:00 
   Oxidation Reduction Reactions 0:26 
    Oxidation and Reduction Overview 0:27 
    How Can One Tell Whether Oxidation-Reduction has Taken Place? 7:13 
    Rules for Assigning Oxidation State: Number 1 11:22 
    Rules for Assigning Oxidation State: Number 2 12:46 
    Rules for Assigning Oxidation State: Number 3 13:25 
    Rules for Assigning Oxidation State: Number 4 14:50 
    Rules for Assigning Oxidation State: Number 5 15:41 
    Rules for Assigning Oxidation State: Number 6 17:00 
    Example 1: Determine the Oxidation State of Sulfur in the Following Compounds 18:20 
   Activity Series and Reduction Properties 25:32 
    Activity Series and Reduction Properties 25:33 
    Example 2: Write the Balance Molecular, Total Ionic, and Net Ionic Equations for Al + HCl 31:37 
    Example 3 34:25 
    Example 4 37:55 
  Stoichiometry Examples 31:50
   Intro 0:00 
   Stoichiometry Example 1 0:36 
    Example 1: Question and Answer 0:37 
   Stoichiometry Example 2 6:57 
    Example 2: Questions 6:58 
    Example 2: Part A Solution 12:16 
    Example 2: Part B Solution 13:05 
    Example 2: Part C Solution 14:00 
    Example 2: Part D Solution 14:38 
   Stoichiometry Example 3 17:56 
    Example 3: Questions 17:57 
    Example 3: Part A Solution 19:51 
    Example 3: Part B Solution 21:43 
    Example 3: Part C Solution 26:46 
III. Gases
  Pressure, Gas Laws, & The Ideal Gas Equation 49:40
   Intro 0:00 
   Pressure 0:22 
    Pressure Overview 0:23 
    Torricelli: Barometer 4:35 
    Measuring Gas Pressure in a Container 7:49 
    Boyle's Law 12:40 
    Example 1 16:56 
   Gas Laws 21:18 
    Gas Laws 21:19 
    Avogadro's Law 26:16 
    Example 2 31:47 
   Ideal Gas Equation 38:20 
    Standard Temperature and Pressure (STP) 38:21 
    Example 3 40:43 
  Partial Pressure, Mol Fraction, & Vapor Pressure 32:00
   Intro 0:00 
   Gases 0:27 
    Gases 0:28 
    Mole Fractions 5:52 
    Vapor Pressure 8:22 
    Example 1 13:25 
    Example 2 22:45 
  Kinetic Molecular Theory and Real Gases 31:58
   Intro 0:00 
   Kinetic Molecular Theory and Real Gases 0:45 
    Kinetic Molecular Theory 1 0:46 
    Kinetic Molecular Theory 2 4:23 
    Kinetic Molecular Theory 3 5:42 
    Kinetic Molecular Theory 4 6:27 
    Equations 7:52 
    Effusion 11:15 
    Diffusion 13:30 
    Example 1 19:54 
    Example 2 23:23 
    Example 3 26:45 
  AP Practice for Gases 25:34
   Intro 0:00 
   Example 1 0:34 
    Example 1 0:35 
   Example 2 6:15 
    Example 2: Part A 6:16 
    Example 2: Part B 8:46 
    Example 2: Part C 10:30 
    Example 2: Part D 11:15 
    Example 2: Part E 12:20 
    Example 2: Part F 13:22 
   Example 3 14:45 
    Example 3 14:46 
   Example 4 18:16 
    Example 4 18:17 
   Example 5 21:04 
    Example 5 21:05 
IV. Thermochemistry
  Energy, Heat, and Work 37:32
   Intro 0:00 
   Thermochemistry 0:25 
    Temperature and Heat 0:26 
    Work 3:07 
    System, Surroundings, Exothermic Process, and Endothermic Process 8:19 
    Work & Gas: Expansion and Compression 16:30 
    Example 1 24:41 
    Example 2 27:47 
    Example 3 31:58 
  Enthalpy & Hess's Law 32:34
   Intro 0:00 
   Thermochemistry 1:43 
    Defining Enthalpy & Hess's Law 1:44 
    Example 1 6:48 
    State Function 13:11 
    Example 2 17:15 
    Example 3 24:09 
  Standard Enthalpies of Formation 23:09
   Intro 0:00 
   Thermochemistry 1:04 
    Standard Enthalpy of Formation: Definition & Equation 1:05 
    ∆H of Formation 10:00 
    Example 1 11:22 
    Example 2 19:00 
  Calorimetry 39:28
   Intro 0:00 
   Thermochemistry 0:21 
    Heat Capacity 0:22 
    Molar Heat Capacity 4:44 
    Constant Pressure Calorimetry 5:50 
    Example 1 12:24 
    Constant Volume Calorimetry 21:54 
    Example 2 24:40 
    Example 3 31:03 
V. Kinetics
  Reaction Rates and Rate Laws 36:24
   Intro 0:00 
   Kinetics 2:18 
    Rate: 2 NO₂ (g) → 2NO (g) + O₂ (g) 2:19 
    Reaction Rates Graph 7:25 
    Time Interval & Average Rate 13:13 
    Instantaneous Rate 15:13 
    Rate of Reaction is Proportional to Some Power of the Reactant Concentrations 23:49 
    Example 1 27:19 
  Method of Initial Rates 30:48
   Intro 0:00 
   Kinetics 0:33 
    Rate 0:34 
    Idea 2:24 
    Example 1: NH₄⁺ + NO₂⁻ → NO₂ (g) + 2 H₂O 5:36 
    Example 2: BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3 H₂O 19:29 
  Integrated Rate Law & Reaction Half-Life 32:17
   Intro 0:00 
   Kinetics 0:52 
    Integrated Rate Law 0:53 
    Example 1 6:26 
    Example 2 15:19 
    Half-life of a Reaction 20:40 
    Example 3: Part A 25:41 
    Example 3: Part B 28:01 
  Second Order & Zero-Order Rate Laws 26:40
   Intro 0:00 
   Kinetics 0:22 
    Second Order 0:23 
    Example 1 6:08 
    Zero-Order 16:36 
    Summary for the Kinetics Associated with the Reaction 21:27 
  Activation Energy & Arrhenius Equation 40:59
   Intro 0:00 
   Kinetics 0:53 
    Rate Constant 0:54 
    Collision Model 2:45 
    Activation Energy 5:11 
    Arrhenius Proposed 9:54 
    2 Requirements for a Successful Reaction 15:39 
    Rate Constant 17:53 
    Arrhenius Equation 19:51 
    Example 1 25:00 
    Activation Energy & the Values of K 32:12 
    Example 2 36:46 
  AP Practice for Kinetics 29:08
   Intro 0:00 
   Kinetics 0:43 
    Example 1 0:44 
    Example 2 6:53 
    Example 3 8:58 
    Example 4 11:36 
    Example 5 16:36 
    Example 6: Part A 21:00 
    Example 6: Part B 25:09 
VI. Equilibrium
  Equilibrium, Part 1 46:00
   Intro 0:00 
   Equilibrium 1:32 
    Introduction to Equilibrium 1:33 
    Equilibrium Rules 14:00 
    Example 1: Part A 16:46 
    Example 1: Part B 18:48 
    Example 1: Part C 22:13 
    Example 1: Part D 24:55 
    Example 2: Part A 27:46 
    Example 2: Part B 31:22 
    Example 2: Part C 33:00 
    Reverse a Reaction 36:04 
    Example 3 37:24 
  Equilibrium, Part 2 40:53
   Intro 0:00 
   Equilibrium 1:31 
    Equilibriums Involving Gases 1:32 
    General Equation 10:11 
    Example 1: Question 11:55 
    Example 1: Answer 13:43 
    Example 2: Question 19:08 
    Example 2: Answer 21:37 
    Example 3: Question 33:40 
    Example 3: Answer 35:24 
  Equilibrium: Reaction Quotient 45:53
   Intro 0:00 
   Equilibrium 0:57 
    Reaction Quotient 0:58 
    If Q > K 5:37 
    If Q < K 6:52 
    If Q = K 7:45 
    Example 1: Part A 8:24 
    Example 1: Part B 13:11 
    Example 2: Question 20:04 
    Example 2: Answer 22:15 
    Example 3: Question 30:54 
    Example 3: Answer 32:52 
    Steps in Solving Equilibrium Problems 42:40 
  Equilibrium: Examples 31:51
   Intro 0:00 
   Equilibrium 1:09 
    Example 1: Question 1:10 
    Example 1: Answer 4:15 
    Example 2: Question 13:04 
    Example 2: Answer 15:20 
    Example 3: Question 25:03 
    Example 3: Answer 26:32 
  Le Chatelier's principle & Equilibrium 40:52
   Intro 0:00 
   Le Chatelier 1:05 
    Le Chatelier Principle 1:06 
    Concentration: Add 'x' 5:25 
    Concentration: Subtract 'x' 7:50 
    Example 1 9:44 
    Change in Pressure 12:53 
    Example 2 20:40 
    Temperature: Exothermic and Endothermic 24:33 
    Example 3 29:55 
    Example 4 35:30 
VII. Acids & Bases
  Acids and Bases 50:11
   Intro 0:00 
   Acids and Bases 1:14 
    Bronsted-Lowry Acid-Base Model 1:28 
    Reaction of an Acid with Water 4:36 
    Acid Dissociation 10:51 
    Acid Strength 13:48 
    Example 1 21:22 
    Water as an Acid & a Base 25:25 
    Example 2: Part A 32:30 
    Example 2: Part B 34:47 
    Example 3: Part A 35:58 
    Example 3: Part B 39:33 
    pH Scale 41:12 
    Example 4 43:56 
  pH of Weak Acid Solutions 43:52
   Intro 0:00 
   pH of Weak Acid Solutions 1:12 
    pH of Weak Acid Solutions 1:13 
    Example 1 6:26 
    Example 2 14:25 
    Example 3 24:23 
    Example 4 30:38 
  Percent Dissociation: Strong & Weak Bases 43:04
   Intro 0:00 
   Bases 0:33 
    Percent Dissociation: Strong & Weak Bases 0:45 
    Example 1 6:23 
    Strong Base Dissociation 11:24 
    Example 2 13:02 
    Weak Acid and General Reaction 17:38 
    Example: NaOH → Na⁺ + OH⁻ 20:30 
    Strong Base and Weak Base 23:49 
    Example 4 24:54 
    Example 5 33:51 
  Polyprotic Acids 35:34
   Intro 0:00 
   Polyprotic Acids 1:04 
    Acids Dissociation 1:05 
    Example 1 4:51 
    Example 2 17:30 
    Example 3 31:11 
  Salts and Their Acid-Base Properties 41:14
   Intro 0:00 
   Salts and Their Acid-Base Properties 0:11 
    Salts and Their Acid-Base Properties 0:15 
    Example 1 7:58 
    Example 2 14:00 
    Metal Ion and Acidic Solution 22:00 
    Example 3 28:35 
    NH₄F → NH₄⁺ + F⁻ 34:05 
    Example 4 38:03 
  Common Ion Effect & Buffers 41:58
   Intro 0:00 
   Common Ion Effect & Buffers 1:16 
    Covalent Oxides Produce Acidic Solutions in Water 1:36 
    Ionic Oxides Produce Basic Solutions in Water 4:15 
    Practice Example 1 6:10 
    Practice Example 2 9:00 
    Definition 12:27 
    Example 1: Part A 16:49 
    Example 1: Part B 19:54 
    Buffer Solution 25:10 
    Example of Some Buffers: HF and NaF 30:02 
    Example of Some Buffers: Acetic Acid & Potassium Acetate 31:34 
    Example of Some Buffers: CH₃NH₂ & CH₃NH₃Cl 33:54 
    Example 2: Buffer Solution 36:36 
  Buffer 32:24
   Intro 0:00 
   Buffers 1:20 
    Buffer Solution 1:21 
    Adding Base 5:03 
    Adding Acid 7:14 
    Example 1: Question 9:48 
    Example 1: Recall 12:08 
    Example 1: Major Species Upon Addition of NaOH 16:10 
    Example 1: Equilibrium, ICE Chart, and Final Calculation 24:33 
    Example 1: Comparison 29:19 
  Buffers, Part II 40:06
   Intro 0:00 
   Buffers 1:27 
    Example 1: Question 1:32 
    Example 1: ICE Chart 3:15 
    Example 1: Major Species Upon Addition of OH⁻, But Before Rxn 7:23 
    Example 1: Equilibrium, ICE Chart, and Final Calculation 12:51 
    Summary 17:21 
    Another Look at Buffering & the Henderson-Hasselbalch equation 19:00 
    Example 2 27:08 
    Example 3 32:01 
  Buffers, Part III 38:43
   Intro 0:00 
   Buffers 0:25 
    Buffer Capacity Part 1 0:26 
    Example 1 4:10 
    Buffer Capacity Part 2 19:29 
    Example 2 25:12 
    Example 3 32:02 
  Titrations: Strong Acid and Strong Base 42:42
   Intro 0:00 
   Titrations: Strong Acid and Strong Base 1:11 
    Definition of Titration 1:12 
    Sample Problem 3:33 
    Definition of Titration Curve or pH Curve 9:46 
   Scenario 1: Strong Acid- Strong Base Titration 11:00 
    Question 11:01 
    Part 1: No NaOH is Added 14:00 
    Part 2: 10.0 mL of NaOH is Added 15:50 
    Part 3: Another 10.0 mL of NaOH & 20.0 mL of NaOH are Added 22:19 
    Part 4: 50.0 mL of NaOH is Added 26:46 
    Part 5: 100.0 mL (Total) of NaOH is Added 27:26 
    Part 6: 150.0 mL (Total) of NaOH is Added 32:06 
    Part 7: 200.0 mL of NaOH is Added 35:07 
    Titrations Curve for Strong Acid and Strong Base 35:43 
  Titrations: Weak Acid and Strong Base 42:03
   Intro 0:00 
   Titrations: Weak Acid and Strong Base 0:43 
    Question 0:44 
    Part 1: No NaOH is Added 1:54 
    Part 2: 10.0 mL of NaOH is Added 5:17 
    Part 3: 25.0 mL of NaOH is Added 14:01 
    Part 4: 40.0 mL of NaOH is Added 21:55 
    Part 5: 50.0 mL (Total) of NaOH is Added 22:25 
    Part 6: 60.0 mL (Total) of NaOH is Added 31:36 
    Part 7: 75.0 mL (Total) of NaOH is Added 35:44 
    Titration Curve 36:09 
  Titration Examples & Acid-Base Indicators 52:03
   Intro 0:00 
   Examples and Indicators 0:25 
    Example 1: Question 0:26 
    Example 1: Solution 2:03 
    Example 2: Question 12:33 
    Example 2: Solution 14:52 
    Example 3: Question 23:45 
    Example 3: Solution 25:09 
    Acid/Base Indicator Overview 34:45 
    Acid/Base Indicator Example 37:40 
    Acid/Base Indicator General Result 47:11 
    Choosing Acid/Base Indicator 49:12 
VIII. Solubility
  Solubility Equilibria 36:25
   Intro 0:00 
   Solubility Equilibria 0:48 
    Solubility Equilibria Overview 0:49 
    Solubility Product Constant 4:24 
    Definition of Solubility 9:10 
    Definition of Solubility Product 11:28 
    Example 1 14:09 
    Example 2 20:19 
    Example 3 27:30 
    Relative Solubilities 31:04 
  Solubility Equilibria, Part II 42:06
   Intro 0:00 
   Solubility Equilibria 0:46 
    Common Ion Effect 0:47 
    Example 1 3:14 
    pH & Solubility 13:00 
    Example of pH & Solubility 15:25 
    Example 2 23:06 
    Precipitation & Definition of the Ion Product 26:48 
    If Q > Ksp 29:31 
    If Q < Ksp 30:27 
    Example 3 32:58 
  Solubility Equilibria, Part III 43:09
   Intro 0:00 
   Solubility Equilibria 0:55 
    Example 1: Question 0:56 
    Example 1: Step 1 - Check to See if Anything Precipitates 2:52 
    Example 1: Step 2 - Stoichiometry 10:47 
    Example 1: Step 3 - Equilibrium 16:34 
    Example 2: Selective Precipitation (Question) 21:02 
    Example 2: Solution 23:41 
    Classical Qualitative Analysis 29:44 
    Groups: 1-5 38:44 
IX. Complex Ions
  Complex Ion Equilibria 43:38
   Intro 0:00 
   Complex Ion Equilibria 0:32 
    Complex Ion 0:34 
    Ligan Examples 1:51 
    Ligand Definition 3:12 
    Coordination 6:28 
    Example 1 8:08 
    Example 2 19:13 
  Complex Ions & Solubility 31:30
   Intro 0:00 
   Complex Ions and Solubility 0:23 
    Recall: Classical Qualitative Analysis 0:24 
    Example 1 6:10 
    Example 2 16:16 
    Dissolving a Water-Insoluble Ionic Compound: Method 1 23:38 
    Dissolving a Water-Insoluble Ionic Compound: Method 2 28:13 
X. Chemical Thermodynamics
  Spontaneity, Entropy, & Free Energy, Part I 56:28
   Intro 0:00 
   Spontaneity, Entropy, Free Energy 2:25 
    Energy Overview 2:26 
    Equation: ∆E = q + w 4:30 
    State Function/ State Property 8:35 
    Equation: w = -P∆V 12:00 
    Enthalpy: H = E + PV 14:50 
    Enthalpy is a State Property 17:33 
    Exothermic and Endothermic Reactions 19:20 
    First Law of Thermodynamic 22:28 
    Entropy 25:48 
    Spontaneous Process 33:53 
    Second Law of Thermodynamic 36:51 
    More on Entropy 42:23 
    Example 43:55 
  Spontaneity, Entropy, & Free Energy, Part II 39:55
   Intro 0:00 
   Spontaneity, Entropy, Free Energy 1:30 
    ∆S of Universe = ∆S of System + ∆S of Surrounding 1:31 
    Convention 3:32 
    Examining a System 5:36 
    Thermodynamic Property: Sign of ∆S 16:52 
    Thermodynamic Property: Magnitude of ∆S 18:45 
    Deriving Equation: ∆S of Surrounding = -∆H / T 20:25 
    Example 1 25:51 
    Free Energy Equations 29:22 
  Spontaneity, Entropy, & Free Energy, Part III 30:10
   Intro 0:00 
   Spontaneity, Entropy, Free Energy 0:11 
    Example 1 2:38 
    Key Concept of Example 1 14:06 
    Example 2 15:56 
    Units for ∆H, ∆G, and S 20:56 
    ∆S of Surrounding & ∆S of System 22:00 
    Reaction Example 24:17 
    Example 3 26:52 
  Spontaneity, Entropy, & Free Energy, Part IV 30:07
   Intro 0:00 
   Spontaneity, Entropy, Free Energy 0:29 
    Standard Free Energy of Formation 0:58 
    Example 1 4:34 
    Reaction Under Non-standard Conditions 13:23 
    Example 2 16:26 
    ∆G = Negative 22:12 
    ∆G = 0 24:38 
    Diagram Example of ∆G 26:43 
  Spontaneity, Entropy, & Free Energy, Part V 44:56
   Intro 0:00 
   Spontaneity, Entropy, Free Energy 0:56 
    Equations: ∆G of Reaction, ∆G°, and K 0:57 
    Example 1: Question 6:50 
    Example 1: Part A 9:49 
    Example 1: Part B 15:28 
    Example 2 17:33 
    Example 3 23:31 
    lnK = (- ∆H° ÷ R) ( 1 ÷ T) + ( ∆S° ÷ R) 31:36 
    Maximum Work 35:57 
XI. Electrochemistry
  Oxidation-Reduction & Balancing 39:23
   Intro 0:00 
   Oxidation-Reduction and Balancing 2:06 
    Definition of Electrochemistry 2:07 
    Oxidation and Reduction Review 3:05 
    Example 1: Assigning Oxidation State 10:15 
    Example 2: Is the Following a Redox Reaction? 18:06 
    Example 3: Step 1 - Write the Oxidation & Reduction Half Reactions 22:46 
    Example 3: Step 2 - Balance the Reaction 26:44 
    Example 3: Step 3 - Multiply 30:11 
    Example 3: Step 4 - Add 32:07 
    Example 3: Step 5 - Check 33:29 
  Galvanic Cells 43:09
   Intro 0:00 
   Galvanic Cells 0:39 
    Example 1: Balance the Following Under Basic Conditions 0:40 
    Example 1: Steps to Balance Reaction Under Basic Conditions 3:25 
    Example 1: Solution 5:23 
    Example 2: Balance the Following Reaction 13:56 
    Galvanic Cells 18:15 
    Example 3: Galvanic Cells 28:19 
    Example 4: Galvanic Cells 35:12 
  Cell Potential 48:41
   Intro 0:00 
   Cell Potential 2:08 
    Definition of Cell Potential 2:17 
    Symbol and Unit 5:50 
    Standard Reduction Potential 10:16 
    Example Figure 1 13:08 
    Example Figure 2 19:00 
    All Reduction Potentials are Written as Reduction 23:10 
    Cell Potential: Important Fact 1 26:49 
    Cell Potential: Important Fact 2 27:32 
    Cell Potential: Important Fact 3 28:54 
    Cell Potential: Important Fact 4 30:05 
    Example Problem 1 32:29 
    Example Problem 2 38:38 
  Potential, Work, & Free Energy 41:23
   Intro 0:00 
   Potential, Work, Free Energy 0:42 
    Descriptions of Galvanic Cell 0:43 
    Line Notation 5:33 
    Example 1 6:26 
    Example 2 11:15 
    Example 3 15:18 
    Equation: Volt 22:20 
    Equations: Cell Potential, Work, and Charge 28:30 
    Maximum Cell Potential is Related to the Free Energy of the Cell Reaction 35:09 
    Example 4 37:42 
  Cell Potential & Concentration 34:19
   Intro 0:00 
   Cell Potential & Concentration 0:29 
    Example 1: Question 0:30 
    Example 1: Nernst Equation 4:43 
    Example 1: Solution 7:01 
    Cell Potential & Concentration 11:27 
    Example 2 16:38 
    Manipulating the Nernst Equation 25:15 
    Example 3 28:43 
  Electrolysis 33:21
   Intro 0:00 
   Electrolysis 3:16 
    Electrolysis: Part 1 3:17 
    Electrolysis: Part 2 5:25 
    Galvanic Cell Example 7:13 
    Nickel Cadmium Battery 12:18 
    Ampere 16:00 
    Example 1 20:47 
    Example 2 25:47 
XII. Light
  Light 44:45
   Intro 0:00 
   Light 2:14 
    Introduction to Light 2:15 
    Frequency, Speed, and Wavelength of Waves 3:58 
    Units and Equations 7:37 
    Electromagnetic Spectrum 12:13 
    Example 1: Calculate the Frequency 17:41 
    E = hν 21:30 
    Example 2: Increment of Energy 25:12 
    Photon Energy of Light 28:56 
    Wave and Particle 31:46 
    Example 3: Wavelength of an Electron 34:46 
XIII. Quantum Mechanics
  Quantum Mechanics & Electron Orbitals 54:00
   Intro 0:00 
   Quantum Mechanics & Electron Orbitals 0:51 
    Quantum Mechanics & Electron Orbitals Overview 0:52 
    Electron Orbital and Energy Levels for the Hydrogen Atom 8:47 
    Example 1 13:41 
    Quantum Mechanics: Schrodinger Equation 19:19 
    Quantum Numbers Overview 31:10 
    Principal Quantum Numbers 33:28 
    Angular Momentum Numbers 34:55 
    Magnetic Quantum Numbers 36:35 
    Spin Quantum Numbers 37:46 
    Primary Level, Sublevels, and Sub-Sub-Levels 39:42 
    Example 42:17 
    Orbital & Quantum Numbers 49:32 
  Electron Configurations & Diagrams 34:04
   Intro 0:00 
   Electron Configurations & Diagrams 1:08 
    Electronic Structure of Ground State Atom 1:09 
    Order of Electron Filling 3:50 
    Electron Configurations & Diagrams: H 8:41 
    Electron Configurations & Diagrams: He 9:12 
    Electron Configurations & Diagrams: Li 9:47 
    Electron Configurations & Diagrams: Be 11:17 
    Electron Configurations & Diagrams: B 12:05 
    Electron Configurations & Diagrams: C 13:03 
    Electron Configurations & Diagrams: N 14:55 
    Electron Configurations & Diagrams: O 15:24 
    Electron Configurations & Diagrams: F 16:25 
    Electron Configurations & Diagrams: Ne 17:00 
    Electron Configurations & Diagrams: S 18:08 
    Electron Configurations & Diagrams: Fe 20:08 
    Introduction to Valence Electrons 23:04 
    Valence Electrons of Oxygen 23:44 
    Valence Electrons of Iron 24:02 
    Valence Electrons of Arsenic 24:30 
    Valence Electrons: Exceptions 25:36 
    The Periodic Table 27:52 
XIV. Intermolecular Forces
  Vapor Pressure & Changes of State 52:43
   Intro 0:00 
   Vapor Pressure and Changes of State 2:26 
    Intermolecular Forces Overview 2:27 
    Hydrogen Bonding 5:23 
    Heat of Vaporization 9:58 
    Vapor Pressure: Definition and Example 11:04 
    Vapor Pressures is Mostly a Function of Intermolecular Forces 17:41 
    Vapor Pressure Increases with Temperature 20:52 
    Vapor Pressure vs. Temperature: Graph and Equation 22:55 
    Clausius-Clapeyron Equation 31:55 
    Example 1 32:13 
    Heating Curve 35:40 
    Heat of Fusion 41:31 
    Example 2 43:45 
  Phase Diagrams & Solutions 31:17
   Intro 0:00 
   Phase Diagrams and Solutions 0:22 
    Definition of a Phase Diagram 0:50 
    Phase Diagram Part 1: H₂O 1:54 
    Phase Diagram Part 2: CO₂ 9:59 
    Solutions: Solute & Solvent 16:12 
    Ways of Discussing Solution Composition: Mass Percent or Weight Percent 18:46 
    Ways of Discussing Solution Composition: Molarity 20:07 
    Ways of Discussing Solution Composition: Mole Fraction 20:48 
    Ways of Discussing Solution Composition: Molality 21:41 
    Example 1: Question 22:06 
    Example 1: Mass Percent 24:32 
    Example 1: Molarity 25:53 
    Example 1: Mole Fraction 28:09 
    Example 1: Molality 29:36 
  Vapor Pressure of Solutions 37:23
   Intro 0:00 
   Vapor Pressure of Solutions 2:07 
    Vapor Pressure & Raoult's Law 2:08 
    Example 1 5:21 
    When Ionic Compounds Dissolve 10:51 
    Example 2 12:38 
    Non-Ideal Solutions 17:42 
    Negative Deviation 24:23 
    Positive Deviation 29:19 
    Example 3 31:40 
  Colligatives Properties 34:11
   Intro 0:00 
   Colligative Properties 1:07 
    Boiling Point Elevation 1:08 
    Example 1: Question 5:19 
    Example 1: Solution 6:52 
    Freezing Point Depression 12:01 
    Example 2: Question 14:46 
    Example 2: Solution 16:34 
    Osmotic Pressure 20:20 
    Example 3: Question 28:00 
    Example 3: Solution 30:16 
XV. Bonding
  Bonding & Lewis Structure 48:39
   Intro 0:00 
   Bonding & Lewis Structure 2:23 
    Covalent Bond 2:24 
    Single Bond, Double Bond, and Triple Bond 4:11 
    Bond Length & Intermolecular Distance 5:51 
    Definition of Electronegativity 8:42 
    Bond Polarity 11:48 
    Bond Energy 20:04 
    Example 1 24:31 
    Definition of Lewis Structure 31:54 
    Steps in Forming a Lewis Structure 33:26 
    Lewis Structure Example: H₂ 36:53 
    Lewis Structure Example: CH₄ 37:33 
    Lewis Structure Example: NO⁺ 38:43 
    Lewis Structure Example: PCl₅ 41:12 
    Lewis Structure Example: ICl₄⁻ 43:05 
    Lewis Structure Example: BeCl₂ 45:07 
  Resonance & Formal Charge 36:59
   Intro 0:00 
   Resonance and Formal Charge 0:09 
    Resonance Structures of NO₃⁻ 0:25 
    Resonance Structures of NO₂⁻ 12:28 
    Resonance Structures of HCO₂⁻ 16:28 
    Formal Charge 19:40 
    Formal Charge Example: SO₄²⁻ 21:32 
    Formal Charge Example: CO₂ 31:33 
    Formal Charge Example: HCN 32:44 
    Formal Charge Example: CN⁻ 33:34 
    Formal Charge Example: 0₃ 34:43 
  Shapes of Molecules 41:21
   Intro 0:00 
   Shapes of Molecules 0:35 
    VSEPR 0:36 
    Steps in Determining Shapes of Molecules 6:18 
    Linear 11:38 
    Trigonal Planar 11:55 
    Tetrahedral 12:45 
    Trigonal Bipyramidal 13:23 
    Octahedral 14:29 
    Table: Shapes of Molecules 15:40 
    Example: CO₂ 21:11 
    Example: NO₃⁻ 24:01 
    Example: H₂O 27:00 
    Example: NH₃ 29:48 
    Example: PCl₃⁻ 32:18 
    Example: IF₄⁺ 34:38 
    Example: KrF₄ 37:57 
  Hybrid Orbitals 40:17
   Intro 0:00 
   Hybrid Orbitals 0:13 
    Introduction to Hybrid Orbitals 0:14 
    Electron Orbitals for CH₄ 5:02 
    sp³ Hybridization 10:52 
    Example: sp³ Hybridization 12:06 
    sp² Hybridization 14:21 
    Example: sp² Hybridization 16:11 
    σ Bond 19:10 
    π Bond 20:07 
    sp Hybridization & Example 22:00 
    dsp³ Hybridization & Example 27:36 
    d²sp³ Hybridization & Example 30:36 
    Example: Predict the Hybridization and Describe the Molecular Geometry of CO 32:31 
    Example: Predict the Hybridization and Describe the Molecular Geometry of BF₄⁻ 35:17 
    Example: Predict the Hybridization and Describe the Molecular Geometry of XeF₂ 37:09 
XVI. AP Practice Exam
  AP Practice Exam: Multiple Choice, Part I 52:34
   Intro 0:00 
   Multiple Choice 1:21 
    Multiple Choice 1 1:22 
    Multiple Choice 2 2:23 
    Multiple Choice 3 3:38 
    Multiple Choice 4 4:34 
    Multiple Choice 5 5:16 
    Multiple Choice 6 5:41 
    Multiple Choice 7 6:20 
    Multiple Choice 8 7:03 
    Multiple Choice 9 7:31 
    Multiple Choice 10 9:03 
    Multiple Choice 11 11:52 
    Multiple Choice 12 13:16 
    Multiple Choice 13 13:56 
    Multiple Choice 14 14:52 
    Multiple Choice 15 15:43 
    Multiple Choice 16 16:20 
    Multiple Choice 17 16:55 
    Multiple Choice 18 17:22 
    Multiple Choice 19 18:59 
    Multiple Choice 20 20:24 
    Multiple Choice 21 22:20 
    Multiple Choice 22 23:29 
    Multiple Choice 23 24:30 
    Multiple Choice 24 25:24 
    Multiple Choice 25 26:21 
    Multiple Choice 26 29:06 
    Multiple Choice 27 30:42 
    Multiple Choice 28 33:28 
    Multiple Choice 29 34:38 
    Multiple Choice 30 35:37 
    Multiple Choice 31 37:31 
    Multiple Choice 32 38:28 
    Multiple Choice 33 39:50 
    Multiple Choice 34 42:57 
    Multiple Choice 35 44:18 
    Multiple Choice 36 45:52 
    Multiple Choice 37 48:02 
    Multiple Choice 38 49:25 
    Multiple Choice 39 49:43 
    Multiple Choice 40 50:16 
    Multiple Choice 41 50:49 
  AP Practice Exam: Multiple Choice, Part II 32:15
   Intro 0:00 
   Multiple Choice 0:12 
    Multiple Choice 42 0:13 
    Multiple Choice 43 0:33 
    Multiple Choice 44 1:16 
    Multiple Choice 45 2:36 
    Multiple Choice 46 5:22 
    Multiple Choice 47 6:35 
    Multiple Choice 48 8:02 
    Multiple Choice 49 10:05 
    Multiple Choice 50 10:26 
    Multiple Choice 51 11:07 
    Multiple Choice 52 12:01 
    Multiple Choice 53 12:55 
    Multiple Choice 54 16:12 
    Multiple Choice 55 18:11 
    Multiple Choice 56 19:45 
    Multiple Choice 57 20:15 
    Multiple Choice 58 23:28 
    Multiple Choice 59 24:27 
    Multiple Choice 60 26:45 
    Multiple Choice 61 29:15 
  AP Practice Exam: Multiple Choice, Part III 32:50
   Intro 0:00 
   Multiple Choice 0:16 
    Multiple Choice 62 0:17 
    Multiple Choice 63 1:57 
    Multiple Choice 64 6:16 
    Multiple Choice 65 8:05 
    Multiple Choice 66 9:18 
    Multiple Choice 67 10:38 
    Multiple Choice 68 12:51 
    Multiple Choice 69 14:32 
    Multiple Choice 70 17:35 
    Multiple Choice 71 22:44 
    Multiple Choice 72 24:27 
    Multiple Choice 73 27:46 
    Multiple Choice 74 29:39 
    Multiple Choice 75 30:23 
  AP Practice Exam: Free response Part I 47:22
   Intro 0:00 
   Free Response 0:15 
    Free Response 1: Part A 0:16 
    Free Response 1: Part B 4:15 
    Free Response 1: Part C 5:47 
    Free Response 1: Part D 9:20 
    Free Response 1: Part E. i 10:58 
    Free Response 1: Part E. ii 16:45 
    Free Response 1: Part E. iii 26:03 
    Free Response 2: Part A. i 31:01 
    Free Response 2: Part A. ii 33:38 
    Free Response 2: Part A. iii 35:20 
    Free Response 2: Part B. i 37:38 
    Free Response 2: Part B. ii 39:30 
    Free Response 2: Part B. iii 44:44 
  AP Practice Exam: Free Response Part II 43:05
   Intro 0:00 
   Free Response 0:12 
    Free Response 3: Part A 0:13 
    Free Response 3: Part B 6:25 
    Free Response 3: Part C. i 11:33 
    Free Response 3: Part C. ii 12:02 
    Free Response 3: Part D 14:30 
    Free Response 4: Part A 21:03 
    Free Response 4: Part B 22:59 
    Free Response 4: Part C 24:33 
    Free Response 4: Part D 27:22 
    Free Response 4: Part E 28:43 
    Free Response 4: Part F 29:35 
    Free Response 4: Part G 30:15 
    Free Response 4: Part H 30:48 
    Free Response 5: Diagram 32:00 
    Free Response 5: Part A 34:14 
    Free Response 5: Part B 36:07 
    Free Response 5: Part C 37:45 
    Free Response 5: Part D 39:00 
    Free Response 5: Part E 40:26 
  AP Practice Exam: Free Response Part III 28:36
   Intro 0:00 
   Free Response 0:43 
    Free Response 6: Part A. i 0:44 
    Free Response 6: Part A. ii 3:08 
    Free Response 6: Part A. iii 5:02 
    Free Response 6: Part B. i 7:11 
    Free Response 6: Part B. ii 9:40 
    Free Response 7: Part A 11:14 
    Free Response 7: Part B 13:45 
    Free Response 7: Part C 15:43 
    Free Response 7: Part D 16:54 
    Free Response 8: Part A. i 19:15 
    Free Response 8: Part A. ii 21:16 
    Free Response 8: Part B. i 23:51 
    Free Response 8: Part B. ii 25:07 

Hello, and welcome to Educator.com; and welcome to the first lesson of AP Chemistry.0000

Today, I'm going to start by discussing the naming of compounds, because we have to know what it is that we are dealing with in order for us to jump into the chemistry.0005

So, let's get started.0013

OK, so this is, of course, the periodic table, and I'm just going to do a quick overview of the periodic table, just to get a sense of where we need to be, because this periodic table is what you are going to be using all the time--it's going to be your primary reference in chemistry--as you already know.0017

Over here, on the left, we have this first column; these are called the alkali metals.0033

Actually, you know what?--let's sort of break it up broadly.0040

Generally, everything from about right here onward--everything that is blue, and this orange right here--so the whole left side--these are all metals.0044

Up here, starting with boron, silicon, arsenic, tellurium--everything here and up to the right--those are going to be nonmetals.0053

So, as you can see, most of the periodic table--most elements (naturally occurring elements) consist of metals.0060

These over here are called the alkali metals; the second column is called alkaline earth metals; these here, in the middle--these are called the transition metals.0066

These over here are called the noble gases, basically because they are unreactive.0078

These are called the halogens.0084

These are the standard nonmetals.0086

The only thing on the left that is a nonmetal is hydrogen.0088

Now, let's talk about some of the diatomic gases.0091

There are certain gases (well, certain compounds; not just gases, because bromine is actually a liquid)--certain compounds that occur...0096

The periodic table lists actual atoms; astatine, iodine, bromine, chlorine, fluorine, oxygen, nitrogen, and hydrogen--they occur as diatomic gases.0105

In other words, there are two atoms that make a molecule of that in the natural, normal state.0116

What you are breathing right now is nitrogen and oxygen gas; not nitrogen and oxygen atoms, but there are two atoms stuck together to form a molecule.0121

That is really all that you have to know.0132

These numbers up on top (the ones right below the symbols): those are the atomic numbers--that is the number of protons and electrons in a neutral atom.0135

This number, right down below it (which, for many things, is almost double the atomic number)--that is the atomic mass; that is an average of all of the masses of the isotopes, depending on the frequency of occurrence of the particular isotope.0144

Again, if any of this is strange--things or words like isotope and stuff--I would definitely urge you to look through it in your books, but for all practical purposes, it won't come up all too often.0162

So, I just wanted you to get a quick overview; and now, we can jump into actually naming the compounds.0174

Because, before we do anything else--many of the problems are not going to actually have equations for us, or symbols for us; they're just going to give us names of compounds.0179

We have to know what it is that we are writing, so we can write an equation--all chemistry begins with a chemical equation.0187

OK; let's start off, quickly, by talking about what ions are.0193

Ions are atoms that have lost or gained an electron--that is it; lost or gained one or more electrons.0201

Your atoms (your naturally-occurring atoms) are neutral; in other words, they have the same number of protons and electrons, which makes them electrically neutral.0223

If I add to those, or if I take away from those, all of a sudden they become charged particles.0231

Positive ions are called cations, and negative ions are called anions.0237

So, an example of that would be something like this: Mg2+ is a magnesium atom that has lost two electrons.0252

Because an electron is the thing that is being added or subtracted, when we add (electrons are negatively charged), it carries a negative charge.0259

So, for example, O2-; that is an atom that has gained two electrons, so now it's carrying two extra negative charges.0268

Magnesium lost two electrons, so now it has two more positive charges than it does negative charges; that is why we put the 2+.0277

Fluorine is -; potassium is +; aluminum is +3; iron(II) is that; there is also an iron(III)--and we'll talk about that in a minute--why it is that certain transition metals actually can lose more than one electron, and they have different oxidation states.0285

That is it--ions, and the first thing that we're going to discuss, of course, is ionic compounds: simple binary ionic compounds, where the cation is made of one element and the anion is made of the other element.0306

Let's go ahead and get started.0319

OK, so now let's go to ionic, and we're going to go from the symbol to the name.0323

We need to be able to go from the symbol and be able to name the compound, and we need to be able to go backwards (from the name to the symbol).0331

So, from symbol to name--we'll do that first.0338

NaCl...now, if you look at your periodic table, you're going to notice that the alkali metals on the left, the alkaline earth metals (which is the second), and then on the right (if you ignore the noble gases), you're going to have the halogens, and then you're going to have the next group and the next group, working to the right.0344

As it turns out, the metals always lose electrons; nonmetals always gain electrons.0365

So, when we're talking about ionic compounds, it's always going to be the metal that is going to be positive, and it's going to be the nonmetal that is going to be negative.0376

Now, what we do when we put together ionic compounds is: we are just putting them together in such a way that we actually cancel the charge.0385

But again, we will get to that in just a minute, when we're dealing with name-to-symbol.0394

So, sodium...this NaCl, the name for this is sodium chloride--that's it.0399

Anytime you see something like this, basically, what you do is you take the name of the metal (sodium); you take the base of the nonmetal, and you add -ide to it; that's it.0404

How about MgBr2; OK, this is just magnesium, and this is bromine, so it becomes "bromide."0422

That's it; it is always consistent--it's the name of the cation first (the positive, or the name of the metal, first), and then the name--the root of the nonmetal, plus -ide.0436

All binary ionic compounds are named like this.0446

How about something like this: Al2O3; don't worry about what the numbers mean yet--we'll get to that in just a minute.0451

This is aluminum oxide.0460

And then one more for good measure: K2S; this is potassium sulfide.0465

That's it; very, very nice.0477

OK, so now we're going to do to ionic, and we're going to go from the name to the symbol.0479

OK, so if I see something like calcium oxide, how do I symbolize that?0489

Well, here is where we have to actually look at the charges; and the charges come from their arrangement on the periodic table.0499

The alkali metals are in the first group on the left; when they react with nonmetals, they actually lose one electron; they're in the first group.0506

The things in the second column (like magnesium and calcium)--they lose two electrons when they react.0517

Well, if we go to the right-hand side of the periodic table (ignore the noble gases), the halogens--they actually gain one electron when they react.0525

The next column over (next to the halogens), the one where oxygen and sulfur are--they gain two electrons.0538

Then, if you go one more, you're going to end up with the nitrogen and phosphorus column; those gain three electrons.0544

So, when you're putting ionic compounds together that consist of a metal from the left and a nonmetal on the right, you look at their charges.0553

In the case of calcium, calcium is a 2+ charge, because it's in the second column.0560

Oxygen has a 2- charge, because it is the second column over, ignoring the noble gases; it has a 2- charge.0566

What we're trying to do here is: we need to combine these in such a way that the charges actually cancel.0574

Here, the positive 2 and the minus 2 cancel, so the symbol is just CaO.0580

In other words, I just need one calcium atom and one oxygen atom to make calcium oxide; the charges balance.0585

OK, now let's do something like sodium sulfide.0594

Sodium sulfide: well, sodium is in the first column, so it has a +1 charge; sulfur has a -2 charge; it's in the second column over from the halogens (again, ignoring the noble gases).0603

Well, this is a +1 and this is a -2; in order to balance this charge, this +1 needs to be +2 to balance this -2.0615

Therefore, we need 2 sodiums; so, the symbol becomes Na2S.0623

I need two sodiums for a total 2+ charge to balance the -2 charge.0628

And again, all ionic compounds have to be neutrally charged--their charges have to balance.0632

Let's do something like aluminum iodide.0641

Aluminum is in the 1, 2, third column over on the periodic table, when you actually skip the transition metals (we're going to get to the transition metals next--they are handled a little differently).0652

So, from your perspective, the first column, the second column...you skip the transition metals, and you go to the other column, where aluminum is.0661

That is a 3+ charge, so everything in that column carries a 3+ charge.0670

Al3+; and iodine is a halogen, so it has a -1; well, how many iodines do I need to balance the 3+ charge? I need three of them.0675

So, this becomes AlI3, aluminum iodide.0684

Let's try calcium phosphide.0690

Calcium phosphide is Ca2+, because calcium is in the second column; phosphorus is in the third column over, so it carries a negative 3 charge when it's reacting with a metal.0701

Well, how do I balance 2+ and 3-? I can't do it directly, so I need the least common multiple of these.0711

The least common multiple is 6.0717

In order to make 6 positive charges, I need 3 calciums; and 6 negative charges--I need 2 phosphoruses.0720

So, this becomes calcium phosphide; that is it.0729

You're just taking a look at the charge on the anion and the cation, and you're arranging them--taking specific numbers of them and putting them together so that the charge is 0--that's all you're doing.0732

We'll finish it off with aluminum sulfide.0743

Aluminum sulfide: OK, we said that aluminum was 3+, and sulfur is in the column with oxygen, so it is a 2-; so again, we have to make 6, so we end up with Al2S3, aluminum sulfide; that is it.0750

OK, now we'll do polyatomic anions.0765

Polyatomic anions: again, you have seen them before; polyatomic anions are anions that are made of more than one atom.0776

Some examples would be something like ClO3-; that is chlorate; SO42-; that is sulfate; maybe PO43-; that is phosphate.0788

Let's see...ClO2-; that is chlorite; and one of the things you are going to notice with the polyatomics--they generally tend to end in -ate or -ite--most of the time.0808

There are some exceptions, like, for example, the NH4+; that is called ammonium.0820

Now, of course, you have seen lists of polyatomic ions; and there is a page, a list, in your book, of polyatomic ions; or, you can go on the Internet, and you can find a list of polyatomic ions--some short, some long--and it lists all of the polyatomic ions that are available.0828

These are treated exactly the same way as the regular atomic anions, in the sense that you are treating them as a whole.0846

When you name them, you just basically use the entire name of the anion; so these actually work out really, really easily.0855

So, let's go from symbol to name first.0863

Symbol to name: for example, if I saw something like NaClO3; well, Na is sodium, and ClO3...when I look it up on the polyatomic ion chart, it is chlorate; so the name is sodium chlorate--very, very simple.0870

If I have something like KClO, well, K is potassium, and if I look up ClO, it is going to be hypochlorite.0891

So, it is potassium hypochlorite.0902

If I had Ca(CN)2, well, Ca is calcium, and CN--when I look it up, it is cyanide; so it is calcium cyanide.0908

So it's very, very simple--you are handling it the exact same way: the name of the metal first, and then the name of the polyatomic anion.0923

Again, these are treated as a whole; it's like--just think of it as one unit that happens to have a specific charge on it: it could be -1, -2, -3...that's it.0929

And again, there is a whole list of these atomic...you will use them enough times so that you'll eventually memorize them, or perhaps you already have them memorized.0940

OK, so now let's go the other way; let's go from the name to the symbol.0949

Now, let's go from name to symbol.0955

Let's do potassium phosphate.0960

If you see "potassium phosphate," well, potassium is in the first column; so, when it reacts as an ion, it is K+.0963

When you look up "phosphate," it is PO43-; we are doing the same thing.0977

We just need to put them together in such a way that the charges balance.0982

In order for the charges to balance, we need three potassiums in order to balance the three negative charges; so we write it as K3PO4: potassium phosphate.0986

How about something like aluminum nitrate?0999

Well, aluminum has a 3+ charge; nitrate, when you look it up--it is NO3, and it carries a -1 charge.1008

I need three nitrates to balance the 3+ on the aluminum, so it becomes Al, and whenever you need more than one polyatomic, you of course put it in parentheses, so it is written as (NO3)3.1016

Now, this three--that means you have three NO3s; so it's three nitrogens, nine oxygens.1030

Now, let's do one more: magnesium hydroxide.1037

OK, magnesium is a 2+; hydroxide, when you look it up--it is a -1; you need two of these to balance the 2+ charge.1047

So, it is Mg(OH)2; good--nice and easy.1054

OK, so now we're going to move on to ionic compounds involving transition metals.1060

Ionic compounds involving transition metals: OK, now let's take a look at two compounds, just to start with.1069

Let's take a look at Fe2O3, and let's take a look at FeO.1084

Well, if I ask you to just name these the way we have been doing, you take the name of the metal; you take the name of the anion, add -ide to it, and you will get iron oxide.1091

Well, it's true--this is iron oxide; however, there is this other compound, that is FeO; this is also iron oxide.1101

These are two entirely different compounds with completely different chemistry; how do we differentiate between the two?1109

As it turns out, transition metals (all of those things in the middle of the periodic table)--they can actually lose different numbers of electrons--the same atom.1114

For example, you can have iron 2+, iron 3+; you can have manganese 2+, manganese 4+, manganese 6+...so these things--we actually have to specify, in parentheses, in the name, how many electrons it has lost--in other words, its charge.1124

So, in this case (and here is how you actually end up doing it), let's take Fe2O3.1140

Fe2O3; well, what do we know about oxygen? Oxygen, when it reacts, always carries a -2 charge.1149

There are three oxygens, so the total charge is -6.1156

Well, this -6 has to be balanced by a +6, because this is a neutral compound, right?--it's neutral.1160

This +6 charge is divided among two irons; that means that each iron is carrying a +3 charge; so what we do is: we write III in Roman numerals, in parentheses, right after the iron.1167

So, it's not just iron oxide; it's iron (III) oxide.1183

This thing in parentheses tells me the charge on an individual iron atom.1186

Now, if you want, you can write it as...it depends on your teacher; I actually use Arabic numerals instead of Roman numerals, so I just write iron (3), like this: iron (3) oxide; it really doesn't matter.1191

Some teachers like Roman numerals--it's more traditional; some teachers actually don't really care; but obviously, depending on your teacher...if they want Roman numerals, give them Roman numerals.1204

You don't want to be losing a point here and a point there for silly reasons like that.1215

Now, this one--how about iron oxide: which iron is this one?1220

Well, oxygen is -2; there is only one of them, so it's -2; that means it has to be balanced by a +2.1224

This +2...there is only one iron here, so this is actually iron (2) oxide.1231

So, when we are naming transition metal compounds, we have to specify the charge on the transition metal.1236

The charge is different, depending on how it has reacted: we express that with these parentheses.1243

Let's do a few of them.1249

Let's do Co2(CO3)3.1252

Now, again, these transition metals--they can react with individual atoms, nonmetals, or they can react with polyatomic ions.1267

Here, we have a cobalt that has reacted with a carbonate.1275

Carbonate--when you look it up, you know the name--it's carbonate; the charge on carbonate is 2-.1279

The charge on carbonate is (you know what, I'm going to do this a little lower, so I can...let me put it over here...Co2(CO3)3) -2.1287

There are three of them, so it gives us a total of -6.1300

This -6 has to be balanced by a +6, because cobalt carbonate is neutral; and this +6 is actually divided by...there are two cobalts that are carrying that total of 6 charge, so each cobalt has a +3.1303

So, we name this cobalt (3) carbonate.1321

That is it--very, very straightforward.1327

Let's do one more: let's do Mn(SO4)2.1330

When I look up SO4)2, or if I know it, I know that SO4 is sulfate; it has a -2 charge; there are two sulfates, therefore the total charge is -4.1336

That is going to be balanced by a +4, because this is neutral.1345

There is only one manganese, so that one manganese is carrying the entire 4 charge.1348

So, this becomes manganese (4) sulfate.1352

That's it--very, very straightforward.1361

OK, so now, let's go from name to symbol.1364

Name to symbol: let's say we have something like palladium (2) acetate.1371

OK, these are really, really simple; they are handled exactly the same way: you are given the charge on palladium--it's right there in parentheses; it tells you that it's a positive 2.1382

Acetate, when you look it up--it is going to be (I'm going to symbolize it as AcO) a negative 1 charge, so palladium acetate...well, this is -1; this is +2; which means you need 2 of these to balance that +2 charge.1390

So, you end up with Pd(AcO)2; that's it.1403

Let's try chromium chloride.1415

Let's give ourselves a little bit more room here; I don't want to use up too much here.1419

Chromium (6) chloride: OK, chromium (6)--that is Cr+6 (or 6+; it doesn't matter--I actually generally tend to write it with the number first).1425

And then, chloride is a -1 charge; well, we need 6 of these to balance that.1440

So, what we get is CrCl6.1445

That's it; that is chromium (6) chloride.1449

If it were chromium (4) chloride, it would be CrCl4; chromium (2) chloride--CrCl2; nice and easy.1451

So now, we have done ionic compounds; now we're going to name covalent compounds.1466

Naming covalent compounds: a covalent compound is a compound that consists of a nonmetal-nonmetal bond.1473

Ionic compound--it always involves a metal with a nonmetal; covalent is nonmetal-nonmetal.1488

Nonmetal-nonmetal bond: OK, so now let's go from symbol to name.1499

If we have something like PCl5; all right, so phosphorus and chlorine--they are both nonmetals.1510

When we name these, the -ide ending stays the same, but what we use is: we need to specify the actual numbers here now--how many (oops, let me get rid of these lines--so PCl5) of each atom there is; and we use the prefixes mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octo- or octa-, nona-, and deca-; so, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.1518

This would actually be called phosphorus (actually, I think that is spelled with a u; you know what, I can't even remember now) pentachloride.1563

Now, notice: when this first atom only has one atom in it, we don't use the mono- prefix; we just say phosphorus pentachloride, carbon dioxide.1582

However, if we had something like this: CO; if the subsequent atoms after the first have just one atom--this is carbon monoxide.1595

We use the mono- for subsequent atoms; we don't use it for the first atom.1608

That is carbon monoxide; and let's see--let's try N2O; N2O is dinitrogen monoxide.1614

You might have something like N2O5, which is dinitrogen pentoxide (you could say pentaoxide, but...it's up to you; I say pentaoxide; you could just drop the a, and do pentoxide).1629

Again, very, very simple--just specify the number of atoms that you have; that is all you are doing.1649

OK, so now, let's go from name to symbol.1654

Name to symbol: let's do nitrogen trifluoride--very, very simple.1662

Nitrogen trifluoride--just put them in: NF3; probably the easiest thing in the world.1671

Let's see; let's try dichlorine monoxide; that is going to be dichlorine, Cl2, monoxide, O: Cl2O; that's it.1677

OK, now let's go ahead and finish up with a discussion of how to name acids.1693

Naming acids: Acid...basically, you are going to have some anion, any anion, balanced by an appropriate number of hydrogen ions.1702

Now, acids are interesting; hydrogen is in the first column; it's right above the alkali metals.1734

When it loses an electron, it loses one electron completely, so it becomes H+; well, H+ reacting with some anion (which is maybe a 1-, 2-, 3-)--they're going to combine.1739

Acids are a little different, because...some people consider them ionic compounds; some people consider them covalent compounds; acidic behavior is more about behavior.1758

We will, of course, talk about acids in detail a little bit later on (in fact, in a couple of lessons), and then again in the middle of the course.1768

But right now, we just want to worry about naming them.1775

An acid is always going to have the H first; so if you see something like this: CH4; that is not an acid.1778

But, if you see something like HCl, that is an acid.1785

Just by convention, again, the cation (which, in this case, is the hydrogen) comes first.1789

Let's go from symbol to name.1794

Let's say that we have something like HBr.1798

All right, now this is called hydrobromic acid.1802

OK, here are the rules.1812

When an acid does not have...well, actually, I'm going to do this a little bit differently.1817

Let me write one more, H2S, so that you can see the pattern: hydrosulfuric acid.1836

When we are dealing with a binary acid...binary means that it's a certain number of H's, and the anion is actually just a single atom--bromide, fluoride, sulfide, iodide, things like that); when it's just those two, it's a binary acid.1849

Now, if it doesn't have oxygen in it--there is no oxygen--then the prefix is hydro-, and then it ends with an -ic; so, hydrobromic, hydrosulfuric...if you saw HCl, this is going to be hydrochloric.1869

Again, notice, we take the root...the root...the root, and we add -ic to it, and that is always consistent.1905

If you have a binary acid that doesn't have any oxygen in it, that doesn't have any polyatomic ions, it is always named the same way.1913

So, let's say HI; this is going to be hydroiodic.1923

It's always a hydro- prefix and an -ic ending, always; there is no exception.1932

Now, let's do something else; let's say if you see an acid like this: H2SO4, or if you see HClO3, or if you see H2SO3, or let's say HCm...no, that's going to be a couple of different things...let's say HNO3.1942

OK, these acids actually have oxygen in them; these are named in a different way.1969

The ones that have oxygen in them are generally going to be associated with a polyatomic ion: notice, this is sulfate; this is chlorate; this SO3 is sulfite; this NO3 is nitrate.1975

You might have HNO2; this is called nitrite.1989

OK, here is how these are named.1995

You don't use the hydro- any time you have an acid that has oxygen in it.2000

This is just the root of the element that makes up the polyatomic ion, and if the name ends in -ate--if a polyatomic ion name ends in -ate--you use the -ic ending; if the polyatomic ion name ends in -ite, you use the -ous ending.2004

This one would actually be called sulfuric acid.2030

Again, it's just a balance of charge: SO4 is just 2-; H is +1; that is why you have 2 hydrogens.2039

That is all you are doing: you need to still balance the charge--this is still treated like an ionic compound.2048

This is sulfuric acid; well, if I look up ClO3, that is chlorate; it ends in -ate, so the -ate becomes -ic; so it becomes chloric acid.2053

This one, SO3, is sulfite; therefore, this becomes sulfurous acid (remember, if it ends in -ite, the -ite changes to -ous; if it ends in -ate, it ends in -ic).2067

Anything with an oxygen in it does not have the hydro- prefix.2082

This right here is nitric acid, because it's from nitrate.2087

This HNO2...NO2 is nitrite; therefore, this is nitrous acid.2095

That's it; that is how acids are named.2106

This is going from symbol to name; if we go the other way around...let's say, for example, you had something like chlorous acid.2108

Chlorous acid...well, this -ous ending is telling me that I am dealing with a chlorite anion; therefore--I know chlorite is ClO2-, so--I know that H is +1; they balance; so the symbol for this is HClO2; that's it.2122

Let's do a couple more: let's say if we said...how about chromic acid?2146

Chromic acid: well, it ends in -ic, so it comes from a polyatomic ion that ends in -ate; as it turns out, it is chromate, CrO4.2155

It is 2-; well, H is +1; how many H's do we need to balance the 2- charge? We need 2 of them.2166

So, the symbol is H2CrO4; that's it.2173

Let's say you had something like...how about phosphoric acid?2182

Phosphoric acid: well, phosphoric acid ends in -ic, so we know that we're dealing with the phosphate anion.2191

The phosphate anion is a 3- charge; H is a +1 charge; so I need 3 of these: H3PO4.2200

That is the symbol for phosphoric acid.2209

OK, so now we have talked about how to name compounds, how to go from the symbol to the name and from the name to the symbol.2211

This is going to be, of course, ubiquitous throughout the course.2217

You absolutely have to be able to understand how to name the compounds.2221

On the AP exam, sometimes the symbol will be given; often, in the free response section, the symbol will not be given.2225

You have to come up with the symbol--not only the symbol, but you have to come up with the equation; in fact, one of the last sections of the free response is a series of reactions, and they're just going to give you the names.2232

They're going to say, "This reacts with this; what happens?"; you have to know the symbol and you have to know the reaction; so naming--if you can't do naming, unfortunately, nothing else will work...so there is that.2246

OK, so I'm going to close off this discussion with a problem in percent composition.2258

Again, this is sort of just a general review of the basic techniques of chemistry, that I'm just going to go over quickly.2264

One sort of standard problem is percent composition; in other words, if I have some compound, like C6H12O6, which is glucose, what percentage of it, by mass, is carbon? What percentage of it is hydrogen? What percentage of it is oxygen?2270

I'm just going to do one of those problems, just so you see the general process--as a review.2289

Percent composition by mass (and generally, they'll be talking about mass; if they talk about anything else, they'll specify what they mean): OK, so calculate the percent composition of each element in Mg(NO3)2.2297

This is magnesium nitrate.2335

Well, let's see what we have.2339

We need...basically, when we are calculating percent composition--as you know, a percent is a part over the whole, so we're going to take the mass of magnesium over the whole, the mass of nitrogen over the whole, and the mass of oxygen over the whole.2344

Let's see what we have: magnesium is 24.31 grams per mole, and then we have 2 nitrogens--that comes to 28.02 grams per mole (and a gram per mole is the unit of molar mass, which I'm hoping you're familiar with), and 6 oxygens gives us 96 grams per mole.2358

OK, so the total mass is going to be 148.33 grams; therefore, the percent magnesium is equal to 2431 divided by 148.33, times 100, and you end up with 16.39%.2388

That means, of this 148.33 grams, about 16 or 16 and a half percent of it is magnesium; that is all that means.2420

Percent nitrogen: it equals 28.02 divided by 148.33, times 100, and you end up with 18.89%.2428

That means, of that mass, about 19% of it is nitrogen.2445

Then, the percent oxygen; you can either calculate it, or you can just add these two and subtract from 100; and you're going to get the same answer.2449

96 divided by 148.33, times 100...and you end up with 64.72%.2459

So, in magnesium nitrate, the majority of the mass is actually occupied by oxygen.2470

OK, thank you for joining us here at Educator.com for our first lesson of AP Chemistry.2478

We'll see you next time; goodbye.2482

Welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we're going to be discussing kinetic-molecular theory and properties of real gases.0003

Up until now, we've been discussing the ideal gas law, PV=nRT; well, the real gases actually behave well at low pressures and high temperatures.0008

When the pressures start to get high (let's say above 3, 4, 5, 6 atmospheres), and the pressures start to drop, volume starts to drop--at that point, the gas behavior starts to deviate from the ideal.0023

Towards the end of this particular lesson, we will discuss about how we adjust for that--how van der Waals adjusted for that.0037

Let's go ahead and get started.0043

We're going to run through the kinetic-molecular theory first.0047

So, again, what we are about to list right here--the four or five axioms of the kinetic-molecular theory--are precisely that: they are axioms; they are observations that we have made, and they are assumptions that we are making.0050

Based on those assumptions, we can start to build a theory, and hopefully have it correspond with what we observe empirically (in other words, what we get from experimentation).0062

Let's just list them out.0073

Let me see: our first axiom: The kinetic energy of an ideal gas (and again, we are talking about ideal gases' behavior--toward the end, we will discuss the deviation from ideal behavior, but the kinetic-molecular theory applies to ideal gases) is directly proportional to the temperature.0076

What that means, in terms of an equation, is: KE (the kinetic energy--the energy of motion of the molecules of the gas) is equal to 3/2 RT.0118

Now, R is the gas constant; however, here it is not equal to .08206 liter-atmosphere per mole-Kelvin.0131

In this particular case, because energy is expressed in Joules, and T here is in Kelvin, this R has to be Joules per Kelvin per mole; so, as it turns out...let's see: 8.31...it's going to be 8.31 Joules per mole-Kelvin.0139

And again, it's just a way of making the units work out; that is all it is; it's the same constant, R, the Rydberg constant; it's just so that it actually works out in terms of the appropriate units.0163

T is the temperature in Kelvin.0172

OK, now, also, what I'll put down here (this is a very, very important relationship--notice, it establishes the relationship between the energy, or the temperature; that is really what is going on here--so temperature is really just a measure of the average kinetic energy of the molecules)...0182

Most people confuse temperature and heat; temperature and heat are not the same thing.0200

Temperature is a measure of the motion, random motion, of the molecules; heat is that energy that actually flows from something hot in the direction of something cold.0204

They are two different things; temperature does not measure heat; it measures kinetic energy--it measures the random motion of the molecules, the vibrations, the bouncing around of each other.0214

OK, let's say there is one other little mathematical thing I want to put down here: the average kinetic energy of a gas particle (this one is not altogether that important, but I might as well just put it down here) equals 1/2 mass, times the average velocity squared.0223

Again, in this particular case, mass has to be in kilograms.0249

When velocity is expressed in meters per second, we get our kinetic energy in Joules.0255

That is why we need mass in kilograms.0261

OK, the second assumption of the kinetic-molecular theory is that the particles are so small (the particles of gas that are bouncing around) compared to the distance between the particles that the volume of the molecules is negligible.0264

In other words, molecules are very, very tiny; and when they're bouncing around in a gas, they are really very far apart from each other; so essentially, you can think of them as volume-less point particles that are just bouncing around.0308

Now, obviously, when we squeeze, when we drop the volume, increase the pressure, lower the temperature--when the volume gets really, really tiny--now the volume of the molecules actually plays an important role.0318

We can't really make this assumption anymore; we have to adjust this; but, for kinetic theory--ideal behavior--we just ignore the volume of the actual particles themselves.0329

They do have volume; they're just not important.0339

Three: the third axiom is: Particles exert no forces on each other.0345

In other words, they are just bouncing around randomly; one has nothing to do with the other; they don't stick together; they don't repel; they just bounce off of each other in perfectly-elastic collisions.0364

Now, that is not the case--we know for a fact that it doesn't happen that way--but for ideal behavior, we can make this assumption.0377

For all practical purposes, it behaves this way at low pressure.0383

OK, now the fourth one: The particles are in constant motion (this makes sense), and the collisions (oops, this random stray line all the way across the page--let's get rid of it here) with the walls of the container are the cause of exerted pressure.0388

So, if I have some gas in a closed vessel, and I measure the pressure, the pressure of that gas is coming from the particles--the gas molecules or atoms--bouncing and hitting the walls of the container.0444

That is where pressure comes from; it's just like a ball that hits a wall--just imagine billions and billions and billions of balls hitting a wall--well, that wall is going to feel it!0457

So, these are the assumptions of the kinetic-molecular theory of an ideal gas.0466

Now, from this, we can go ahead and deduce some things.0473

I'm not going to go ahead and derive any of these equations; I'm just going to throw them out, because, again, we're concerned with using the equations--not necessarily where they come from.0477

You can go ahead and follow the derivation in any one of your textbooks--they are either in the appendix or in the actual text itself--so I'll let you look at them if you want to; it certainly helps if you do.0484

If not, no harm--no foul--we're just going to be able to use them; that is what we're going to do.0496

The root is equal to (let's see) velocity and 3R...actually, that's going to be for the whole thing...3KT over m: so, the root mean square speed--just think of it as the average speed at which a particle is moving--is equal to 3 times K times T times m.0502

Now here, K is something called Boltzmann's constant, a very, very, very important constant--probably the single most important constant in physics (at least in my opinion--other people would say that the Planck constant is--they are closely related, in fact).0540

Boltzmann constant: that is 1.38x10-23 Joules per Kelvin.0560

T is temperature in Kelvin, and m is mass in kilograms.0573

It is the mass of an individual particle in kilograms; that is why we use the small m instead of the capital m.0582

Now, let's go ahead and write this same thing as 3RT/M, where R is, as we said before, the Rydberg constant: 8.31 Joules per mole-Kelvin--and that actually happens to equal Boltzmann's constant times Avogadro's number, 6.02x1023.0587

So, this is also a really interesting relationship to keep in mind--that R, the Rydberg constant, equals the Boltzmann constant times Avogadro's number.0623

8.31 Joules per mole-Kelvin equals 1.38x10-23, times 6.02x1023.0632

This is where you end up getting R from.0640

M, the capital M, is the molar mass.0643

This just gives me average--the root mean square speed of gas particles.0649

OK, so let's throw a couple of other definitions out there.0657

Essentially, what I'm going to be doing is just sort of laying out these things--what they are--defining them, and then, once I have them on the page, I'm going to go ahead and use them to solve some problems.0660

But, I just wanted to lay them out as they are, as opposed to laying one out, doing a problem...I'm going to save the problems until the end.0669

There is something that we call effusion, and effusion is nothing more than the passage of a gas through a tiny opening--that is it.0677

So, if I have a balloon and I poke a little needle hole in it, the gas from the inside of the balloon effuses out.0696

The rate of effusion is just the rate at which it actually comes out.0702

So, it's going to be a certain volume per unit time, like, let's say, 10 milliliters per second.0708

That means 10 milliliters of air are escaping for every second; that is all it is--a rate is just an amount over a unit time.0713

But, you know this already.0721

Well, somebody by the name of Graham discovered that the rate of effusion is inversely proportional to the molar mass of the particular gas.0723

Or, we can write it as: The rate times the molar mass is equal to a constant.0736

Well, for two gases under similar circumstances--for the same temperature and pressure--they are going to equal the same constant.0750

Therefore, what you have is something like this: you have: The rate of the first gas, times its molar mass, equals the rate at which the second gas effuses, times its molar mass.0760

This is how I like to use it; however, it is totally equivalent to (and more often than not, you will see it written like this): The ratio of the rates--rate 1 over rate 2--is equal to the molar mass of 2 over the molar mass of 1.0774

It really doesn't matter: you can write it this way; you can write it this way--it's a totally personal choice; most books will write it this way, because they like to have ratios of the same thing.0794

The rate of one over the rate of the other equals the square of the molar mass of one over the square of the molar mass of the other...absolutely the same thing.0801

OK, now let's define something called diffusion.0812

Diffusion just means one gas mixing with another.0816

If I open up a vial of pure ammonia, and if I put it on the table, let's say 5 feet away from me, it's going to take a little while, but eventually, I'm going to actually smell the ammonia.0826

Well, the ammonia is mixing with the air--the oxygen and nitrogen--the 70%...air is nitrogen and oxygen; it's mixing with the air, and diffusion is just the extent to which it mixes.0837

The rate of diffusion is how quickly it actually mixes.0851

Effusion and diffusion are actually closely related; you can actually use the same equation for both.0855

Again, we will talk more about this when we actually do a problem; it will make more sense; but I just want to throw out the meanings.0861

So, effusion: how quickly a gas escapes from a tiny opening; diffusion is how a gas actually mixes with another gas--how quickly it penetrates that other gas.0866

OK, now let's go ahead and talk about real gases, as opposed to ideal gases.0879

Well, we know that the ideal gas law is PV=nRT; I am going to rearrange this, and I am going to write it as nRT/V, and here is the reason that I am going to do this.0888

We said that the ideal gas law works for low pressures--low pressures where the gas molecules are flying around; they're really far apart from each other; the individual volumes don't matter; but, if I were to all of a sudden take a volume, where the individual gas particles really fall apart, and they are floating--they don't matter--and I increase the pressure, by increasing the pressure, I reduce the volume.0902

So now, I have dropped it down to something like this, where the same number of particles...now the particles don't have as much space as they did before.0930

So now, the volumes of the individual atoms and molecules start to matter.0944

What ends up happening is that this initial volume...when we use this particular volume, now, because there are so many particles, and the volume of the particles is actually a fair percentage of the total volume--now, the volume available for the particles to move around in is not the same as--is actually lower than--the ideal volume, by assuming that volumes don't matter.0952

So, if I assume I have a point--if I have a particle and another point--there is a certain volume that is available to it; but now, if this point is an actual volume--occupies volume--it's taking away volume, so there is less room for this other particle to move around.0979

We have to make an adjustment to this side by reducing the volume.0993

We write this as P=nRT/V-nb; and I'll talk about what b is in just a second (n is just the number of moles).0997

This is a volume adjustment; it is saying that, as we increase the pressure or decrease the volume or lower the temperature (which also decreases the volume), now the volume of the individual particles matter, and I have to make an adjustment for the volume.1011

There is less volume available for the particles of gas that are initially there to move around in.1024

It is not V; it's less than V--that is why it is V-nb.1030

OK, now, the next adjustment is this: we assumed, in the kinetic-molecular theory for ideal gases, that the particles exert no forces on each other.1034

They have no attractive force to each other.1042

Well, as it turns out, particles do have an attractive force to each other, and in fact, the more polar the particular molecule (like, for example, water molecule)--they're going to stick together.1044

Because they stick together, the collisions that they experience are not elastic, so the pressure that we actually measure is going to be less than the pressure that it would be under ideal conditions, because now you have fewer particles actually bouncing around and hitting the containers, because more of these particles are actually sticking together.1055

There is loss of energy, if you will, in some sense; the total energy of the system is conserved, but individually, there is sort of a loss; so the pressure that we measure is actually going to be less than the pressure...I'll say that the pressure observed is going to be less than ideal pressure.1078

That is why I put the "obs" here.1095

So, this factor is going to be a, times n over V squared; this is the pressure adjustment.1097

This is the volume adjustment; this is the pressure adjustment.1107

Now, I'm going to rearrange this again, and I'm going to write it as P+a times (n over V)2, times V-nb, equals nRT.1110

This is PV=nRT, but because of real gas behavior, I have made adjustments to the pressure and the volume for real gas behavior.1126

Now, this a and b are called van der Waals constants, and we have calculated different constants for different gases.1135

You can see them in any chapter on gases in a chemistry book.1154

They have them for (most of them list maybe 10 or 12) the most common gases--methane, oxygen, nitrogen, hydrogen...things like that.1157

That is all it is; these are just numbers that you put in there; n is just the number of moles of particles that you are dealing with, and this is a better representation of how gases behave at high pressures and low temperatures--in other words, small volumes, because volume matters now.1165

That is all this is; this is just PV=nRT, adjusted for real gas behavior.1183

OK, let's go ahead and jump into our examples, and I think a lot of this will start to make sense.1189

So, our first example is going to be: Calculate the root mean square velocity of atoms in a sample of methane gas (which is CH4) at 40 degrees Celsius.1194

So, calculate the root mean square velocity of atoms in a sample of CH4 at 40 degrees Celsius.1227

Basically, how far is your average atom flying around at?1231

OK, well, let's (let me see--we know what we're going to deal with...so...) just use our equation: root mean square speed is equal to 3 times R times T, over the molar mass; square root.1236

Well, T (temperature) is in Kelvin; 40 degrees Celsius becomes 313 Kelvin; we have to make sure to work in Kelvin.1253

Also, remember molar mass: molar mass has to be in kilograms per mole--not grams per mole, so we know that oxygen is 16 grams per mole, but oxygen is going to be .016 kilograms per mole; that is what is really important here.1264

These problems are not difficult; the difficulty is going to be remembering to work in the appropriate unit, so that we actually get our answer in meters per second.1280

OK, so methane is CH4; C is 12; there are 4 H's--that is 16; so, 16 grams per mole becomes 0.016 kilograms per mole.1287

And now, when we put these numbers in, we get 3, and we said R is 8.31, not .08206, so 8.31 Joules per mole-Kelvin, and the temperature is going to be 313 Kelvin, and we have 0.016 kilograms per mole; all of this under the square root sign.1304

Now, when we do the mathematics, as far as the numbers, we're going to get 699 meters per second.1335

Now, I want to show you where the meters per second comes from.1341

Kelvin cancels with Kelvin; mole cancels with mole; now, what we end up with, as far as units, is Joules over kilograms.1344

All right, here we go: J over kg; well, the Joule is kilogram-meter2 per second2; that is the unit--force times distance, a newton times a meter, gives me a Joule.1355

So, the unit of a Joule is a derived unit; it's kilograms-meters2 per seconds2, over kilograms.1374

Well, kilograms cancels with kilogram, leaving us meters2 per seconds2; then, when I take the square root of that, I get meters per second.1382

So, the units work out--very, very important.1391

R has to be 8.31 Joules per mole-Kelvin; temperature has to be Kelvin; molar mass has to be in kilograms per mole--very, very important.1394

OK, let's move on to a second example here.1403

We have...let's see...the problem says: The effusion rate of an unknown gas is found to be (we can measure this) 32.50 milliliters per minute.1407

In other words, 32.50 milliliters are leaking out of a hole every minute.1438

That is all that is; it's a rate--an amount per time.1444

Now, under identical conditions, the effusion rate of oxygen gas (O2) is found to be 31.50 milliliters per minute.1448

Is the gas (the unknown gas) methane, carbon monoxide, nitrogen monoxide, carbon dioxide, or nitrogen dioxide?1482

Well, let's use what we know.1500

We know that Graham's law says that the rate1, times the molar mass of 1, equals rate2, times the molar mass of 2.1502

Well, we know the rate of the unknown gas--we measured it--that is 32.50 milliliters per minute (I'll go ahead and leave the unit off--it's not that important).1514

I don't know what its molar mass is; I would like to know that, because, when I know that, I can just compare it to the molar masses of my choices and pick the right one.1526

Well, I know the rate for the second one is 31.50 milliliters per minute--that is the oxygen--and its molar mass is going to be...it's O2, so it's not 16; it's 32--32 grams per mole.1534

In this case, we can use the 32 grams per mole; it doesn't have to be in kilograms per mole, because, again, the ratios cancel out.1550

It is OK, as long as the units are the same.1557

That is it; so we have that the square root of the molar mass is going to end up being 5.48, and then, when we square both sides, we get a molar mass of about 30 grams per mole, and when we compare it, it looks like nitrogen and oxygen--nitrogen is 14; oxygen is 16; that is 30.1561

So, our gas is nitrogen monoxide.1584

That is it: Graham's law; rate is related this way; you could put it in that other form, where you have rate1 over rate2 = molar mass of 2 over molar mass of 1.1588

I like it this way, because everything is consistent--1, 1, 2, 2--but it is your choice.1599

Let's do another problem here.1607

This one says: Calculate the pressure exerted by 0.600 moles of nitrogen gas in a 2.0 liter vessel at 35 degrees Celsius, using a) the ideal gas law, PV=nRT, and b) van der Waals equation (the van der Waals equation was that adjusted one, the P-a over n/V squared--that one).1614

We want to compare them to see what we are looking at--to see how closely, actually, ideal behavior and non-ideal behavior is for this particular situation.1664

Well, 35 degrees Celsius--that is a pretty high temperature, in terms of Kelvin.1673

2 liters volume is a reasonable volume; and it is .60 moles.1678

Well, let's just sort of see what happens.1683

OK, so a) PV=nRT: we rearrange; we get nRT/V, and we just plug the values in.1686

We get 0.600 (is our number of moles); R is .08206.1695

And remember, when you are dealing with the ideal gas law, you have to use the .08206 for R; when you are dealing with issues of Joules and things like that, root mean square speed...that is when you have to use the 8.31 Joules per mole-Kelvin.1701

The problem itself--if you just stop and take a look at what units you want--they will tell you which R you are actually going to use.1719

Then, 308 Kelvin should be for 35 degrees Celsius, and our volume is 2.0 liters.1726

We end up with 7.58 atmospheres; that is pretty high pressure.1735

OK, now let's use pressure + a, times (let me write it as n2 over V2...no, you know what...I'm just going to leave it as (n/V)2).1743

Notice this n/V, by the way: it's the number of moles over a volume--it's a concentration.1759

The pressure is actually contingent on the concentration, which makes sense; it has to do with: the more molecules you have, the more heavily concentrated, in a given volume--the pressure is going to increase.1764

So, just recognize that; that is why I didn't do n2 over V2.1774

I left it that way so that you could see that it is a concentration term.1779

...V-nb=nRT.1782

OK, and then, when we rearrange, we get P=(nRT/(V-nb))-(a(n/V)2).1788

Now, the pressure adjustment constant (van der Waals constant) for nitrogen gas is 1.39, and (the units are irrelevant; you can certainly look up the units if you want--they are not altogether important) BN2 is equal to...the b, the volume adjustment parameter, is 0.0391.1807

When we put all of these values in here, we end up with 7.67 for this first term, and we end up with 0.1251 for this second term, which gives us a total pressure of 7.56 atmospheres.1833

So notice, under the ideal gas law, we have 7.58 atmospheres; using the real gas behavior, we have 7.56 atmospheres.1853

7.56 and 7.58 are virtually the same, so under these conditions, we are welcome to go ahead and use the ideal gas law.1862

Which--just by looking at this--35 degrees Celsius is a pretty high temperature; 2 liters--it's a pretty big volume, actually, for .6 moles of gas, so you can pretty much say to yourself, "You know what? I'm just going to go ahead and use the ideal gas law; I don't want to use the van der Waals equation--it's not important."1870

This confirms that that is the case.1889

If the conditions were different, you might get a lot of deviation.1892

OK, so we talked about the kinetic-molecular theory; we have talked a little bit about root mean square speed, average kinetic energy of a gas sample...we have talked about effusion and diffusion and done some problems.1896

In the next lesson, we're going to actually sort of tie it all together and do some regular gas problems.1908

Thank you for joining us at Educator.com, and we'll see you next time--goodbye!1914

Hello, and welcome back to Educator.com.0000

Last lesson, we finished off the discussion of gases with real gases and the kinetic-molecular theory.0002

This lesson, we're just going to sort of tie it all together; we're just going to do some practice problems.0008

We're just going to work on the entire range of gas problems.0013

We're not going to do 1,000 problems, but we're going to do a fair sampling, and it will be the type of problems that you will see on the AP exam, and the type of thinking that you're going to have to do, and a little bit of the manipulations.0017

So, let's just go ahead and get started--that is the best way to make sense of anything.0029

The first example that we're going to do is the following (let's see...go ahead and write 1 here)--the problem is as follows.0036

The density of a gas was measured to be 1.25 grams per liter (now, notice: we have grams per liter--it's still a mass per volume; the actual, individual units don't really matter; for gases, grams per liter is pretty standard, simply because gases don't weigh very much).0046

So, we measured this density at 29 degrees Celsius and 1.75 atmospheres.0073

OK, we want to know: What is the molar mass of this gas?0082

A pretty standard problem--it's often how we find the molar mass of a gas.0089

What is the molar mass of the gas?0093

Now, I can guarantee you that this problem, in one form or another, will show up on the AP exam--perhaps even twice; I guarantee it; it always does.0098

So, let's see; let's talk about how we're going to approach this.0106

Well, let's just look at the definition of molar mass: they are asking for a molar mass, so what is molar mass?0110

Let me use red ink here.0115

Molar mass is equal to the mass in grams, over the number of moles; that is it--that is all molar mass is.0118

Let me rearrange this equality; let me write it as mole equals mass over molar mass.0129

Now, let me use a capital M for molar mass, but I'll go ahead and leave mass itself as...0139

And you know this already--you know, when you're finding the number of moles, you take the mass of something and you divide it by the molar mass; you get the number of moles.0145

Well, mol is n in the ideal gas law, so let's go ahead and put--wherever we see n, let's put mass over molar mass; how is that?0152

We write PV=nRT; that means PV=mass, over molar mass, times RT; it equals--and then we drop the V down; we solve for P; we get P=mass, times R, times T, over molar mass, times volume.0163

Now, let me switch some things around here; I'm going to bring the V and put it under the m, under the mass, and I'm going to write mass over V, volume.0190

I'm going to separate it out from everything else, and I'm going to write RT over (no, wait...PV over mass...oh, I'm sorry; I made a little mistake here); I drop down the m; I drop down the V; I actually want to drop...I'm looking for molar mass, so that is what I want to solve.0201

I want to bring the M up here, drop the V and the P down here...there we go: it looked like there was something wrong.0229

And then, I'm going to combine the mass and the mass over V; I'm going to write that as a unit and write RT/P (so this is the same thing--we just played with some variables).0237

Well, what is mass over volume?0247

You know that it is density; so, there you go--the molar mass is equal to the density (which--I used a Greek small letter delta), RT, over P.0250

That is how you find the molar mass.0266

If you had it the other way--if you had the molar mass and you wanted the density--you just rearrange it; that is what is nice about this--you just rearrange your variables, based on the units that they are made of, and you can come up with all kinds of different variations of the particular equation at hand.0268

That is often how you do it in laboratories.0282

So, let's go ahead and put our values in.0285

We have the density, which is 1.25 grams per liter; we have R, which is .08206 liter-atmosphere per mole-Kelvin, and then we have the temperature, which is 29 degrees Celsius, which is 302 Kelvin (again, we always work in Kelvin).0287

Then, our pressure, it says, is 1.75 atmospheres.0312

OK, that is it: so we get: atmosphere cancels atmosphere; Kelvin cancels Kelvin; liter cancels liter; what we'll end up with is g on top, mole on the bottom--grams per mole--molar mass; our units work out.0317

Our final answer is 17.7 grams per mole; that is it.0333

We took the PV=nRT; we solved the molar mass thing for moles; we put it in; we rearranged, and we get the molar mass equals the density, times R, times T, divided by P.0339

This equation right here will show up on the exam, in various forms; you will see it in the multiple choice, and more often than not, chances are pretty good (I would say 80 to 85%) that you will see it in the free response section or the essay section (the part that asks you for some descriptive--where you are not really doing anything mathematical, but you're describing what is going on).0350

So, it will more than likely show up.0370

OK, so let's go to another example; that is what we are here to do.0375

2: we have: Equimolar quantities of nitrogen gas and nitrogen dioxide gas are put into a closed vessel at constant temperature (so, closed vessel--fixed volume; constant temperature; we're good).0382

This is going to be an example of a qualitative problem that you see in the free response section, but they are not going to ask you to do any calculations, but they are going to ask you to reason things out.0419

So, this is a typical, typical example of what you will see on there.0428

Part A (and it comes in multiple parts): A: Which gas has the larger partial pressure?0432

These kind of questions want to test your qualitative understanding of what is going on.0445

In other words, you might be able to solve a problem mathematically, because you have sort of seen it a thousand times, and you know how to fiddle with it; but if you don't actually know what is going on, you won't be able to answer these questions.0449

So, these questions are probably the hardest, in the sense that they are testing whether you can reason things out, based on what you know chemically.0459

So, you have to know the chemistry; it is not just the math.0467

The larger partial pressure: OK, well, all right; we know that the partial pressure is equal to the number of moles, times R, times T, over V; it's just the ideal gas law for a mixture of gases.0471

We know that the total pressure is equal to the sum of the partial pressures; each partial pressure is equal to the number of moles of that particular gas, times RT, over V.0487

Well, b (a and b are just the nitrogen and O2)--that equals the number of moles of b, times RT, over V.0496

It is equimolar--that means the same number of moles.0504

That means the number of moles of nitrogen gas equals the number of moles of nitrogen dioxide gas.0508

Therefore, the partial pressures are equal.0513

The partial pressure of nitrogen gas is equal to the partial pressure of nitrogen dioxide gas.0516

You have reasoned it out; there it is.0523

Part B: here it is interesting: Which has the greater density?0526

Which has the greater density: well, in the problem that we just did, we said that the molar mass is equal to the density, times RT, all over P; let's just solve for the density now, instead of the molar mass.0538

So, let's go ahead and convert this to: Density equals P times M, over RT; in other words, P, R, T...those are fixed.0553

Molar mass; if the molar mass is higher, the density is higher--that is what this equation tells me; I just rearrange an equation.0565

Well, of these two--the nitrogen gas and the nitrogen dioxide gas--which has a higher molar mass? Nitrogen dioxide.0572

Therefore, that implies that NO2 has the higher density.0579

This equation tells me so.0585

This is the kind of stuff that you have to do.0589

Much of the multiple choice section is also going to be stuff just like this: reasonably simple--you can reason it out in a number of steps--no real math involved (or if there is math, it's simple numbers--2, 4, 6, 8); but there it is.0593

It's qualitative understanding; that is what is important.0605

Any of your professors will tell you that, if they had a choice between you understanding things qualitatively or quantitatively, your qualitative understanding is actually a little bit more important.0611

Math will always come; qualitative understanding doesn't always come; or, if it does, it comes a lot slower.0620

OK, C: let's see: Which has a greater concentration?--interesting.0628

Which has a greater concentration: Well, what is concentration?0638

Concentration--the unit of concentration is defined as (when you see :=, it means "is defined as") number of moles per liter.0649

Well, they have the same number of moles--their equimolar amounts were put in--and they are in the same volume flask, so concentration of 1, of nitrogen gas, equals the concentration of NO2 gas--nice!0661

OK, let's see what is next.0678

Which has the greater average kinetic energy?0682

OK, which has greater average kinetic energy: in other words, on average, which collection of atoms or molecules are actually moving faster?0687

Well, kinetic energy--remember what we said?--3/2 RT is average kinetic energy.0702

It is directly proportional to temperature.0716

Temperature is constant; therefore, the average kinetic energy of the two gas samples is the same.0721

That is kind of extraordinary: different gases, different molar masses--their average kinetic energy is actually the same.0727

It is a function of temperature; that is it; it's not a function of anything else.0734

OK, E: Which has the greater average molecular speed?0740

Energy and speed: not necessarily the same here.0745

Which has the greatest average molecular speed: well, Urms is equal to 3RT over M; so now is where that molar mass thing comes in.0748

So, nitrogen--the higher the molar mass on average, the slower the molecule.0779

Nitrogen dioxide is heavier than nitrogen gas; therefore, nitrogen gas is lighter; therefore, nitrogen gas is faster.0786

So, that implies that nitrogen gas is faster--moving around faster.0792

OK, now, F: Which gas will show greater deviation from ideal behavior?0803

This one is a bit challenging.0825

Think about these two molecules: nitrogen gas, N2, is nonpolar; NO2 is a bent molecule--it is polar.0830

Because it is polar, there is greater charge distribution, which means that the individual molecules of NO2 are going to attract each other more.0847

Because they attract each other more, the effect that they have on the pressure, the volume, the relationship that we had to adjust in the van der Waals equation, is greater.0856

So, NO2 will demonstrate greater deviation from ideal behavior, precisely because it is more likely to experience that deviation because of the pressure adjustment--the volume adjustment.0864

That's a typical problem that you will see.0881

OK, let's see: number 3: A mixture of neon and argon gas at a total pressure (PT) of 1.4 atmospheres--if there are twice as many moles of neon as argon (so we have twice as much neon as we have argon gas), what is the partial pressure of argon gas?0884

So, we have a mixture of neon and argon gases; the total pressure in there is 1.4 atmospheres; if there are twice as many moles of neon as argon, what is the partial pressure of the argon?0940

OK, this is a mole fraction/partial pressure kind of thing.0951

We said that the mole fraction...remember, we said it is the number of moles of the part, over the total number of moles of the whole, which is also equal to, as far as pressures are concerned, the partial pressure of argon, over the total pressure.0956

Well, let me rearrange this.0973

The partial pressure of argon is equal to the mole fraction of argon, times the total pressure; I know the total pressure--it's 1.4--I should be able to figure out the mole fraction here, and that should give me my answer.0976

OK, well, let's see if we can't figure this out.0988

Let me use this version of it here; yes, I think I will use this version of it here, instead.0996

The mole fraction is also equal to the moles of argon, over the total number of moles, and it says that the moles of argon, let's say, is x.1004

OK, so the moles of argon is x; well, the moles of neon--it says that I have twice as many, so it's 2x.1022

Therefore, the mole fraction of argon is going to equal x over the total (which is x + 2x); or, if I wanted to do just 1 over 1+2, that is fine.1029

That is equal to the partial pressure of argon, divided by 1.4.1041

OK, well, x over (x + 2x) is equal to x over 3x; the three x's cancel; that leaves me with one-third, equals the partial pressure of argon over 1.4; I multiply 1.4 times one-third, and I get that the partial pressure of argon is equal to 0.47 atmospheres--which makes sense.1049

It's twice as much; there are 3 total; so therefore, it's going to be one-third of that 1.4.1077

Again, the mathematics sort of bears it out, based on reasoning it out.1085

Mole fraction--just plug it in; everything will fall out.1089

Let's see what we have: let's see, #4: we have: 2 moles of hydrogen gas, plus 1 mole of oxygen gas, gives 2 moles of water gas at a constant pressure and volume.1097

If Ti is the initial temperature before the reaction takes place (meaning here on the left), what is the final temperature after the reaction takes place (in other words, here on the right)?1125

Before and after: those are the important words here.1156

Be very careful; read the question very carefully.1159

Well, there is a change in the system here, right?--so we're going to use the P1V1/T1n1= P2V2/T2n2.1162

But, they say that pressure and volume are constant, so they basically drop off.1175

That leaves me with 1/T1n1= 1/T2n2.1180

Another way of rewriting this is just T1n1=T2n2.1188

Again, it is important to be able to see where all of this comes from.1196

The equation doesn't just drop out of the sky; you can rearrange it; you can fiddle with it to get what it is that you need.1199

Now, you just sort of put them all in.1204

We solve this for T2; we want our final temperature--that is T2.1207

That is equal to the initial temperature, Ti, times n1 over n2--the total number of moles to begin with, the total number of moles that you end with.1211

Well, Ti...the total number of moles that you begin with is 2+1: 3 moles.1224

The total number of moles that you end with is 2; so our answer is...that is it; that is the expression.1233

The initial temperature, times 3, divided by 2, will give me the final temperature--and it's nothing more than an application of the ideal gas law, with pressure and volume completely ignored.1240

All it is is the temperature and the number of moles.1250

Before the reaction, the number of moles is 3 moles of gas particles; after the reaction, the number of particles is 2 moles; that is what we have to watch out for.1253

Let's see: we'll do a quick multiple choice one here.1267

Which of the following conditions would be most likely to cause deviation from ideal behavior in a gas?1272

We have 1) Low pressure, 2) Low volume, third choice is Low temperature, fourth choice is High temperature, and our (oops, can't have these random lines here; let me put temp there) 5)--we have High pressure.1306

OK, so our choices are: 1 only; 2 and 3 only (let's see); 2, 3, 5; 1 and 4; and the final choice is 5 only.1347

Well, which of the following conditions would be most likely to cause deviation from ideal behavior?1380

So, deviation from ideal behavior has to do with high pressure; high pressure or low temperature causes the volume to be small.1386

A small volume, a high pressure, low temperature: these things tend to induce deviant behavior.1396

1 only (low pressure): no, that is not it.1406

B: 2 and 3 only--well, low volume: yes, that will do it; low temperature: yes, low temperature will induce a low volume; that will do it.1414

High temperature: no; but there is also high pressure--high pressure will also cause a low volume, so this says "2 and 3 only," but 2 and 3 and 5 also work, so it isn't B.1425

We look at C (2, 3, and 5): yes, that works.1437

Let's check our others, just to be on the safe side.1441

1 and 4: 1--no (low pressure); 4 (high temperature)--no, that is definitely not it.1443

5 only: well, 5 works (high pressure)--yes, that is certainly viable--but it's not that one only; it's also 2 and 3 (low volume, low temperature, high pressure).1450

The idea is the "low volume" part.1460

Low volume comes from high pressure; low volume comes from low temperature.1463

So, that is what you have to keep in mind; so deviant behavior comes when you have low volumes.1468

OK, so that gives you a sampling of the type of problems that you are going to see on the AP exam.1475

Some of them are from a multiple choice section; some of them are from what they call the essay section; some from the free response section.1479

This is pretty typical of the type of thing that is going on.1487

They are not altogether difficult mathematically; it's just that they require a qualitative understanding.1489

It is very, very important that you know the chemistry.1494

Being able to do the math is nice, but, as you will discover as we go on in the course, a lot of the mathematics tends to be very, very long; it's not a one- or two-step thing.1498

If you understand the chemistry, the math will follow; but just being able to do the math doesn't mean that you will be able to reason out the chemistry, because the math can lead you astray.1506

As you see, if you miss some of the units, you are going to be all over the place; numbers are not going to make sense.1515

Qualitative understanding is what we seek in science.1520

Math will always be there; qualitative understanding will not always be there.1523

OK, thank you for joining us here at Educator.com; we will look forward to seeing you next time--goodbye.1528

Hello, and welcome back to Educator.com.0000

Welcome back to AP Chemistry.0002

We're going to start, today, on our unit discussing thermochemistry, and there are going to be about three or four lessons discussing thermochemistry--chemistry related to heat.0004

Today, we're going to talk about energy, heat, and work--some of the basic, fundamental concepts--and eventually, we will get more into the chemistry of it.0015

Let's just jump in and get started.0023

The first thing that we want to definitely make sure you know is that temperature and heat are not the same thing.0026

So, temperature and heat are not the same thing; this is probably the single biggest misconception.0032

OK, so temperature is a measure of the average kinetic energy of the particles in the sample that you are taking the temperature of.0048

Heat is the thing that actually flows when there is a temperature differential.0078

In other words, heat is the thing that flows from a hot object to something that is colder.0085

Or, it doesn't necessarily need to be an object; any time there is...let's say you have some hot thing, or something that is at a higher temperature; it doesn't have to be hot; there just has to be a difference in temperature.0089

So, if something is 30 degrees Celsius, and let's say the air surrounding it is 10 degrees Celsius, heat is that energy, if you will, that flows from the higher-temperature object into the lower-temperature object.0099

Heat is the energy that actually flows; what temperature does is actually measure the random motions of the molecules.0113

There is a difference between the two; they are related, but they are not the same thing.0121

Heat is the thing that flows from hotter to colder--we'll say objects (objects could also just be the air)--due to the temperature differential.0126

Differential, as you know, just means the difference.0162

Heat is a form of energy; it is a very disordered form of energy.0166

OK, so heat--profoundly important; thermochemistry is going to be all about the heat.0179

Work is another form of energy.0187

Clearly, energy can come in different ways; you can have light energy; it could be heat energy; work is another form of energy; all of these things are energy.0199

They are represented by the same unit, and the idea, essentially, of major areas of science, is: How do we convert one form of energy to another form of energy?0208

That is really what is going on.0217

Energy...work is another form of energy, and we define it as Work=Fd, which is force times distance.0221

So, the best way to think about that is: if I apply a certain force to an object, and if I move it a distance d, well, the work is going to be the force that I apply, times the distance that I actually move it.0232

Now, notice: work, the way we discuss it in science, is not the same sort of way of everyday usage of the word "work."0243

I can stand and I can push against a wall, and I'll definitely start sweating, and I'll definitely start getting tired, eventually.0252

But, the wall isn't moving; I'm applying a force, but the wall isn't moving.0258

Force is a positive value, but distance is 0; the wall isn't moving.0261

Scientifically, there is no work actually being done; yes, I am doing work, in the sense that I am actually expending energy, but no actual work is being done by the force.0267

So, there is a difference: when we say "work," in chemistry, in physics, in engineering, we're talking about a force that causes something to move.0277

OK, so let's look at some units.0287

Force equals mass times acceleration; that is Newton's second law.0290

So, if you want to give a certain body of a certain mass a given acceleration, you have to apply a certain amount of force to it, and the relation between these three variables is F=ma.0297

Well, mass is in kilograms; that is the standard unit of mass.0306

Acceleration is in meters per square second.0313

Well, this unit right here--the kilogram-meter per square second--this is called a newton.0317

A newton is a unit of force, symbolized with an N.0323

Now, we just said that work equals a force times a distance; well, force, we just said, is a newton; and distance is in meters; so now, we have this unit of a newton-meter.0327

This unit, which is a (well, let me go ahead and actually write out everything so you see it)...0346

We said that a newton is a kilogram-meter per square second, and then we multiply it by a meter, which is there; so this is the force; this is the distance; and we end up with kilogram-meter squared per second squared, and this is called a Joule.0355

This is our standard unit of energy, the Joule.0377

If you want, you can think of it as kilogram-meter squared per second squared, or you can think of it as kilogram-meter per square second, times meter--a newton-meter.0381

So, a newton-meter is a Joule, or if you want to break down the newton and sort of put some things together, you can get kilogram-meter squared per second squared.0390

It depends on the problem you are working on; sometimes you will just deal with newtons; sometimes, you will need to break it up.0399

As you go on in science, you will realize that you will need to sort of break up more of these things, in terms of units, to deal with certain constants.0404

Constants--depending on the problem, you might have to use different constants, which are the same constant, but in different units, they have different values.0412

OK, so the biggest piece of advice I can give is: watch your units.0421

If we're talking about a mass, make sure that kilogram matches with gram; or, if something is in two different units, make sure that kilograms, meters, seconds, Joules, newtons...that everything matches; otherwise, your values will be wrong.0426

OK, so let's talk about the two ways to actually transfer energy.0442

The two ways to transfer energy--to transfer energy from one object to another, from one thing to another, from one environment to another, from the outside to an inside, to transfer energy: we have two ways of doing it.0448

We can transfer it as heat or work, or both.0465

That is it; so, in some sense, this is really simple.0473

We're going to have some sort of a system, and we want to either give energy to it or take energy away from it; well, there are only a couple of ways we can do that.0477

We can either put heat into it or take heat out of it, or we can do work on it, or it can do work on us.0485

So, let's be a little more specific here; we'll draw a picture, and we're going to start talking about systems and surroundings.0491

OK, so I have a system; a system is defined as the thing that we are interested in.0500

I know that that is kind of a vague definition--"thing that we are interested in"--it will make more sense as we do some problems and as we talk about it more.0513

Well, for chemistry, most systems are going to be the reaction: the reaction, and maybe the vessel that is containing it...a solution of hydrochloric acid plus sodium hydroxide--that is your system...the beaker plus the water; that is the system.0525

The surroundings--that is everything else.0540

That is everything else; so, very, very intuitive--nothing strange is going on here.0545

We said that we can transfer energy in two ways: we can do it via heat, or we can do it via work.0550

This is our system; this is (I'm sorry, I should put the system in here)...I'll call this sys, and out here is the surroundings.0557

So now, this is the boundary, and it's always going to be something; there is going to be some sort of a boundary that separates your system from your surroundings.0575

Now, we're going to transfer heat in and out; we're going to transfer work in and out.0583

We have to pick a point of view.0588

In chemistry, we always look at the system's point of view; it's not always the case; there might be other circumstances, maybe in engineering, where you need to look at the surroundings' point of view; but in chemistry, we always look at it from the system's point of view.0591

From the system's point of view, when heat is added to a system (OK, so we say heat in), that heat is positive.0602

When heat is given off of a system, that heat is negative; in other words, the system is losing heat--it's going from a certain value of heat, it's losing heat, and its final heat is actually less than that.0617

So, energy and heat is flowing out; so heat is negative.0631

That is the sign convention; if heat is put into the system, that means more energy; so it has a certain amount of energy; we put more heat in; it goes up; that is positive.0635

We are looking at it from the system's point of view; notice, it would be the reverse if it were the surroundings' point of view.0643

The other way to transfer energy is work.0648

Now, the system could do work on the surroundings; the surroundings could do work on the system.0651

We will be a little more specific about what that means, but the same sign convention applies.0657

If I do work on the system--work on the system--that is positive work.0662

That means I have put energy into the system via the work that I have done to it.0676

If the system does work on the surroundings, that is going to be negative work.0680

This is called work done by the system.0687

So, in this case, these prepositions are very, very important; so, when you read your questions, makes sure you read them carefully, because it will say "work done on the system" or "work done by the system."0693

It won't tell you whether it is positive or negative; you have to decide.0701

So, work done by the system--that means the system used its energy to do something to the surroundings, and in the process of doing it, it transferred its energy to the surroundings.0706

Now, energy is depleted.0715

Work is a little more abstract than heat; with heat, you have a pretty good intuitive sense--you know when something gets colder or gets hotter.0718

Work is a little different, but again, we will do some problems, and we'll get used to it.0725

So, two ways to transfer energy: heat and work; work into the system is positive; heat into the system is positive; heat out of the system is negative; work out of the system is negative.0730

Again, from the system's point of view--we are always looking at it from the system's point of view.0744

Whenever a process actually gives off heat (the system gives off heat), we call it exothermic.0749

It makes sense, right?--exo, outside.0755

And then, heat in is called endothermic--an endothermic process; that means the system is taking in heat.0758

These are very, very important words, and we will use them in the context of our problems.0767

OK, so let's look at just a couple of quick examples.0773

If I take methane gas, and if I combust it (burn it in oxygen), I produce two moles of CO2, 1 mole of CH4; I need 2 moles of oxygen; I produce 2 moles of CO2 and I produce 2 moles of water, and I produce heat.0776

As you know, any time you burn something, it gives off heat; you feel the heat radiation; that is giving off energy as heat.0797

This is an exothermic process.0804

Burning, combustion, is an exothermic process.0810

If I take nitrogen gas, and if I react it with oxygen gas, in order to produce a couple of moles of nitrogen monoxide gas, as it turns out, I have to add heat to the system.0813

So, what happens: if I have a vessel, and if I actually ran this reaction, what would happen is: the vessel itself would get cold--I would feel the vessel get cold--and the reason is, now the reaction--in order for it to work--needs energy put into it.0826

So, it takes heat away from the glass (let's say it's in a glass container); it sucks heat out of the glass container, and because the heat from the surroundings is not actually going into the glass fast enough, when I touch the glass, I feel cold.0843

So, it isn't that the glass got cold; it is that the heat was pulled out of it, bringing it to a lower temperature.0856

That is what is going on.0863

This is an endothermic process.0864

The system evolves heat and gives it up.0870

This system pulls in heat--needs the energy in order to perform its functions (in this case, the systems are the reactions).0873

OK, now let's go ahead and put some mathematics to this.0883

We said that there are two ways to transfer energy: heat and work; well, the change in energy of a system is going to be the heat that goes in or out of the system, plus the work that goes in or out of the system.0890

Again, heat and work are just two different types of energy.0909

So, if a system starts in one state of energy, and it goes to another state of energy, a couple of things have happened.0912

It has either lost or gained heat, lost or gained work, or some combination of both.0920

This is our fundamental equation; this is actually a statement of the first law of thermodynamics--that energy can neither be created nor destroyed; that any energy change is going to take one form or the other: heat, work, or both.0926

But, energy is conserved; that is what this equation is saying.0942

OK, so these heat and work thermodynamic quantities--they have two parts.0947

They have a magnitude, and they have a sign.0957

Well, again, we dealt with the sign issue here; with thermodynamics, oftentimes we have to stop and think about the physics of what is going on, in order to decide what the sign is going to be.0963

Often, we know what the magnitude is, but we have to make sure to keep the sign straight; otherwise, the mathematics will not work out.0974

All right, now let's fiddle with this equation; let's talk a little bit more specifically about this thing called work.0982

OK, so a common type of work in chemistry (in fact, the most common in general) is work done by a gas.0991

So, it's either going to be...well, work done by a gas: notice, here is that "by" again; that is expansion; and when I draw the picture in a minute, it will show you what I mean.1015

Or, work done to a gas, which we call compression.1030

Let's say I have this container, and it has a moveable piston that can move up and down; this is a chamber that has the volume.1044

So, I can push this piston down and squeeze this volume--make it smaller--or I can pull the piston up and make the volume bigger.1056

In this particular case, if there is a gas in here, and it expands, that means the gas is actually pushing against the external pressure of the atmosphere.1064

It is doing a certain amount of work; if I actually push the piston down, and the volume gets smaller, this pressure is higher than this pressure; I am actually pushing this down--I am doing work on the system here.1073

Let's take a look at what this looks like mathematically.1088

We just said that the change in energy is equal to q + w, the heat that goes in and out of a system plus the work that is done on or by the system.1091

OK, q: well, we said that work is equal to force times distance; great--simple substitution.1101

We'll leave q alone, and now, here is what I'm going to do: I know that there is a relationship between force and pressure and area; as it turns out, if I take the force of something, and if I divide it by the area (remember, back when we talked about gases, we said that pressure is equal to the force per unit area, so...) I have this F here; if I divide by A, in order to retain this, I have to multiply by A, so I basically have multiplied this F/d by A/A, right?1111

I did this in order to manipulate this F a little bit.1144

Now, I have F/A, and I have distance times area.1147

Well, this thing right here--I see it in the profile, but really, it is just a cylinder.1153

Let's see what we get here: we have q; well, force over area is pressure, and distance times area is...well, distance is...let's say that this force here is a pressure that is being applied to this piston.1160

The area is, of course, the area of this circular part; so there is a certain pressure being applied; well, the area times the distance...the distance is the distance that this piston actually goes down or up, right?1186

So, this is the distance; well, area times the distance, as far as the circular cylinder is concerned, is just volume.1204

You end up with: Change in energy equals q; P is constant, so the only thing that is changing here is, from this to this, whether we either press down or pull up.1215

Pressure...PΔV; this is pressure times the change in volume.1233

This is what we mean by pressure, volume, and work.1238

So, when we push down on a gas, we are actually doing work on the gas; we are doing work to the system.1242

When the gas expands and pushes the piston up, then the gas is actually pushing against an external pressure, and it is actually doing work on the surroundings.1251

This final equation...this is often how we will do it: we're going to be dealing with some gas in a container, and this is going to be the expression that we use: The change in energy of the system is going to be the heat that flows in or out of the system, plus the change in volume, based on a certain amount of pressure.1266

This pressure is always going to be the external pressure.1291

It is an external pressure that pushes down, and squeezes it, and makes the volume smaller; or, it's going to be the pressure against which the gas pushes in order to make it bigger.1294

In either case, it is a pressure that is pushing against the boundary here.1306

So, it is always an external pressure--that is what this P is, not the pressure of the gas.1311

OK, so now let's talk about some signs.1318

All right, this is going to be important; so now, we're going to actually talk about this term right here, and we're going to relate...so we know that work and pressure times volume are equivalent in terms of magnitude whenever we are doing something like this.1322

So now, we want to talk about what sign convention we are going to use.1339

All right, so let's just write "Work = PΔV"; so, work equals pressure times the change in volume.1345

If I start at a certain volume, and if the gas expands, that means the gas inside is pushing against an external pressure; well, the final volume is bigger than the initial volume.1351

Final volume, minus initial volume, is going to be bigger than 0.1369

Well, in this particular case, it is bigger than 0; the pressure is bigger than 0; here, this work is going to be positive; this number is going to be positive.1375

Pressure times the change in volume, for an expansion, is positive; but, because it is expanding, it is the system that is doing work on the surroundings; therefore, the work is negative, because again, we said we are looking from the system's point of view.1383

If the system is pushing against this piston and expanding, it is doing work on the surroundings.1398

In other words, work is leaving the system.1404

Therefore, work is negative; so, by putting this negative sign, we also account for the other direction.1406

If the outside pressure is pushing down on the piston, and making the volume smaller, that means the final volume is smaller than the initial volume.1413

Therefore, ΔV, which by definition is final minus initial, is less than 0.1425

Well, if ΔV is negative, and P is positive; this negative sign--negative times negative--makes the work positive, which works with the fact that now the surroundings are doing work on the system.1430

The work is positive, because the system is now taking in energy in the form of work.1444

It is positive from the system's point of view.1451

The energy of the system is increasing.1453

So, this is the actual relationship: work is negative pressure times the change in volume (change in volume always being final volume minus initial volume--which is the definition of Δ--final minus initial).1456

OK, so with this little bit of a background, let's just jump into the problems, and I think we can start to make sense of some of this.1473

Let's do Example 1: So, we want to calculate the change in energy for an endothermic process in which 29.6 kilojoules of heat flows (and notice: we said endothermic), and where 13.7 kilojoules of energy is done on the system.1482

OK, so we want to calculate: we say that, for an endothermic process, 20.6 kilojoules of heat is flowing.1540

Endothermic means heat is coming into the system; that means heat is positive.1545

"Where 13.7 kilojoules of energy is done on the system": on the system means work is flowing into the system--that means work is also positive.1551

So, we write our equation: change in energy equals q + w; q is heat; w is work.1559

We said an endothermic process, so it's positive, so it's going to be positive 20.6 kilojoules; so notice, the problem will give you a magnitude; it won't necessarily tell you what direction it is going.1569

If it said exothermic, then I would have to use -20.6.1582

If this said "by the system," I would have to use -13.7.1585

But here, it's endothermic, so heat is positive, and then the work is also positive 13.7, because it is done on the system.1589

Kilojoules...this is a simple arithmetic problem.1597

So, we end up with 34.3 kilojoules of energy.1600

I put heat into it in the amount of 20.6 kilojoules; I did work on the system by 13.7 kilojoules; the total amount of energy that I imparted to the system is a positive 34.3 kilojoules of energy.1608

Oh, by the way, I should let you know: oftentimes, another unit of energy is something called the calorie; it's an older unit of energy, and it is defined as the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.1626

The calorie is a unit of heat; it's a unit of energy, because that is what heat is--it's just energy.1640

So, as a conversion factor, one calorie equals 4.184 Joules.1646

We are probably not going to run into calorie too much; we may or may not; we'll find out; but just so you know...1653

You have heard of the word "calorie"; you hear about it all the time; that is what it is: 1 calorie is that many Joules--it's a unit of heat; it's a unit of energy.1658

OK, so Example 2: What is the work associated with the expansion of a gas from 0.750 liters to 1.760 liters at constant external pressure of 13 atmospheres?1667

The gas is in a container; there is a pressure of 13 atmospheres on the outside; and again, pressure is always going to be external.1724

On the outside, 13 atmospheres is pressing on it.1732

But, the gas is actually expanding from the inside, and it goes from .750 liters to 1.760 liters.1736

In other words, it is pushing out against the surroundings.1746

It is doing work on the surroundings, which means that energy is flowing out of the system and into the surroundings, which means that it is actually going to be negative.1750

Well, work equals negative pressure, times change in volume, equals negative 13 atmospheres; change in volume--we have 1.760 liters, minus 0.750 liters, and when I do this, I end up with -13.13 (I don't know if you can read that) liter-atmosphere.1761

OK, so notice: atmosphere, liter; this is not exactly a unit of energy that you have seen, although it is a unit of energy; it just happens to be different units, and I will show you what this is.1809

The conversion factor here is this: 1 liter-atmosphere is equal to 101.3 Joules of energy.1822

I'll show you in a minute where this actually comes from.1833

But again, it just goes back to the point: units are very, very important; just because you end up with some unit that seems strange doesn't necessarily mean that it is strange.1835

So, all of the math that we have done is correct; work is equal to pressure times change in volume; pressure is in atmospheres; volume is in liters; this is actually a unit of energy--it's a unit of work.1844

It is actually equivalent (by a conversion factor) to a Joule.1858

So, don't let this throw you off.1863

Now, let's go ahead and do the conversion.1865

When we multiply this out, we end up with 1330.4 Joules, or (can't have these stray lines floating around--let's erase this) 1.33 kilojoules.1867

There we go.1887

OK, so let me see what I have here.1890

OK, so really quickly...you know what, let's skip the unit conversion part; if you can, you are welcome to look it up, but the liter-atmosphere--one liter-atmosphere is equal to 101.3 Joules, so that conversion will always be there for you.1898

Let's just keep going on with these examples.1918

All right, another problem here: Now, a balloon is inflated from 4.00x106 liters to 4.50x106 liters by the addition of 1.4x108 Joules of heat energy.1922

So, I add some heat to the system, and I add 1.4x108 Joules of heat.1972

Now, if the balloon expands against a constant pressure of 0.95 atmospheres, what is the ΔE for the process?1981

OK, so let's just draw a quick picture here: we have this balloon--obviously, we are talking about a really, really large balloon; we pump in heat--pumping in heat energy--so heat is going to be positive.2019

This balloon is going to expand, which means that it is going to push out against the .95 atmospheres; so now, the balloon is actually doing work on the surroundings.2034

Work is going to be negative.2043

Now, let's work it out.2049

ΔE is equal to q plus w; we have worked out our sign convention--heat is going in; work is going out, because it is expansion.2052

We also know that work is equal to -PΔV, so let's go ahead and calculate our work.2062

The work is equal to -0.95 atmospheres (that is the external pressure; and again, pressure, volume, work; PΔV; that P is the external pressure against which you are pushing, or that is pushing on you, or pushing on the system), times 4.50x106, minus 4.00x106.2076

We end up with (when we do this mathematics) -475,000 liter-atmospheres; and then, when we multiply by 101.3 (oops, that is not kilojoules--the conversion factor is 101.3 Joules per liter-atmosphere), we end up with a total of -4.8x107 Joules of work.2103

Well, we just (oh, no, we definitely don't want that...OK) said that ΔE equals q + w; we calculated w--that is -4.8x107 Joules--so now ΔE equals q plus that.2141

Well, q, positive--it is 1.4x108 Joules (you know what, I am always confused about whether I should write it out or do the letters; for years it has been like this--OK, I'll just do the letter), minus 4.8x107 J, so our change in energy is 9.2x107 J.2163

Notice what has happened here: energy flowed in as heat.2196

Work ended up expanding the gas; the gas ended up pushing against the external atmosphere; so energy came in as heat; energy left as work.2202

The total energy change for this process was 9.2x107 Joules; this is a positive value, so there is now more energy in the system that stayed as heat than went out as work.2214

That is all this says--the fundamental equation; this is the first law of thermodynamics.2231

The change in energy is equal to the heat that goes in and out of a system, plus the work that goes in and out of a system.2235

Energy itself can be transformed, but it cannot be created or destroyed; it always has to be accounted for.2241

OK, thank you for joining us here at Educator.com.2247

We'll see you next time for some more thermochemistry; goodbye.2250

Welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we're going to continue our discussion of thermochemistry.0003

We're going to talk about enthalpy and Hess's Law.0007

I have to, before I begin...I wanted to discuss--just take a couple of minutes to talk about--thermodynamics and thermochemistry, and a lot of the terms that are sort of being thrown around, and some of the equations.0010

Back in the early...well, not early days; in the turn of the century, there is a saying about thermodynamics; a very famous thermodynamicist says this; he said, "None of us really understands thermodynamics; we just get used to it."0025

Now, of course, that is not completely true; we understand it; but to a large extent, a lot of it really is true.0037

Thermodynamics is a very, very strange thing; heat behaves in very difficult ways--very unusual ways.0043

These terms, like enthalpy and heat and energy and work, and pressure-volume...I know that a lot of that is very, very odd; it's difficult to wrap your mind around; it's difficult to sort of get a good, intuitive feeling for what is going on.0049

My recommendation for dealing with that is: don't really worry about it too much.0065

A lot of the comfort that comes from dealing with thermochemistry and thermodynamics, aside from some of the stuff like today's (it's actually not that bad)...a lot of it just comes from familiarity.0069

You will be seeing it over and over and over again, and you will be doing the problems over and over again, so that you will get more of a sense.0079

So, if you don't understand it the way you do other things, I wouldn't worry about it too much; that is just the nature of thermodynamics.0085

Again, just to throw something out there to make you put your mind at ease: it's like that for everybody.0092

With that, let's go ahead and get started.0099

OK, so let's start with a definition.0103

We're going to define enthalpy: enthalpy (which we use the symbol H for, and you will understand in a minute, when we actually equate it to heat) is equal to the energy of a system, plus the pressure of the system, times the volume of the system.0108

So, let's say I had some container of gas; it's going to have a certain energy that is associated with it; it's going to have a certain pressure that is associated with it and a certain volume that is associated with it.0132

By definition, that is the enthalpy; if I take the energy, plus the pressure, times the volume, I get the enthalpy.0142

Now, like most thermodynamic properties, we don't really know what absolute enthalpies are; the only thing we can actually measure (which is science: in science, we measure things) are changes in enthalpy.0149

So, ultimately, we are going to be concerned with ΔH; and, in fact, all of thermodynamics is going to be concerned with the Δ of some properties.0162

Later on in the course, we will talk about ΔS; we will talk about ΔG, which is free energy entropy; and H is enthalpy.0169

We are concerned more with Δ, and, a little bit later in this lesson, we will show why the Δ is important...in any case, just so you know...0177

The definition of enthalpy is the energy plus the pressure times the volume of the system.0189

Now, let's recall what we did last time: we said that the change in energy of a system is equal to q, plus the work, which was -PΔV.0194

Now, let me rearrange this; let me bring this -ΔV over to this side; q equals the change in energy, plus PΔV--now, let me just set that aside for a minute.0209

Now, let me come over here, and let me take my definition of H as equal to E, plus PV, and now I'll do ΔH.0222

Well, ΔH, which is final minus initial, equals Δ of this, which is the ΔE plus Δ of PV.0236

Well, in this particular case, if we keep the pressure constant, that means we can pull this out of the Δ; we end up having ΔH=ΔE+PΔV.0247

Notice what we have: we have that q, which is the heat that is transferred, is equal to the change in energy plus the pressure times the change in volume (constant pressure).0266

And we have that ΔH, just based on the definition, equals ΔE, plus pressure, times Δvolume.0276

These are the same; q equals ΔH at constant pressure.0282

So, this is a constant pressure situation here, OK?--at constant pressure (which is pretty much what we are doing--what we do most of our chemistry in, at a constant atmospheric pressure), q equals ΔH.0293

In other words, the enthalpy is nothing more than the heat that is transferred--the energy that flows as heat.0307

In other words, if a certain reaction takes place, and it gives off a certain amount of heat or it takes in a certain amount of heat, that heat is equal to the enthalpy.0314

So, I can talk about the enthalpy by just referring to the heat; all I have to do is deal with the heat--it happens to be the same as the enthalpy, under constant pressure conditions.0325

That is why constant pressure is really, really nice, because under those conditions, I don't have to deal with enthalpy; I just deal with the heat; they are equivalent.0335

Heat is a very easy thing to measure; you just run a reaction, and you measure how hot something gets or how cold something gets; that is your enthalpy.0343

It is a thermodynamic quantity that is related to heat; under constant pressure conditions, it is the heat--it is equivalent to the heat; so that is really nice.0351

This is the important thing to know.0362

Heat and enthalpy are equivalent in magnitude under constant pressure conditions.0366

OK, so when we talk about the enthalpy of a reaction, ΔHrxn, well, that is equivalent to the heat of the reaction.0370

We will often talk about the heat of the reaction.0380

OK, now, ΔH of the reaction is equal to...actually, you know what, no; I'm not going to...I'll save this for the next time; there is no need to confuse us with any information that we don't need right away.0392

OK, so let's just do a quick example to get a feel for what is going on--a little bit of stoichiometry.0408

Example 1: When 1 mole of C6H12O6, which is glucose, is fermented to ethanol at a constant pressure, 67 kilojoules of heat is released.0415

The system releases heat; heat is flowing out of the system; it is negative.0458

How much heat is released when 7.6 grams of glucose is fermented?0467

1 mole of glucose is fermented to ethanol at constant pressure; 67 kilojoules of heat is released--exothermic.0490

The enthalpy is negative; now heat and enthalpy is the same--constant pressure.0499

How much heat is released when 7.6 grams of glucose is fermented?0505

OK, so let's write out what this looks like to get a sense of how we sort of start these problems, and what it looks like, notationally, for a chemist.0509

So, C6H12O6 is fermented to 2 C2H5OH (ethanol--this is regular drinking alcohol), plus carbon dioxide gas.0518

And we often write the ΔH over here as -67 kilojoules; so we see that heat is released; ΔH is negative; it is an exothermic process.0534

That is what it means: exothermic--ΔH is negative; ΔH is negative--it is exothermic; it is giving off heat, which means, in addition to this product and this product, one of the other products is that much heat.0543

That is why we write it; imagine heat as that third product that also comes out of the reaction.0556

This heat comes from the bonds in the carbon, hydrogen, and oxygen; that is where it is coming from.0563

OK, let's draw a little energy diagram, so you see what is going on here.0569

C6H12O6: this is H, enthalpy (heat at constant pressure), and this is just the reaction coordinate.0576

The reaction coordinate just means that the reaction is proceeding in that direction.0588

Well, there are some...here we have the C2H5OH, ethanol, plus our CO2.0594

Now, what this means--this -67 kilojoules--as it turns out, thermodynamically, the reactants--there is more heat in these bonds; when going from glucose to ethanol as CO2, the amount of energy in these bonds is actually 67 kilojoules less.0605

So, this difference right here, from this point to this point, is the 67 kilojoules.0627

Because, again, energy cannot be created or destroyed, the energy in these bonds is returned into the energy of these bonds.0634

But now, I have an excess amount of energy; what am I going to do with it?--well, the reaction just releases it as heat.0642

That is what this says; it is going to a lower heat--this excess heat is what is given off.0647

That is why it is -67; this is what it actually looks like.0655

The products are thermodynamically at a lower energy.0658

OK, so let's go ahead and calculate: well, C6H12O6 is 180 grams per mole; we have 7.6 grams of it, times 1 mole--180 grams; that gives us 0.0422 mol, and I hope that I did my arithmetic correctly.0663

Well, I have 0.0422 mol, and it is telling me that it releases 67 kilojoules per mole (that is what the problem says).0694

It is a simple arithmetic problem.0705

-2.8 kilojoules of heat is released.0708

2,800 Joules of heat is released with 7.6 grams of glucose.0723

7.6 grams of glucose is not very much; it's a handful--not even a handful; it is 2,800 Joules; that is a lot of heat.0729

There is a lot of heat in those bonds; that is why the human body metabolizes glucose--it breaks it down, not into alcohol and CO2--it breaks it down completely into carbon dioxide and water, and all of the energy that is released--the body uses that energy to produce a molecule called adenosine triphosphate, and it is the adenosine triphosphate that runs the body.0743

That is our energy currency, and it all comes from the energy that is stored in the bonds of C6H12O6.0767

Well, just 7.6 grams produces 2,800 kilojoules of energy!0773

You can imagine the amount of glucose we actually take in, in the form of carbohydrates and other things; the body requires a lot of energy to run.0778

All right, now let's talk about something called a state function.0790

Let me...all right, let's define a state function.0798

A state function is a property (you could call it a state property; I don't know why they call it a state function, but it is a property) that does not depend on the path taken to achieve that state.0803

OK, so let's say I have something here and something there; these are two states; I can get from this state to this state--I can either go this way, directly, or I can go this way and come back; I can go this way, this way, this way, this way, this way, this way.0846

Now, as it turns out, there are certain properties that are not state functions, like heat and work.0862

So, for example, if I went from here to here, I would have to do a certain amount of work.0872

Clearly, if I went from here to here to here to here to here to here, I am doing more work.0876

But, as it turns out, energy (which, as we said, is equal to heat plus work, neither of which is a state function)--as it turns out, energy is a state function.0880

As it turns out, it doesn't matter how I get to the final state--all that matters is where I started and where I ended up.0894

That is why we are concerned with ΔS: ΔH, ΔG--those are state functions; enthalpy is another state function.0900

So, we said that enthalpy is equal to the change in energy, plus the pressure, times the change in volume, right?0907

Well, this is a state function; pressure is a state function--it doesn't matter how I get there; at a certain point, and at the pressure that I start and end up with, it is just a certain pressure; it doesn't matter how I get there.0916

Volume: it doesn't matter how I go from 3 liters to 5 liters; I can go up to 18 liters, drop down to .1 liter, and then go up to 5 liters; I have still just gone from 3 to 5--the net effect is the same.0929

So, ΔH is also a state function.0942

It is a state function because it is the sum of two state functions.0944

Energy is a state function despite the fact that neither of these is a state function.0948

This is also quite extraordinary, that that is the case.0953

OK, so as far as chemistry is concerned, now: chemistry--if we start with certain reactants, and we want to end up with certain products, well, as far as the enthalpy is concerned, it doesn't matter how I get there.0957

I can get there in 2 steps; 15 steps; 147 steps.0974

Now, yes, there are areas of chemistry where we are concerned about the steps, but as far as a thermodynamicist is concerned, all he cares about is the enthalpy at the beginning and the enthalpy at the end.0978

It is a state function; it doesn't matter how you get there; all that matters is that you get there.0988

It is the two states that matter; that is all that matters.0993

Because of that, we can actually take a reaction that we are interested in, and perform it, and if we want to find the enthalpy of that reaction, well, if we have enthalpies of other reactions that we can use to get to our final reaction, we get our final enthalpy.0996

OK, so now, this is the idea of Hess's Law.1014

And, rather than talking about it or writing about it, the best thing to do is just do a problem, and of course, it will make sense.1021

So, let me write out Hess's Law here.1026

Well, I won't write it out; we'll just do an example, and it will make sense.1032

OK, we want to find the ΔH for the reaction of sulfur, plus oxygen gas, going to sulfur dioxide gas.1036

In other words, there is a certain heat of reaction associated with this; either it absorbs heat to create SO2, or in the process of creating SO2, it releases heat.1059

I want to find the enthalpy--the heat of the reaction.1069

Well, how do I do that?1072

OK, so we want to find ΔH for the reaction; well, as it turns out, we just happen to know that we have a couple of reactions at our disposal.1074

We know that, if I take sulfur plus three-halves oxygen gas, in the process of creating sulfur trioxide gas, we happen to know that the ΔH of that is -395.2 kilojoules.1088

We also happen to know that sulfur dioxide gas, 2 moles of that, plus oxygen gas, goes to 2 sulfur trioxide gas, and we happen to know that the ΔH of that equals -198.2 kilojoules.1110

So, we have this reaction that we know the ΔH for; we have this reaction that we know the ΔH for.1134

Hess's Law says it doesn't matter how we get to our final reaction, if we can come up (oops, no, we don't want these stray lines here) with a way of manipulating these equations (switching them, multiplying them by coefficients, a lot like you do for linear equations in linear algebra or algebra courses that you have taken).1140

If we can fiddle around with them and add the equations to come up with a final equation--this one that we want--well, we will just add the ΔHs, and we will get the final enthalpy of the reaction, because again, ΔH is a state function; it doesn't matter how you get there, as long as you get there.1166

So, let's see what we are going to do.1184

So, how can I fiddle with these equations in order that, when I add them vertically, I end up with this equation?1186

OK, well, let's see; let's reverse...let's see, what can I do?1194

I am trying to create SO2 gas; so notice that SO2 here is on the right-hand side, but in these equations, SO3 is on the right-hand side.1201

The only equation that has SO2 in it is over here; I want to reverse it, and there is one SO2 here, but there are 2 SO2, so I am going to flip equation 2, and I'm going to divide it by 2.1212

OK, so let me...actually, let me rewrite everything again, so that we have it on one page.1228

We have S + O2 going to SO2; that is the reaction that we want; and we are given S + 3/2 O2 goes to SO3; ΔH equals -395.2 kilojoules.1241

Our second equation (we'll call that #1) is 2 sulfur dioxide gas, plus O2, goes to 2 SO3, and the ΔH for that is -198.2 kilojoules.1262

OK, so we said we want this equation from these equations.1279

We are going to flip this equation, #2; so, we are going to reverse #2 and divide it by 2.1284

When we reverse it and divide it by 2, this 2 SO3 comes to the left, and it ends up becoming just SO3; this 2 SO2 ends up on the right, but becomes SO2; and this O2 also ends up on the right, but it becomes 1/2 O2.1299

Now, what happens to the ΔH?--well, exactly what you think.1315

If you flip a reaction--if you flip an equation--you change the sign of ΔH; if you divide an equation by 2, you divide the ΔH by 2.1318

So now, the ΔH is no longer -198; it is going to equal +99.1 kilojoules.1327

And now, we leave the...here, our other equation is S + 3/2 O2 goes to SO3, so S is on the left--there is one S on the left--so let's leave that one alone.1337

We have S + 3/2 O2 going to SO3; that ΔH stays the same: -395.2 kilojoules.1351

Now, we just add straight down.1367

Everything that is on both sides--if there is something on the left and something on the right, they cancel.1369

SO3 on the left; SO3 on the right; it cancels.1375

S comes down; that is taken care of.1379

I have 3/2 O2 on the left; I have 1/2 O2 on the right; 3/2 minus 1/2 is equal to...well, there you go: O2 (two halves)...+ O2.1383

O2 is taken care of; now, the only thing left is the SO2.1400

There you go; I have the final equation that I wanted by messing around with the equations for which I did have information.1405

Now, all I do is: I just add the ΔHs.1413

When I add the ΔHs: +99.1, minus 395.2; we get -296.1 kilojoules.1416

I used reactions that I knew, manipulated the equations, made the appropriate changes to the enthalpy, and I added straight down, added straight down; now I have the enthalpy for this reaction.1426

This is Hess's Law; I can use reactions that I do know to find a reaction that I want.1439

OK, let's do another example here.1447

Let's do this on a new page.1454

Calculate the ΔH for the synthesis (which means formation) of di-nitrogen pentoxide gas, N2O5, from its elements.1461

OK, elements--so, the reaction that we want is nitrogen gas, plus oxygen gas, goes to N2O5 gas.1489

Now, we have to balance this: so, we have 5 and 2, so we'll put a 5 here; we'll put a 2 here.1505

We'll put a 2 here; now it's balanced.1512

This is the equation that we want; OK, now here are the equations that we have at our disposal.1515

Equation #1: we have H2 + 1/2 O2 goes to H2O; the ΔH of that is equal to -285.8 kilojoules.1521

Our second equation is: N2O5 + H2O goes to 2 HNO3, which is nitric acid; the ΔH of that is -76.6; exothermic, exothermic.1540

3: we have 1/2 N2 + 3/2 O2 + 1/2 H2 goes to HNO3; the ΔH of this is equal to -174.1 kilojoules.1565

So, our task is to take these three equations, manipulate them by multiplying by coefficients, reversing them, and then adding them straight down and arranging them in such a way so that, when they add, they add to the final equation.1585

OK, so let's see what we have; what can we do?1604

I notice I have N2O5 on the right, and here I have N2O5 on the left, so let me just leave that one alone for now.1608

Let me see: I have 2 N2 on the left, 5 O2; I have 2 N2, and the only equation here that has N2 in it is this one, so I'm going to multiply equation #3 by 4.1623

So, multiply #3 by 4, and what I end up with, when I multiply the third equation by 4--I get 2 N2 + 6 O2 + 2 H2 goes to 4 HNO3, and the ΔH also gets multiplied by 4.1638

Negative...so this is multiplied by 4 minus 696.4 kilojoules; so whatever I do to the equation, I do the same thing to the enthalpy.1673

OK, now let's see what I have; HNO3...I have 4 HNO3, but I don't have any HNO3 here, so I need 4 HNO3s on the left.1683

I have 2 HNO3s here, so I'm going to flip this equation and multiply it by w.1695

So, we'll flip 2 and multiply by 2.1699

When I flip this and multiply by 2, I get 4 HNO3 goes to 2 N2O5 + 2 H2O, and now the ΔH for this...I have flipped it, and I have multiplied it by 2, so now this is a positive 153.2 kilojoules.1711

Now, #1--let me see: #1 equation: H2; I have...what do I have?...I have 2 H2s on the left; I have an H2 here; I need to cancel the H2s.1741

I have 2 H2Os on the right; I have an H2O on the right here; so I have used equation 3 and used equation 2; I need to get this on the left, and I need to multiply it by 2, so I'll do the same thing.1755

I'll flip #1 and multiply by 2, and I end up with the following equation.1765

I get 2 H2O goes to 2 H2 + O2; and again, the ΔH is going to be +571.6 kilojoules.1776

And now, I should end up with what I have; so let me see here; let me...we are concerned with this equation, this equation, and this equation; so let's see what cancels.1797

H2O; 2 H2O, 2 H2O; 2 H2, 2 H2, right?--this is on the left of the arrow; this is on the right of the arrow.1815

They are on top of each other, but it is where they are on as far as the arrows are concerned.1824

I have 4 HNO3, 4 HNO3; 2 N2, so I'll bring that down; that takes care of the 2 N2.1830

I have 6 oxygens on the left; I have 1 oxygen on the right; so, 6 minus the 1 leaves 5 oxygens on the left.1841

So, that is taken care of and that is taken care of; now, the only thing left is the 2 N2O5, which is on the right, so I am not adding it.1853

That is on the right-hand side of the arrow; so I get 2 N2O5, which is exactly what we want: 2 N2 + 5 O2 goes to 2 N2O5, exactly what we found.1866

Now, let's just add the ΔHs straight down, and when we add them, we end up with 28.4; again, you might want to check my arithmetic; I'm notorious for bad arithmetic.1880

So, this reaction: nitrogen gas plus oxygen gas to form di-nitrogen pentoxide: it is positive 28.4 kilojoules, so this is endothermic.1893

That means, in order for this reaction to go forward, I actually have to add heat to it.1906

Or, if I leave it alone, and there are other circumstances where there is something happening--there is enough heat that it can pull from the surroundings--it will pull that heat from the surroundings in order to make this go.1911

So, there you have it--Hess's Law.1924

We want to find the enthalpy, the heat of a given reaction; well, if we have other reactions at our disposal that we know the heat for, and we can rearrange those equations in a certain way--we can fiddle with the enthalpies appropriately--add the equations; we will get the equation that we are looking for, and we will get the enthalpy of that equation.1926

OK, thank you for joining us today at Educator.com for the discussion of enthalpy.1947

We will see you next time; goodbye.1953

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

In the last couple of lessons, we have been talking about enthalpy and energy and heat--things like that.0004

Today, we're going to continue our discussion of enthalpy, and we are going to be talking about standard enthalpies of formation; and we are going to be able to use these standard enthalpies of formation to help us bypass Hess's Law.0011

When we are given a reaction that we need to find the enthalpy for, the heat of reaction for (whether it is given off or absorbed), we can just do it by reading off of a table of enthalpies that have been prepared for a whole number of compounds.0024

It's very, very convenient, because Hess's Law, although it is very convenient in the sense that we can manipulate equations and just add them--still, it requires a little bit of work.0039

With standard enthalpies of formation, we have a standard, and then we have some things tabulated, and we can just do some really simple arithmetic and get the answer that we want--very, very powerful.0049

So, with that, let's go ahead and get started.0060

As often with these, we are going to start off with a definition, because I think it always makes it a little bit more worthwhile.0064

So, definition: the standard enthalpy of formation (and, if you have not figured it out by now, when we say "enthalpy," we just mean heat--you can use heat of formation, enthalpy...I actually prefer to use the word "heat"; I don't really care for the word "enthalpy" myself; but there it is) of a compound is the change in enthalpy that accompanies (let me spell this a little bit more appropriately here) the formation of 1 mole of the compound from its elements, with all substances in their standard states.0071

OK, so the standard enthalpy of formation of a compound is the change in enthalpy that accompanies the formation of 1 mole of the compound from its elements, with all substances in their standard states.0184

That means, if I wanted to talk about the enthalpy of formation of liquid water, it would look like this.0195

H2 gas (because hydrogen is a gas in its standard state); and standard state means 25 degrees Celsius, 1 atmosphere pressure--normal; the way you would find it on an average day.0205

Plus O2 gas (these are its constituent elements); they would form liquid, and the ΔH standard of formation--this is the symbol for it, ΔHf for ΔH of formation--this little degree sign at the top means that it is standard conditions, 25 degrees Celsius, 1 atmosphere pressure; equals...some number.0218

This is what it would look like: we calculate the ΔH for the formation of a compound, 1 mole of that compound.0244

So notice, this is not balanced; now it is balanced.0251

1 mole of the product...it is always like this; this is always going to be 1, so we don't really balance this equation in terms of whole-number coefficients.0256

Because this is 1, you will end up with fractions over here.0265

So again, it is a standard; we pick a standard, and in chemistry, the standard is the mole; so that is what it is.0269

Now, let's actually list the standard states.0277

We have listed a couple of them.0279

What we mean when we say "standard states": for a gas, it is 1 atmosphere; that is standard for a gas.0281

For a solution--if we are talking about a solution in standard state--we mean 1 Molar.0291

And again, this is an older notation, an m with a line over it; you are more accustomed to seeing the capital M; Molar, mole per liter.0297

OK, an element...the state of an element is the state the element takes at 1 atmosphere and 25 degrees Celsius, which is roughly room temperature.0305

An element...the standard state of aluminum is metal; the standard state of bromine is liquid; the standard state of mercury--liquid; the standard state of oxygen is a gas.0328

The standard state of sodium is the metal; that is it--an element.0341

OK, now you should notice: standard states--this 1 atmosphere, 25 degrees Celsius, 1 Molar solution--this is not the same as standard temperature and pressure that we talked about with gases.0346

That is 0 degrees Celsius and 1 atmosphere pressure; so don't mix the two--when we are talking about a standard state, we are talking about a particular state that that compound, that element, is taking.0360

We have listed that standard as: a gas under 1 atmosphere pressure; a solution, 1 Molar; an element, the state that it assumes under roughly room temperature (25 degrees Celsius) and 1 atmosphere pressure--in other words, your average day at sea level.0373

OK, now, once again (I can't reiterate this enough): enthalpies of formation are always given, always given per mole of product, because the product is the thing that we are finding the enthalpy of formation for.0389

The enthalpy of formation is the amount of heat either generated (either given off) or absorbed when you form that product from its constituent elements.0418

We just did this one up here.0431

H2 + 1/2 O2 goes to H2O.0435

Now, here is the nice part: the ΔH of the final reaction, the standard ΔH, the enthalpy of reaction (or the heat of the reaction--standard heat of reaction) is equal to the ΔH of formation of the products, minus the ΔH of formation of the reactants, when you add them together.0440

So, I didn't want to use a summation symbol; and this also includes...well, actually, you know what--let me go ahead and write out...instead of using the summation symbol, I'm going to write it out.0481

You are going to notice that I am not a really big fan of symbolism; I think it's nice, but I often think it gets a little bit in the way.0497

I will put "Sum of the ΔH formations of the products" on the right-hand side of the arrow, minus the sum of the ΔHs of formation for the reactants.0504

That is it; this is the standard definition of ΔH.0529

We use the tabulated...at the end of your book, in the appendix, you are going to see a list that is called "Standard Thermodynamic Data."0532

It is going to list a whole bunch of compounds, and you are going to have three columns in there.0540

The first or second--it is usually the first or second column--one of the columns, anyway--is usually the enthalpy, the ΔH.0546

It has a ΔH; it has a little degree--this zero degree--on it; and it has a little f.0553

It is given in kilojoules per mole.0559

Well, another column is going to be ΔG, which is free energy, which we will talk about later.0562

The other one is going to be S, entropy, which is in Joules per mole per Kelvin.0567

But again, those are other thermodynamic quantities that we will talk about a little bit later in the year, when we talk about equilibrium, spontaneity, and things like that.0576

Right now, we are just concerned with enthalpy.0584

That table in the back--that column that lists the ΔHs of formation--that is what you are going to use to tabulate this.0587

It will make more sense in a minute, when we do a problem.0593

OK, a couple of things to remember: let's see, the first thing to remember is that the ΔH of formation for elements is 0, because you are not forming the elements; the elements are already there.0597

It's very, very convenient that it is 0; it comes in very handy.0615

OK, #2: ΔHs of formation are per mole of product formed.0619

So, when you come across a reaction, make sure the reaction is balanced--balance the equation, so that you have the appropriate stoichiometric coefficients.0643

Balance the equation, because again, we are using the ΔH values of formation, but they are per mole of product; but when you are using it in an actual reaction, you may have 3 or 4 or 5 moles, which means you have to multiply the ΔH of formation by 3 or 4 or 5.0659

And again, it will make more sense in a minute, when we do an example, which we are actually going to do right now.0678

Example 1: OK, ammonia is burned in air to form nitrogen dioxide and water; what is the ΔH of this reaction?0686

OK, so let's see what this says; write out the equation.0721

Always start with an equation; this is chemistry; there is always an equation somewhere.0726

OK, so ammonia is NH3; it is burned in air, which means it is going to take oxygen; and it forms nitrogen dioxide, NO2, plus H2O.0732

When we balance this, we end up with 4; with 7; with 4; and with 6.0746

Now, this is our reaction; we want to find the heat of this reaction; we want to find the enthalpy of this reaction--ΔHrxn, the standard ΔH.0754

We are going to use the ΔHs of formation that are tabulated in the back for this, for this, for this, and for this.0767

Don't mistake the two: the final heat of reaction is for the entire reaction; what we are doing is we are using the heats of formation that are tabulated for the individual pieces of this reaction--products and reactants.0774

So, we said that the ΔH of the reaction is equal to the sum of the ΔHs of formation for the products, including coefficients, minus the sum of the ΔHs of formation for the reactants, including coefficients.0787

All right, that equals--well, let me put a circle around the (oops, I actually wanted to use red; there we go--so that we have it here) equation; ΔH.0824

We are going to take the ΔH of formation of H2O, multiplied by 6; add it to the ΔH of formation of NO2, multiplied by 4.0843

From that, we are going to subtract the ΔH of O2, multiplied by 7.0851

And then, subtract the ΔH of formation of NH3, multiplied by 4.0856

Products minus the reactants--the sum of the ΔHs of formation of the products, minus the sum of the ΔHs of the reactants.0864

All right, so now, when we look up 34, 4...so which one shall we do first?...let's do...it doesn't really matter, so let's do the NO2 first.0875

I always like to do it this way; so we have 4 moles (that is a 4) times the ΔH of formation of NO2 from its constituent elements.0888

We look in the back at our thermodynamic tables; it is 34 kilojoules per mole; again, it is per mole of compound.0900

That is what a ΔH of formation is; it is a standard.0907

Plus 6 moles--that is where this comes from--that is where the 6 comes from; this, times the ΔH of formation of water, which is -286 kilojoules per mole.0910

From that, we are going to subtract 4 moles, times the ΔH of formation of NH3, which is -46 kilojoules per mole; plus 7 moles--that is where that comes from--times 0 kilojoules per mole, because again, this is an element--oxygen gas.0926

We said that oxygen gas's ΔH of formation (for elements) is 0.0953

So, it's nice; these are not elements--these are compounds--so they have actual, finite values.0958

Well, when we do this, we end up getting -1396 kilojoules; is this exothermic, or is it endothermic?0965

ΔH is negative; that means it is giving off heat; that means, when you mix NH3 and O2 gas, you produce three things: you produce nitrogen dioxide; you produce H2O, and you produce heat.0976

That is what this negative means; this is exothermic--it gives off heat.0988

Now, notice: this is in kilojoules; moles here have canceled; ΔH of formation is given in kilojoules per mole--we are multiplying by the stoichiometric coefficient in the balanced equation to give us a total for the reaction, as written in the balanced equation.0994

Moles cancel; the ΔH of reaction--standard ΔH of reaction--is equal to -1396 kilojoules.1012

ΔH of reaction is in kilojoules.1024

The ΔH of formation for the individual compounds--that is in kilojoules per mole of compound formed; that is why it is standard--we have standardized it to one mole.1027

OK, now let's see: let me show you what happened in terms of Hess's Law.1040

1/2 of N2, plus 3/2 H2, goes to NH3.1054

The ΔH of formation is equal to -46 kilojoules per mole.1063

H2 + 1/2 O2 goes to H2O; the ΔH of formation is -286 kilojoules per mole.1071

1/2 N2 + O2 goes to NO2; the ΔH of formation is equal to 34 kilojoules per mole.1083

These are just the numbers that I used, that I looked up; and O2--that is just 0; the ΔH of formation equals 0.1095

If I used Hess's Law, I would take these equations; I would manipulate them, flip them, turn them, multiply by things, and add straight down to get the final equation that I wanted.1104

Then, I would add the appropriate changes in the ΔH.1114

Well, a lot of that has already been done for you; again, the process--the steps don't matter.1117

All that matters is where you start, where you begin; because these values have been tabulated, I can just take the products, minus the reactants; that gives me my ΔH of reaction from the ΔHs of formation that are tabulated in the thermodynamic data in the back.1122

That is it; OK, let's do another example, and I think this will suffice to explain the use of ΔH of formation.1140

Calculate the ΔH of reaction, standard, for the thermite reaction (thermite--for those of you who don't know about thermite, it is the very, very, very powerful, very...well, you'll find out in a minute, as far as the ΔH, whether it is positive or negative).1151

The reaction is: 2 moles of aluminum metal, plus 1 mole of iron (3) oxide, goes to aluminum oxide, plus 2 moles of iron.1176

So, when you mix aluminum metal with iron oxide, you end up producing aluminum oxide; you end up releasing iron metal.1201

It's kind of amazing, actually.1212

So now, let's see what we have.1214

Let's go ahead and list some ΔH of formation values: the ΔH of formation of aluminum equals 0 (aluminum is an element); the ΔH of formation of Fe2O3 is equal to -826 kilojoules per mole.1218

The ΔH of formation of Al2O3 is equal to -1676 kilojoules per mole: wow, incredibly exothermic!1245

The ΔH of formation for iron--well, iron is an element: 0.1260

Good--so now, let us take 1 mole times -1676 kilojoules per mole, plus 0.1268

That is the aluminum plus the (yes, that is right...the products) iron, minus the reactants, which is 1 mole times -826 kilojoules per mole, which is the Fe2O3, plus 0.1286

When we do this, we get -850 kilojoules; that is 850,000 Joules.1311

Exothermic is an understatement.1324

Exothermic releases a lot of heat; in fact, it releases so much heat that the iron that comes out is not solid iron; it is actually liquid iron; the iron is melted.1327

OK, so let's see what we have here.1341

Do we need to do anything more?--no, that is it.1347

So, we found our ΔH of the reaction; it is equal to -850 kilojoules; notice, ΔH of reaction is in kilojoules, not kilojoules per mole.1349

Kilojoules per mole is ΔH of formation; that is what we used to get the ΔH of the actual reaction.1361

That is it; that is standard enthalpies of formation--nothing too difficult; I hope that made sense.1369

With that, I will go ahead and stop there.1374

Next time, we are going to talk about calorimetry, the transfer of heat and transfer of energy directly from one object to another.1376

Until then, thank you for joining us here at Educator.com; we'll see you next time; take care.1385

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we're going to be talking about calorimetry; we are going to be finishing off our discussion of thermochemistry.0004

We have talked about enthalpy, and a little bit about Hess's law, and things like that; we are going to finish it off with a discussion of calorimetry today.0011

So, let's just jump on in and see what we can do.0018

OK, so let's recall that, at constant pressure, the ΔH, the enthalpy, is actually equal to the heat of the reaction--which is very, very convenient for chemistry.0022

We do things at constant pressure so, when we talk about the heat and the enthalpy, they are the same.0032

Let's just write that down to begin with: recall that, at constant pressure, ΔH is equal to the q.0038

Remember energy: you can transfer energy to and from a system...or the change in energy of a system--the internal energy of a system, ΔE...there are two ways that energy can be changed.0053

It can either transfer as heat in or out of the system, or as work; in other words, you can do work on the system, or work out of the system.0066

So, q was heat; w was work, and it was usually pressure-volume work; and at the constant pressure, as it turns out, the enthalpy is equal to the heat of reaction.0074

We speak about the heat of a particular reaction: 2 H2 + O2 goes to 2 H2O; there is a certain heat that is given off--it is the enthalpy; it's written as ΔH=...negative for exothermic, positive for endothermic, things like that.0085

Now, let us define something called the heat capacity.0100

No, I'm not going to write "definition"; I just said that it is a definition.0106

Heat capacity: the heat capacity is basically the heat absorbed by a material, divided by the increase in temperature.0110

For example, let's say you had absorbed 20 Joules of heat--some material, we don't know what it is--and the rise in temperature was, let's say, 4 degrees Celsius.0134

Well, of course, you know, when you do division, basically what you are doing with all division is taking the denominator to 1; so, this is going to be Joules per 1 degree Celsius.0147

That is the whole idea behind units; so 20 over 4--you get 5 Joules per degree Celsius.0156

That means, for every 5 Joules of energy that you put into this thing, the temperature of that thing is going to rise 1 degree Celsius.0163

That is a general definition of heat capacity: heat absorbed, divided by the increase in temperature.0170

OK, now, temperature is going to be expressed sometimes in degrees Celsius, degree Kelvin--the scales for degree Celsius and degree Kelvin are the same, so the actual gradation is the same; they just differ by 273; so it's not a problem.0176

Sometimes, you will see J/K; sometimes you will see J/ΔC.0193

All right, now different substances have different heat capacities, and this is what is kind of interesting.0198

Now, as a standard, in order for us to have something to discuss...as a standard, we choose 1 gram of a substance.0204

When we choose 1 gram of a particular substance to apply a certain amount of heat to, to see how much the temperature rises, it is called the specific heat--so not just the heat capacity, but the specific heat capacity.0217

So, as a standard, we choose 1 gram of a substance, and call it the specific heat capacity.0228

Now, heat capacity is Joules per degree Celsius; well, when we choose one gram of that, it's going to be Joules per degree Celsius per gram.0252

The unit for specific heat capacity: the unit is Joules per degree Celsius per gram of substance, which is often written as Joules over gram-degree Celsius; that is the unit of heat capacity.0263

OK, there is also molar heat capacity, and it is exactly what you think it is.0283

It is going to be the amount of energy that I need to...the amount of energy that one mole of something absorbs, and rises by 1 degree Celsius.0295

Or, if I want another way of saying it, if I want to raise the temperature of 1 mole of something by 1 degree Celsius, or 1 degree Kelvin, how much heat do I have to put into it?0305

1 gram; 1 mole; we can work with both; so the unit of that is, of course, Joules per degree Celsius per mole, which is equivalent to (the way you will see it is) mole-degree Celsius.0314

And again, this can be a K instead of a C; it just depends on what the problem is asking for.0333

Specific heat capacity/molar heat capacity: the amount of heat required to raise a particular amount of something 1 degree Celsius.0337

OK, now let's talk about calorimetry.0347

There are two types of calorimetry: there is constant-pressure calorimetry, and there is constant-volume calorimetry.0354

We're going to start with constant-pressure calorimetry.0359

Constant-pressure calorimetry measures heats of reaction (you remember that little rxn; it's a shorthand for "reaction"), or the enthalpies of reaction, for reactions occurring in solution--so in aqueous solution or some kind of solution; it doesn't necessarily need to be water as the solvent--it could be anything else--but something that happens in solution.0366

If I have hydrochloric acid, and I mix it with sodium hydroxide, and I stir it all up to...you know that it's an acid-base neutralization reaction...it is going to give off some heat.0409

Well, in calorimetry, what you are doing is: that reaction is taking place in solution, so the heat that that reaction either gives off or absorbs is going to be reflected in the temperature of the solution.0422

We actually use the solution to measure decrease in temperature, and because we know the particular heat capacity of water, it is an indirect way of measuring the actual amount of heat that is being given off, or released, from the particular reaction.0436

It will make more sense when we do a little bit of a problem.0452

Now, recall that energy is conserved; so, if a reaction gives off heat (exothermic), the solution gets warmer.0456

You know that from experience; if a particular reaction is taking place in solution, the solvent that is containing it is like some sort of a container, if you will.0489

The reaction itself gives off heat; well, the water absorbs that heat, so the heat given off is equal to the heat absorbed.0497

That is sort of the fundamental idea behind calorimetry.0504

Calorimetry is a device that measures heats given off or absorbed, using a particular medium that we know the properties of; and that gives us the heat of a particular reaction.0507

OK, so we measure the rise in the temperature or the decline in the temperature, and then we calculate heat absorbed or released.0521

OK, now, energy released equals energy absorbed.0529

If a system is releasing heat, the surroundings are absorbing that heat.0544

If a system is absorbing heat, it is the surroundings that are giving off that heat; it's a trade-off, and it happens at the boundary.0548

That is why we talk about systems and surroundings and universes.0556

OK, now, this is the particular equation that we are interested in: mCΔt.0561

q is the energy or the heat; so energy or heat...heat and energy are interchangeable.0571

This right here--this is the mass of the solution, or the mass of the particular thing that you are discussing--it could be a solid; it could be a liquid--its mass, generally in grams.0582

This C right here is the symbol for heat capacity--specific heat capacity.0594

Heat capacity: I'll say it's heat capacity, and I'll put "specific," because more often than not, it is specific heat capacity.0601

Remember, we are talking about a gram of something; this is why this mass is in grams.0610

This is going to be specific heat capacity, and this ΔT (I'm sorry, this shouldn't be a small t) is a change in temperature.0614

So, that is the change in temperature; now, mind you, it is not the beginning or the ending temperature--it is the actual change in the temperature.0625

It is the final temperature, minus the initial temperature--that is the definition of Δ.0633

It is the change in temperature; so, if a particular problem says the temperature went from 25 to 30, our ΔT in this case is 5.0639

Or, if they say (a problem could say) there was a temperature increase of 4.7 degrees, our ΔT is 4.7; so this is the change.0648

OK, so once again: mass, times heat capacity, times the change in the temperature, is equal to the energy; let's make sure that the units match here.0658

Mass is going to be in grams; the heat capacity is in Joules per gram per degree Celsius; and the change in temperature is going to be in degrees Celsius.0667

Degrees Celsius cancels degrees Celsius; gram cancels gram, leaving you Joules--sure enough, Joules: heat or energy is in Joules.0679

This equation, right here, is our fundamental equation for calorimetry.0688

Anything having to do with heat capacity, work, the heat absorbed or released by a substance (liquid, solid, whatever) is equal to the mass of that substance, times its heat capacity, times the change in temperature that it experiences.0692

It's fantastic: 1, 2, 3, 4: you have 4 variables--given any three of them, you can find the fourth.0711

Let's say you have the heat, the mass, and the change in temperature; you can calculate the heat capacity.0718

Let's say you have the heat capacity, the temperature, and the q; you can calculate the mass.0723

Whatever it is--like any equation, you can rearrange it to get what it is that you need.0728

You will run into different problems where you have to solve for each one of those, so it isn't just solving for one or the other; this is just how it is arranged.0733

You will rearrange it accordingly.0742

OK, well, let's just go ahead and do a problem, and I think it will start to come together.0745

Let's do an example: we have: 50 milliliters of a 1 Molar hydrochloric acid solution (and this m with a line on it--that is just an older symbol for molarity; it is just something that I'm used to; you will see the capital M--just as a reminder) is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution, NaOH.0755

OK, temperature rises from 25 degrees to 31.9 degrees (this is Celsius).0794

Calculate the heat released...let's see...and the molar heat released.0811

OK, so 50 milliliters of a 1 Molar hydrochloric acid solution is mixed with 50 milliliters of a 1 Molar sodium hydroxide solution; the temperature rises from 25 to 31.9 (let's make this 31; sorry, it looks like a 37 here); we want you to calculate the heat released in this reaction when I mix these, and the molar heat released (in other words, the heat released per formation of 1 mole of whatever it is that is actually reacting here).0836

Let's just write a reaction; this is chemistry--you always want to start with a reaction, if you can, which...more often than not, you are going to be starting with some reaction.0867

We are putting together hydrochloric acid, and we are putting together sodium hydroxide.0875

We have H and we have OH; this is a neutralization reaction.0880

Any time you have an H and an OH, they are going to seek each other out, and they are going to form liquid water.0885

Of course, what you have left over is the sodium chloride.0892

Once you actually write it out, then you balance it; in this particular case, everything is balanced, so our reaction is HCl + NaOH → H2O, which I am actually going to rewrite a little differently, simply because I like to.0895

It's up to you; you can write H2O; I have always been a big fan of writing it HOH, because I know that this H (let me do this in red)--this H comes from here, and this OH comes from here.0908

I like to keep them separate, especially when I am balancing things that aren't 1 to 1; it makes it a little bit easier to balance.0921

So, HOH is perfectly acceptable--unless your teacher says otherwise, but hopefully, they will be OK with it.0928

50 milliliters of a 1 Molar--all right, so let's see what we have here.0935

Well, they want the heat released, so q=mCΔT.0942

Well, we need to find the mass of the solution; so the heat that is going to be released in this reaction (let me draw a little picture here)--this reaction takes place throughout this solution.0952

The solution is 100 milliliters of solution; well, the heat released into it is going to go into the water.0963

The water is going to rise in temperature; that is what is happening here.0970

The mass of the water--well, 100 milliliters; if you have 100 milliliters, times...water is 1 gram per milliliter, so you have a mass of 100 grams.0976

This mass has to be expressed in grams.0988

We have 100 grams of water; and again, heat released is heat absorbed.0991

Heat released by the reaction is heat absorbed by the water, by the solution.0995

Well, we have 100 grams of water; now, the heat capacity of water happens to be 4.18 Joules per gram per degree Celsius.1000

What that means is that, if I have 1 gram of water, and if I want to raise it by 1 degree Celsius, I have to put in (I have to heat it by) 4.18 Joules.1010

It is a very high heat capacity; that is what makes water the extraordinary substance that it does--one of the things is the fact that it has such a high heat capacity.1019

The change in temperature, now: 25 to 31.9--what is that--6.9 degrees Celsius?1027

OK, so now, Celsius cancels Celsius; gram cancels gram; we do our multiplication, and, if I am not mistaken, we end up with (if I have done my arithmetic right) 2884.2 Joules.1034

Now, you can express this as kilojoules if you want; that is going to be 28.9 kilojoules; but I'm just going to go ahead and leave it as Joules here--it's not a problem.1054

That is how much heat is released, based on everything that they gave us.1062

Rise in temperature: the heat released by the reaction is absorbed by the water; this is what calorimetry is about; usually, water is the particular...because we know the heat capacity of water, we can measure it, so this reaction released this much heat.1069

Now, this was part A; now we want to find the molar heat; OK.1086

This is: now we want to find the amount of Joules per mole of whatever it is we are discussing.1093

In this particular case, this particular reaction is a neutralization reaction.1100

The reaction that is taking place is: H+ is getting together with OH- to form H2O.1104

What we want to find is the amount of heat that is going to be released per mole of water formed.1112

That is what this is--as is: with 50 milliliters of a 1 Molar HCl, 50 milliliters of a 1 Molar NaOH, what we have is 28.9 kilojoules.1122

Now, we want to know what that is per mole of water formed.1131

OK, here is where we have to calculate how much is actually formed.1135

Well, 50 milliliters is 0.050 liters; times 1 mole per liter--that gives me 0.050 moles of HCl, which gives me 0.050 moles of H+, right?1141

One molecule of HCl releases 1 hydrogen ion.1159

Now, we do the same thing for the hydroxide.1164

We have 0.050 liters of the sodium hydroxide, and it is also 1 Molar; liters cancel; that gives me 0.050 mol of OH-, because 1 mole of sodium hydroxide releases 1 mole of hydroxide.1167

Well, this is nice: the equation is 1 to 1, so there is nothing limiting here; so every H is going to join with an OH--nothing is going to be left over, and we have 1:1:1.1184

We are not going to have .1 here; we don't add these; it says 1 mole of H+ and 1 mole of OH- goes to 1 mole of water.1197

We're going to form 0.050 moles of H2O.1204

OK, so now, we can go ahead and finish the problem.1213

We said that we ended up releasing 2884.2 Joules in this reaction.1219

Well, the amount of water formed was 0.050 moles; so again, when you divide, you are taking the denominator, and you are turning it to 1.1229

So, this becomes 57684 Joules per mole, which is equivalent to 57.7 kilojoules per mole(when the numbers start to get this large, generally they will express it in kilojoules)--per mole of water formed.1244

So, a particular set of conditions gives you the amount of heat, just in Joules.1270

Then, if you adjust that by putting the number of moles of the thing that you actually created, or that you are talking about, in the denominator, that gives you the molar heat released.1276

There you have it; OK.1289

Now, that was constant-pressure calorimetry; and again, constant-pressure calorimetry takes place in solution, because you are not really changing the pressure of the environment when you are pouring--when you are working in a solution.1295

The pressure doesn't change; the pressure is the atmospheric pressure on that solution, so everything works out really nicely.1307

Well, there is also something called constant-volume calorimetry, and so let's talk about that a little bit.1314

Constant-volume calorimetry: this takes place in something called a bomb calorimeter.1324

So basically, what a bomb calorimeter is (I'll give you a quick...): it is this container, and it is full of water, and inside the container, there is another container.1336

That is connected to the outside; and what we do is: we put some compound in there; we ignite it; and we burn it--we combust it completely--we blow it up, basically.1353

Now, because it is constant volume--the volume doesn't change--we're keeping a constant volume, the pressure rises; the heat rises; the heat escapes into the surrounding medium.1365

Well, what we have done with a standard: we have actually already calculated the heat capacity for the entire calorimeter, which involves the metal, any parts, the solution, the water--everything in there.1382

So, for constant-volume calorimetry, it isn't q=mCΔT; it is actually CmΔT; this Cm actually includes (so I'll put it over here--q=mCΔT)...this Cm is the heat capacity of the entire bomb--of the calorimeter itself.1396

It includes--it has already taken into account--the heat capacity of the water, the mass of the water, the heat capacity of the metal, the mass of the metal...everything is taken into account; so this one is actually going to be Joules per degree Celsius, or Joules per degree Kelvin.1422

This is slightly different; so, for constant-volume calorimetry, this is our equation.1441

The mass is accounted for, which is why they ended up putting it as a subscript on the heat capacity; so now, it's just a symbol reminding you that we are talking about constant-volume calorimetry.1448

The principle is the same: so you are just taking something--in this case, you are blowing it up; you are completely combusting it, seeing how much heat it releases or...well, yes, it's going to release heat.1460

The heat passes through the bomb into the solution; the temperature of the solution rises; we measure that temperature change, and then, we just use this.1470

Same thing; so, let's do an example.1480

OK, let's do this one in red.1484

Example (let's see): 0.1964 grams of quinone (which is C6H4O2) is placed in a bomb calorimeter; the bomb calorimeter is known to have a heat capacity of 1.56 (let me actually write this over here, so I can include the units)...Cm = 1.56 (not 7) kilojoules per degree Celsius.1488

So here, they are actually expressing the specific heat in kilojoules per degree Celsius.1553

Now, the rise in temperature is (I don't know where these lines are coming from; OK) 3.2 degrees Celsius.1559

Now, the rise in temperature--this is the ΔT, right here.1575

Calculate (whoa, that was interesting; we don't want that; OK) the heat of combustion per gram and per mole of quinone.1584

We want to calculate the heat of combustion; we want to calculate the enthalpy of this combustion reaction.1619

Quinone--so, just to let you know, we have this thing quinone--I'm just going to write out the reaction with names.1625

Combustion: Quinone + O2 → CO2 + H2O.1632

All combustion reactions are the same; O2 is the other reactant; they produce carbon dioxide, and they produce water; this is a standard combustion reaction.1638

We want to find out what the heat is per gram of quinone, and per mole of quinone.1647

OK, so we use our constant-volume equation, which is CmΔT.1652

Well, we know what Cm is; it is 1.56 kilojoules per degree Celsius.1665

We want to multiply that by 3.2 degrees Celsius.1673

What you end up with, when you multiply this out: you get 4.99 kilojoules.1678

OK, 4.99 kilojoules is released; now, A: per gram--so 4.99 kilojoules was released, and we said that there were .1964 grams of this quinone, so we take the 4.99 kilojoules, and we divide it by the mass, 0.1964 grams.1686

We want Joules per gram; when you divide by something, you are making the denominator 1.1718

So, you end up with 25.4 kilojoules per gram; that is how much heat is released.1725

If you have a gram of quinone--not a lot--and you combust it, it releases--the bonds in quinone release--25.4 kilojoules of heat; that is a lot of heat.1734

Now, per mole: well, we just have to now change this to moles.1748

Well, we know how to do that: we use the molar mass, grams per mole.1752

We go to 25.4 kilojoules per gram, times grams per mole, the molar mass; and now, we look up quinone--the molecular formula.1755

It ends up being 108 grams per mole; we end up doing the multiplication; we get 2743 kilojoules per mole.1771

108 grams of quinone--that is 1 mole--it releases 2,743 kilojoules (in other words, 2 million, 743 thousand Joules) of heat.1783

That is a very, very, very powerful molecule; there is a lot of energy stored in those bonds.1796

So again, basic calorimetry: constant-pressure calorimetry--let's just see--constant P: you have that the heat comes from mass, times the heat capacity, times the change in temperature.1802

And you have constant volume--Constant V--which is q=CmΔT.1819

The mass has been accounted for; it is just one calorimeter; you have the bomb; everything is accounted for.1828

Now, this equation could be used for anything; as far as calorimetry is concerned, the heat capacity--notice, we have been using this for water, 4.18 Joules per gram-degree Celsius.1834

But, water doesn't have to be the thing that I am heating up; this equation is perfectly valid for anything that you heat up, whether it is iron, concrete...anything at all; water, mercury...1845

Each different substance has its own heat capacity, so there you have it.1857

OK, let's close off with one more problem, and it will sort of tie everything together.1864

OK, a 28.2-gram sample of nickel at 99.8 degrees Celsius is placed in--is dropped in, if you will--150.0 grams of water at 23.5 degrees Celsius.1872

So, the nickel is at 99.8 degrees Celsius; we drop it in water that is at 23.5 degrees Celsius.1909

That is what this says.1915

After thermal equilibrium is reached (in other words, after we allow everything to settle down, where now no more heat is being transferred between the metal and the water)--after thermal eq... (oh, let me write it out, actually--after thermal equilibrium is reached), the temperature of the solution is 25 degrees Celsius.1919

What is C--what is the heat capacity for nickel?1952

So here, we are taking hot nickel and dropping it in water; the heat is released into the water, so the water rises in temperature.1959

Now, from that, we want to be able to measure the heat capacity of the nickel.1969

OK, let's draw a little schematic of what is actually going on.1974

OK, I have the nickel here at a temperature of 99.8; this is actually a good way to do this, because you can actually see it pictorially--what is going on.1978

We have the water, which is at 23.5 degrees Celsius; this is degrees Celsius; now, this one drops in temperature; this one rises in temperature; and at some point, when thermal equilibrium is reached, they are both going to be at 25 degrees Celsius.1988

That is where we ended up; that is the thermal equilibrium.2005

OK, now, energy that is lost by the nickel is energy that is absorbed by the water.2010

So, in other words, the heat that the nickel gives off is equal to the heat that the water absorbs.2023

They are equal to each other; OK.2037

I'm going to run this calculation one way, and then I'm going to do something slightly different--at the end, I'm going to talk about this little negative sign that is going to show up out of nowhere.2041

So, q of nickel is mCΔT, and q of H2O is mCΔT; so now, let me see; let's go ahead and put our values in.2049

We have 28.2 grams of nickel; we are looking for C--that is what we want--and ΔT is 25-99.8 (I don't know if I should do it this way or not; OK, that is fine); 25 minus 99.8--that is equal to the mass of the water, which was 150, and the specific heat of water is 4.18, and the change in temperature is 25 minus 23.5.2077

25 minus 23.5: that is the ΔT for the water--it goes from 23.5 to 25, and this one goes from (the final minus initial) 25 from the 99.8; that is why we have set it up this way.2124

OK, so let's do 28.2C, times (25 minus 99.8), equals 940.5 Joules; when we run through this math, we multiply this by this, divide through; we should end up with .446.2141

OK, this is the heat capacity for the nickel; this is in Joules per gram per degree Celsius.2177

OK, we said that the (so let's stop and take a look at this) heat capacity is not a negative number; this is just a mathematical artifact here, based on how we did the problem.2187

When you are doing this particular kind of problem, what you end up having to do is--because the heat absorbed is equal to the heat released, well, as it turns out, because one is releasing heat, it is losing, and the other one is gaining that heat, so it's positive.2200

So, as it turns out, the sum of those is actually equal to 0; that is the whole idea.2227

From one perspective, the loss is equal to the gain; therefore, to actually make them equal, you really have to stick a negative sign in front of it.2234

Or, you can just run this particular calculation as-is, just leaving everything: mCΔT, mCΔT.2245

Because one is dropping in temperature and one is rising in temperature, the Δ (because it's defined as final minus initial)--one of these is going to be a negative number.2254

That is where this little negative sign shows up; this is an artifact of the mathematics, based on how you did it.2264

If you want to run this same problem and not have to worry about this sign at the end, you can just say the q of one equals negative (the q of the other), because that is what it is.2273

One of them is a positive quantity--it is being absorbed; the other one is a negative quantity--it is being released.2287

It just depends on your perspective.2294

As far as the mathematics is concerned, one is the negative of the other.2296

So, you can go ahead and put the negative sign in here, and then run this calculation; or you can just run this calculation, doing final minus initial for nickel, final minus initial for water--leaving everything as it is, based on the definition of Δ.2300

And then, if you end up with a negative sign, just go ahead and drop that negative sign; the negative sign is a physical feature of the problem--it doesn't necessarily reflect the actual mathematics of the problem.2317

Heat capacity is a positive capacity, so we will just take the .446.2329

As long as you keep that in mind with problems of this nature, where something is dropping in temperature and something is rising in temperature, the final is going to be--one of them is going to be--negative (the ΔTs).2334

We have to account for that negative; you can account for it here, in the beginning, or you can account for it at the end.2345

I hope that that makes sense; we may actually be returning to a particular problem like this, maybe, to talk about it a little bit more, because sometimes there is a little bit of confusion regarding this.2353

OK, thank you for joining us for AP Chemistry, and thank you for coming to see us here at Educator.com.2362

Take good care; goodbye.2367

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

We just finished talking about thermochemistry, and now we are going to move on to discuss a series of topics that are going to form, really, the most important part of chemistry.0003

Not that what we talked about wasn't important--I shouldn't say important; I should say the real core of what it is that you are going to be using in your future work.0015

We are going to be discussing--the first thing we are going to be starting with is kinetics.0025

We are going to talk about kinetics; after that, we are going to talk about equilibrium; after that, we are going to talk about acid-bases and some more aspects of acid-base chemistry.0030

And then, we are going to talk about thermodynamics in greater detail--not just enthalpy, but we are going to talk about free energy and entropy, and then we are going to round it out with a discussion electrochemistry.0040

Afterward, we will actually come back and discuss some of the topics like bonding and solutions and things like that.0052

But, I wanted to get to the heart and soul of what chemistry is--the things that you are going to be using--because we want to spend a fair amount of time with it and get really, really comfortable with it, over and over and over again.0060

Our first jump into this is chemical kinetics.0073

Chemical kinetics is the study of how fast something happens.0078

As it turns out with chemistry, you have thermodynamics, which tells you whether something can happen; and then you have chemical kinetics, which is to tell you how fast something happens.0083

You might have this wonderful, wonderful reaction that works thermodynamically, and it is the greatest thing you have ever seen, but it is just so slow that maybe it comes to completion after four thousand years.0096

There are plenty of reactions like that...so both are important: you have to know that a reaction actually works--will go forward without too much effort on your part--and that it will actually go forward in a reasonable amount of time, because there is the practical issue.0106

So, let's just jump in, and today we are going to talk about reaction rates, and we're going to introduce this idea of rate laws.0119

Much of our discussion today is just going to be getting you familiar with the context--that is what it is going to be about today.0126

OK, let's get started.0136

Let us start with a definition, like we often do.0139

No, I am not going to write the word "definition"; I am just going to write the rate.0147

Well, you all know what a rate is; it is the change in something over the change in time.0153

In this particular case, let's write ΔA over ΔT, where A, in the square brackets, is the concentration of a particular species (it could be any species--molecule, atom, whatever we happen to be on...ion...more often than not, it is going to be something in solution).0160

So, it's the change in concentration; so Δ is the concentration in moles per liter: molarity--pretty standard: M, and you have also seen it as m with a line on it; that is my particular symbol for it.0183

The rate is the final concentration of something, minus the initial concentration of something, divided by the time it took for it to go from the initial concentration to the final concentration.0198

Let's just take, as an example, a particular reaction: we'll write the balanced reaction, 2 moles of nitrogen dioxide decompose into 2 moles of nitrogen monoxide, plus a mole of oxygen gas.0210

Of course, all of these are gases, so let's go ahead and write that, just to be absolutely complete.0227

Now, I'm going to give a table: it's going to be a timetable, a table of time increments.0233

Then, I'm going to give you the concentrations of this, this, and this.0239

Then, we are going to see what is going on here.0242

Let's draw this out; so we have time t: 0, 50, 100, 150, 200, 250, 300, 350, and 400.0245

Now, we have NO2; and again, square brackets always means concentration in chemistry, in moles per liter.0265

NO2, concentration of NO, concentration of O2...so, 0.0100, 0.0079 (it's going to take me a couple of minutes to write this; I apologize), .0065 (I just want to give you as broad a picture as possible), 0.0055, .0048, and then these are 0 and 0...1, 2, 3, 4, 5; you know what, let's just stop with 5 of them.0274

0.0021, .0011, 0.0035, .0018, 0.0045, 0.0023, .0052, 0.0026, and it goes on; but we have a pretty good idea here, so let's talk about what is happening here.0312

This is a particular reaction that was run, and it is a decomposition of nitrogen dioxide to nitrogen monoxide plus oxygen gas.0347

At different time intervals (at 50 seconds, then 100 seconds, then 150, and so on), I actually measure the concentration.0354

Well, at time 0, where we started, I started off with (let me use red) 0.0100 moles per liter of the NO2.0362

Notice, this is 0 and this is 0; in other words, nothing has happened yet.0373

This is right at the beginning of the reaction.0377

So, 50 seconds later, I measure the concentration in the flask; I notice that I have 0.0079 moles per liter; there is 0.0021, 0.0011 of the oxygen.0379

That is all this is saying; I am just tabulating different concentrations of all three species at different times--50-second intervals.0392

100 seconds, 150...at 200, I have .0048--just about half; .0052, .0026.0399

Something you should notice here: notice that this column tends to be half of this column.0410

Well, it makes sense; if you go up here and you look at the coefficient, this coefficient is 1; this coefficient is 2.0416

It is telling me that, for every 1 mole of oxygen produced, 2 moles of nitrogen monoxide is produced.0422

These numbers confirm the stoichiometry.0429

This is a standard sort of kinetic data.0432

You are measuring concentrations, and you are taking time.0438

Well, let's go ahead and see what the graph of this looks like.0441

OK, so let me move over here; now, let's go ahead and draw this graph: it will be something like this.0445

This is the time axis, and this is the concentration axis; I'll just put brackets for concentration.0453

Let's see, let's do 1, 2, 3, 4...let's do 0.0100 here, and let's...so these are in increments of...0.0025, 0.0050, 0.0075; that is right--I know how to do math.0464

Then we have 1, 2, 3, 4, 5, 6, 7, and 8; OK, so now, we are going to take the data that we have, and we are going to graph it.0491

One of the graphs that we have looks like this: it goes down to about something like that; we have another one that goes up to about like that.0507

And then, we have another one which is about halfway--something like that.0527

OK, so this right here--this is the NO2.0533

It starts off at 0.0100 concentration, and as time passes, it decreases; the concentration goes down, down, down, down, down.0539

Now, notice what is interesting to notice about this: this is not a straight line; this is a curve--it starts at a certain thing, but it actually starts to slow down--it starts to stabilize and become...OK, so that is interesting.0547

This up here--this graph--is for the NO.0558

As the reaction starts, there is none of it; the concentration increases as time goes by: boom, boom, boom, boom, boom, boom, boom.0564

Because the stoichiometric coefficient is 2 NO2, 2 NO, this line and this line are actually just copies of each other.0573

As this one drops, that one goes up, and the crossover is going to be completely symmetric.0581

Now, this line over here--this graph--that is the O2.0587

Notice, it is half of the NO, which works with the stoichiometry; for every 1 mole of O2, 2 moles of NO is actually produced.0591

This gives a graphical representation of what we have.0601

Now, let's just find an average rate from, let's say, from t0 to t50; so, average rate that equals Δ...in this particular case, let's do ΔNO2 over Δt.0604

Let's take from 0 to 50--the rate from 0 to 50; it is going to be...Δ is final minus initial, so it is going to be 0.0079 (after 50 seconds, that is the concentration of the NO2--it has gone down), minus the initial, which is 0.0100.0630

The time difference is 50 minus 0; what you end up with is -4.2x10-5 moles per liter per second.0657

So, remember these numbers right here: this is concentration--it's moles per liter.0672

That means that the nitrogen dioxide is actually decreasing, on average, between 0 and 50 seconds--the first 50 seconds of the reaction, it is decreasing at 4.2x10-5 moles per liter per second.0676

That is what this negative sign tells me--it tells me that it is decreasing; it is confirming this.0689

You also know this from the slope; the slope along that curve is negative.0695

OK, now, in kinetics, it is customary to express all rates as positive.0700

So, this negative 4.2x10-5--we will usually express it as a positive rate, so we will actually just negate it, so it will be something like this.0710

Let me go back to blue here.0721

The rate is equal to -ΔNO2/Δt, equals -(-4.2x10-5) moles per liter per second.0725

Now, again, this is just a customary thing; when you get a negative rate, you know that something is decreasing; by now, chances are that you have had calculus or you are in calculus--you have a sense of what a rate is.0744

A negative rate means something is decreasing; a positive rate means that something is increasing.0760

Again, I myself don't particularly care for using positive rates--I like to keep things...I like to use the negative and positive signs, but it is customary, so I just thought that you should know that.0764

Symbolically, it looks like this.0773

This is the symbol that you will see: -ΔNO2 when something is a reactant that is actually decreasing as you move forward in a reaction.0776

OK, now let's see some average rates for different time intervals.0786

Let's do 0 to 50, 50 to 100, 100 to 150; so, let's put together another table here: the time interval and the average rate (which is -ΔNO2/Δt); we are just going to calculate rates for different time intervals.0791

OK, and again, this is just a symbol; that is why this negative is here.0812

From 0 to 50, we already calculated as 4.2x10-4; again, we are keeping it positive.0815

From 50 to 100, when we use the data, we end up with 2.8x10-4.0825

So again, we take the concentration at 100 minus the concentration at 50, divided by 50, because 50 seconds passed.0835

That is what we are doing; and then, from 100 to 150, you have 2.0x10-4.0843

This is going to be in (by the way, the units are) moles per liter per second.0852

If we go from 150 to 200, and then we will go from 200 to 250; we will get 1.4x10-4, and we will get 1.0x10-4.0858

So, at different times along the reaction (from 50 to the next 50, next 50, next 50), notice: the rate of the reactions, the rate of concentration decline--it is not constant.0871

That is confirmed from the graph, of course; if it were constant, you would have seen a straight line, but the graphs don't actually make a straight line--they actually slope down a little bit, so...different values.0886

OK, so not the same--the rate actually decreases.0899

Now, these are average rates: averages...sometimes we want the instantaneous rate at a given time.0904

Instantaneous rates: for an instantaneous rate, what you do is: you take the graph; you draw a tangent line at a particular point that you are interested in; and then you take two points on that line, and you calculate the slope.0913

The derivative is what you are calculating, but it is just the slope of the tangent line; so those of you who have had calculus--it is just--you know that that line...the slope represents the derivative of that graph, if you were to have an equation for it.0936

More often than not, you will not, because you are dealing with raw data.0951

You are going to take the data, make the graph, draw the tangent yourself, and then estimate.0954

So, for an instantaneous rate, we form the tangent line to the graph at (t, f(t)), or (t, concentration)--that is what f(t) is--it's the concentration--so maybe I should write that.0961

Time t, and the concentration of whatever species you have to be dealing with...then, calculate the slope.0991

That will give you an instantaneous rate: calculate the slope.0999

OK, so let's do something like that.1007

Let's take our graph, and we have a particular curve, and then we are going to draw a tangent line to that; let's say we choose 100--that is, we want to know what the instantaneous rate is; what is the rate of the reaction at t=100?1011

OK, well, you draw the tangent line; you pick a couple of points on that tangent line; you see where they hit the y-axis; and then, of course, you see where they hit the x-axis.1034

You take Δy/Δx (in other words, that is x2; that is x1; y2, y1); you just take y2-y1/x2-x1 from points that you pick from the line that you drew (not other random points).1050

The point that it passes through--the tangent--is at the point that you are interested in; so, you will get (when we do this calculation based on the data that we have earlier--this is just a small portion of the other graph around 100)...the rate ends up being 2.4x10-5.1072

And again, it is going to be moles per liter per second.1094

So, at 100 seconds, that is the rate of decrease of nitrogen dioxide, based on the particular data that we collected.1097

OK, well, notice: we did the reactants; we had NO2 going to NO + O2; balance it.1106

Well, what if I wanted to express the rate with respect to the products, either nitrogen monoxide or oxygen gas?1118

I can do that, and it ends up looking something like this.1126

Rate of NO2 consumption: well, I know that that equals the rate of NO production; it is also equal to twice the rate of O2 production.1130

That is what the stoichiometry tells me--that the rate of this is twice as fast as this, because the ratio is 2:1.1158

Well, the symbols are -ΔNO2/Δt; this is a symbol for the rate.1165

Again, we put a negative sign here so that everything is positive, because this Δ is going to be negative; so negative, negative makes this positive.1173

Well, that is equal to ΔNO/Δt; that is equal to twice the ΔO2/Δt.1182

That is it; it doesn't really matter; more often than not, we tend to work with reactants--the depletion of the reactants.1193

If you wanted to, you can express it as the increase of the products, but again, most of the time, we'll be talking about (it's just sort of standard, at this point, to talk about) the depletion of reactants.1202

OK, now, let's talk about what it is that is actually going on here, and how it is that we can derive some sort of a rate law, some sort of a mathematical expression.1217

This is a symbol, and we have some data; how do we take some of this data and convert it into an actual, mathematical formula that expresses just how fast the rate is changing?1228

These right here give me rates of change of concentrations.1241

OK, if we take the reaction 2 NO2 → NO + O2, well, when you start the reaction, there is no nitrogen monoxide, and there is no oxygen gas.1248

So, the reaction is just going to fall forward as fast as it can.1265

Well, in chemistry, many reactions...some go to completion all the way, but at some point, you actually do reach a point called equilibrium (which we will talk about later in more detail), but understand that this reaction is actually reversible.1268

At some point, enough nitrogen monoxide has formed, and enough oxygen has formed, so that now the reverse reaction starts to take place.1282

Well, so now you have the forward reaction of the decomposition of nitrogen dioxide, and you have the backward reaction, the reverse reaction, of the formation of nitrogen dioxide.1290

Things can get kind of complicated here.1300

So, what we do when we run these kinetic experiments is: we try to measure concentrations as close to t=0 as possible.1303

In other words, start the reaction, and immediately take all of the measurements that you need, in order to find the rate of the reaction, before this reaction has had a chance to get to a point where the reverse reaction becomes significant--where it starts to change the actual depletion of the nitrogen dioxide.1314

If I measure the rate of the depletion of nitrogen dioxide right at the beginning of that calculation, I don't account for how fast...I don't have to worry about when the reaction has run a little bit, and now the reverse reaction is going to affect the rate at which this depletes.1333

It changes things; it is very, very complex to deal with that mathematically and physically.1353

What we try to do is: we try to measure reaction rates right at the beginning of reactions.1358

When we do that, as it turns out, under these conditions, the rate is actually proportional to the concentration of the reactants, raised to certain powers.1366

So, in the case of NO2, as it turns out, the rate of the reaction--the actual rate of the reaction (this is just a rate of consumption and depletion; that is not the rate of the reaction) actually ends up being proportional to...raised to some power.1384

The rate is equal to some constant, K, times the concentration of nitrogen dioxide, raised to some power.1404

Now, this K and this n have to be determined experimentally.1416

I can't just read it off from the equation.1422

Let me say this again; let me move to another page and write this again.1425

Now, as it turns out, the rate of a reaction is proportional to some power of the reactant concentrations.1429

If I have some species, A, going to B + C, the rate of the reaction, which is symbolized as -ΔA/Δt, is equal to some constant, K, times the concentration of A raised to some power.1465

This right here is what we call a differential rate law.1486

Now, what if you had something that had 2 reactants, A + B, going to...I don't know, C + D.1491

Well, now the rate law (and you can choose any one of these to actually symbolize it; let's just go with A again): -ΔA/Δt (this is just the symbol for the rate) is equal to some constant, K, times A to some power, times B to some power--because now, both reactants are involved.1498

Again, if you have 3, you will have 3 terms; it is just like that.1528

So again, the rate of a reaction is proportional to some power of the reactants' concentration.1532

Depending on whether you have one reactant or two reactants...if you have 1 reactant, you have 1 term like this; if you have 2 reactants, you have 2 terms like this...1537

This is called a rate law; specifically, it is called a differential rate law.1549

Now, again, this K, this n, and this m--they have to be determined experimentally; we have to run the reactions, and we have to work this out.1554

We will actually do this in the next lesson; we will actually talk about how to derive this rate law; it is actually quite nice--quite easy.1562

OK, now, let's write: these are differential rate laws.1571

We also have something called integrated rate laws, which we will also talk about in, not the next lesson, but the lesson afterward.1591

These are called differential rate laws.1596

It is when the rate of a reaction is expressed as some constant, times the concentration of the reactant, raised to some power.1600

If there is 1 reactant, that is 1 thing; if there are 2 reactants, you include those; if there are 3 reactants, you include all of those.1609

You would have An, Bm, Cs power...something like that, times some constant.1615

What is important is: the K, the n, and the m are determined experimentally.1621

You can't just read them off from the stoichiometry.1625

These coefficients in the actual equation have nothing to do with these numbers.1628

OK, now, let's see: let's actually go through a sort of basic example of determining a particular rate law for a particular reaction.1634

Let's start with a reaction: 2 N2O5 → 4 NO2 + O2.1647

OK, now let me just put in carbon tetrachloride; so this happens in carbon tetrachloride solution.1661

I'm actually going to...you know what, let me actually write the equation again up here.1670

2 N2O5 → 4 NO2 + O2, in CCl4.1679

OK, so now we have some time measurements, and we have some concentration measurements: N2O5...1692

0, 200, 400, 600, 800, 1000, 1200, 1400; we have 1.00, 0.88, 0.78, 0.69, 0.61 (not 71), 0.54, 0.48, 0.43.1702

OK, so we have this set of data that we collected at different time increments; we measure the concentration of the N2O5, and sure enough, it is decreasing.1750

Now, when we graph this (I'll just do the graph right here), we are going to end up with something like...if this is 1, start off with 1; let's go ahead and put 0.2 down here, and let's put 400, 800, 1200, 1600...so, it's going to look something like this.1763

This data is just plotted on a graph: 400, 800, 1200, 1600--nice and simple; nothing particularly complicated.1791

This is time; this is concentration of the species; OK.1805

So now, what we want to do is: we end up actually...so now, I'm going to leave this here, and I'm going to move to the next page...when I take that graph, and I pick a couple of points, I can actually measure the concentration of N2O5 at particular points.1810

When I do that, I have the following (I can actually measure the rate, the instantaneous rate; remember, you draw tangent lines at a given point; you measure the rate at that point; you measure the rate someplace else)...1836

So, at a concentration of 0.90 moles per liter, the rate ends up being 5.4x10-4.1857

0.45: 2.7x10-4; now, notice what has happened here.1871

At a certain concentration, I measured the rate of the reaction with a tangent line.1884

At another concentration, I measured the rate; but notice the relationship between these two--I did this specifically.1891

This is half of that; well, this is half of that; what happens here--the correlation--this also happens to be half of that.1899

So, when you have something like that, where when the concentration is halved, the rate is also halved (it won't always happen like this; this just happens to be this particular data that we did)--we arranged it so we can check the rate at concentration that is easily comparable.1911

.9; .45; this is twice that; this is half of that; as it turns out, when I halve the concentration, the reaction rate is also cut in half; that is a linear relationship--that is the whole idea of it.1946

If you double something, and the thing that you are measuring also doubles, that is a linear relationship.1963

If you double something, and the thing that you are measuring goes up by a factor of 4, that means it is a quadratic relationship.1970

2, 4; 3, 9; here, you have halved it or doubled it, depending on which direction--how you want to look at it.1977

Here, you have doubled something; the difference has doubled; so it is a 1:1 relationship--that means it is a linear relationship: linear means an exponent of 1.1987

So, our rate law is: this is the symbol when we write the rate; in fact, oftentimes, you don't really need this symbol; I know that this symbol actually confuses a lot of students--it doesn't need to; I don't particularly care for it very much.1998

It is just the symbol--it means rate; for all practical purposes, you can actually leave this off and just write "rate =."2018

We said that the rate equals some constant (K), times the concentration of the reactant (N2O5), raised to some power (n).2026

Well, here, because the relationship is linear and I read it straight off, when a concentration is halved, the rate is halved; or when a concentration is doubled, the rate is doubled.2037

That is a linear relationship; that means n equals 1.2049

So, I don't need this; so this is the rate, equals some constant (K), times N2O5.2055

That means the rate of this reaction, based on the data, is actually some constant (K), raised to the concentration of the di-nitrogen pentoxide, raised to the first power.2071

This is the differential rate law for this particular reaction.2083

The most important thing to take away from this is that, under conditions where we actually measure rates, near points where the reaction is just starting, where we have a concentration and then another concentration: we have the rate at one concentration and the rate at the other concentration.2101

When we see how the concentration changes, and then we compare it to how the actual rate changes, that gives us the exponent of the particular concentration term of the reactant.2128

That n is called the order--the order of the reaction.2140

This is a first-order reaction; so the decomposition of di-nitrogen pentoxide is a first-order reaction.2148

Now, I'm going to go ahead and stop this lesson here; I just wanted to give you sort of a brief run-down about how this works.2155

Next lesson, we are actually going to get deeper into this, and we are going to calculate differential rate laws; we are going to calculate the exponents; and we are going to calculate the rate constants more systematically.2161

But, I just wanted you to see and get a sense of what this looks like and how things operate.2173

Thank you for joining us at Educator.com; we'll see you next time for a greater discussion of kinetics; goodbye.2178

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to continue our discussion of reaction kinetics, chemical kinetics.0004

In the last lesson, we introduced rate laws--in particular, the differential rate law.0010

Let's just recap what that is, and we'll get into the actual method (or one of the methods) of determining that differential rate law for a particular reaction, given a particular set of data.0015

The method is called the method of initial rates; let's go ahead and get started.0029

We said that (you know, I think I'm going to use black ink now) the rate of a reaction, which is how fast it is going, symbolized by -Δ some species (reactant species), where this negative comes from...Δt is equal to some constant, times the product of the reactants (however many there are), raised to (A, B...let's say there are three reactants) certain powers.0035

These powers are called the order of that reactant.0079

n, m, s...the order of the reactant; this m, s, and n, and K, are experimentally determined.0085

The idea is this (we talked a little bit about it last time, but let's be more specific about it)--the idea is: we want to start a reaction, and we want to measure the rate of that reaction, the rate of depletion of this particular thing (the reactant) as quickly as possible, before any of the products have had a chance to build up and start working in reverse, because as product builds up, the reaction goes backwards.0103

That is the thing; chemistry, as it is going one way, is also going the other way; so, before it becomes too complicated, we want to see if we can measure the rate of the reaction--before anything else gets in the way.0131

Let's just write this down--so, the idea is this: we run several experiments with different initial rates...not initial rates; I apologize--initial concentrations.0144

We pick specific concentrations to start with--reactants and products--and we measure it immediately.0171

We start with different initial concentrations and measure the initial rates (that is what we are measuring--initial rates) before any products have an opportunity to build up and complicate matters.0179

OK, we then compare the rates of the different experiments...we then compare the rates among experiments...to see how the rate depends on the concentration (or specifically, the concentration change from one experiment to the other).0228

It depends on the concentration of a given reactant.0272

This will make a lot more sense when we actually do our first example, which we are going to do right now.0277

We are going to run a series of experiments--2, 3, 4, 5 of them--and we are going to modify concentrations (initial concentrations).0283

So, if I have an Experiment 1 and Experiment 2, and initial concentration 1 and initial concentration 2, I'm going to measure the rate; and depending on the concentration changes, I am going to compare the rates, and I'm going to see what the relationship is (if it doubles; does it go up by 4? does it go up by 6? does it go up by 19? 14.3?), and that is how I get those exponents.0295

That is how I get these numbers first.0318

Then, when I am done with that--once I get the orders of the reactions--then I can take any one of the data in the experiment and put it in here to find K.0320

Let's go ahead and do that; it will make a lot more sense.0329

OK, so let's take (you know what, I think I'm going to start with a fresh page here)...let's do the following reaction.0332

Let's take ammonium ion, plus the nitrite ion; when I mix those together, I am going to end up with nitrogen dioxide gas, plus 2 moles of liquid water.0341

Ammonium and nitrite ion forms nitrogen dioxide gas, which bubbles off; and I am left with liquid water.0361

Notice, I have two reactants--two reactants; I did three experiments on this; here is my data.0372

Experiment #1, Experiment #2, Experiment #3 (you know what, I probably don't need to make them this big--leave myself some room here--Experiment #2, Experiment #3)...0390

Now, I have NH4+ concentration; NO2- concentration; and I have initial rate, which is what I am measuring.0403

Now, my first experiment: I started (remember, we are dealing with reactants here; we are measuring the rates before any products have a chance to build up)...0.100 moles per liter, and here I started with an NO2- concentration of 0.0050 moles per liter.0418

When I ran this experiment, right as it started, I measured the rate to be 1.35x10-7.0441

Now, I ran a second experiment: 0.100; I left this concentration the same, but I doubled this one: 0.00...no, I doubled this concentration: 0.0100.0450

When I make a change from experiment to experiment, I change one species--I don't change everything.0467

The reason is because I need two experiments to compare; later, what I'm going to do is...notice, I have left this the same, and I have left this the same, but I have changed this; I'm going to use experiments 1 and 2 to compare, to find out how it's related to nitrite ion concentration--how the rate depends on this concentration.0473

I change one species at a time; I don't change all of them.0493

I left this the same and changed this one; when I do that, I ended up with a rate of 2.70x10-7.0496

Now, Experiment 3: 0.200; now, I doubled the ammonium concentration, and I left this concentration the same.0505

When I do that, I ended up with a rate of 5.4x10-7.0517

So, my first experiment; second experiment; third experiment: .1, .005, .1 (wait, I'm sorry, this is .1, not .01--I knew I had too many...I was going to say it looks a little odd); .100, and this is .100; that is right, .2.0523

OK, so .1, .1 initial; initial rate; and then, I do another experiment, and I start with .2 mol of this, .1 mol of that, and I measure an initial rate, 5.4.0548

Now, I can do my comparison.0561

OK, so here is where the good stuff starts.0563

Now, we said that the rate, which we will symbolize as -Δ (and I can pick any one--I am just going to go ahead and pick this one, so this is just a symbol); it's equal to some constant (K), times the NH4+ raised to some power (m--actually, let me use n, because I want to...); and the other is NO2-, raised to the power of m.0566

So this is it; this was our assumption: under these conditions of initial rate, as it turns out, the rate is dependent on the product of the concentrations of the reactants, raised to different powers.0601

I write this, and now I am going to use this experiment to determine m, to determine n and determine K.0615

OK, now, Experiment 1 says that the rate is equal to 1.35x10-7.0622

Well, the rate is equal to K, times the concentration of NH4+, which in Experiment 1 is 0.100 to the m power, and the concentration of NO2-, which is 0.0050 to the n power.0641

I just literally plug in these numbers into what I wrote down.0662

Experiment 2 says (I'll say this is rate 1, and this is rate 2, of course): rate 2 is 2.70x10-7, and that is equal to K times...now I use the values for Experiment 2: .100 to the n power, times 0.100 to the m power.0669

OK, now I compare the two rates.0698

Rate 1, divided by rate 2, is equal to 1.35x10-7, divided by 2.70x10-7, equals--this over that equals--K times 0.100 to the n, times 0.0050 to the m (there is a lot of writing here, but it is important to see everything), over K times 0.100 to the n, over 0.100 to the m.0704

Now, look what happens: 1.35x10-7, over 2.7x10-7 (let me make this a little more clear here--this looks a little confusing...10-7)...this is equal to one-half.0742

K cancels K; .100 to the n, .100 to the n--those cancel; what I end up with is .0050 to the m, over .100 to the m.0758

This is just .0050, divided by .100, to the m power.0772

Well (oh, I don't know where these lines are coming from!), it equals...this is just one-half to the m power.0780

Well, 1/2 equals 1/2 to the m power; that implies that m is equal to 1.0796

So, because m is equal to 1, that means the nitrite ion concentration, m, is 1; that means it is an order 1 in nitrite.0807

So now, I will do the same to find n; so I have already found m--now I am going to do the same thing for n.0829

This time, I am going to compare Experiment 2 and Experiment 3.0836

Let's go back to black.0841

OK, Experiment 2: the rate is equal to 2.7x10-7; it is equal to K times 0.100 to the n, times 0.100 to the m, over (oh, actually, no, not yet).0845

OK, and then we will do Experiment 3: the rate is equal to 5.40x10-7, is equal to K times 0.200 to the n, times 0.100 to the m.0880

Well, rate 2 divided by rate 3 (and you can do it in either order--you can do rate 3 over rate 2; it doesn't really matter) equals 2.70x10-7, over 5.40x10-7, equals K times 0.100 to the n, 0.100 to the m, divided by K times 0.200 to n, 0.100 to the m.0900

K cancels K; that cancels that; that is equal to 1/2; that is equal to 1/2, this time to the n power.0939

Well, 1/2 to what power is equal to 1/2?--that implies that n equals 1.0949

I found my differential rate law.0956

OK, now my differential rate law is this: my rate is equal to some constant (K), times the concentration of ammonium ion raised to the first power, times the concentration of nitrite raised to the first power.0960

The rate is first-order in each; the total order of the reaction is 2--you add up the orders.0978

So, m is equal to 1; n is equal to 1; their sum is equal to 2, so the overall order of this reaction is 2.0986

That is what I have done--I have compared rates to find this--but I am not done.1001

Now, I want to find K.1004

Now that we have our rate law, that the rate is proportional to the concentration of ammonium raised to the first power, times the concentration of nitrite raised to the first power, I can take any one of those experiments, 1, 2, or 3, plug in the values (I have a rate; I have this concentration; I have that concentration), and I can solve for K, and that is exactly what I am going to do.1005

I'm just going to go ahead and choose Experiment 2.1030

Experiment 2 says: the rate is 2.70x10-7, equals K times (oops, I don't have to actually do the brackets when I'm putting numbers) 0.100 raised to the first power, and the 0.100 raised to the first power.1034

And now, I just go ahead and do my division.1070

OK, let me see here; .1; .01; minus and minus; and I end up with K is equal to 2.70 (please check my arithmetic; my arithmetic is notorious for being wrong) times 10 to the -5.1079

There we go; and the units, in this particular case, is liters per mole-second.1098

The reason is because we have a rate, which is moles per liter per second, and then you are dividing by moles per liter, dividing by moles per liter, so this K ends up with liters per mole per second.1106

K--this is not the unit for all K's; this is just the unit for K for this particular reaction, which is second-order--first-order in both of these.1125

OK, method of initial rates: we found the rate law, and then we used one of the experiments to find K, and then we plug it back in, so our final answer is: the rate equals 2.70x10-5 times the concentration of NH4+, times the concentration of NO2-.1137

That is our differential rate law for this reaction.1164

We are done.1167

OK, let's do another example.1168

This time, we are going to use three reactants.1172

This time, we will do bromate ion, plus 5 bromide ion, plus 6 hydrogen ion, forms 3 molecules of bromine plus 3 molecules of water.1182

3 reactants: our rate law is going to have 3 things in there.1202

OK, you want to determine the order of each reactant; you want to determine the overall order; and we want to determine the rate constant.1207

That is our task, OK?1219

Let's write it down: determine the order of each reactant, overall order, and rate constant.1221

OK, well, here is our experimental data...well, let's go ahead and write the rate law first--the general one.1242

The rate is equal to some constant (K); it is going to be that raised m power, Br- raised to n power, H+ raised to s power.1249

This is the general form; we are going to find m, n, and s, and we are going to find K, based on the following data.1268

So, we have four experiments; let's see...1274

Experiment: BrO3-, Br-, H+, and initial rate...1284

All right, Experiment 1; 2; 3; 4; 0.10, 0.10, 0.10; initial rate is 8x10-4.1296

0.20, 0.10, 0.10; notice, I have only changed one thing in this case--the concentration of the bromate: bromate, bromide, hydrogen ion.1314

We get 1.6x10-3--that is interesting; OK.1328

0.20, 0.20, 0.10; this time, from here to here, we have changed the bromide.1336

We end up with 3.2x10-3.1344

The fourth one: 0.10, 0.10, 0.20; we have gone back--that is the same; that is the same; that is what is different.1349

We end up with 3.2x10-3.1362

OK, so in order to find m, which goes with the bromate ion, I'm going to use Experiments 1 and 2, because bromate changes from Experiment 1 to 2.1369

Let's go ahead and do that.1384

I'm just going to go ahead and write it out without writing absolutely everything.1387

So, rate 1, divided by rate 2, equals 8x10-4, divided by 1.6x10-3, equals...well, the rate is equal to that whole expression--is equal to K times (and now, I'm going to stop using brackets; I'm just going to use parentheses; but I do mean concentrations) concentration 0.1 to the m; 0.1 to the n; 0.1 to the s, over K times 0.2 to the m, 0.1 to the n, 0.1 to the s.1390

We have a whole bunch of cancellations; the only thing that doesn't cancel is this.1444

So, this number is equal to 1/2, equals 1/2 to the m power; that implies that m equals 1.1450

Therefore, the bromate power is 1; we are done with that.1465

Now, let's compare rate 2 and rate 3; in other words, Experiment 2 and Experiment 3.1473

Rate 2, divided by rate 3, equals 1.6x10-3, over 3.2x10-3, equals K times .2 to the m, .1 to the n, .1 to the s, over K times .2 to the m, .2 to the n, .1 to the s.1480

Cancel, cancel, cancel, cancel, cancel, cancel; we end up with 1/2 equals 1/2, this time, to the n power, which implies that n equals 1.1517

Well, n equals 1; that is the bromide ion concentration.1530

So, bromide is also order 1.1533

OK, now we will do rate 1 compared to rate 4.1537

Rate 1 is 8x10-7 (is that right?); no, it's 8x10-4, not -7.1545

8x10-4; I was going to say: that is a little bit too much.1559

8x10-4, divided by 3.2x10-3, equals K times .10 to the m, .10 to the n, .10 to the s, over K times .10 to the m, .10 to the n (not s yet--.10 to the n--wow, symbols everywhere), .2 to the s.1565

Cancel, cancel, cancel, cancel, cancel, cancel; we end up with 1/4 equals 1/2 to the s power.1602

Well, 1/2 raised to what power gives me 1/4?--that implies that s equals 2, so the hydrogen ion concentration has an order of 2.1613

There we have it: we have our rate, which is equal to (I'm going to go ahead and use one of them as a symbol) -ΔBrO3-, over Δt, just a symbol, is equal to some constant, K.1625

BrO3- to the first power, Br- to the first power, H+ to the second power: this is our differential rate law.1648

You can just...that; this is just a symbol that says "the rate is."1659

The rate of depletion of bromate is equal to some constant, times this, this, this, squared.1666

That is what is going on.1673

Now, I take any one of the experiments; I put the values of the experiments in--that one, that one, that one--and I have the rate, because I calculated the rate--it's part of the data--and I solve for K.1674

So, I'm going to go ahead and take...oh, let's just take Experiment 1; it's not a problem.1686

OK, so I get 8.0x10-4 equals K times 0.10, times (that is the bromate concentration) the bromide concentration--is 0.10, and this other one is 0.10 squared.1692

When I solve for K, I get 8; in this particular case, it is liters cubed, over moles cubed, second.1711

And again, the unit doesn't really tell you all that much; it just tells you what is going on up here.1722

But, the real thing--this number is what is important.1727

The order of bromate is 1; the order of bromide is 1; the order of hydrogen ion is 2.1731

The overall order is 1 plus 1 plus 2; the overall order is 4.1739

The rate constant is 8.1743

OK, that is the method of initial rates: you write down "equals constant K, times the particular species, raised to..." and so on; it's usually not going to be more than 2 or 3; I think 3 is about the most that you are really going to get, as far as reactants are concerned.1749

Then, you take the data that has been collected: Experiment 1 has initial concentrations for each of the reactants, and a measured rate.1768

You compare the rate of one with the rate of the other by putting in values based on this equation, just like we did to derive those numbers.1778

Once you have the orders, you use them, plus one of the experimental values, concentration, concentration rate, to find K.1787

And now, you have your final differential rate law.1798

So, in this particular case, we get that the rate is equal to 8 times that, that, that: the rate of the reaction.1802

Now, I can plug in any concentration I want, randomly: OK, .26, 19, whatever; and based on this that I derived from experiment, I can tell you what the rate of the reaction is going to be--how fast it is going to be.1818

OK, so that was the method of initial rates; we did a couple of examples; next lesson, we are going to talk about the integrated rate law.1836

This was the differential rate law.1843

Thank you for joining us here at Educator.com; we will see you next time; goodbye.1845

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

In today's lesson, we are going to be talking about the integrated rate law and reaction half-life.0004

Last time, we discussed the differential rate law that says that the rate of a reaction is proportional to the concentration of reactant, to some power.0009

We used some concentration-time data to actually work out what that rate law was.0021

Now, what we are going to do is: we are going to use calculus; we are actually not going to do the derivations, but using the techniques of calculus, we can take that differential rate law; we can convert it to an integrated rate law; and then, instead of the differential rate law (which expresses rate as a function of concentration of reactant), we are going to express concentration as a function of time.0027

Let's jump in and see what we can do!0049

OK, so, as we said, the differential rate law expresses rate as a function of reactant concentration.0053

So, we had something like this: the rate equals (and there is this interesting symbol)--let's just use di-nitrogen pentoxide, N2O5/Δt.0059

And again, this is just a symbol expressing rate; and this negative sign is because, in chemistry, they prefer to have rates being positive.0073

Because a reactant is diminishing, the final minus the initial concentration is going to give you a negative value here, so a negative and a negative gives you a positive.0080

I wouldn't worry too much about this; what is important here is this thing--this K, times the N2O5; and in this particular case, it would be some value of n, and we actually calculated it before in a previous lesson: n=1.0088

So, this is a first-order reaction.0105

The decomposition of di-nitrogen pentoxide is a first-order reaction.0107

Well, the integrated rate law (this is the differential rate law) expresses concentration as a function of time, which is very, very convenient.0112

Given a certain amount of time--10 seconds, 50 seconds, 2 minutes--what is the concentration at that moment?--very, very convenient: as a function of time.0139

Now, the differential in differential rate law (or what we call just the rate law--when you hear "the rate law," they mean the differential rate law--they mean this thing)--the "differential" part comes from this symbol, this Δ; in calculus, they become differentials.0149

They become something that looks like this: d...let's say dA/dt.0165

When you fiddle around with this expression, and then you perform an operation called integration on it (the symbol for which is something like that), that is why it is called the integrated rate law; that is where these names come from.0170

If you go on in chemistry, you will understand--you will actually do the derivations; it is actually quite simple, but we will skip it for our purposes.0181

OK, so given a differential rate law, we can find an integrated rate law.0187

Or, if we have an integrated rate law, we can actually go backwards and find a differential rate law.0193

So, since we are going to deal with first-order reaction, let's write a reaction: aA decomposes into products; and the products themselves are not altogether important--what is important is that we have a single reactant on the left side of the arrow.0198

So, the rate is equal to (I'm going to skip the symbol; I'm just going to say) K, times the concentration of A, to the first power.0216

When we fiddle with this thing (actually, you know what, let me go ahead and write...it is actually pretty important: -ΔA/Δt; we might as well be consistent): K times the concentration of A to the first power; when we integrate this expression, we end up with the following.0229

We end up with: the logarithm of the concentration of A equals -K, times t, plus the logarithm of the initial concentration.0257

This (oops, try to get red ink here...red ink) is our integrated rate law.0272

For a first-order reaction, this is our differential rate law, and this is our integrated rate law, OK?0281

This says that the logarithm of the concentration at any given time, t, is equal to minus the rate constant, K (which is the same K here), times the time, plus the logarithm of the initial concentration.0287

Now, take a look at this equation here: this here (let's just call it y)...let's call this -K m; this t is your x, plus b.0303

So, this equation--the integrated rate law actually takes the form y=mx+b.0315

As it turns out, if you have some time and concentration data (which we will do in a minute), if you have a certain time and you are measuring concentrations at different times--if you actually plot, not the time and the concentration data, but if you plot the logarithm of the concentration, versus the time (the time on the x-axis, and the logarithm of the concentration); if you end up with a straight line, that actually tells you that this is a first-order reaction.0322

Just given some straight kinetic data (time, concentration), you plot the time on the x-axis and the logarithm of the concentration on the y-axis; if you get a straight line, that tells you that you have a first-order reaction.0350

That tells you you could write this and this; it automatically gives you the integrated and the differential rate laws.0363

That is what is so great about this particular one.0370

It is very practical, because again: at any given time, t, you can measure the concentration of your particular reactant.0372

OK, so let's go ahead and do an example, and I think it will make more sense, as always.0381

Let's go back to...no, let's keep it red; not a problem.0388

Our example will be: The decomposition of di-nitrogen monoxide...di-nitrogen pentoxide; I'm sorry...at a constant temperature (at a constant t) gave the following kinetic data.0394

OK, so we have--let's see: t is going to be in seconds, and then, of course, our concentration...0426

Now, if you don't mind, I actually tire of doing the whole brackets for concentration; I just like put parentheses for concentration; I'm going to sort of go back and forth between them--I hope you understand what it means.0433

When we are talking about kinetics, a parentheses around a species (like di-nitrogen pentoxide) is actually going to mean concentration.0445

Normally, it is true: a square bracket is concentration; I shouldn't be so reckless, but I am.0453

OK, so the concentration of N2O5 (and this is in moles per liter, as always--concentration is in moles per liter when you see it like that)...0460

OK, so we are going to do 0 seconds, 50 seconds, 100 seconds, 200 seconds, 300 seconds, and 400.0470

The data we have is: we start with an initial concentration of 0.100--four decimal places; 0.0707; 0.0500; 0.0250; 0.0125; and 0.00625.0480

This is the kinetic data that we collected.0505

At different time intervals, we measured the concentration of reactant left in the flask, and these are the concentrations that we got.0509

Now, we want to know what the order of this reaction is, and we want to know what the rate constant is.0515

Let's go ahead and start.0521

Now, what we are going to do in order to find the order: we said that we are going to plot the logarithm of the concentration, versus the time (y versus x).0524

Whenever you see something versus something, it is always y versus x.0534

So, this is going to be the y-axis; this is going to be the x-axis...actually, the logarithm of this is going to be the y-axis.0537

Let's make a new table and draw a little line here; and this time, I'll do the table in blue.0543

It is going to be the same thing; the time is going to be the same thing--it's going to be 0, 50, 100, 200, 300, and 400.0550

Now, we are going to take logarithms of these numbers; so, when we take the logarithms of those numbers, we end up with the following data.0560

This is going to be ln of the N2O5 concentration in moles per liter.0569

We end up with -2.3, -2.6, -3, -3.7, -4.4, and -5.0575

Now, we are going to plot this.0589

We have our y-axis and our x-axis: this is our time; this is going to be our logarithm of the N2O5 concentration.0592

We will just put some numbers up here: so, -6, -5, -4, -3, -2; so, -6, -4, -2, and we will go with 100, 200, 300, 400; this is 100 seconds (oops, all of these lines are showing up again--this is always happening down at the bottom of the page--it's kind of interesting).0604

All right, so 100, 200, 300, and 400 seconds: when we plot this data, we actually end up getting something which is (let's go down to about right there; I guess that is pretty good), believe it or not--we actually do end up with a straight line, with all of these different points at various points.0633

Now, I should let you know: this is kinetic data; kinetic data is not always going to line up in an exactly straight line--it's not going to be that the line is going to go through every single point exactly.0661

But, it is going to be pretty good; you are going to get a linear correlation.0672

So, you should be able to draw a line; in other words, you are not going to get points all over the place.0676

You are going to be able to draw a line through many of the points.0681

Don't think that is has to be exact; real kinetic data, like real science, is not...doesn't fall into nice, perfect, clean square boxes.0685

There is a little bit of deviation; don't let that confuse you.0697

The idea is: when you plot this, you get something that is a straight line--pretty much a straight line.0700

OK, so because it is a straight line--because this kinetic data gives you a straight line--that implies that you have a first-order reaction.0706

That means that your rate law is K times the N2O5 to the first power, and that means your integrated rate law is (let's write it down here): the logarithm of the concentration of N2O5 equals -K, times t, plus the logarithm of the initial concentration of N2O5.0722

That is what is happening here: the initial concentration was 0.100; so we have identified that it is first order.0757

We can go ahead and write a differential rate law for it; we can write an integrated rate law for it; everything is good.0769

Now, the second part is: how do we find the rate constant?0776

OK, well, we have the straight line; the rate constant is (let me go ahead and move to the next page) the slope, the negative slope, of that line.0779

So, the slope equals -K.0791

So now, all we have to do: we take 2 points on that line--it doesn't matter which 2 points; now, when you are taking points on a line (on a straight line for a graph), you can only use points on that line.0799

We just said that real kinetic data doesn't always give you an exactly straight line; so you are going to have a best fit line, and you are going to do a least squares fit on that line.0813

Well, when you pick 2 points randomly on there, you don't pick 2 data points, unless those data points actually happen to be on the line.0825

If they are not, you pick points on that line.0833

There you go: you have your 2 points; you do Δy/Δx; set it equal to -K; and you solve for K.0836

When we take a couple of points on the graph that we just did in the previous slide, we end up with the following.0847

We end up with: -6.93x10-3, and the unit is per second, equals -K.0853

Slope equals -K; negative equals negative; so K, the rate constant, equals 6.93x10-3 per second.0867

That is our rate constant.0879

What we did: given kinetic data (time and concentration), we plotted time on the x-axis, and the logarithm of the concentration on the y-axis.0881

If you get a straight line, it's a first-order reaction; if you don't get a straight line, it is not a first-order reaction; so it works both ways.0892

If you do get a straight line, the slope of that straight line is negative of the reaction constant, K.0902

OK, so now, let's move on to our second example, or actually a continuation of this first example.0914

Using the data and results from the previous example, find the concentration of N2O5 at t=150 seconds.0926

OK, well, so now we have K; we have the integrated rate law, which is: the logarithm of the concentration is equal to -K, times t, plus the logarithm of the initial concentration.0958

What we are looking for is the concentration of N2O5 at 150 seconds.0979

Well, if we look at our graph, or look at our data table, we have t at 100; we have t at 200.0984

We can't just split the difference to find the concentration in between there at 150.0991

You will see that in just a minute; but let's just solve our equation.0996

We end up with logarithm of the concentration of A (because we are looking for the concentration of A--this is the dependent variable here--this whole thing), equals -K (K was 6.93x10-3--let me make this a little more clear), times t (well, at 150 seconds), plus the natural logarithm of the initial concentration (the initial concentration was 0.1000).1000

OK, so let's go ahead and do this: equals...well, when you do the math, you end up with: ln of the concentration of A equals -3.3425, and in a logarithmic equation, you solve by exponentiating both sides.1036

So, let's go ahead, and I'll just write "exponentiate," which means e to the ln of A equals e to the -3.3425, and you end up (well, the e and the ln go away) with a concentration of A=0.0353 moles per liter.1059

There you go: we used the kinetic data to find the integrated rate law, to find the reaction constant...the rate constant; and then we used that equation to find the value of a concentration at any given time (in this case, 150 seconds).1093

Now, let's take a look at some values.1111

The concentration of N2O5 at t=100 was 0.500; that is from the data table.1119

The concentration of N2O5 at t=200, also from the data table, equals 0.250.1133

OK, now, the concentration of N2O5 that we found, which was at t=150--it doesn't equal...it's between 100 and 200; it isn't this plus this, divided by 2; it isn't .375; it isn't right down the middle.1146

That is the whole idea--that this relationship is logarithmic, so we found this to be equal to 0.0353, not 0.0375.1165

This is not right; hold on a second--I think we have some values wrong here; this is .1; this is .05; 0.0500; 0.0250; .0375.1185

It is not half of that; you actually have to use the integrated rate law to find the rate--it's this--the concentration--not that.1204

Be very, very careful; don't just think you can look at the data and say, "OK, well, half the time between 100 and 200, at 150, is half the concentration between .5 and .25, which is .0375"; that is not the case.1214

You have to use the integrated rate law, which gives you .0353; the concentration is actually less--more diminished than you would have expected.1226

OK, so let's go back here; now that we have taken care of the integrated rate law, we also want to talk about something called the half-life.1236

Let's give a definition.1247

The half-life of a reaction is the time required for the concentration of a reactant (in our case, the reactant) to reach 1/2 its original value.1256

In other words, when does A equal A0/2?1296

When does the concentration of A equal A0/2?--when is half the reactant used up?1304

That is sort of an important point; it is a line of demarcation; it gives us a standard--a reference, if you will--a reference point: half-life; it's pretty common.1309

We also speak about half-life in radioactive decay: how much time does it take for half of a particular nuclide, radioactive isotope, to completely disappear by half of its original amount?1318

So, in kinetics, we talk about the same thing: the half-life of a reaction is the time required for the concentration of a reactant to reach 1/2 of its original value.1331

In other words, concentration of A equals 1/2 of A0, the initial concentration.1339

OK, well, the symbol is t1/2...half-life; the time.1345

Now, let's go ahead and derive our half-life equation.1359

So, we have our equation, our integrated rate law: the logarithm of A equal -Kt, plus the logarithm of A0.1363

Well, let's see: how shall we do this?1378

OK, let me go ahead and bring this over here; so we have ln of A, minus ln of A0, equals -Kt.1381

Now, let me rewrite this logarithm thing as (yes, that is fine) logarithm of A, divided by A0, right?1397

It is the property of a logarithm; the logarithm of something, minus the logarithm of something else, is the logarithm of that first something divided by the second something.1408

Equals -Kt; and now, we will use the fact that A=A0/2, right?1416

We will go ahead and put this A0/2 in here.1424

We get the logarithm of A0/2, over A0, equals -Kt.1428

The A0s cancel; we end up with ln of 1/2 = -Kt.1439

We get that t is equal to ln of 1/2, over K.1447

So, the half-life--when half of the concentration has completely been depleted, the time that it takes for that to happen--is equal to the logarithm of 1/2, over the rate constant.1454

Take a really, really close look.1469

We can also write this another way: we can also write it as logarithm of 2 (I'm sorry, this is -K, right?--because the minus sign...) over K, and that just has to do with how we derive the equation.1471

Either one of these is fine.1488

This one or this one: you can use either one; it is just a question of: logarithm of 1/2 is a negative number, negative over negative is positive, logarithm of 2 is positive, so either one is absolutely fine; it's just that, the way I derived it, we came up with this one.1490

Now, notice what is important about this: the time it takes for a first-order reaction to be halfway complete, to be 50% depleted, does not depend on the initial concentration.1507

It doesn't depend on concentration at all; it just depends on a number, the logarithm of 1/2, and it depends on the rate constant.1518

So, it is a function of the rate constant; for a first-order reaction, the half-life does not depend on concentration; that is very, very important.1525

But, we will summarize this towards the end.1534

OK, so now that we have that, let's go ahead and do another example.1537

OK, a certain first-order reaction has a t1/2 equal to 20 minutes--has a half-life of 20 minutes.1550

That means it takes 20 minutes for a certain reaction to be halfway complete.1568

A: We would like you to calculate the rate constant.1571

B: We would like you to find out how much time is required for the reaction to be 75% complete.1583

OK, so we have a certain first-order reaction; it has a half-life of 20 minutes.1610

We want you to calculate the rate constant, and we want you to tell us how much time is required for the reaction to become 75% complete.1614

All right, well, part A: we have: t1/2=...let's just use the ln 2/K.1623

We are solving for K here; we know the t1/2, right?1642

So, K is equal to ln 2 over the t1/2.1645

Well, ln 2 is a certain number, and t1/2; this is just ln 2 over 20 minutes; and what we end up with is 0.0346 per minute; that is the rate constant--nice and simple.1654

So, you can find a rate constant from the half-life, or you can find a rate constant from the kinetic data--the slope.1672

Either one is fine; it just depends on what you have at your disposal.1679

Now, part B: well, 75% complete of the reaction--what does "75% complete" mean?1684

That means that your final concentration is only 25% of your original.1693

75% means that your final concentration of A is equal to 0.25 A0 (that means you only have 25% of the initial concentration left; 75% of it is gone).1701

We just plug that in to our integrated rate law.1714

So, let's write our integrated rate law again; we write it over and over again, until it becomes second nature.1720

-Kt + ln of A0 (and you notice, I am getting, actually, very, very lazy; I am not even writing the parentheses anymore).1726

We are talking about kinetics; we are talking about rate laws; we are talking about concentrations in moles per liter; I don't want to hammer certain things that are self-evident.1735

OK, so now, we said that A=.25 A0, so we write ln of 0.25 A0 = -0.0346t + ln, and our initial concentration was...actually, it doesn't even matter; it is just A0.1746

We know that it is 75% complete, so we don't actually have to have a number; we just know that this is .25% of that, because it is 75% done.1776

OK, so now, we move this over; we put them together; it looks like this.1784

The logarithm of 0.25 A0, over A0, equals -0.0346t.1792

These A0s cancel; we end up with ln of 0.25, divided by -0.0346, equals t.1803

When we solve this on our calculator, we end up with 40 minutes; very nice.1815

Now, this another way to actually do this, based on the percentage that they asked.1824

They said 75% complete; 75% complete--that means 50% is gone (half-life), and then to 75% means that another half-life (half of the half) is gone.1829

So, you can think of it as the first 20 minutes, and then another 20 minutes, because again, the half-life of a first-order reaction is constant.1842

It doesn't depend on concentration; it only depends on the K.1851

In this case, they say it's 20 minutes; the half-life of any amount is always going to be 20 minutes for this reaction.1856

It means 20 minutes must elapse before half of what is left over goes away.1864

Then, half of that left over goes away--it takes another 20 minutes; half of that--another 20 minutes.1869

But, I think the best thing to do is just to use the equations, instead of sort of reasoning it out that way.1874

When you get a little bit more comfortable with it, that is fine; you can look at 75%, and you can say, "Oh, that is 2 half-lives; 2 times 20 is 40."1879

Just go ahead and plug it in, and everything will work out fine.1887

OK, so we have taken care of the differential rate law for a first-order.1890

We have talked about integrated rate law for a first-order.1894

We have talked about the half-life of a first-order reaction.1898

All of this, basically, comes from standard kinetic data, where you, at different points of time, measure the concentration of the reactant; you plot it, and then the plot will actually tell you whether you are looking at a first-order or a different kind of order reaction.1902

So, next lesson, what we are going to do is talk about second-order and zero-order reactions.1919

We will just continue with the same theme: new rate laws, new integrated rate laws, new formulas for half-life; and then, we will summarize everything.1924

Thank you for joining us here for AP Chemistry, and thank you for joining us at Educator.com.1931

We'll see you next time; goodbye.1936

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to continue our discussion of integrated rate laws.0004

Last lesson, we introduced the integrated rate law for a first-order reaction, and also the half-life formula for a first-order reaction.0007

Today, we are going to talk about second-order reactions and zero-order rate laws.0015

So, let's just jump right on in.0019

OK, so again, we are talking about (in order to simplify matters for ourselves, just so we get a good sense of the kinetics) a single reactant that decomposes to products.0024

That is what we have been doing this entire time.0035

Our fundamental reaction that we have been working with is the following products.0037

OK, now, the rate law for a second-order is the following--this is the differential rate law; and again, when we say "rate law," they mean differential rate law.0043

When they mean integrated rate law, they will specifically say "integrated rate law."0055

So, the rate is equal to -Delta;A/Δt; this rate symbol, which is the differential part in differential rate law, is equal to some constant, K, times the concentration of A to the second power.0060

That is the differential rate law for a second-order reaction.0079

Now, when we integrate this particular function, we end up with the following.0084

Let me write; this is the differential rate law; now, the integrated, which relates concentration as a function of time.0091

It is going to be 1 over the concentration of A, is equal to Kt, plus 1 over the initial concentration.0112

For a second-order reaction, your integrated rate law says 1 over the concentration of A is equal to the rate constant, times the time, plus 1 over the initial concentration of A.0124

So again, in this particular case, we can set it up as y=mx+b, but this time, the y is 1 over the concentration--the data that we take.0138

That is what is important; and the slope of the line that we get is the rate constant.0151

Notice, this time it is a positive slope--positive K instead of negative K.0158

What we are going to do is: we are presented with some standard raw kinetic data; we are going to find...we have the time; we have concentration; we are going to also change those concentrations to 1/concentration, and we are also going to do logarithm of concentration, because now, we have to check; now, we have 2 different types of equation, 2 different types of reaction orders.0161

Now, we have to check to see whether it is first-order or second-order.0185

We have to have 2 columns: one with the logarithm of A, one with the reciprocal of the concentration of A.0188

And again, you will see that in just a minute, when we do the example.0195

So, now, this is the integrated rate law; let's go ahead and work out the half-life for a second-order reaction.0199

And again, remember: half-life means that the concentration of A is equal to the initial concentration, over 2; it is when half of it is gone.0209

We will go ahead and write; that implies using this equation with those values: 1 over A0 over 2 equals Kt, plus 1 over A0; this ends up being 2 over A0, minus 1 over A0.0219

I flip this and move this over; it equals Kt; 2 over A0 minus 1 over A0 is 1 over A0, equals Kt.0247

Now, I will divide through by K, and I get that t of 1/2 is equal to 1/K, times the concentration (the initial concentration).0257

This is the formula for the half-life of a second-order reaction.0269

Notice, in this particular (you know what, this is not too clear; let me write this out; I need to erase this, and let me write it bigger over to the side; I think it will be a little more clear; let me get rid of this thing; A0...so, we get t is equal to 1/K, times the initial concentration)...0274

Now notice: in this case, the half-life of a second-order reaction not only depends on K, but it also depends on the initial concentration.0302

That was not the case with the first-order reaction; a first-order reaction does not depend on the concentration--it just depends on K; it's constant.0310

This is not constant; so the half-life--if you start with a certain amount, half of it is going to take a certain amount of time to get rid of.0317

Well, now that you have gotten through half of it, now that is a new initial amount.0325

In order to release half of that, that half-life is going to change; it is actually going to end up being longer.0328

So now, it depends on two things: the rate constant and the concentration.0334

There you go: we have that equation, which is the integrated rate law; we have this equation, which is the half-life; and, of course, we have this equation, which is the differential rate law.0339

These are the things that matter in kinetics: the differential rate law, the integrated rate law, and the half-life formula.0354

Pretty much everything else can be worked out from this information, and all of this, of course, comes from raw kinetic data: time, concentration.0360

OK, now, let's go ahead and do an example, because that is the best way that these things work.0368

Example: we have butadiene; butadiene dimerizes (and dimerization means that two molecules of something stick together); butadiene dimerization was studied, and the following kinetic data were obtained.0377

OK, so let's go ahead and write out the formula.0417

2 C4H6 turns into C8H12.0420

OK, so now, we have our time value; I'm going to go ahead and write it as one big plot.0429

Our time value; we have our concentration of C4H6, which is our reactant.0437

Now, what we want to do here is: we want to find the rate constant; we want to find the order of the reaction; and we want to find the half-life (I'm sorry about that; I should have actually told you what it is that we are actually going to be doing here).0446

So, the first thing we want to do is: we want to find the order of this particular reaction, based on the data that we are given.0459

Then, we want to find out what the rate constant is; and then, we want to find out what the half-life of the reaction is, based on this data.0467

So, t (let me go ahead and write this down) is 0, 1000 seconds, 1800, 2800, 3600, 4400, 5200, and 6200.0475

And what we have is 0.0100, 0.0625, 0.0476...no, that is not right; that is 0.0100; so this is going to be 0.00625--the butadiene is diminishing--0.00476, 0.00370, 0.00313, 0.00270, and we are almost done--no worries--0.00241; and 0.00208.0498

OK, given this raw data, find the order of the reaction; find the rate constant; and find the half-life of this reaction, based on the fact that we started with .0100 moles per liter of this butadiene.0552

All right, so we need to check: is it first-order or second-order?0569

We need to plot both the logarithm of C4H6 versus time, and the reciprocal of C4H6 versus time, to see which one of these gives us a straight line.0574

If this gives us a straight line--the logarithm--it is first-order; if this one gives us a straight line, it's second-order.0591

That is how we do it.0597

OK, so here is what the data looks like.0599

Let's see; let's do logarithm first.0602

We get -4.605, -5.075, -5.348, -5.599, -5.767, -5.915, -6.028, and our last one is -6.175; reciprocals are a lot easier.0604

What we have is 100; 160; 210; 270; 320; 370; 415; and 481.0639

OK, so now we are going to plot (I'll do it in blue): we are going to plot this versus time, and we are going to plot this versus time, to see which one gives us a straight line.0654

Let's go ahead and take a look at what these plots look like.0669

I'm going to go ahead and put them next to each other; so, I'm not going to give you too much detail on these graphs--you can actually go ahead and plot them yourself on a piece of graph paper, or using a software like Excel or something like that.0671

What you are going to end up with, as it turns out--let's see here, this is going to be 200...yes; so let's do the logarithm of the butadiene versus time here; and let's do the 1 over the reciprocal of butadiene (which I'll just call A) over time here.0687

When I do this, the logarithm graph is something like this--definitely not a straight line.0714

When I do the reciprocal, I get a straight line; so I will have you confirm this, but this is exactly what happens.0721

So, because it is the reciprocal that gives us a straight line, it is a second-order reaction.0729

Second-order: so, we can go ahead and write the differential rate law: C4H6/Δt equals the constant times C4H6 to the second power.0737

We know it now; it is a second power; we derived it from the graph now, so we have an order of 2.0754

Now, the next thing we want to know is: what is the rate constant?0763

OK, well, the rate constant--the best thing to do is to go ahead and (let's see, how shall we do the rate constant)...let's just go ahead and use the integrated rate law.0767

We know that the second order is: 1, over the concentration of A, equals -K (actually, you know what?--what am I saying?--let me just...write it out, so we have it)...equals positive K times t, plus 1 over A0.0780

This is y=mx+b.0808

We just graphed the reciprocal over that; we got a straight line, so now, what we want to do is: we want to pick two points on that straight line, take the Δy over the Δx of those two points...0811

And again, the points need to be on the line; if the data points that we have in the original data, when we calculated it, are on the line, that is fine--you can use those data points.0825

But, if they are not, make sure that you are using points that are on the line.0835

When we calculate Δy/Δx from this graph, based on the kinetic data that we derived, we end up with the following.0838

Let's see: let's use a couple of points, actually; so let's use Δy over Δx; it equals...as it turns out, a couple of the data points do fall on the line.0848

I pick the first and the last; so it's going to be 481 minus 100, over 6200 minus 0, which gives me 0.06145, and the unit is going to be liters per mole-second.0862

I know the units for the rate constants don't really make sense; they tend to change, depending on what order it is; it is actually the number that matters.0882

This is our rate constant; and now, we want to find t1/2.0889

OK, well, t1/2 of a second-order reaction is 1/K, times the initial concentration.0897

It is equal to 1 over 0.06145, and the initial concentration was 0.0100; so, when we do that, we get 1627 seconds.0909

There you go: raw data: time and concentration; we calculated the logarithm of concentration; we calculated the reciprocal of the concentration; we plotted ln of concentration versus time for one graph; we plotted reciprocal of concentration versus time for a second graph.0926

We saw which one of those graphs gave us a straight line; in this particular case, it was the reciprocal versus time that gave us a straight line: second-order.0947

Because it is second order, I go ahead, and I can write the differential rate law, if I need it.0955

We need the rate constant; so I just found a couple of points on that straight line; I took the slope, y2-y1 over x2-x1.0961

I ended up with the rate constant; and then, because I have the half-life formula, and I have the rate constant which I found, and I have the initial concentration (which is the initial concentration from raw data), I was able to calculate the half-life of this reaction--very, very straightforward.0969

Nothing altogether too complicated: again, differential rate law, integrated rate law, half-life--those three things are what is important, as far as the kinetics of a reaction are concerned.0985

OK, that was second-order reaction.0996

Let's talk about a zero-order reaction.0999

All right, so a zero-order reaction is exactly what you think it is; it says that the rate (well, let me actually write it in the top here)...1001

Zero-order reaction: well, the rate (which is equal to -Δ of some reactant/Δt) equals KA to the 0 power, which is K times 1, which is K.1011

In other words, the rate itself is constant; the rate doesn't change.1032

The integrated rate law: when we integrate this particular function, you end up with the following.1039

You end up with the concentration of A (you know what, I am not even going to write these anymore; they are driving me crazy), equals -Kt, plus A0.1045

The concentration at any given time is equal to minus the rate constant, times time, plus the initial concentration.1060

Or, it equals the initial concentration, minus rate constant, times time; it's diminishment.1067

Again, y=mx+b.1072

So now, if you had some raw kinetic data, and you just plotted the concentration versus the time, and you got a straight line, that is a zero-order reaction; there you go.1077

Now, half-life: A=A0/2; so let's put it in here.1088

A0/2=-K times t, plus A0.1097

A0/2 minus A0=-Kt.1111

-A0/2=-Kt.1117

Therefore, t1/2 equals A0 (these lines--they always show up in the most inopportune moments), divided by 2K.1125

Let's do it in red.1143

We have this; we have this; and we have this.1145

For a zero-order reaction, the differential rate law says that the rate is equal to K, the rate constant; it is a constant.1155

The integrated rate law says that the concentration at any time A is equal to -Kt, plus A0.1163

The half-life formula says that the initial concentration, divided by twice the rate constant, will give you the amount of time it takes for half of whatever you started with to disappear.1169

Notice, in this case: again, it depends on the initial concentration, and it depends on K.1181

Now, you might be wondering what a zero-order reaction is; where would you run across a zero-order reaction?1187

When would you have a reaction that doesn't actually depend on the concentration?1192

Zero-order reactions show up in places where...usually in things that involve catalysis; and, extended to the biological realm, that means enzymes.1197

So, for example, if you have--let's say just 10 enzymes, catalyzing a reaction; well, if all of those 10 enzymes are busy (meaning if they are full, doing what they are supposed to do), it doesn't matter how much more reactant you actually put in there.1210

It is called a substrate--the thing that the enzyme grabs onto and fills with; so it doesn't matter--once those 10 enzymes are full, you can't make the reaction go any faster, because it doesn't matter how much concentration--how much more of the reactant--you actually put in; the reaction rate is controlled by how many enzymes are actually doing the work and how fast they are doing it.1229

So, under catalysis conditions, the reaction rate actually depends on the catalyst, not so much the concentration of the reactant.1252

Now, this isn't always the case; we're just saying that, when you run across a zero-order reaction, that more often than not, it is going to be in a catalysis situation.1260

That is just a qualitative bit of information that you should know, if it happens to come up.1268

OK, so now, let us take a global view of what it is that we have done with zero-, first-, and second-order reactions.1273

Let's summarize what we have, and it will give us a good, nice one-page summary of reaction kinetics and how to deal with them.1280

OK, so this is going to be a summary for the kinetics associated with the reaction aA going to products.1288

So again, this entire time we have been talking about a single reactant decomposing into products.1314

It diminishes, and it forms something.1320

How can we figure out the rate?1322

All right, so here, let's go ahead and make ourselves a little table; and we will go ahead and put a line here, and we will say first-order, second-order, zero-order reaction.1324

OK, the differential rate law for a first-order reaction is: the rate is equal to K, times the concentration of A to the first power.1345

The rate is equal to K, times the concentration of A to the second power.1360

And, the rate equals just plain old K.1366

That is the differential rate law.1369

OK, our integrated rate law: we have, for a first-order reaction--we have: ln of A is equal to -Kt, plus (oops, I always do an h for ln; I don't know why) ln of A0; again, concentrations, kinetics--we are talking about concentrations.1371

Now, the rate law for a second-order is reciprocal: 1 over A is equal to Kt, plus 1 over A0.1395

And, this one is: A is equal to -Kt, plus A0.1408

These are the integrated rate laws.1414

OK, a plot giving a straight line is going to be ln of A versus t, y versus x.1416

So, the plot of ln(A) for a first-order reaction--the plot of ln(A) versus t gives a straight line.1440

Here, it is 1/A versus t that gives a straight line.1445

Here, it is A versus t that gives a straight line.1450

And again, it is just the left-hand side of the integrated rate law: ln(A), 1/A...you check the data.1456

Slope: well, the slope is -K, K, -K of the particular line, if you happen to get one.1464

In other words, the slope is that; so you can find K; that is how you find K.1478

And, half-life: the t1/2 is equal to ln(2) over K; notice, for a first-order reaction, it is kind of interesting; it does not depend on concentration at all; it is constant.1483

The half-life for a second-order reaction is equal to 1/K times the concentration of A0.1503

It depends on the initial concentration.1512

And, t1/2 of this one equals the initial concentration over 2K.1515

So, this is everything that we have done: first-order, second-order, zero-order reaction rates.1521

When we are given kinetic data, which involves time and concentration, we have to plot the logarithm of the concentration versus time, the concentration versus time, and 1 over the concentration versus time.1528

Whichever one of these graphs gives us a straight line, that is the order of the reaction.1542

When we have the order of the reaction (let's say, for example--boom!--we come up with second-order), we know the differential rate law; we can write the integrated rate law; we can find K from the slope; and we have the half-life formula.1547

We can tell you what the concentration is at any time; we can tell you what the concentration is going to be at any percentage that we want along the way, instead of the time--all of this information right here.1561

This is a summary of standard kinetic operating procedure.1573

We'll go ahead and leave it at that; next time, we will talk about temperature dependence of reaction rates, and talk about something called the Arrhenius equation, and activation energy, and things like that.1581

But until then, thank you for joining us for AP Chemistry and kinetics, and thank you for joining us at Educator.com.1592

Take care; goodbye.1599

Hello, and welcome back to Educator.com0000

Today, we are going to continue our quick review of the basics of chemistry before we jump into some solid AP work.0002

We're going to go over stoichiometry today.0010

I'm going to talk about several varieties of stoichiometry problems.0013

Stoichiometry, essentially, is about amounts; that is what it is.0018

In chemistry, the basic unit of amount is the mole.0026

This is something that I know you all have heard before, so really quickly: a mole is 6.02x1023 particles.0031

"Particles," because in chemistry, we can talk about atoms; we can talk about molecules; we can talk about ions, electrons, protons...whatever.0042

I tend to use just the generic term "particles"; and the problem itself will specify what particle we're talking about.0049

In one mole (so this is a conversion factor) of something, we have 6.02x1023 particles.0057

Now, the periodic table is arranged based on the mole; a molar mass, that you see on the periodic table--that is one mole of oxygen atoms, one mole of magnesium atoms, one mole of iron atoms--those weigh those particular numbers.0065

So, the mole is the standard unit that we use in chemistry--again, because we can't count individual atoms (they're too tiny), so we have to choose a unit, kind of like a dozen or a century--things like that.0079

Stoichiometry basically concerns conversion factors: setting up a bunch of conversion factors based on what you know versus what you want.0093

There is a series of steps to that.0103

So, I'm going to introduce this notion of how to deal with stoichiometry problems.0106

Because the mole is always what we're talking about, the mole is the central hub; you start from the mole and you go to the mole, or you pass through the mole getting to where you want.0111

One way or the other, you are going to have to use a mole.0120

The reason is: the equations in chemistry (like, for example, 2H2+O2 → 2H2O, the reaction of hydrogen and oxygen gas to produce water)--well, these numbers here--the coefficients--the 2, 1, and the 2--they represent moles.0122

What this says is, "Two moles of hydrogen gas need to react with one mole of oxygen gas in order to produce two moles of H2O."0143

So, all equations--balanced equations--represent the standard that we use to work from. 0150

It's all in moles.0158

Now, in any stoichiometry problem, any species that you're dealing with always has (this is the best way to think about it) moles (you know what, I'm going to actually draw this up here) of X.0160

Every species that is mentioned in a stoichiometry problem--something like that; from the mole, we can go to grams of X, or we can go to the number of particles of X.0179

Again, X might be a molecule; it might be an atom; it might be an ion; it depends.0198

Later on, we might need to go to liters of X.0202

But, they all have to pass through the mole.0209

Again, every species under discussion has one of these little things associated with it.0213

If a stoichiometry problem has two species that are mentioned, and there is some relationship among the two species (which we will get to in a minute), that species is also going to have something like this.0218

Let's call it mol y, and again, you can go to grams of y; number of particles of y; or maybe liters of y.0231

Now, here is what is interesting: the reason that I say that each species--you should write something like this: when you do something like this, this will give you a pathway to actually find out how to write the problem, stoichiometrically; how to solve it, mathematically.0250

This gives you a solution path.0264

Any time you are moving from one species to another--like, for example, in a minute we're going to sort of choose a couple of reactants and a couple of products and stoichiometric problems--movement this way or that way is based on the mole ratio, and these mole ratios are precisely the coefficients in the balanced reaction.0267

Now, this will make more sense in a minute, when we actually do a problem, and I'll show you the way to actually do this.0286

Hopefully, you are already good at this; but if not, this will be a good review and a little bit of practice.0293

Having said that, let's just jump into a problem, and I think it will make sense.0299

OK, so example number 1: How many Cl atoms are there in 18.50 grams of phosphorus pentachloride?0304

OK, let's see: so now, let's identify the species: one of the species is chlorine; one of the species is PCl5; and so, I'm going to write that.0333

I'm going to write mol PCl5, and I'm going to write mol Cl; those are the two species.0344

Now, what is it that I'm interested in?--I'm actually interested in the number of atoms, so I need the number of Cl atoms.0354

OK, what am I given?--I'm given grams of PCl5.0365

I'm given grams of PCl5; I'm supposed to find the number of chlorine atoms.0375

Here is how you do it: you go...this is the path that you follow, and each arrow represents a conversion factor.0380

I need to go from grams of PCl5 to moles of PCl5.0390

From moles of PCl5, I need to go to moles of Cl; this is going to be the mole ratio.0394

From moles of Cl, I can go to the number of chlorine atoms.0400

Here is the setup (actually, let me draw this a little bit higher, so that I can make some room): grams PCl5...0405

This is why we say each species has this little setup here: when you set it up like this, you can actually go ahead and draw the pathway; so here, I'm going from grams of y, to moles of y, to moles of x, to number of particles of x.0418

That's it; I'm following a pathway, and I can go anywhere I want, but I'm always passing through moles; there is a mole ratio.0434

OK, you always write what you are given: 18.50 grams of PCl5.0441

This is one conversion factor, two conversion factors, three conversion factors: I know I'm going to have three conversion factors--I don't even need to fill in the numbers yet.0451

I can just automatically do this.0461

Now, the thing that you are coming from is going to be the unit on the bottom; the thing that you are going to is going to be the unit on the top.0463

So, I'm going to have grams of PCl5, moles of PCl5; well, how many grams in a mole of PCl5?0471

Well, the molar mass is 208.22 in one mole of PCl5.0481

Now, my second conversion: what is the relationship between moles of PCl5 and moles of Cl?0487

One mole of PCl5 releases 5 moles of Cl; that is just based on the molecular formula--it is PCl5.0493

All right, PCl5: one of these produces five of these.0501

So, again, the thing that you are going to is on top--mole of Cl; the thing you are coming from is down at the bottom--mole of PCl5.0507

Well, the relationship is: one mole of this produces five moles of that; that is the mole ratio; that is this step.0524

Now, I'm looking for number of Cl atoms in one mole of Cl.0532

Well, one mole of Cl--one mole of anything contains 6.02x1023 (let me actually write "atoms").0544

So again, here, the particle we're talking about is atoms.0558

When I do this multiplication, I end up with (OK, I can write it down here) 2.67x1023 atoms of Cl; that is my final answer.0561

Again, it's just based on this relationship; every species has a relationship; from mole, you can go to grams, to particles, to liters.0578

Then, among the species, there is a mole ratio--a relationship.0588

In this one, it was 1 mole of this produces 5 moles of that.0592

That is how you proceed through this network.0595

You have something that you are given; you have another thing that you want; you find a path through there.0598

If you're given this and you want that, you find a path through there, and it always passes through moles, because the mole is the basic unit of chemistry.0603

OK, we'll get to some more stoichiometry in just a minute.0614

The next part of the review: I'm going to cover an empirical formula problem.0618

Oftentimes, you are given a certain amount of data, and you need to find the empirical formula of a compound.0623

The empirical formula is the formula that gives the lowest number ratio of those things.0630

So, for example, if I had something like C6H12O6, which is glucose; and let's say I had another sugar, C12H24O12; well, these are two different sugars.0636

But notice, the base formula--I can actually divide it by 6 here; I end up with CH2O, and I end up with CH2O, right?0656

That is the lowest number ratio of atoms to each other.0666

In other words, all sugars have a ratio of one carbon to two hydrogens to one oxygen.0670

That is the empirical formula; it's a general umbrella formula that covers all of the molecules of a given compound.0676

Each individual compound has its own molecular formula.0684

These are molecular formulas; they are specific to the compound.0687

But, an empirical formula covers both; that's it.0690

So, let's go ahead and do an empirical formula problem.0695

Let's see...let's write this as example 2: all right, I have a compound that is 24.5% sodium, and 14.9% silicon, and 60.6% fluorine.0698

I want to know what the empirical formula is.0722

OK, here is how you handle it: because you are given percentages...percentage is based on 100; so, since they give us the percentages, let's just presume that we have 100 grams of something.0724

So, we can convert these percentages: 24.5% of 100 grams is 24.5 grams.0735

What I'm going to do is: I'm going to convert these to moles.0741

So, sodium: I take 24.5 grams, and I convert to moles: well, one mole of that is about 23 grams, so I get 1.07 moles of sodium.0744

Now, I do the same for silicon: silicon--I have 14.9%, which is 14.9 grams if we take a 100-gram sample; we're just doing this so we can make our math easy.0760

Well, one mole of silicon is 32.07 grams, so I get 0.46 moles of silicon.0771

Then, when we do fluorine, we have 60.6 grams of fluorine; one mole of fluorine is 19 grams, and I end up with 3.2 moles of that.0782

Well, now that I have it in moles, I divide by the lowest number of moles; so I divide all of them by 0.46, 0.46, 0.46, and when I do that, I end up with 2.3 here; I end up with 1 here, of course, and here I end up with 7.0796

I need whole-number ratios; so 2.3--in order to make this a whole number, I have to multiply by 3, which means I have to multiply everything else by 3.0822

This becomes 21; this becomes 3; and this becomes 7.0832

So, this is my empirical formula: I have: Na7Si3F21.0837

That is my empirical formula.0851

It could be Na14Si6F42, or any other multiple, but the basic relationship, the ratio, of sodium to silicon to fluorine is 7:3:21.0853

This procedure always allows you to find the empirical formula.0866

You change the percentage to gram, gram to mole for all of them; you divide by the lowest number of moles to standardize, so that you have at least one of them with a 1; and then you multiply by an appropriate constant to make sure that all of them give you whole numbers, because we can't leave it as 2.3; you can't have a fraction of an atom.0872

That is a standard empirical formula problem.0891

OK, so now we're going to take the next step, and I'm going to show you a molecular formula problem.0894

Once we find the empirical formula, if we happen to have the molar mass of a particular compound, we can find the actual molecular formula.0901

Let's do that one.0908

Let's go ahead and make this example 3: Menthol is composed of carbon, hydrogen, and oxygen.0913

1.005 grams is combusted to produce 2.829 grams of CO2 and 1.159 grams of H2O.0932

If menthol is 156 grams per mole, what is its molecular formula?0969

Here, we're going to end up having to find the mass of the individual components--the carbon, the hydrogen, the oxygen.0994

We're going to use the empirical formula, and then, from there, we're going to use the fact that they give us the molar mass.1003

We're going to divide that molar mass, the 156, by the mass of the empirical formula to see how many empirical units actually go into the whole formula.1010

That is how we get the molecular formula.1018

Let's just jump right on in.1021

Notice, this didn't give you the percentages straight on; this is a slightly different problem--slightly more complicated.1023

Let's see: 2.89 grams of CO2; let's go ahead and...let me write this equation down: menthol, when you combust it, it's with O2, and it's going to produce CO2 + H2O.1029

All right, now: 2 CO2...how many grams of carbon...OK, so now we have 2.829 grams of CO2.1054

One mole of CO2 weighs 44 grams; now we're going to do...so we know that all of the carbon ends up in CO2; all of the hydrogen in menthol ends up in H2O; it's the oxygen that is split between the two.1077

We need to find the number of grams of carbon, the number of grams of hydrogen; they give us the total number of grams of menthol that were combusted, so the balance of that is going to be the oxygen.1092

Then I can start my problem.1104

2.829 grams of CO2) times...again, this is a mole ratio: 1 mole of CO2 produces one mole of carbon, right?1107

One mole of CO2 produces one mole of carbon, times 12 grams of carbon per mole, and I end up with 0.7715 grams of carbon.1119

Now, I have 1.159 grams of water times...1 mole of water is 18 grams: 1 mole of H2O produces 2 moles of hydrogen, right?1135

Erase that: 2 moles of hydrogen--1 mole of this produces 2 moles of hydrogen.1159

Hydrogen is 1 gram per mole, so I get 0.1288 grams of (oops, this isn't carbon; this is) hydrogen.1165

Now, if I want to find the mass of oxygen, I take the 1.005 grams of menthol, and I subtract the amount of carbon, 0.7715, plus 0.1288, and I end up with a mass of 0.1047 grams of oxygen.1182

Now I have my masses of each, based on a combustion analysis.1211

I hope you understand how it is that I came up with that.1216

Now, I can start my empirical formula problem.1218

I have 0.7715 grams of carbon, times 1 mole over 12 grams; that gives me 0.06429--often, with these empirical formula problems, you want to carry out the decimals as far as you can, to get a good number.1222

Then, if I do hydrogen, I have 0.1288 grams, times 1 mole, which is 1 gram, and I get 0.1288--this is in moles.1242

Oxygen: I have 0.1047 grams; 1 mole of that is 16 grams of oxygen (again, we'll use the oxygen atom), and this is going to equal 0.0065438 moles.1258

Now that we have moles, we need to divide by the smallest, and it looks like the smallest here is this one.1284

When we divide by that, when we divide by 0.0065438, and we divide by 0.0065438, we end up with 10 here; we end up with 20 here; and we end up with 1 here.1289

So, the empirical formula for menthol is C10H20O1.1312

Now, (OK, so here is where we have to...) what we do is: we take the molar mass of the compound, the 156 (let me actually write this out), and we divide it by the molar mass of the empirical formula.1324

When we do that, again, the empirical formula is the lowest number ratio, so in this particular case, we had C10H20O1; that means the carbon to hydrogen to oxygen ration is 10:20:1.1363

Well, it might be C20H40O2; we don't know.1380

So, we divide the mass of the empirical formula into the molar mass of the compound, which we do know.1385

This ends up being...they said it was 156 grams per mole, and the molar mass of the empirical happens to also be 156 grams per mole; therefore, the ratio is 1.1391

Therefore, menthol is C10H20O1.1405

If this number were 3, it would be C30H60O3; that's it.1411

This is the actual...not only empirical formula, but it happens to be the molecular formula for menthol.1416

This process will always work for molecular formula.1422

You run through it; you find the empirical formula; and then you divide the molar mass of the compound by the molar mass of the empirical formula, and that will give you the ratio--the number by which to multiply all of these little subscripts in order to get your final molecular formula.1425

Let's do a stoichiometric calculation based on an equation.1443

The fermentation of glucose produces ethyl alcohol and carbon dioxide according to the following equation: C6H12O6, and fermentation (so no oxygen involved), goes to 2C2H5OH + 2 CO2.1450

The question is, "How many grams of glucose are required to produce 5.2 grams of ethanol, C2H5OH?1474

So, how many grams of glucose do I need in order to produce 5.2 grams of C2H5OH?1502

OK, so let's take a look: the two species that we are talking about are glucose and C2H5OH; we need to pass through moles.1508

They want...we are given grams of C2H5OH that we want; we want to find the number of grams of glucose.1519

Here is the process: based on this equation (I'll write it like this), I need grams of glucose.1526

Grams of glucose comes from moles of glucose.1538

Moles of glucose I can get from moles of ethanol.1543

Moles of ethanol comes from grams of ethanol.1551

That is the process: I am given grams of this--I am given 5.2 grams of this; I have to go to the moles--from grams to moles, mole to mole (I use mole ratio--that one), and then from mole back to gram.1559

This is my path: I go here to here to here to here; that's it--I have 1, 2, 3 arrows; that is three conversion factors.1574

Let's go ahead and write this down; we always start with what we're given.1586

5.2 grams of EtOH (EtOH is just a shorthand for ethanol); that is 1, that is 2, that is 3; grams of EtOH here; I'm going to moles.1590

Moles of EtOH--notice, I'm not going to do the numbers until afterward--I have moles of EtOH here, because, again, they have to cancel, and it's also the thing that I'm coming from.1605

I'm going to moles of glucose, and this is moles of glucose; grams of glucose...1617

5.2 grams of ethanol; ethanol happens to be 46 grams per mole; the mole ratio--how many moles of ethanol to moles of glucose?1629

Moles of ethanol to moles of glucose: 1; that is why you need a balanced equation.1638

Glucose is 180 grams per mole; therefore, my final answer is 10.17 grams of glucose.1645

10.17 grams of glucose will produce 5.2 grams of ethanol--standard stoichiometry.1655

The two species were glucose and CH3OH; there is my 1 mole of glucose; mole of ethanol; these are the relationships: I need grams of glucose coming from grams of ethanol; this is my path.1664

If you can't just do it automatically, draw it out; see the solution path, and the path will tell you what the mathematics looks like.1681

Let's finish off this lesson with a limiting reactant problem.1692

Limiting reactant problems are going to come up very, very often, so basically, it's just...a balanced equation is such that it says, "All of this will react with all of that; nothing is left over."1697

But in real life, it doesn't work like that; often, when we run a reaction, one of the reactants is going to be left over; one of them is going to run out.1709

The thing that runs out--when it runs out, the reaction stops.1717

It's a limiting reactant--it limits the outcome of the reaction.1720

It controls how much product you actually get, because when it runs out, again, the reaction stops.1725

Let's take a look at this equation here: 2 aluminum hydroxide, plus 3 moles of sulfuric acid, produce 1 mole of aluminum sulfate, plus 6 waters.1732

I'm given 0.350 moles of aluminum hydroxide, and I'm given 0.450 moles of sulfuric acid.1756

My questions are: a) which is the limiting reactant? part b) how many grams of aluminum sulfate (Al2(SO4)3) can be recovered, theoretically?1769

That is what we are calculating; we are calculating a theoretical yield, as if everything went right, theoretically.1805

And part c) is: How many moles of excess reactant are left over?1811

Let's do part a: what is the limiting reactant?1831

We need to find out which one of these runs out first.1834

You can choose either one to start with: basically, what you're going to do is...here is what it looks like: I'm going to go ahead and choose the aluminum hydroxide.1837

0.350 moles of Al(OH)3; the mole ratio between this and this is 2:3.1847

Moles of Al(OH)3 (this is the problem with chemistry--there is just so much writing as far as these symbols are concerned--the names!) to moles of H2SO4; I need to do a mole ratio calculation.1863

It is 2 moles of this to 3 moles of that; this actually tells me that I require 0.525 moles of H2SO4.1885

In other words, if I have .350 moles of aluminum hydroxide, this calculation tells me that I actually need .525 moles of H2SO4 to completely react with it.1897

Well, this is required; the question is: I require .525--do I have .525? I do not (let me do this in red).1908

I don't have .525--I only have .50; therefore, H2SO4 is limiting.1921

Because it's limiting, it controls the rest of the reaction; this is the number that I use to find out how much of this I'm going to get; not this.1932

When this runs out, the reaction stops; there is going to be excess of this.1940

Therefore, this is what controls how much I have.1944

Let's go ahead and do part b, which is, "How many grams of aluminum sulfate can be recovered?"1947

All right, well, since this is my limiting reactant, that is the one that I take: 0.450 moles of H2SO4 times...well, what is the mole ratio here?--it's 3:1; 3 moles of H2SO4...this is not going to work...1954

3 moles of H2SO4...oh, these crazy lines; OK, let me try this one more time.1987

Let's start from the top: 1 mole of aluminum sulfate comes from 3 moles of H2SO4, and then we have 342.7 grams of aluminum sulfate per mole.2002

There we go; this cancels that; that cancels this; my final answer is going to be in grams, and it's going to be =51.32 grams of Al2(SO4)3 (it's tedious writing all of those symbols, isn't it?).2035

There you go; no, I'm not going to have...I do not want those lines to show up; therefore, I'm going to write this one more time.2061

=51.32 grams of Al2(SO4)3; there we go!2073

Based on .450 moles of H2SO4, the most aluminum sulfate I can recover is 51.32 grams.2088

The reason is: because that is what controls the reaction.2096

OK, now: this final problems says, "How many moles of excess reactant will be left over?"2101

This is our limiting, so this is our excess: we need to find out how much is used, and then subtract it from the amount that we started with, and that will give us the amount left over.2107

The amount used is, again, going to be based on the limiting reactant; so, part C: 0.450 moles of H2SO4...2117

Well, I have that 3 moles of H2SO4 requires 2 moles of aluminum hydroxide.2132

I end up with 0.30 moles of aluminum hydroxide used.2145

0.350 moles-0.30 moles used (let me actually write these out: moles of Al(OH)3 minus 0.03 mol used up) leaves me with 0.05 moles of aluminum hydroxide to end up with.2156

You see what we did: once you know the limiting reactant, that controls the rest of the reaction.2191

.450 moles of H2SO4 uses up .3 moles of aluminum hydroxide.2196

I started with .350; I subtract, and I'm left with .05 moles of aluminum hydroxide; that is my excess reactant.2201

OK, so in this lesson, we have done some basic stoichiometry coverage.2212

We have talked about empirical formula; we have talked about molecular formula; and we finished off with a limiting reactant problem.2215

That, and a previous lesson with naming, should provide sort of a good, general foundation for the rest.2222

Starting with the next lesson, we're going to jump straight into some good, solid AP chemistry.2230

Until then, thank you for joining us here at Educator.com; we'll see you next time; goodbye.2234

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

The last couple of lessons, we have been talking about reaction kinetics.0004

We have talked about the differential rate law; we have talked about the integrated rate law; and today, we are going to close off the discussion of kinetics with a discussion of activation energy and something called the Arrhenius equation.0010

You know from your experience that, when you raise the temperature on a given reaction, that the reaction tends to proceed faster; or, if you drop the temperature, that somehow things seem to go slower.0022

Like, for example, the whole idea of refrigeration is based on that fact: you drop the temperature, and the reactions that cause spoilage actually slow down.0034

So, not only does the rate depend on concentration, but the rate also depends on temperature.0043

The Arrhenius equation actually accounts for that dependence on temperature.0048

Let's just jump right on in.0053

OK, now again, we have seen that the rate depends on concentration; so, if we have some general equation like aA + bB → products, well, we know that the rate is going to be equal to some rate constant, times the product of A raised to some power, times the product of B raised to some power.0056

And, these n and m and n could be integers; they could be numbers; they could be anything.0083

Then again, let's reiterate that the m, the n, and the K are determined experimentally; these are not things that we can read off from the equation, the way we will do later on, when we discuss equilibrium.0090

Well, again, as it turns out, the rate is not only dependent on concentrations, but it is also dependent on the temperature.0101

So, as it turns out, the rate constant itself shows an exponential increase with temperature.0108

This constant of proportionality--yes, it is true that the rate is contingent, is dependent on A and B raised to some power; but in some sense, this K is a measure of that dependence.0132

As it turns out, for the temperature itself, the rate constant shows an exponential increase with temperature.0143

The rate constant depends on the temperature.0149

So now, we are going to introduce something called the collision model, and it is exactly what you think it is.0153

It accounts for this temperature dependence the same way that it accounts for the concentration dependence.0160

Let me just write this collision model; and the collision model basically says (I don't really need to write it down) that, in order for things to react, they basically need to come into close contact with each other.0166

And, since you know that molecules and atoms are sort of flying around, especially in the gas phase, at very, very high speeds--they don't just come close to each other; they collide.0181

And that is what it is; in order for a reaction to take place, they have to collide.0191

Well, so we know that, of course, the kinetic-molecular theory says that, as you raise the temperature, the average velocity of the molecules--of the atoms--of the particles increases.0196

Therefore, you have a higher frequency of collisions.0210

A higher collisions--more things collide; therefore, there is a greater chance of the reaction actually taking place.0213

And, that is all that is really going on.0221

Let's take a look at this.0224

Well, let's see: shall we write...let's write something for the collision model.0231

So, in order for a reaction to proceed, particles of reactants must come into close contact by colliding.0236

Now, I know this is not the most precise, rigorous definition; but it gives us a sense of what is going on, and that is really what is important; we want to understand the chemistry--we want to understand where these equations come from, if it actually makes sense--and this is perfectly fine.0267

OK, now, what is interesting is that, even though the temperature rises, and more particles are colliding--as it turns out, the frequency of collision goes up; but as it turns out, the rate itself seems to be not as high as it would be otherwise.0286

There are some other things going on; so, before we discuss those other things, let's just talk about what this man, Arrhenius, actually proposed.0302

Arrhenius proposed the idea of an activation energy.0312

The activation energy, if we want to define it: it is the threshold energy that the reactants must overcome in order to go to products--or in order for the reaction to actually move forward.0318

And now, we're going to draw a picture, and I think the picture is going to make it a lot more clear.0353

So, this is a standard energy diagram, and on the y-axis we have energy; and this thing is called the reaction coordinate.0358

The reaction coordinate is just a measure of...sort of time moving forward; it isn't time itself--it is saying how far forward the reaction is proceeding.0368

Let's go ahead and put the reactants over here; we'll put the products over here; as it turns out...0378

Now, thermodynamically, notice that the products over here are actually lower than the reactants.0386

Thermodynamically, this reaction is spontaneous; it will actually move forward.0392

But, that doesn't mean that it is going to move forward; there is this energy hill that has to be overcome--it has to get over this hump--in order for it to actually move forward.0397

This is what the activation energy is.0408

The reactants, the molecules...whatever it is--they have to slam into each other; they have to collide with enough energy to actually overcome this barrier.0412

Now, at a given temperature, not all of the velocities of the particles are uniform; it is not like all of the particles have one velocity.0422

There is a distribution of velocities, and only those velocities of the particles that are involved, that have enough kinetic energy for that kinetic energy to be transformed and equal to this activation energy (it's E with an a) will actually move forward.0430

That is why, when you raise the temperature--yes, it is true, you are actually causing more things to move faster and more things to collide; but not everything is going to move at enough speed to actually overcome this threshold barrier.0450

So, let's go ahead and draw a little...so this thing up here--we call it the transition state; we also call it an activated complex--the transition state.0466

So if we were to take something like...let's take a particular reaction; let's take 2 BrNO → Br2 + 2 NO; so here, basically, two molecules of BrNO need to collide, and when they do collide, they end up releasing a bromine molecule and 2 nitrogen monoxide molecules.0478

Well, in order for that to happen--it's true that they need to slam into each other with enough energy; if they actually overcome it--if they actually have enough energy to get over that activation energy and to form a transition state, well, you might think that it actually looks a little bit...something like this.0514

In order for this to happen, a bromine and a nitrogen bond has to break, and a bromine-bromine bond has to form; that is what is happening here.0532

So, we could think of it as BrNO and then another BrNO, and these dots are used for bonds that are breaking and bonds that are forming.0544

This thing would be the transition state; they would slam together, and as this bond is breaking (the BrNO bond), the Br-Br bond is forming.0556

This was sort of what it might look like; now, we don't know exactly what it would look like, but that is what we postulate.0567

We call this thing the transition state; there is a transition from reactants to products.0574

That is all that is going on here.0579

And, we will often see these energy diagrams in any number of contexts.0581

OK, now, let us talk a little bit about this distribution of velocities, so you see what is going on, exactly.0589

Now, we said that, at a given temperature, there is a distribution of velocity; some particles are moving very slowly; some particles are moving very, very fast; in general, they sort of average out, and the distribution looks something like this--like any other distribution.0601

It will be...OK, so this is the energy axis; and now, let's see...this distribution--let's call this at temperature 1; so let's say, at a certain temperature, there is a distribution of speeds; certain things are slow; certain things are fast.0617

Well, as it turns out, if the activation energy is here, that means only those particles have enough energy, have enough speed, to surmount that barrier.0638

All of the others do not: which is why, despite the fact that we raised the temperature, and we have a whole bunch of collisions--as it turns out, the sheer number of collisions that actually lead to a reaction is actually very small.0652

This is why it is very small: because, among the distribution of speeds (very slow, moderately slow...on average, where most molecules fall, and the very fast), it is only the very fast that have enough energy to overcome that activation barrier--the activation energy.0665

That is what this is.0682

Well, if we raise the temperature, we raise, on average, the velocities, and the distribution changes; the distribution looks like this now.0683

So now, notice that this is temperature 2; and the temperature 2 is higher than temperature 1.0694

Well, temperature 2 is higher than temperature 1, which means, on average, you have more things that are fast.0701

Well now, we have basically pushed, on average, all of the particles to a faster speed, but they still occupy a distribution: some are still slower than others; most are in the middle; some are fast0708

But notice, now you have a lot more (I'll do this in red) particles that have enough energy above the activation energy to actually move forward with the reaction, as opposed to (let's do this other one in black) this right here, the original temperature.0721

So, as you raise the temperature, you provide more particles with enough energy to overcome that activation energy barrier.0750

This is a qualitative description of why raising the temperature increases the reaction rate.0762

OK, now, Arrhenius proposed this: he proposed that the number of collisions with enough energy equals some percentage of the total collisions--right?0770

If you have 1,000 collisions, and let's say, of those 1,000 collisions, only 100 of them have enough energy--well, it's a certain percentage.0798

He quantified this; he actually came up with a mathematical formula: so, the total collisions (we won't worry about where these came from or how he derived this) e to the -Ea/RT.0805

So, this ea/RT is the fraction of the total collisions that have enough energy to move forward.0823

And notice: this thing right here, the activation energy, shows up in this exponential.0837

R is the (let me rewrite this, actually: e to the -Ea/RT) gas constant, which is not .08206--we are dealing with energy in Joules, so it is actually 8.31 Joules per mole-Kelvin.0843

T is the temperature in Kelvin; so it's very, very important--you might be given degrees Celsius, but when you use this equation (and we'll sort of make it a little bit better in a minute), the temperature has to be in Kelvin, and R has to be 8.31, not .08206.0868

This Ea--that is the activation energy, the energy that has to be overcome in order for the reaction to move forward--the energy that takes you up to the transition state--the energy that takes you up to the activated complex for it to go over the hump; it is the top of the hump.0888

OK, now, as we said, even those particles with enough energy to overcome the barrier--still, we don't see the kind of rate change that we would expect.0905

Well, that is because there is something else going on.0917

Not only do these particles have to have enough energy in order to overcome the barrier--they actually have to be oriented properly.0920

In other words, they can't just slam into each other this way or that way or that way, and a reaction will take place; there is a specific way that they have to slam into each other in order for a reaction to take place.0928

So, there are two requirements (I'm going to put this down at the bottom here, and I'm going to do this in blue) for a successful reaction.0939

Two requirements for a successful reaction: the first is, of course, that the collision energy has to be greater than or equal to the activation energy, right?--we need to get over the hump.0955

Well, not only do we need to get over the hump, but the molecular orientation (well, I don't want to say "molecular"--well, yes, that is fine) must be such that it allows the reaction to take place.0977

We need an activation energy--the collision energy has to be greater than the Ea--and the orientation of the molecules has to be such that it actually allows for the reaction to take place.1015

That is why the number that we actually see, even though we have a bunch of particles that certainly have enough energy to overcome the barrier--still, only a fraction of those actually makes it past the reaction point.1026

So, we have a bunch of particles; only a fraction of them have a certain velocity, in order to activate, in order to get over the hump.1039

Of the amount that actually have enough to get over the hump, only a fraction of those have the right orientation.1049

So as it turns out, really, it is kind of surprising that reactions proceed the way they do, because really, ultimately, very, very few molecules actually satisfy these things.1057

But again, since we are talking about so many, we actually do see reactions take place.1066

OK, now, I am going to write a mathematical expression that quantifies everything that we have talked about.1072

Now, the rate constant is going to equal z, times p, times e to the (-Ea, divided by RT).1084

Now, let's talk about what these mean.1096

z is a factor that accounts for collision frequency--in other words, how often things collide.1099

p accounts for orientation, and it is always going to be less than 1; but that is fine--we are not really worried about that.1114

And of course, e to the (-Ea/RT)--that accounts for the fraction of collisions having enough energy--in other words, greater than or equal to Ea.1127

These three things: the frequency of collisions, the orientation, and those that actually have enough energy when they collide: all of these things contribute to the rate constant.1161

And, as we said, the rate constant increases exponentially with temperature.1177

All of these things are a function of temperature.1182

This is a constant; this is a constant; all of these things are essentially constants and factors.1184

It is this temperature that changes.1189

Now, we are going to write this as: K (which is the rate constant) equals Ae to the (-Ea/RT).1192

This A is called the frequency factor, and it accounts for these things--it accounts for the frequency of the collisions and the frequency of collisions that actually have the right orientation.1203

OK, this is called...this is the Arrhenius equation, right here.1219

This is the Arrhenius equation.1231

Now, like all things that tend to involve exponentials, let's go ahead and take the logarithm, and see if we can't come up with some linear relationship.1233

When I take the logarithm of both sides, I get the logarithm of K is equal to the logarithm of this whole thing; and the logarithm of a multiplication is the logarithm of one plus the logarithm of the other, so it is going to equal the logarithm of A, plus ln of e to the (-Ea/RT).1240

Well, ln(K)= ln(A), plus--the ln and the e go away, right, so you end up with -Ea/RT.1264

Now, let me rearrange this: I get ln(K)=...I'm going to pull out the -Ea/R, because these are constants, times (oops, no, we want this to be very, very clear) 1/T, plus the logarithm of A.1283

Now, notice what we have: we have (oh, these lines are driving me crazy) RT ln(A); let me actually do it over here.1314

We have y=mx+b.1331

The logarithm of the rate constant--that is the x; let me draw it down here: y=mx+b.1336

When we do y versus x, when we plot the logarithm of the rate constant against 1 over the absolute temperature in Kelvin, what we end up with is a straight line.1350

The slope of that straight line is equal to the negative of the activation energy, over R.1362

So, this gives us a way of finding the activation energy for a given reaction, when we measure the temperature and the rate constant.1368

The y-intercept actually gives us ln(A); well, when we exponentiate that, it gives us a way to find the frequency factor, A.1375

There we go; this is our basic equation that we are going to use when we are dealing with temperature dependence of the rate constant.1386

OK, now, let's see: let's go forward, and let's rewrite the equation again.1397

We have: logarithm of the rate constant equals minus the activation energy, over R, times 1 over the temperature, plus the logarithm of A (which is the frequency factor).1405

Thus, for a reaction which obeys the Arrhenius equation, the logarithm of the rate constant, K, versus 1 over the temperature, gives a straight line with slope equal to -A/R.1422

And again, R equals 8.31 Joules per mole-Kelvin, not the .08206.1464

Now, what is really, really nice is that most rate constants actually do obey the Arrhenius equation.1472

So, because that is the case, it actually ends up lending support to the collision model.1477

We derived this from the collision model; the fact that most things obey this is supporting the collision model--that is supporting evidence; so, our model is actually very, very good.1484

Now, let's go ahead and see if we can do an example; I think it's the best way to proceed.1499

Our example: the following data were obtained for the reaction 2 N2O5 decomposes to 4 NO2, plus O2.1507

The following data was obtained: the temperature, which was in degrees Celsius, and the rate constant, which is in per second.1537

We have 20, 30, 40, 50, and 60; and these are rate constants, OK?--so this is per second, so we are looking at a first-order reaction.1549

2.0x10-5; we have 7.3x10-5; 2.7x10-4; 9.1x10-4; and 2.9x10-3.1562

Notice, as the temperature increases, the rate constant increases; the rate is getting faster.1582

A higher rate constant means a faster rate.1586

OK, now, let's see: what is it that we want to do here?1591

We want to find the activation energy for this reaction.1597

In other words, how much energy do the N2O5 molecules have to have in order for this reaction to proceed properly?1602

OK, well, let's go ahead and graph this data.1610

Now, I'm going to give a rough graph; it's going to be reasonably accurate, but the idea is to see what is going on here.1614

This is going to be 3; this is going to be 3.25; this is going to be 3.50; and then we have 6, 7, 8, 9, 10, 11, 10, 9, 8, 7, 6, so -6, -7, -8, -9, -10, -11.1623

OK, so these are...this axis, the vertical axis, we said, is the logarithm of the rate constant; and this is going to be 1/temperature.1649

So, let's go ahead and actually...so what we do is (now, notice: we were given K, and we were given T; what we have to do is) calculate 1/T and ln(K).1667

OK, so let me go ahead and just redraw everything here; let's do temperature in degrees Celsius; let's do temperature in Kelvin; let's do 1/T; and let's do ln(K)--so we can actually see the data, instead of just throwing it out there.1681

We have 20, 30, 40, 50, and 60; then, we have 293, 303, 313, 323, and 333.1701

Now, we have 3.41x10-3; 3.30x10-3; 3.19x10-3; 3.10x10-3; and 3.00x10-3.1718

Then, we have -10.82; -9.53; -8.22; -7.00; -5.84.1739

So, again, our equation is 1/T and ln(K), which is why we took that data that we got, and we calculated the 1/T and the ln(K).1753

Now, this is what we plot: this on the x-axis, this on the y-axis.1764

When we do that, we end up with some line that looks like this.1771

Let's go over here; I'm just going to pick a couple of points: 1, 2, 3, 4, 5--we have 5 points, but I'm just going to pick a couple of them, because it is the point that I want to make that is ultimately important.1776

We go ahead and we draw a line through those; and again, this is kinetic data; everything is not going to lie on a straight line, but you are going to get a linear correlation.1787

You are doing the "best fit" line.1798

When you do that, you pick a couple of points on that line, and you calculate Δy/Δx.1799

Now, Δy/Δx, which, in this case, is equal to Δ of ln(K) (that is the y) over Δ of 1/T.1808

When you do this, you end up with -1.2x10-4 Kelvin.1820

Now, we said -1.2x10-4; that is the slope--well, the slope is equal to the activation energy over R, which is equal to the activation energy (which we are seeking) over 8.31 Joules per mole per Kelvin.1832

When we multiply this through, we end up with an activation energy equal to 1.0x104 Joules per mole; there we go.1856

We were given kinetic data, which consisted of temperature and rate constants; we used the logarithmic version of the Arrhenius equation; we found 1/T and ln(K); we wrote those values down.1872

We plotted that: ln(K) versus 1/T; we got ourselves a straight line; we picked a couple of points on that line; we found the slope; and we know that the slope is equal to negative of the activation energy, over the gas constant.1885

We solved basic algebra, and we ended up being able to find the activation energy of 1.0x104 Joules.1901

That means that the particles--a mole of particles--need to have enough activation, and need to have 1.0x104 Joules (in order for this to actually proceed) per mole.1908

That is all this is; that is all we did.1921

OK, let's see: what else can we do?1926

Let's do another example here.1933

This time, I will go ahead and...actually, before we do the example, let me give you a little preface to the example.1935

Now, instead of kinetic data, the activation energy can also be gotten from the values of the rate constant at only 2 temperatures.1944

In other words, we don't need a whole table of rate constants and their corresponding temperatures...or, temperatures and their corresponding rate constants.1980

As long as we have two temperatures and two rate constants, we can actually find the activation energy.1989

So, let's see how that is done.1994

Now, at temperature T1, we have rate constant K1.1996

The relationship is: the logarithm of K1 is equal to minus the activation energy over R, times 1/T1, plus the ln(A).2009

Again, this is based on the equation: K is equal to A, times e to the (-Ea/RT).2022

This is the Arrhenius equation, and, when we take the logarithm of both sides, we get this version.2032

Well, for any K and T, that is the relationship.2037

And now, let's take another temperature; how about at T2?2042

Well, at T2, we have a K2; we have another rate constant, K2.2047

The relationship is: ln(K2) is equal to -Ea, over R (it doesn't change; it's the same reaction), this time times 1/T1, plus ln(A).2053

And again, A doesn't change; A is the frequency factor--it's a constant for that particular reaction.2066

Well, let me take this equation minus this equation.2072

So, I write: ln(K2)-ln(K1), which equals ln(K2/K1), is equal to...when I take this side and subtract this side, the ln(A)s cancel, and I end up with the following.2077

-Ea/R, times 1/T2, minus 1/T1.2097

Now, in your chemistry books, you will often see this flipped, and you will see the sign change here.2106

All they have done is factor out a -1 from here, in order to get rid of this -1.2112

So, you will also see it as: Ea/R, times 1/T1, minus 1/T2; it's up to you.2117

I personally prefer this, because it doesn't change anything.2128

I think it is important that equations be written exactly as how they were derived, and that signs be left alone, simply so you can see everything, so everything is on the table.2133

When you start to simplify things, yes, you tend to make them look more elegant--and this is generally true of science; we tend to like our equations to look elegant.2143

But, understand something: that in science, and in mathematics, the more elegant something looks--that means the more that is hidden.2153

That is the whole idea: when something looks elegant--when something looks clean and sleek and simple--that means something is hidden.2160

In this particular case, what they have done is: they have hidden the negative sign, and they have flipped this around.2165

That is fine; this is reasonable straightforward--it is not going to confuse anybody too badly.2170

But, I think that if we are going to take something and subtract something else, we should leave things exactly as they are.2174

So, don't let this minus sign confuse you because it looks different than what you see in your books; it's the same equation; they just don't like minus signs--which is generally true of chemists; they tend not to like minus signs; I don't know why.2180

OK, so here we have this equation; if you are given two temperatures and two rate constants, you can calculate the activation energy.2193

Let's do an example.2204

We have the following reaction: we have: methane (CH4), plus 2 moles of diatomic sulfur, forms carbon disulfide, plus 2 H2S gas (hydrogen sulfide gas).2208

OK, now, we take a couple of measurements: temperature in degrees Celsius, and we calculated some rate constants for that.2233

That ended up being...let me see: we did it at 550 degrees Celsius, and we also did it at 625 degrees Celsius; we got 1.1, and we got 6.4; it makes sense--higher rate constant-faster rate; higher temperature-faster rate; so everything is good.2248

We want to find the activation energy.2267

Well, great; we have two constants, and we have two temperatures; let's use our equation that we just derived.2271

And again, you don't have to know that equation; as long as you know the Arrhenius equation, everything else you can derive from there, because you are just taking logarithms and fiddling with things.2282

That is why we are showing you the derivation--to show you that you don't have to memorize the equation; it is where the equation came from and what you can do with it.2290

OK, so let's take the logarithm of K2/K1 (well, you know what, let me write it again), is equal to (-Ea, over R)(1/T2 minus 1/T1).2299

And again, temperature is in Kelvin.2317

This is equal to the logarithm of 6.4 (see this number), over 1.1; is equal to -Ea/8.31, times...now, 1/T2; T2 is 625 degrees; that is 898 Kelvin...so it's 1/898, minus...T1 is 550; that is 823 Kelvin.2321

When we solve, when we do this and this, we end up with the following: =1.7609, equals -Ea × -1.22x10-5.2354

We end up with an activation energy of 144,195 Joules, or 144 kilojoules.2372

I am not a big fan of significant figures one way or the other, which is why it looks like this.2386

Numbers are probably going to be a little bit different, as far as you are used to in your book, but it is the process that is important, ultimately--not the significant figures.2393

Later, if you become an analytical chemist, then significant figures will be a bigger issue.2400

That is it; we have used the Arrhenius equation, which, again, says that the rate constant is equal to some constant (called the frequency factor), exponential, minus activation energy over RT.2404

From this one equation, we can do multiple things.2422

This expresses that the rate is dependent on temperature.2427

This is the independent variable; this is the dependent variable.2431

When we are taking kinetic data, we will often have temperature and a rate constant, and we can do things to that, based on how we fiddle with this equation.2435

OK, with the Arrhenius equation, this concludes our discussion of kinetics.2447

I want to thank you for joining us for this discussion, and thank you for joining us here at Educator.com.2453

We'll see you next time; goodbye.2457

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

In our last lesson, we concluded our discussion of kinetics, but I thought it would be a good idea to actually spend some time doing some practice problems, directly from the AP exam, and just to give you an idea of what it is that you are going to be doing on the exam, because ultimately, again, this is an AP chemistry course, and the final thing is the AP Chemistry test.0005

There is nothing altogether different as far as the example problems that we did, but I thought it would be nice to have some real, specific examples of what it is that you are going to see.0025

So, we're going to do a couple of multiple-choice and a couple of free-response questions, and things like that.0035

Let's just jump right in and see what we can do.0040

OK, so our first series of questions is going to be three multiple-choice questions, and it is going to be based on the following hypothetical reaction and the following choices.0044

OK, so let's see: we'll say: Consider the following hypothetical reaction.0055

Now, mind you, these multiple choice questions--you can't use your calculator in these, so the numbers are going to be very, very simple.0072

Often, you will be able to just look at something and see what it is that is going on, so if you find yourself having to do multiple strange computations, or if you think that you can't do it because of a computation, chances are that there is something wrong.0078

These multiple-choice questions--they test basic understanding: can you follow a logical train of thought, where the numbers are completely secondary?0091

It is the free-response questions where you can use your calculator to actually find specific answers.0102

Hypothetical reaction: we have: our reactant R, plus reactant S, goes to product T.0109

Now, the following are our choices.0117

A) Rate = K times concentration of R to the first power.0124

B) we have: The rate = K times the concentration of S to the second power.0134

Our C) choice is: The rate is equal to K, times the concentration of R to the first, the concentration of S to the first.0143

D) Our choice is: The rate is equal to K, times the concentration of R to the second power, the concentration of S to the first.0152

And E), our final choice, because you have ABCDE to bubble in: The rate is equal to K times R squared, S squared.0163

So, we have a hypothetical reaction, and we have these choices to choose from, based on the questions that we are going to ask.0176

The first question: #1: OK, it says: When R and S (in other words, when the concentrations of R and S) are doubled, the rate increases by a factor of 8 (so they are telling you this; they are telling you that they double the concentration of R; they double the concentration of S; the rate increases be a factor of 8).0183

The question is: What is the rate law?0219

A through E are our choices.0228

OK, so how do we deal with this?0232

Again, we should be able to do this reasonably quickly.0235

Well, they are telling us that they are doubling R and they are doubling S, and that the rate is increasing by a factor of 8.0239

Well, there are a couple of ways that we can do this.0248

I could say to myself, "Well, if I double the rate..." I can say by doubling something, I am just making it twice that; well, if the rate increases by a factor of 8, I can say, "Two to what power equals 8?"0249

But, I think that may be a little bit confusing, so let's go ahead and do this a more natural way.0268

Let's just go ahead and stick the doubled rate into these to see what happens.0274

Well, notice what happens: if I put in twice R into the first one, I get K2R; the 2 comes out--I get 2 KR.0279

Let me just do that; let's try choice A.0293

Try choice A: I'm a little bit longer on this first one, but it shouldn't take this long--I just want you to see the process.0297

Try choice A: Well, I get that the rate is equal to K, times twice the rate of R.0304

2 just comes out; so twice the KR.0313

Well, KR is the rate; so by doubling the rate, all I have done is multiply the rate by a factor of 2; that is not the one.0319

I do the same thing with S; so, if I try choice B, they are telling me that the rate is equal to K2S--just double the rate and put it in there; and this time, it's 2S2, right?--because our choice is S to the second power.0328

Well, this is equal to 4 times KS2.0345

So now, by doubling the concentration of S, I put it in the choice, and I run the mathematics, and I get that my final answer is 4 times K to the S squared.0352

Well, K to the S squared was the original rate; 4 times that--that is not it, either.0363

So, as it turns out, the only one of these, when I put it in, where I double R and I double S and I run the mathematics--the only thing that gives me an actual increase by a factor of 8--is choice D.0369

The answer is D, and the best way to see that--just in terms of by looking at it without having to run through this--is just: take the number 2--they double the rate--and put it in here.0386

2 squared is 4; times 2 is 8.0398

You get 8 on this side.0402

That is how you solved this problem: just put it in and see which one actually gives you what it is that they are asking for.0404

I hope that makes sense.0411

OK, under the same circumstances now, problem #2: When R and S are doubled (again, the concentrations are doubled), the rate increases by a factor of 2.0414

Well, they say that both R and S are doubled; so, if you go back to your choices, and if you look through your choices--if you put 2 in for R, 2 in for S, well, the only one that actually gives you double the rate is choice A.0448

So, A is the answer.0469

And again, if you don't see that: the rate they told me was K, times R to the first power.0477

Well, if I put in 2, they tell me R and S are doubled; in this case, this choice A is not dependent on S at all--it only depends on R; so when I put 2 in there, I get K times 2R equals 2 times K times R.0492

Well, 2 times R--that equals 2 times the rate; that means the rate is doubled; that is the only choice.0513

If I put it into the other choices, I get that the rate is increased by a factor of 4, by a factor of 8, by a factor of 16...so the only one that is doubled is choice A, so our answer is A.0520

That is how we do it; these are very, very quick--you just need to be able to know what to put where and what to multiply.0534

OK, well, let's take a look at #3.0539

This time, they say, R is doubled; so this time, they are only doubling R.0543

R is doubled; S is unchanged.0552

Well, let's write this a little differently.0562

This time, what we have is: When R is doubled (the concentration of R), and S is unchanged, the rate is unchanged.0569

OK, well, they tell me that R is doubled, and they tell me that S is unchanged, and they tell me that the rate is unchanged.0608

Well, if I go back to my choices and if I take a look--well, if R is doubled--notice that choice A, choice C, choice D, and choice E all have R in them.0618

So, if I stick a 2 in there, I am going to get an increase; there isn't going to be no change; but they tell me that S is unchanged.0631

Well, if S is unchanged, the only choice (which is choice B)--the rate on this one is equal to K, times the concentration of S squared.0641

Well, they are telling me that S is unchanged; well, if I don't change S, I don't change the rate.0658

S unchanged; rate unchanged; my only choice is B, because everything else has an R in it, and if the rate is doubled, that means there is going to be some sort of a change to the rate.0663

In this case, my only choice is choice B, under these circumstances.0675

I hope that makes sense.0682

OK, let's try a free-response question here.0686

No, actually, I'm sorry; let's try one more multiple-choice question.0697

OK, so now: For the hypothetical reaction A + B → C, based on the following data, what is the rate law?0700

OK, so we are going to give you some data here; and they give you some data: experiment, and then we have the concentration of A; we have the concentration of B; and then we have the rate.0737

Experiment 1, 2, 3; 0.20, 0.20, 0.40, 0.10, 0.20, 0.10; and we have 2.5x10-2; we have 5.0x10-2; and we have 5.0x10-2.0752

Our choices are the following: K times the concentration of A; K times the concentration of A squared; C is K times the concentration of B; D is K times the concentration of B squared; and E: K times the concentration of A, times the concentration of B.0778

OK, so now, let's take a look at this data.0804

Whenever we compare two experiments, when we are given concentration data and rate information, whenever we compare two experiments, we need to find something where only one of the concentrations changes--because again, we are only dealing with one independent variable for anything that we compare.0811

So, in this case, when we change B from point 1 to point 2, when we double it, we ended up doubling the rate.0828

OK, so that means that it is going to be first-order in B.0841

The rate is K, so so far we know that it is first-order in B, because again, when you double something and you double the rate, that is first-order in that reactant.0848

Well, now let's take a look at Experiment 1 and Experiment 3.0862

Let me do this in red; now, we look at 1, and we look at 3.0866

Here, the B concentration stays the same, but we double up A and A.0870

Let me see here: .2, .4, .2, point...oh, wait; I'm sorry; our data is wrong.0880

It is not .1; this is .2; that is .20; OK, let's try this again, shall we?0891

Let me erase this, and let me put in blue again: this is .10, .20; this is .20; OK.0900

Yes, I think that is going to change some things.0911

OK, so now let's take a look at Experiment 1 and 2 again; that is fine; so .2, .2; that is the same; we double up the concentration of B, and then we end up doubling the rate; so yes, not a problem.0913

Our rate is...that part is the same; it is going to be first-order in B.0927

And now, let's see; now, let's look at Experiment 2 and Experiment 3.0934

.2, .2; B is the same; when I end up going from changing the concentration of A from .2 to .4, when I double that, the rate goes from 5.0x10-2 to 5.0x10-2; I double the concentration, but the rate didn't change.0939

Therefore, the rate is not dependent on A at all; so we are done.0956

Rate is KB; our choice is C; that is it.0961

Any time you have data like this, you take a look at some doubling, some changing, concentration, and you see how the rate changes.0969

That will tell you the order of that particular concentration term in the differential rate law; and then, you just look through your choices.0977

Good; OK, let's see; let's try another one here.0987

#5: Reactant P underwent decomposition, and the concentration was measured at different times.1001

Now, based on the data, what is the rate law?1038

OK, so we have time data, and we have concentration of P data.1055

0 time (this is in hours, and the concentration is in moles per liter, as always): 0, 1, 2, 3; so we started off at 0.4, and an hour later, we measured the concentration; we have 0.2; an hour later, we have 0.1; an hour later, we have 0.05.1064

Now, our choices are the following.1088

First-order in P (I'm getting a little sloppy here); second-order in P; twice; D--we have half KP; and E--we have K.1094

OK, so they are saying that some reactant, P, underwent decomposition, and the concentration was measured at different times.1132

Based on the data, what is the rate law?1140

OK, take a look at the time: 1 hour, 1 hour, 1 hour.1142

The time increment is uniform; it's 1 hour.1148

Now, notice the concentration: after 1 hour, there is half of the initial concentration.1151

After another hour, there is half of this concentration.1157

After another hour, there is half of this concentration.1160

So, every hour, it is diminishing by half.1163

That should immediately tell you something; we are talking about half-life here; and what is interesting is that the time--the half-life for it to diminish by half, diminish by half, diminish by half--is uniform.1168

The t1/2 is clearly 1 hour, and it is constant.1184

Any time you have a constant half-life, you are talking about a first-order reaction.1189

Remember that the t1/2 of a first-order reaction is equal to the logarithm of 2, over K, or the logarithm of 1/2, over -K.1199

It is a constant; it doesn't depend on the concentration.1217

Because it doesn't depend on the concentration, it stays constant.1220

That is what this data is telling me.1225

One hour, one hour, one hour; half-life is constant in order for it to diminish by half, diminish by half, diminish by half.1228

So, that automatically tells me that I am talking about a first-order rate law; that is my answer.1237

My choice is A; I hope that makes sense.1247

OK, let's see what we can do here.1254

Let's see: OK, let's try a free response question.1260

This is #6: We have the following reaction: we have 2 nitrogen monoxide, plus chlorine gas, forming 2 moles of NOCl.1264

OK, we have some kinetic data; we run a series of experiments--we run 3 experiments--1, 2, 3.1278

We measure the concentration of nitrogen monoxide (initial concentration); we measure the initial concentration of Cl2; and we also measure...this time, we are measuring the rate of appearance of NOBr.1286

That is fine; we can measure the rate of appearance--a rate is a rate; rate of appearance, rate of disappearance--it's the same; it's just that one is negative and one is positive.1309

We have: the following data were obtained: .02; 0.04; 0.02; 0.02; 0.02; 0.04.1317

We have 9.6x10-2; we have 3.8x10-1; and we have 1.9x10-1.1333

Our first question to you is: what is the rate law?1346

What is the rate law?--OK, it should be easy enough.1352

We have some concentration data; we have rate data; we hold one of them fixed; we check the other one to see what is going on.1357

When we check Experiment 1 and 2, we notice that the chlorine concentration is the same.1366

We doubled the nitrogen monoxide concentration; when we doubled the concentration, we quadrupled the rate.1372

OK, I hope you see that: double, from .02 to .04, the concentration of nitrogen monoxide; this 3.8x10-1 is four times 9.6x10-2.1383

OK, I hope that you see that; watch these exponents very, very carefully.1395

So now, double NO concentration means quadruple the rate.1401

This implies that it is second-order in NO.1413

2 squared is 4; that is where this 2 comes from here.1423

OK, now, let's take a look at Experiment 2 and 3.1430

When we look at 2 and 3, we notice that we have...actually, not 2 and 3; we will have to look at 1 and 3; I'm sorry.1435

1 and 3; and the reason is because .02...now, we're going to leave the NO concentration the same; we are going to double the Cl2 concentration.1448

So, in this particular case, when we double the chlorine concentration, we end up doubling the rate: 1.9x10-1 is twice the 9.6x10-2.1459

We double the rate.1476

Well, that means--when you double something and the rate doubles, that means it is first-order.1479

Therefore, our rate law is equal to some K, times NO to the second power, times Cl2 to the first power; that is our answer.1487

That is it--nice and simple.1504

OK, let's do our next phase.1507

Part B: what is the value of the rate constant? Include units.1513

OK, what is the value of the rate constant? Include units: I do my units and my numbers separately; that is just something that I like to do.1517

You are welcome to do it any way that you like.1545

OK, so let's see: we have "What is the value of the rate constant? Include units."1548

OK, well, we just said that the rate is equal to K times NO squared, times the Cl2 concentration.1553

Well, just pick any one of the experiments; you have the rate; you have the NO concentration; you have the Cl2 concentration; just solve for K.1565

So, K is equal to the rate, divided by the NO concentration squared, times the Cl2 concentration.1575

Let's just take (I think I'm going to use Experiment 3) 1.9x10-1, divided by 0.02 squared, times 0.04; and when we do that, we end up with 1.2x104.1587

That is the numerical value; now, let's do the units.1615

The rate: OK, the rate is in moles per liter per second, because that is how much was showing up, or that is how much was disappearing.1619

That is this up here; it is in moles per liter per second.1635

These are concentrations; you get moles per liter squared, times moles per liter; you get moles per liter cubed.1638

You get moles cubed over liters cubed.1648

Well, let's see what we have: this becomes 2; this becomes 2; that cancels that and cancels that; you flip that over, and you get liters squared over moles squared-seconds.1652

That is your unit; so your answer is 1.2x104 liters squared/moles squared-second.1672

That is your answer.1684

So again, my recommendation, as far as kinetics is concerned, is: in one of the lessons (not the last lesson, but the one right before that), we did a summary of first-order reaction, second-order reaction, zero-order reaction.1687

On that summary, we discussed the differential rate law; we discussed the integrated rate law; we discussed the half-life; and we discovered the quality of the graph that gives us a straight line (whether it's the logarithm versus time; whether it's 1 over the concentration versus time; or whether it's just concentration versus time).1704

That is what you want to know, as far as kinetics is concerned.1728

If you understand that summary, if you understand that table and where each thing is coming from, all of these problems will fall out naturally.1731

OK, thank you for joining us here at Educator.com, and thank you for joining us at AP Chemistry.1741

See you next time; goodbye.1747

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to start on what I consider to be probably the most fundamental, the most important, topic of the entire chemistry curriculum.0004

It is the one thing that shows up absolutely everywhere, in all areas of science.0012

It is the concept of equilibrium.0016

Starting from now, we are going to spend...I am certainly going to talk about it in great detail, and in great depth, and we are going to do probably more than the usual number of problems.0018

The reason is because these equilibrium problems (and, as we move on, acid-base and further aqueous equilibria and thermodynamics and electrochemistry and things like that)--the nature of the problems is such that there is no real algorithmic procedure to attack the problems with.0031

I mean, there is--there is certainly a set of things that you can do--but each problem is different, and it is very, very important to actually understand the chemistry.0049

We want the chemistry to lead the mathematics, not the other way around.0058

We don't just want to jump into a problem and start throwing equations together; we want to know what is going on.0061

So, I just wanted to give you that heads-up before we begin; this particular set of topics that we are going to talk about today is going to be more of just introductory, getting you comfortable with the notion of equilibrium, trying to wrap your mind around the concept of equilibrium.0067

Then, with the next lesson and subsequent lessons, we will really dive in and really get into the problem-solving aspect of this.0081

With that, let's go ahead and get started.0088

OK, so let's take a look, initially, at a reaction like this: the reaction of hydrogen and oxygen to form water.0094

2 H2 + O2 → 2 H2O.0102

Now, notice: we wrote this arrow as just one arrow to the right.0107

Well, when this reaction runs forward (so when we put hydrogen and oxygen in a container and then we ignite it), all of the hydrogen and all of the oxygen end up turning to water.0110

We say that the equilibrium of this thing is very, very far to the right.0122

It is true that, at the end, there is a little bit of hydrogen left, and a little bit of oxygen left, but for all practical purposes, there is none left.0125

All of it is completely reacted, which is why the arrow goes in one direction.0132

Most reactions, however--remember, when we talked about kinetics, we said that as a reaction starts to move forward, some of the products start to show up, and some of those products start to break down, and they go back the other way to form reactants.0137

Well, let's take a look at a reaction like this.0150

Nitrogen plus 3 moles of hydrogen gas goes to 2 moles of ammonia.0154

So, if we put nitrogen and hydrogen gas in a container, and we let it start to react under a given set of conditions, it is going to move forward, and it is going to form the ammonia.0161

But, what is going to happen is: after enough ammonia starts to form, there is a certain point at which the ammonia starts to break down, and it starts to decompose into hydrogen gas and nitrogen gas.0171

Well, there comes a point where the forward reaction and the back reaction happen at the same rate.0181

Because they are happening at the same rate, we can no longer measure which one is really moving forward or backward.0186

The reaction is still talking place, so it is still a "dynamic" equilibrium (and I will write that in a minute).0192

But, the idea is that now, the forward and the reverse reaction are taking place at the same rate; that is what we call the equilibrium condition.0198

The system has reached a point, a natural point, for that reaction.0207

So, an equilibrium condition is something that is sort of like a fingerprint for that particular reaction.0212

In a minute, we will give a mathematical expression for that fingerprint, which is specific to that reaction.0218

That is it; it is that simple; equilibrium is when the forward and the reverse reactions are happening at the same rate, and we represent it as a double arrow.0226

Most reactions fall under this category.0234

In fact, all reactions do, but some reactions are so heavily forward, or so heavily in the back, that we just go ahead and, for all practical purposes, we assume that everything has reacted or nothing has reacted.0236

We say the equilibrium is far to the right (meaning all products), or we say it is to the left (all reactants--nothing has happened).0249

So, let's go ahead and actually write this down.0257

Equilibrium is the state where the forward and reverse reactions (rxns) happen at the same rate.0265

They happen at the same rate, so that there is no net reaction movement.0294

It is as if the reaction has come to a stop; there is no more change.0310

In other words, no more NH3 forms; no more nitrogen diminishes; no more hydrogen diminishes.0313

They reach values that are stable--constant.0319

Let's sort of see what this looks like graphically.0322

We'll make an extra-long graph here; and this axis--let's say we have concentrations; so we'll start with, let's say...no, let's go up here.0328

Let's start with the H2 concentration; so, it's going to look something like this.0342

OK, and this is just the reaction coordinate--time, in other words--so we'll just go ahead and put "time" here.0350

As time proceeds, the hydrogen concentration drops, drops, drops, drops, and then it levels off; it stays some place.0356

Well, now the nitrogen concentration--it is going to drop, drop, drop, drop, and then it is going to level off.0362

Then, of course, there is the ammonia concentration; if, after a certain amount of time, we take measurements of the ammonia concentration, it is going to rise, rise, rise, rise, and then it is going to level off.0372

Well, this point--where it levels off--that is the equilibrium point; so this is a graphical representation of what equilibrium looks like.0384

Let's stay we start with a flask; hydrogen and nitrogen are in there; and notice, the hydrogen diminishes (right?--it's being used up); the nitrogen diminishes; it's...0394

Let me label these; I'm sorry--so this first one is the H2gas; this second one is the NH3 gas; and this third one is the nitrogen gas; OK.0405

The nitrogen diminishes, diminishes, diminishes; and notice that the hydrogen--the slope here is actually steeper than the slope here, which makes sense, because hydrogen is being used up three times as fast as the nitrogen.0424

At any given point, initially, the slope of the hydrogen is going to be three times the slope of the nitrogen.0439

Here, the slope is positive, because there was no ammonia to begin with; but ammonia is forming, so again, this is concentration (that is what this axis is).0446

We are taking measures of concentration; the concentration of ammonia is increasing, increasing; but there comes a point where all three of them reach a value that no longer changes.0455

They just sort of stop; everything is constant at this point; that is the equilibrium condition.0465

The reaction is still going, forward and backward, but it is happening at the same rate, so there is no net movement of the reaction.0469

So, let us write here: Equilibrium is a dynamic process; in other words, it means that it is a dynamic equilibrium.0476

"Dynamic equilibrium" means that what is happening is still happening; it isn't that the reaction has come to a stop, except that it is happening at the same rate, so we don't notice a net process taking place.0492

It is different than sort of a static equilibrium, where...for example, like a rock that is poised some place, let's say out in Utah or something; the rock is not moving, so all of the forces on that rock are balanced.0505

But that is it; there is no...everything is balanced, but nothing is actually happening.0520

In chemistry, in chemical equilibrium, the reaction is still moving forward and backward; it always has to, because molecules are slamming into each other all of the time; but there is no net notice of actually any change.0524

That is what this represents graphically.0538

OK, now, we come to the good part: we have a way of actually representing this equilibrium condition mathematically.0540

So, let's go ahead and describe, for a general reaction that looks like this: aA + bB (and again, from now on, most of the time, we're going to be using the double arrows, because we are going to be talking about equilibrium conditions) to cC + dD.0550

Well, as it turns out, most of science works like this; you run an experiment; you collect a bunch of data; and hopefully, the data--the relationship among the numbers--the data that you have collected--is sort of clear mathematically.0569

But, often that is not the case.0588

Often, what you have to do is use just trial and error--and this is exactly how they did it, back in the old days.0590

They collected a bunch of data; so, for example, they will put a certain concentration of nitrogen, a certain concentration of hydrogen, and they will come back to it after a certain amount of time (once it has reached equilibrium), and they will measure the concentrations, and they will just collect data for different concentrations.0594

Sometimes they will start with all nitrogen and hydrogen--no ammonia; sometimes they will start with only ammonia--no nitrogen and hydrogen; sometimes they will start with a whole bunch of one and a little bit of the other two--all kinds of different things.0613

And then, they measure, at equilibrium, what the concentrations are.0625

Well, once you have these numbers, you have to try to find a way to see if there is some mathematical formula, some equation, that actually fits those numbers.0628

When it comes to data, often we try to find a constant.0638

In other words, is there a relationship among the different values at equilibrium that stays the same?0641

And, as it turns out, there is, and it is called the equilibrium constant--the equilibrium constant expression.0649

So, a reaction of this type, where you have these reactants--A, B, C, and D capitalized are the species; a, b, c, and d small are the coefficients--the equilibrium expression looks like this.0654

We write Keq; we will also write K; we might not always write "eq," but any time you see a K something, with some subscript, it is an equilibrium kind of constant.0667

It is equal to the concentration of C, raised to the c, the concentration of D, raised to the d; over the concentration of A, raised to the a, and the concentration of B, raised to the b.0678

What that means is that, at equilibrium, if I measure the concentrations of A, B, C, and D, and if I take the concentration of C, raise it to its stoichiometric coefficient, multiply it by the concentration of D, raised to its stoichiometric coefficient (and remember, concentrations--these brackets always mean moles per liter--always), over the concentration of the reactants (A raised to its stoichiometric coefficient and B raised to its stoichiometric coefficient)--as it turns out, this number stays constant.0696

Later, towards the end of this lesson, actually, we will see a wonderful example of that, where we start with a bunch of different concentrations, and the concentrations are different at equilibrium.0727

In other words, the equilibrium position is different, but the constant is the same for every single particular experimental set of conditions.0738

The equilibrium constant is a mathematical representation of the equilibrium; it is a fingerprint for that reaction, at a given temperature.0746

The K is temperature-dependent; so, when you see these problems, it will often give you a specific temperature.0756

It changes with temperature, which makes sense--because, if you remember the Arrhenius equation, the Arrhenius equation says the rate of a reaction is dependent on temperature. 0760

Oftentimes, the equilibrium will change as rates change.0769

So, that is it; this probably the single most important thing; if you don't take anything away from chemistry, take this away from chemistry.0774

We wait for a system to come to equilibrium; we measure the equilibrium concentrations and the relationship of the concentrations of the individual species at equilibrium.0782

That is what is important: this is at equilibrium; we cannot emphasize that enough.0791

We will talk about another quotient, which has the same form, but it's not at equilibrium; it is called the reaction quotient; but we use that to decide which way the reaction is moving at a given moment.0795

But, this is at equilibrium; so when you see a Keq or a K (oops, we don't want these random lines)...you know what, let me just leave it as Keq for the time being (and let me correct these lines; I don't want you to think that I am cancelling anything out here; this is D...), this is at equilibrium; very, very important.0807

OK, so let's just do a couple of examples.0838

Well, actually, before we do that, let's do some rules regarding the equilibrium constant expression, the Keq.0841

So, some rules--very, very simple; #1: Pure liquids and solids: if you have a pure liquid or a solid as part of the equation, they don't show up in the equilibrium expression: "as reactants or products, are not included in the Keq expression."0849

OK, so if you have a solid or a liquid in reactants or products, it is not included in the equilibrium expression.0898

2: Reactants or products which are aqueous (in other words, basically, ionic compounds)...I'll put aqueous ionic, because I think it's important, because we can certainly have covalent compounds that are aqueous, like sugar, which actually do show up in equilibrium expressions, because they are not ionic; so, aqueous ionic...must be expressed in free ionic form.0906

"Must be expressed"--and this is often where kids get into the most trouble, when they are putting together equilibrium expressions.0943

They don't really stop and realize that, when they are looking at a molecular formula, the molecular formula gives the combined ion, like silver nitrate or sodium sulfate.0950

Well, silver nitrate and sodium sulfate, in solution, are free ions, silver ion and nitrate ion.0961

It isn't silver nitrate that shows up in the expression; it's actually the silver ion, the nitrate ion, the sodium ion, the sulfate ion; so it's very, very important.0967

We'll do an example in just a minute to make this perfectly clear.0975

So, reactants or products which are aqueous ionic must be expressed in free ionic form.0978

But you know this already, because you know that, when you are dealing with ionic compound mixtures, you want to write the net ionic reaction, not just the molecular equation, because you have some spectator ions.0984

Spectator ions are ones that don't participate in the chemistry; we want only the things that participate in the chemistry.0993

In other words, we want the net ionic reaction...free ionic form.1000

OK, so let's do an example.1007

Example 1: Write the equilibrium expression (Keq) for the following.1012

OK, we have A: so, we have: N2 + 3 H2 goes to NH3.1034

And we say "goes to" simply out of habit; again, we are talking about equilibrium now--from now on, it is always going to be some kind of an equilibrium that we are talking about, unless otherwise stated.1046

So, this is gas, and this is hydrogen gas, and the formation is NH3 gas.1056

Well, the Keq expression for this is going to be (this is a gas--it's not a liquid or solid--so it does show up in the equilibrium expression)...we have: the concentration of NH3 (oops, this is a 2 here) raised to the 2 power, because that is the stoichiometric coefficient.1063

It is products over reactants; again, the Keq is always products over reactants.1082

Then, we have (in the reactant side) the nitrogen gas concentration, raised to its stoichiometric coefficient (which is 1); and then we have the hydrogen gas concentration, H2, raised to its stoichiometric coefficient.1088

This is the equilibrium expression.1104

At equilibrium, if I measure the ammonia concentration, the nitrogen concentration, and the hydrogen concentration; and if I plug them in to this number; no matter what set of conditions I start off with, the ratio (this expression) will be constant.1107

It is a fingerprint for that reaction at a given temperature; it will not change.1123

OK, B: Here we have an example of an ionic situation: If I take a solution of potassium iodide, and if I mix it with a solution of silver nitrate, the question is: what is going to be the equilibrium expression when I have silver iodide (solid, because that is going to be the precipitate) and potassium nitrate?1129

So again, we are looking at an ionic situation; we are looking at this double-displacement reaction, where the potassium goes with the nitrate and the silver goes with the iodide.1153

The silver iodide drops out and falls to the bottom of the flask as a solid--that is the precipitate--and these stay aqueous.1163

In other words, they are dissolved as free ions.1170

Let me go ahead and write the total ionic equation for this; it's a nice review.1174

Potassium iodide is soluble (and this is balanced, by the way--we have to balance it before we do anything else--standard procedure in chemistry: always balance)1179

K+ + I- (free ions), plus silver+, plus NO3- (free ions) goes to silver iodide (that is a solid; so a solid--the whole idea behind a precipitate is that it sticks together--water doesn't dissolve it; the bonds between the silver and the iodide are stronger than the bonds that the water might have...the solvation), plus K+, plus NO3-.1188

Well, K+ cancels K+ (oh, again with the lines!); NO3- goes with NO3-; and we are left with a net reaction of the following.1224

We are left with: silver ion, plus iodide ion, going to silver iodide, solid.1234

This is aqueous; this is aqueous.1244

Now, here is the interesting thing about it: we said that solids don't show up in the equilibrium expression; therefore, in the numerator, we just put a 1; that is it.1247

This iodide is in solution; it is aqueous; this silver ion is aqueous; so the equilibrium expression for this (let me actually rewrite the equation, since we are on a new page here: so...): I have: Ag+ + I- goes to AgI, solid.1259

The Keq expression, equilibrium constant expression--there is nothing over on the product side; so again, it's products over reactants; this is solid--it doesn't show up.1279

It is the only product; so nothing shows up at all in the numerator.1289

On the bottom, we have Ag+ to the first power, and we have I- to the first power.1293

So, it is the net ionic reaction that you use for your equilibrium expression.1300

It's very, very important to do this in any kind of situation that involves some sort of soluble ionic compound.1305

Now, over on the product side, you might have gas formation; you might form water, and you might form CO2, for example, if you end up mixing a carbonate with an acid.1311

You have to account for all of this; it is the net ionic reaction that plays a part, that gives you the particular equilibrium expression.1324

OK, let's do: 4 NH3 gas + 7 O2 gas forms 4 moles of nitrogen dioxide gas and 6 moles of H2O; and let's just say that this is liquid.1333

So again, we have gas, gas, gas, liquid; liquids don't show up in the equilibrium expression.1363

Therefore, the Keq for this is going to be...1368

Oh, here is another thing: now that we are talking about equilibrium, I personally get a little...there is a lot of notation involved in this, as you can tell.1373

You have brackets; you have symbols; you have charges; you have lines separating things; you have Keq; there is a lot of writing.1382

I personally...concentrations are always written with brackets, but since we are talking about equilibrium, and often we will continue to talk about equilibrium, and we are always going to be writing some kind of K: Keq, K, KP, whatever it is--we are always going to be talking about some equilibrium expression--instead of brackets, I prefer to use parentheses.1391

I hope that that doesn't confuse you.1414

Again, my parentheses here, in our discussion of equilibrium, always represents concentrations.1416

It is just so I can sort of make the writing a little bit faster, because my brackets tend to be a little sloppy.1423

So, Keq is going to be the NO2 concentration, raised to the fourth power, over the NH3 concentration, raised to the fourth power, times the O2 concentration, raised to the seventh power.1427

Notice, this did not show up in the numerator; it is a liquid; it does not show up.1446

Pure liquids, pure solids--pure liquids and solids--a pure liquid or solid is different that an aqueous solution.1452

You just show that in an aqueous solution, you have free ions floating around; that is not pure--that is just some ions floating around in solution; they are dissolved.1459

By "pure liquid," I mean like, for example, if the reaction said NaCl liquid; that doesn't show up...because NaCl--you can melt salt, and you can actually have a pure liquid salt, but that doesn't mean it actually shows up in the equilibrium expression.1467

So, differentiate pure liquid from aqueous; aqueous--yes; pure liquid--no.1485

OK, let's see what else we have here.1491

D: let's see another example--we have: Calcium carbonate, CaCO3--this time, this is going to be a solid, and it is in equilibrium with calcium oxide, which is also solid, plus--interestingly enough--CO2 gas.1497

You don't often see this--solids and gases in the same thing.1514

So, calcium carbonate decomposes into calcium oxide and CO2.1517

It is a reversible reaction.1521

Well, the Keq for this: well, this is a solid; we don't care about it; this is a solid--we don't care about it; we have just this, so as it turns out (and this is a balanced equation, so it's not a problem), it just equals the CO2 concentration.1523

That is it--nice and simple.1541

This equilibrium expression only depends on the CO2 concentration at a given temperature.1543

In other words, at a given temperature, no matter where I start, with this and this and this, at some point, the CO2 concentration will always be the same; it will always end up at the same point.1550

It is a constant; it is a fingerprint for this reaction at a given temperature.1563

OK, now, in case you are wondering why it is that solids and liquids actually don't show up in equilibrium constant expressions, it is the following.1572

I can take a gas, and I can change the concentration of it by changing the pressure, volume, temperature...things like that.1580

I can condense it or expand it; the concentration changes; I can change the volume.1586

A solid, whether it is, let's say, 1 gram of sodium chloride or 150 grams of sodium chloride--well, concentration-wise, concentrations of liquids and solids don't change, because the concentration is a measure of the amount per volume.1592

Well, in a solid and liquid, the amount per volume is a constant; so, when something is a constant, it doesn't show up in an expression because it is constant.1609

It is part of the equilibrium constant, if you will; it is part of it...it doesn't show up.1619

That is the reason why: gases and aqueous ions--yes; their concentrations can change.1624

For example, I can take one liter of a liquid; I can drop in a certain amount of salt and dissolve it; now, I have a certain concentration of sodium ions--let's say it's .5 moles per liter.1630

Well, I can drop in more salt; I haven't changed the volume--I have changed the amount.1646

So, now, the concentration has changed; this is why free ions and gases show up in the equilibrium constant expression; their concentrations are not constant.1650

But, for solids and liquids, I can't actually change the concentration of them--they are constant.1661

OK, let's do another example here.1666

Let's continue with using our nitrogen and hydrogen to form ammonia.1669

We have Example 2: we have N2 + 3 H2 going to 2 NH3.1677

OK, for this reaction, the following equilibrium concentrations were recorded at 128 degrees Celsius.1691

So again, most of these equilibrium things--they take place at a certain temperature, because Keq is temperature-dependent.1718

All right, so for this reaction, the following equilibrium concentrations were recorded at 128 degrees Celsius.1723

What this means is that, at 128 degrees Celsius, we measured the amount of ammonia, hydrogen gas, and nitrogen gas that there was in a flask, and here is what we discovered.1729

We discovered that the ammonia concentration is 3.1x10-2 Molar (moles per liter--actually, you know what, let me actually write out "moles per liter").1740

I have never really cared for that capital M sign; I actually like to have all of my units absolutely apparent.1755

OK, the nitrogen concentration was measured to be 8.5x10-1 moles per liter.1761

The hydrogen ion concentration was 3.1x10-1 moles per liter.1772

This is an equilibrium condition.1781

In a given flask, I measured the concentration of the three species involved in this reaction, and these are the concentrations that I measured.1784

This is at equilibrium--"the following equilibrium concentrations."1792

Now, we want to calculate the Keq, OK?1796

Well, we know what the Keq is; we know that it is the products over the reactants, raised to their respective stoichiometric coefficients.1802

So, we have the NH3 concentration squared, over the N2 concentration times the H2 concentration cubed.1811

Well, that equals...the NH3 concentration is 3.1x10-2 moles per liter squared (I'm going to leave off the units; I hope you don't mind--units are not altogether that important for equilibrium concentration calculations, because they are going to change, depending on these exponents; so we don't really worry about them all that much).1821

N2 concentration is 8.5x10-1.1841

And then, 3.1x10-1 cubed.1847

When we multiply all of this out and divide, we end up with 3.8x104; that is the Keq at that temperature for that particular reaction.1853

That is it--nice and simple: measure the concentrations and put it in there--simple, basic math--it's only arithmetic.1873

OK, now, let's calculate...this is...so now, we'll do Part B.1880

Now, let us calculate the following equilibrium: Calculate Keq for 2 NH3 going to 3 H2 +N2.1889

Notice, this is the same reaction, except it's reversed.1906

So now, I'm going to calculate the equilibrium based on ammonia being the reactant, and hydrogen and nitrogen being the products.1909

Well, again, it is just the Keq--by definition, it is always the same; it is always going to be products over reactants.1917

So now, we have the H2 concentration cubed, times the nitrogen concentration, divided by the ammonia concentration squared.1924

Well, you notice that this is just the reciprocal of your other Keq that we found.1936

Let's call this Keq...let's call it K prime...how about Keq prime?1940

That is this one; well, this is equal to 1 over the Keq; it's just the reciprocal, which makes sense--when we reverse a reaction, we just take the reciprocal of it to find...1949

OK, so it's going to be 1 over 3.8x104, and this one ends up being...let's see what we get; we get 2.6x10-5.1960

OK, and now we will do one more.1976

Part C is: Calculate the Keq for this one: 1/2 N2 (now we're going to change the coefficients) + 3/2 N2 goes to NH3.1980

Basically, we have taken the standard balanced equation, the one with the integral coefficients--the 1, the 3, the 2--and we have multiplied everything by 1/2.2002

We have divided everything; it is still balanced, but now, because the equilibrium expression is dependent on the stoichiometric coefficients, it is going to change a little bit.2014

So now, the Keq expression is equal to N2...I'm sorry, it's going to be NH3, divided by the concentration of N2 to the one-half power...2023

OK, this is not going to work; these lines all over are getting in the way.2043

All right; it always seems to happen at the bottom of the page--it's very interesting.2046

OK, N2 raised to the 1/2 power, times (oh, this is H2) the concentration of H2 raised to the 3/2 power.2051

Well, if you notice what this is--essentially, what we have done...so let's call this K double prime...this is equal to our initial Keq that we found up there for the standard, balanced reaction--this is equal to K raised to (oh, what is wrong with these lines!?) 1/2.2067

It is as if we have taken...not as if; we actually have; we have taken the equilibrium expression for the original balanced equation, the ones with the integral coefficients, and we have raised it to the power of the thing that we multiplied the equation by.2095

Going from the original equation to this equation, we have multiplied everything by 1/2.2112

Well, that means take the original Keq and just raise it to a power of 1/2.2117

In other words, take the square root of it; that is what these fractional powers mean.2122

Something to the 1/2 power means to take the square root of it.2126

And, when we do that--when we take 3.8x104 to the 1/2 power, we end up with...oh, you know what, I didn't even do the calculation; that's OK--don't worry about it; just put it in your calculator and plug it in.2130

So that is it; this is the original that we got; we multiplied the equation by 1/2 to get a new equation with new coefficients--still balanced; the Keq changes accordingly.2150

The Keq changes by simply taking the original and taking the square root of it.2160

So now, let's write down the take-home lessons of all of this.2165

If you reverse a reaction, that means the new Keq is the reciprocal of the original Keq, which makes sense.2167

If you reverse a reaction, you switch products and reactants; now what is on the bottom comes on top, and what is on the top goes on the bottom.2196

Now, if you multiply a balanced equation by a factor m, then your new K is equal to K raised to the power of m; that is it.2203

We are just saying that if you change the equation--do anything to it--the equilibrium expression also changes.2238

That is all that is going on here.2243

OK, let's do our final example, and we will sort of wrap up this basic introduction to equilibrium.2245

OK, Example #3: The following data were collected for N2 + 3 H2 going to 2 NH3 at a particular temperature--the temperature is not specified.2254

OK, so here is what it looks like: let's go ahead and do (no...yes, that is fine...you know what, I am actually going to start the data on another page).2288

So, the following data were collected for N2 + 3 H2 going to 2 NH3 at a particular temperature.2305

And now, I will go ahead and give you the table: this is the experiment; this is the initial concentration; the equilibrium concentration; and the actual K, which we know the expression for.2310

So, we have three experiments: we have one experiment: the N2 concentration was 1.000, and of course, concentration is in moles per liter.2334

The H2 concentration is at 1.000, and the NH3 concentration was 0.2346

Our equilibrium concentration, when we measured it, is 0.921, 0.763, and 0.157.2353

And the K ends up (so this is...now the K is equal to) NH3 concentration squared, divided by the N2 concentration times the H2 concentration cubed; that is our equilibrium expression.2366

The K ends up equaling 6.02x10-2.2386

When we did Experiment 2, we found the following.2393

We started with an N2 concentration, an H2 concentration, and NH3 concentration, and this time we did 0, 0, and 1.000.2397

I'll go over and discuss what all of this means in just a moment.2410

We ended up with 0.399, 1.197, 0.203.2413

And, when we calculated K, we get 6.02x10-2.2424

Notice, they are the same, even though these are not.2431

And we did one more experiment, and we started off, this time: nitrogen, hydrogen, and ammonia.2434

We did 2.00; we did 1.00 and 3.00.2444

We ended up with (oops, no, we don't want these stray lines)...we did 2.59; we did 2.77; and 1.82.2451

And, when we...6.02x10-2.2465

OK, so here is what is going on: we did three experiments.2470

The first experiment: what we did is, we measured initial concentrations; so we put in a flask 1 mole per liter of nitrogen gas, 1 mole per liter of hydrogen gas, and no ammonia.2474

We let the system come to equilibrium; we came back once it reached equilibrium--once we realized that these values are not changing; everything is constant--and we measured them.2486

It turns out that the nitrogen concentration was .921; hydrogen concentration was .763; and the ammonia concentration was 0.157.2497

We put these values, because they are equilibrium concentrations, into the equilibrium expression: ammonia squared over nitrogen times hydrogen cubed (let me rewrite the nitrogen to make it a little more clear here--we are dealing with N2).2506

And what we did was got the number 6.02x10-2.2520

That is fine; equilibrium concentrations--we stick it into the expression; we get a certain number.2525

Now, a whole different experiment: this time, the initial concentration of the nitrogen and hydrogen was 0.2529

We just put in ammonia in the flask, and we let the system come to equilibrium.2536

In other words, the ammonia started decomposing; now, nitrogen and hydrogen are forming.2542

At equilibrium, when the concentrations were constant and nothing was changing anymore, we measured the concentrations.2546

Well, now, the nitrogen is .399; hydrogen is 1.197; .203; completely different equilibrium conditions--completely different; we started differently, and we ended at different concentrations.2552

But notice, when we put these numbers into this expression, 6.02x10-2: the same number shows up.2564

Hmm, interesting pattern.2572

Experiment #3: This time, we start with a little bit of everything.2574

We start with 2 moles per liter of nitrogen, 1 mole per liter of hydrogen, and 3 moles per liter of ammonia.2578

Well, we let the system come to equilibrium; all the concentrations are now constant--no longer changing.2585

We measure it; we get this, this, this; we put this into the equilibrium expression, and what do you know--6.02x10-2.2591

This is proof that these things right here--these are equilibrium conditions; equilibrium conditions depend on the particular experiment that you are running at any given moment.2599

These individual numbers are different among the experiments, but the relationship (based on the equilibrium expression) of the products raised to their coefficients (stoichiometric coefficients), divided by the reactants raised to their stoichiometric coefficients--that number stays constant.2611

This is very profound: any time you have some data, and data which should not have anything to do with each other, when you mix and match and when you keep getting the same number over and over again, you have something very, very special there.2629

That is the whole idea; much of science is based on trying to elucidate some sort of a constant--what stays?--and this is what it is for the equilibrium condition.2642

At this temperature, this reaction--this is a fingerprint of that reaction.2652

Equilibrium does not change; equilibrium conditions change; the equilibrium constant doesn't change.2661

That is why we call it a constant; it is a fingerprint for that reaction.2667

OK, so last but not least, let's distinguish between the equilibrium constant and the equilibrium position.2672

These are equilibrium positions that vary, depending on the experiment.2681

This is the equilibrium constant, based on this expression; it does not change--it is not variable.2685

So, we speak about constants, and we speak about conditions--very, very simple; very important.2692

OK, so hopefully this has given you a sense of what equilibrium is.2698

Again, a profoundly important topic: those of you that go on into biology, biochemistry, biophysics...things like that--the equilibrium condition (or physics--any kind of science)...all systems tend toward some kind of an equilibrium.2704

It is the reason we are alive, because our bodies are maintaining--constantly are trying to maintain--a certain equilibrium.2722

In fact, if you want to give sort of a broad definition of disease, a broad definition of disease is a deviation from equilibrium; that means something is going wrong.2729

The body is trying to bring things back to equilibrium; equilibrium is where everything wants to be--the point of lowest energy.2738

We can express that mathematically, with chemical reactions, with these expressions.2746

It is actually quite amazing that it is this simple.2750

Thank you for joining us here at Educator.com and for AP Chemistry.2753

We will see you next time for a continuation of equilibrium; goodbye.2756

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

We are going to continue our discussion of equilibrium--in my opinion, absolutely the single most important concept, not just in chemistry, but in all of science.0004

Equilibrium is the thing that all systems tend to--this notion of balance; they don't like being off balance (too much of this, too much of that).0013

All systems tend toward equilibrium, and chemical systems are no different.0023

Last lesson, we introduced the notion of an equilibrium expression, this Keq where, beginning with any sort of concentrations of a particular reaction, once the system has come to equilibrium, we measure those concentrations at equilibrium.0027

We put it into the equilibrium expression, and it ends up being the constant.0043

Different equilibrium conditions=different values; but the relationship among those values is a constant.0047

Today, we are going to continue that discussion, and we are going to introduce a variation of the expression for reactions that involve a gas.0054

As it turns out, when you have a reaction that involves a gas, you can either work in moles per liter (like we did last time--you are certainly welcome to do so by taking the number of moles in a flask and the volume of the flask, and dividing to get your concentration in moles per liter) or, as it turns out, pressure--you can work with pressures, because pressure is actually a measure of concentration.0062

We will show you how, mathematically; it is actually very, very nice.0087

Let's get started.0090

OK, so now, let's just go ahead and write, "For equilibriums involving gases, the K can be expressed in terms of P, pressure."0112

What that means is that I can either express my equilibrium expression with moles per liter concentrations or with pressure.0135

I'm going to write the two expressions, and then we will see how they are actually related, because there is a relationship between the two, which allows us to go back and forth, depending on what the problem is asking.0142

Because sometimes, it might be easier to deal with pressure--sometimes, easier with concentration; it just depends.0153

So again, we are going to stick with our tried-and-true nitrogen, plus 3 moles of hydrogen gas, going to 2 moles of ammonia.0158

We know that the actual Keq expression in concentrations is: the concentration of ammonia squared, over the concentration of nitrogen times the concentration of hydrogen cubed.0170

Well, as it turns out, I can also express it this way: and now, I put a P down as a subscript to the equilibrium expression--P for pressure.0189

KP--it's the same thing, and it equals the partial pressure of NH3 (partial pressure means the pressure just of that gas in the container, because you have three gases in a container: one gas has one pressure, on has another pressure--we call those the partial pressures), squared (so again, everything is the same; it's products, over reactants, raised to the stoichiometric coefficients; the pressure of NH3, squared...), because this is a 2, divided by the pressure of nitrogen gas, raised to a 1 power, because this stoichiometric coefficient is 1, times the partial pressure of H2 cubed, because its stoichiometric coefficient is 3.0198

So now, let's talk about how these two are related.0240

Well, whenever we talk about gases, the one equation that we talk about is the PV=nRT expression.0244

Let me just erase a little bit of this and give me a little more room.0251

Pressure times volume equals (let me make my V a little more clear) the number of moles, times the gas constant, times the temperature in Kelvin.0254

Let's rearrange this a little bit.0267

I'm going to write P=nRT/V.0269

Now, I'm going to take two of these and combine them: n/V times RT: just a little bit of mathematical manipulation--I can do this.0275

I can take the denominator and put it with one of the numbers in the numerator.0286

Now, notice what I have: what is n/V?--n is the number of moles, and V is volume in liters; so, as it turns out, P and n/V are related by a constant, RT.0289

As it turns out, pressure is an alternative form of representing concentration in moles per liter.0308

n/V is just moles per liter, so this says, if I have something which is a certain number of moles per a certain number of liters, if I multiply that by RT, I actually get the pressure.0314

Therefore, instead of concentration, I can just put the particular pressure.0326

That is all this is: in dealing with gases, it is often difficult to...not difficult to deal in concentrations; it tends to be easier, simply by nature of gases, by measuring pressure.0330

That is all this is; so, whenever we have an equilibrium condition that involves some gases, we can use its pressure, because pressure is just an alternative form of concentration; that is it.0342

Heads, tails--it's just another way of looking at concentration.0353

Now, let's see what the actual relationship is between these two values.0357

OK, so P is equal to n/V(RT); so P is equal to concentration times RT (we will just use C as concentration, instead of writing it as n/V) or, we can write C=P/RT: concentration is equal to pressure over RT.0361

This is just standard mathematical manipulation.0383

So now, this is concentration; we are going to rewrite the expression here, and we are going to put concentration back in to see how it is related here.0386

We will do: Keq is equal to...it's pressure over RT, the concentration: so we get the partial pressure of NH3, over RT, squared.0397

So all I have done is: I have just taken this expression, put it into here, and divided by pressure of N2/RT to the first power, times the pressure of H2/RT (that is concentration squared, concentration raised to the first, concentration cubed, right?)--just basic math.0413

And then, I have partial pressure of NH3, squared, times 1/RT, squared, over the partial pressure of N2 times 1/RT, times the partial pressure of H2, cubed, times 1/RT, cubed--so far, so good.0443

I have: partial pressure of NH3, squared, over partial pressure of N2, partial pressure of H2, cubed, times 1/RT, squared, over 1/RT to the fourth power.0474

All right, and now, I am going to have...this 2 is going to cancel that; I'm going to end up with (let me write it one more time--well, two more times, actually): PNH3, squared, over PO2, times the PH2, cubed, times 1/(1/RT, squared) (because this cancels two of those, leaving that) and 1/(1/RT squared) is RT squared.0497

So, it equals the partial pressure of NH3, squared, over the partial pressure of N2 and the partial pressure of H2, cubed, times RT, squared.0542

Well, this is the Keq; that is this expression.0562

This expression right here is the KP (RT)2.0567

So as it turns out, Keq equals KP, in this particular case, times RT squared.0572

There is a relationship between expressing it with concentrations and expressing it with pressures.0584

The relationship is that the Keq equals KP (for this particular reaction--we'll do the general one in a minute), times RT squared.0590

I can go back and forth between the two; so, if I'm working with KP and I want Keq, in concentrations, I can do that.0599

If I have Keq and I want KP, I can do that by multiplying by the square of RT.0605

OK, so now, let's do the general version; I just wanted to show you where the math came from.0612

For the general expression, aA + bB in equilibrium with cC + dD, our relationship is the following.0618

From now on, I'm not going to put the Keq; I'm just going to put K; whenever you see K with no subscript on it, it just means moles per liter, concentration; K equals Keq.0630

So, K is equivalent to Keq; when we speak of P, we will write KP; that means we are dealing with pressures.0641

When we just see a K, without this eq, they are the same--simply to avoid writing the eq over and over again.0651

As it turns out, the K is equal to KP, times RT to the negative Δn, where Δn equals the sum of the coefficients of the products, minus the sum of the stoichiometric coefficients of the reactants.0659

We can write it that way; or, another way that it is written is in terms of KP.0681

KP is equal to K times (RT)Δn; either one of these is fine.0686

If you have K, you can find KP; if you have KP, you can find K by just using this expression; that is it--nothing more than that.0696

Δn is this coefficient plus that coefficient, minus that plus that; that is all.0705

Let's just do an example, and everything will make sense.0711

Example 1: OK, at 427 degrees Celsius, a 2.0-liter flask contains 40.0 moles of H2, 36.0 moles of CO2, 24.0 mol of H2O, and 11.8 mol of CO, carbon monoxide, at equilibrium.0718

So, at equilibrium, we measure 427 degrees; the system has come to equilibrium; we measure in a 2-liter flask; we find that we have this many moles of each of these species.0766

Now, the reaction is as follows: CO2 gas + H2 gas goes to carbon monoxide gas + H2O gas.0777

All of these are gaseous species: that means all of them are involved in the equilibrium expression.0797

Now, our task here is to find K and KP; so, find the equilibrium constant expression, in terms of moles per liter, and find the KP constant, in terms of partial pressure.0804

Well, we have the relationship, so we can find K, and then we can find KP, based on this--no problem.0824

OK, so now, this is our first example of a real equilibrium problem, in the sense that we really have to watch everything that is going on.0833

No two problems are going to read the same way; we can't follow an algorithmic procedure for solving every problem.0842

Physical systems--now, it just depends; different things can happen--you can have different data.0850

It can be worded in a certain way; you have to be able to extract the information that is necessary.0858

Yes, there are certain things that are universal, that you can always count on, but you have to watch every single little thing.0865

Notice here: they are giving you the volume of a flask, and they are giving you the amounts in moles; so you actually have to calculate the moles per liter, before you put it into the equilibrium expression, because the equilibrium expression requires that it be in concentrations and not in moles.0872

You have to sort of watch for that.0888

Let's write the equilibrium expression, K; we will do K first.0892

It is going to equal the products over the reactants, raised to their stoichiometric coefficients.0895

This is a balanced reaction, so everything is 1:1:1:1.0900

We have the concentration of CO, times the concentration of H2O (everything is a gas), over the concentration of CO2, times the concentration of H2 gas.0904

Well, the concentration is the number of moles per volume; our volume is 2 liters (I'm going to go ahead and do this in red), and our moles are 40, 36, 24, and 11.8.0915

So, CO--the concentration of CO, moles per liter of carbon monoxide--is 11.8; so this is going to be 11.8 moles per 2 liters.0932

I take the number of moles and divide by the liters; it has to be concentration.0945

Times the concentration of H2O: H2O is 24 moles, so it's 24.0 divided by 2.0950

The concentration of CO2: well, CO2 is 36 moles, 36.0 moles; it is sitting in a 2-liter flask, so its concentration is 36/2, or 18 moles per liter.0959

And now, the H2 is 40 moles in a 2-liter flask.0973

Now, you might think to yourself, "Well, wait a minute; 2, 2, 2, 2: don't the 2's cancel?"0977

Yes, in this case they cancel; that is because all of these coefficients are 1, 1, 1, 1.0981

But, you can't guarantee that all of the coefficients are 1, 1, 1, 1, so you can't just use the mole values in here.0989

The equilibrium expression explicitly requires that you use concentrations, moles per liter, so let's just use moles per liter.0996

Yes, they will cancel, but at least we know what is going on; we won't lose our way.1003

That is what is important: write down everything--don't do anything in your head; don't cut corners; don't take shortcuts.1007

I promise, it will go badly for you.1013

OK, when we calculate this: 0.197; that is what we wanted--we wanted the K.1015

Now, we want the KP, because that is the other thing that we have to find.1026

Well, we know that KP is equal to K times RTΔn.1029

Well, what is Δn?--Δn is adding the product coefficients and subtracting the other coefficients.1036

Well, Δn is (let me go ahead and write it here) equal to 1+1 (is 2), minus 1+1 (2), is equal to 0.1045

So, in this case, Δn is 0.1056

KP is equal to K, times RT, to the Δn, equals K, times RT to the 0 (anything raised to the 0 power is 1); is equal to K.1058

So, in this case, KP is equal to 0.197...in this case; and the only reason it is this way is because everything is in a 1:1:1:1.1079

2 reactants, 2 products: 1+1 is 2; minus 1+1 (is 2) is 0; that is it.1091

Watch what you are doing very, very carefully; do not cut corners on equilibrium--do not cut corners ever, in any problem that you do.1098

Write everything out; it's very, very important.1104

OK, let's see: now, let's do a slightly more complicated problem, and again, this is going to be an example of taking information that is written in a particular problem and reasoning it out.1107

It isn't just the math--the math is actually pretty simple once you know what is going on.1125

That is the biggest problem with chemistry, or physics, or anything else: it's not how to turn it into math; it is, "What is going on, so that I know which math to use?"1130

That is the real issue, and no two problems are the same; no two problems will be worded the same; you cannot count on that, especially at this level.1139

OK, so Example 2 (and again, we are going to do a lot of problems, like I said, from here on in--from equilibrium all the way through at least electrochemistry, because this is the heart and soul of chemistry, not to mention the free-response questions that you are going to face on the AP exam)...1148

OK, so the question is a bit long: A sample of gaseous PCl5 (phosphorus pentachloride) was placed in an evacuated flask so the pressure of pure PCl5 would be 0.5 atmospheres.1172

So, a sample of gaseous phosphorus pentachloride was placed in an evacuated flask so the pressure of the pure PCl5 would be .5 atmospheres.1221

We stuck it in there so that it would be .5 atmospheres.1229

But, PCl5 decomposes according to: PCl5 goes to PCl3 + Cl2.1232

The final total pressure in the flask was 0.84 atmospheres at 250 degrees Celsius.1259

Calculate K at this temperature.1287

OK, so let's make sure we understand exactly what this problem is asking.1298

Problems are going to be very, very specific; do not read into it anything that is there--read exactly what is there--very, very important.1302

Don't cut corners.1308

A sample of gaseous PCl5 was placed in an evacuated flask so the pressure of pure PCl5 would be .5 atmospheres.1310

In other words, they filled it up with gas, and the pressure of the PCl5 gas was .5 atmospheres.1316

The problem is: "But, PCl5 decomposes"--in other words, the PCl5 they put there at .5 atmospheres--all of a sudden, it starts to come apart.1322

It decomposes into PCl3 and Cl2.1331

Well, now there is an equilibrium that exists; now, you not only have PCl5 at .5 atmospheres; now, you have also produced some phosphorus trichloride gas and some chlorine gas, and there is also some PCl5 gas left over.1334

This is an equilibrium expression; so now, it is becoming a little bit more complicated.1347

Now, you have three things in the flask, when you only introduced one.1351

We measure the final pressure of the flask, and it comes out to .84 atmospheres at 250 degrees Celsius.1354

Calculate the K at this temperature (K, we said, is K equilibrium, moles per liter).1363

They have given this to us in pressures; so, the first thing we have to do is: we have to find KP and then calculate K.1369

So, we want to find (well, if it's OK, I'm not going to write out everything; we know what we are doing)...so notice: they didn't write and say "Find KP."1379

They said, "Find K"; but the problem as written, since we are dealing with atmospheres and gases (PCl5 gas, PCl3 gas, Cl2 gas)...we are going to deal with pressures, and then from pressures, we are going to use the RT and Δn expression that we just worked with to find the K.1387

Make sure that you understand what it is that they are asking for.1404

Don't just stop by getting the KP.1407

OK, let's see how we are going to do this.1411

We are going to introduce something called an ICE chart; and this ICE chart is something that we are going to use for absolutely the rest of the time that we discuss this.1414

All equilibrium problems--acid-base problems, further aspects of acid-base equilibria, solubility-product equilibria, electrochemistry--we are going to deal with these things called ICE charts.1422

This is sort of an introduction to them, and it is a way of dealing with what is going on in the problem.1433

If you understand these, everything should fall out naturally.1438

OK, so let's write the equation.1442

PCl5 is in equilibrium with PCl3 + Cl2.1445

ICE stands for Initial concentration, before anything happens; C stands for Change--what changes take place; and E stands for Equilibrium concentrations.1452

Well, it is these equilibrium concentrations that go into the equilibrium constant expression, right?1464

That is what we said: the Keq: those values that we put in there are concentrations at equilibrium.1469

I means Initial; C means change.1475

Even if you are not sure what to do in a problem, just start, and just write down what you know; write down what is happening; eventually, the solution will fall out.1482

The single biggest mistake that kids make is: they think that they are supposed to just look at a problem and automatically know what is going on.1491

Even I don't just look at a problem and know what is going on!1497

After all of these years of experience, I have to sit there and stare at it, and sometimes just see where I am going.1500

ICE chart is a great place to start with equilibrium problems.1506

It will give you a sense of what is happening; then, you can put the math together.1511

Don't ever feel that you have to just know what is happening; you are extracting information.1514

And...sorry about that--E stands for equilibrium.1520

OK, so our initial concentration of PCl5 was .5 atmospheres.1528

Remember what we said: pressure and concentration--they are the same; they are just different sides of the same coin, so we can deal in pressures the same way we deal with concentrations.1534

We start with 0.5 atmospheres, where, before anything happens--before PCl5 decomposes--there is no PCl3, and there is no Cl2; so this is our initial condition.1544

It's nice; well, a certain amount of PCl5 decomposes.1555

Well, look at our equation; it's 1:1:1.1561

For every 1 mole that decomposes, 1 mole is produced, and 1 mole is produced of the Cl2.1563

We can say that, if -x is the amount that disappears (or, again...concentration and pressure are the same thing)...so, if the pressure drops--PCl5--by a certain amount, that means it has to increase here by that same amount for each.1571

For every 1 atmosphere that drops, that means a certain amount has been used; well, that means that a certain amount has been produced of the PCl3 and the Cl2.1591

That is what the stoichiometry tells me.1598

That is why I have -x, +x, +x; I hope that makes sense--because, again, when this decomposes, this is forming; that is what is going on.1600

So, if 1 mole of this decomposes, 1 mole of this is formed; 1 mole of this is formed.1610

If 5.2 moles of this decomposes, 5.2 moles of PCl3 is formed; 5.2 moles of Cl2 is formed.1614

We don't know how much is formed yet, so that is why we use x: -x here, +x, +x.1621

Now, we add: the initial plus the change gives us the equilibrium condition.1627

At equilibrium, I have 0.50-x.1633

Here, I have 0+x is x; 0+x is x; so at equilibrium, this is how much I have.1637

Now, they want us to find K; well, what other information do they give us?1645

They give us the fact that our total pressure is equal to .84 atmospheres.1651

Well, our total pressure is the pressure at equilibrium: this plus this plus this.1657

The total pressure is the sum of the individual pressures; so, we write: 0.50-x+x+x (write it out; don't do it in your head first) =0.84 atm.1665

Well, this -x cancels with that x, and I'm left with 0.50+x=0.84; this is simple arithmetic--there is nothing hard about this.1682

x=0.84-0.5; it equals 0.34 atmospheres; look at that.1693

I have just solved for x, x, x; I did it; that is nice.1701

I found the value of x: .34 atmospheres of PCl3 show up; .34 atmospheres of Cl2 show up; and .34 atmospheres of PCl5 is lost.1710

So now, we can actually find our K.1724

That is the best part, which is ultimately what we want; so we are solving for KP here.1727

So now, I have partial pressures: I have the partial pressure of chlorine gas, is equal to x, which is 0.34 atm; I have the partial pressure of the PCl3 gas, which, again, is x, which is 0.34 atm; and I have the partial pressure of the PCl5 gas, which is 0.50-x (that is the equilibrium) minus 0.34, which is 0.16 atm.1732

Now, I can put these values into my equilibrium expression.1767

I know what my equilibrium expression is; my equilibrium expression is KP equals the partial pressure of Cl2, times the partial pressure of PCl3, divided by the partial pressure of PCl5, each raised to the first power, because the stoichiometric coefficients are 1.1772

Well, that equals (let's go down here) 0.34, times 0.34, divided by 0.16; I do the multiplication, and I get 0.72; this is my KP; KP equals 0.72.1792

They didn't ask for KP; they asked for K, which means they asked for Keq.1815

Well, the relationship between K and KP is the following.1822

K is equal to KP times (RT)Δn, if that is correct; let me double-check; KP equals...-Δn.1826

OK, so we have...let's see...yes, OK; so now, what is Δn, first of all?1847

Δn is equal to...well, we take the...again, let's write out the equation, so we have it on this page.1862

We have PCl5 in equilibrium with PCl3 plus Cl2; Δn is equal to 1+1-1; 2-1=1.1871

So, K is equal to KP (which is 0.72), times RT...OK, so here is where we have to be careful; R is not...well, the R that we are going to be using here is 0.08206; that is liter-atmosphere/mole-Kelvin; times temperature in Kelvin; it has to be in Kelvin, because the unit of this is liter-atmosphere...1887

You know, let me write out the units here, so you can see it.1920

This is liter-atmosphere per mole-Kelvin; so the temperature has to be in Kelvin.1925

At this particular temperature, add 273; you get 523 Kelvin to the -1 power (negative Δn; that was the relationship).1935

You put in KP; R is .08206; temperature in Kelvin is 523; negative Δn...Δn was 1; negative 1.1948

We do the math, and we end up with 0.017.1957

There you go; in this particular problem, there was a lot going on; we handled it by just sort of stopping, taking a look, and making sure we wrote everything out.1966

We introduced this thing called an ICE chart; we write out the equation on top; underneath, we write the initial concentrations, we discuss the changes that take place, and we add to get our equilibrium.1976

From there, we take a look at what the problem is asking.1988

Sometimes, they might give us a K, and they might ask for a particular concentration.1991

I take those equilibrium concentrations, and I put them into my expression, and I solve that way.1996

In this case, they gave me a total pressure so I could find x.2001

I used that to find the K; it just depends on what they are asking.2005

This is why we are going to do a lot of different types of problems for these equilibrium and so on...at least through electrochemistry.2009

OK, so let's do another example.2018

Example 3: We have: Solid ammonium chloride, solid NH4Cl, was placed in an evacuated chamber, then heated; it decomposed according to: NH4Cl, solid, decomposes into ammonia, NH3, gas, plus hydrogen chloride gas.2027

It's not hydrochloric acid; it's hydrogen chloride gas.2086

Now, after heating, the total pressure was found to be 4.4 atmospheres.2090

Our task is to calculate the KP, the equilibrium constant with respect to pressures.2119

OK, well, let's write out the equilibrium expression first; it's always a great thing to do--write everything out.2125

Write out the equilibrium expression; write out the ICE chart; and then, just see where you go from there.2132

So, the equilibrium expression here, based on this equation--well, we have a gas; we have a gas; and we have a solid.2137

Solids don't show up in the equilibrium expression, so in this case, it is just these two.2143

The coefficients are 1, so we have: the KP is equal to the partial pressure of NH3 gas, times the partial pressure of HCl gas.2148

That is it; if we find the partial pressures, we are done--we plug them in, we multiply them, and we are done.2157

OK, now let's do our ICE chart.2163

Do our ICE chart: always do it this way.2165

NH4Cl (write out the equation, and do it underneath; don't do it separately--you want to be able to keep everything straight) goes to NH3 + HCl.2169

Our initial concentration; our change; and our equilibrium concentration...2183

OK, solid NH4Cl--we don't care; it doesn't even matter, so we just put lines there; it doesn't show up in the equilibrium expression--it doesn't matter.2188

Our initial NH3 and HCl concentration--well, we started off with solid NH4Cl; so, initially, there is none of these.2198

The change--well, a certain amount shows up; that certain amount is what we want.2208

So, +x+x, right?--so again, you have to be able to see what the question is saying.2214

It is telling you that you start off with NH4Cl; it decomposes--when something decomposes, that means it is going away; the products are showing up.2220

That is why you have a +x and a +x here.2229

This + goes here; it's not that +.2235

0+x is +x; 0+x is +x; well, they are telling me that the total pressure in this flask is 4.4 atmospheres.2239

Well, which gases are in the flask?2251

Well, the solid is a solid; that doesn't matter--that doesn't do anything for the gas.2256

The gases in here are NH3 gas and HCl gas.2261

So, I basically have: x+x=4.4; they are telling me that the total pressure in there is 4.4 atmospheres; that has to be made up of the amount of NH3 gas and the amount of HCl gas.2265

Well, that is even, because they are forming a 1:1 ratio.2283

So, 2x is equal to 4.4; x is equal to 2.2; well, 2.2--there you go; that is the partial pressure of HCl and the partial pressure of NH3.2287

I have 2.2 and 2.2 (2.2, not 2.4...oh, numbers and arithmetic!); these numbers are the one that I put back in here.2307

So, KP is equal to 2.2, times...OK, now watch this: even though 2.2 and 2.2 is the same, please don't write 2.22.2319

They are different species; I promise you, if you write 2.22, and if somewhere along the way you get lost and you have to come back, you will spend five minutes trying to figure out what happened, because again, stoichiometric coefficients--they show up in the equation, so write them separately; write 2.2 times 2.2.2332

Even though they are the same, they represent different species.2352

Don't mix them up.2355

We multiply that; we get 4.84; 4.84--that is the KP, and we double-check: "Calculate KP"--that is what we wanted; we're done; that is it.2357

So, they gave us a certain amount of information; they gave us an equation to work with; we wrote down the KP expression; we wrote down the equation.2370

We wrote an ICE chart (Initial, Change, Equilibrium--ultimately, it is the equilibrium that we are concerned with, because the system has come to equilibrium).2380

We followed it; we get x and x.2388

They tell us that the total pressure is 4.4; well, the total pressure is the sum of the individual pressures.2392

There are only 2 gases in here (the NH3 and the HCl); each one is x.2397

2x=4.4; x=2.2; at equilibrium, 2.2 atmospheres is hydrogen chloride gas; 2.2 atmospheres is ammonia gas.2401

You plug that into the equilibrium expression; we multiply, and we get 4.84: a standard equilibrium problem.2413

OK, this sort of gets us going with the types of problems we are going to be dealing with with equilibrium.2419

We are going to systematize this, and this whole idea of the ICE chart--our problems are going to start to become a little bit more complex, but this whole idea of using an ICE chart, writing the equilibrium expression, and then seeing what they want, based on what they give us--the ICE chart itself is going to be different.2425

That is what is going to change.2443

The approach does not change.2445

OK, so thank you for joining us here at Educator.com for AP Chemistry and equilibrium.2448

We will see you next time; goodbye.2452

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

We are going to continue our discussion of equilibrium, and today we are going to introduce something called the reaction quotient.0005

As you will see in a minute, the reaction quotient is just like the equilibrium expression, except it's used at any given time during the reaction.0011

It actually will tell us in which direction the reaction must go in order to reach equilibrium.0021

Sometimes it's too far to the left; it wants to go to the right; sometimes the reaction is too far to the right; it wants to go to the left.0027

Sometimes, it is exactly at equilibrium.0032

So, this reaction quotient is going to be one of the fundamental things that we use, and we will actually see it over and over again as we continue to decide where our reaction is and in what direction it needs to proceed.0035

And, as you will see when we start doing the problems, it is going to describe the mathematics--what are we adding? What are we subtracting?--things like that.0045

So, anyway, let's go ahead and get started.0055

OK, so let's do a quick review; so let's do this in blue.0059

We'll take our equation that we have been dealing with, which is nitrogen gas, plus 3 moles of hydrogen gas, in equilibrium to form 2 moles of NH3 gas--ammonia.0065

OK, now we said that the equilibrium expression, K...we are no longer writing Keq, remember--if we just do K, that will mean that we are talking about moles per liter, as opposed to KP, which means we are definitely dealing in partial pressures, either in torr or atmosphere, as long as the units are consistent.0079

OK, so K is products over reactants, so it is going to be the concentration of NH3 squared, over the concentration of N2 (and again, I am using parentheses for concentration instead of the standard brackets, simply to make this a little cleaner, because my brackets tend to be a bit messy) and H23.0098

So, this is our equilibrium expression; and now, just as a review for what this equilibrium expression tells us: depending on what this number is--if it's a really, really big number, well, notice: if it's a big number, that means the numerator is a lot bigger than the denominator.0121

That means that there is more product than there is reactant.0137

That means that the equation favors the products; that means that, at equilibrium, most of what is in that flask is going to be products.0142

We say it is really far to the right.0151

If this is a really small number, this ratio, that means that the numerator is small and the denominator is big.0153

That means that the reactants actually are favored in the equilibrium; so at equilibrium, you are going to find most of the stuff on the left-hand side; it's going to be mostly nitrogen and hydrogen, instead of the ammonia.0162

That is what this equilibrium constant is a measure of; it is a measure of the extent to which a reaction moves forward or doesn't move forward.0173

That is all it is; it is just a numerical value expressing that.0181

So, we can speak about it qualitatively--"It's far to the right; it's far to the left"--or we can be precise and quantitative--"It is far to the right, because the Keq is 500," or "The Keq is .001."0185

This tells us specifically how far to the left or right; so that is all that that is.0197

OK, now, as we said, there is also something called the reaction quotient, which we can use to tell us in which direction a reaction must go in order to reach equilibrium--in other words, where it is at that given moment.0202

OK, so let's go ahead and define our Q; this is called the reaction quotient.0216

It is the same expression as the equilibrium constant; so, in other words, it's the same thing as this (products over reactants, raised to their stoichiometric coefficients).0229

It's the same expression as Keq, but concentrations/pressures are taken at any given moment.0241

So you know that this constant--equilibrium constant--it is the ratio of these concentrations, once the system has come to equilibrium.0275

Well, we can measure the concentration any time we want (of the NH3, the N2, the H2, or the products, or the reactants), and we can put those into this expression, and we can see how far away from the equilibrium constant it is, and that will tell us whether it is too far to the left or too far to the right.0283

That is all we are doing here; so, for a general reaction, aA + bB going to cC + dD, we have that the reaction quotient, Q, is equal to the concentration of C raised to the c power, the concentration of D raised to the d power, over the concentration of A raised to the a power, the concentration of B raised to the b power.0304

It is exactly the same as the equilibrium expression, except these concentrations are at any given moment.0330

That is all; now, here are the criteria by which we decide where a reaction is--how far from equilibrium.0337

If Q is bigger than K, then the reaction will proceed (let's say...yes, you know, I think "proceed" is a good word; I wanted to use "shift," but I think "proceed" is better, or "move") to the left to reach equilibrium.0346

In other words, if Q is bigger than K, that means this is bigger than that, that means it has gone too far to the right; or it has not gone too far to the right--it is too far to the right.0381

In order for it to reach its equilibrium point (which, as we said, is a fingerprint for that particular reaction at a given temperature), it needs to move to the left; that means it needs to decompose product to form more reactant.0396

It needs to decompose this to form more this; it needs to move to the left--that is what that means.0408

So now, if Q is less than K, well, it's just the opposite: then, the reaction will proceed to the right to reach equilibrium.0413

At any given moment, if we take a bunch of concentrations of products and reactants and we stick it into this expression, we solve it, and it ends up being less than the K, that means there is too much of this and it needs to move in this direction to reach equilibrium.0438

That means this denominator is too big; Q is too small.0452

It needs to go this way, so it's going to proceed to the right to reach equilibrium.0456

So, in this case, reactants are depleting; products are forming.0461

And of course, last but not least, if Q equals K, well, you know the answer to this one: then the reaction is at equilibrium (here we go again with the stray lines; OK, that is nice).0467

That is it--nice and simple; a mathematical way to see where a reaction is and to see where a reaction is going.0487

OK, let's go ahead and do an example.0496

Example 1 (let's see): For the equation H2O gas (you know, that's OK...well, I don't know; I want to sort of skip writing the gas, but I guess it's pretty important) + Cl2O gas (oops, can't have a double arrow going--I need it to go that way and that way) forms 2 HOCl gas at 25 degrees Celsius, the equilibrium constant equals 0.0900.0510

OK, so: for the reaction H2O gas + dichlorine monoxide, that goes to 2 HOCl gas (hydrogen hypochloride gas, OK?--this is not hypochlorous acid; it's in the gaseous state, so it's not aqueous, so it's not the acid--this is the hydrogen hypochloride), the equilibrium constant for this reaction at 25 degrees Celsius is .0900.0561

Notice, there is no unit here; we will get to that in just a minute.0586

Here is what we want to do: For the following concentrations, determine the direction (I don't want to run out of room over there) the reaction must go in, in order to reach equilibrium.0590

OK, so: For the following concentrations, determine the direction the reaction must go in order to reach equilibrium: a standard reaction quotient problem.0637

It is going to be basic; we are going to do it for most equilibrium problems, because we want to know where equilibrium is.0645

OK, so the first one: we have: A partial pressure of H2 O is equal to 200 torr, and the partial pressure of Cl2O is going to equal 49.8 torr, and the P of HOCl is equal to 21.0 torricelli.0651

Now, our Q is equal to the partial pressure of HOCl squared, over the partial pressure of H2O times the partial pressure of Cl2O.0680

Well, that equals 21.0 squared (that is the HOCl), divided by the partial pressure of H2O, which is 200, and it is 49.8; there we go.0694

Now, in this particular case, this is torr and this is torr squared; so the unit on top is going to be torr squared; this is torr; this is torr; it's going to be torr squared, so it's going to end up without a unit.0715

That is why this equilibrium constant doesn't have a unit.0725

It is because the torricelli, torricelli cancels with torricelli, torricelli, down at the bottom.0729

So, now, we get that Q is equal to 0.0443; so Q is equal to .0443; K is equal to .0900; clearly, Q is less than K, which implies that the reaction will move to the right to reach equilibrium.0733

In other words, reactant will deplete; product will form; it will move to the right to reach equilibrium.0766

It hasn't reached equilibrium yet; it's still moving to the right to reach eq.0774

That is it; that is all you are using the reaction quotient for--to tell you which direction it is going in.0780

OK, all right, let's see: so, let's do another one--another set of conditions.0786

This time, we will do: A 3.0-liter flask contains 0.25 moles of HOCl, 0.0100 mol of Cl2O, and 0.56 mol of H2O.0796

OK, notice: they give us moles, and they give us the actual volume of the flask.0828

Well, again, when we deal with these reaction quotients and equilibrium expressions, they have to be in moles per liter--in concentrations.0832

Let's just take each one and find the concentrations before we put them into our reaction quotient.0841

So, our concentration of HOCl is equal to 0.25 mol, divided by 3.0 liters; that is going to equal 0.0833 Molar (that m with a line over it means molarity; it's an older expression; you are accustomed to seeing it: capital M).0846

The Cl2O: that is equal to 0.0100 moles, again divided by 3 liters, because it is in the flask; so, its concentration is 0.00333 Molar.0875

And finally, our H2O concentration is equal to 0.56 mol, divided by 3.0 liter; and this concentration is 0.1867 Molar.0895

Now that we have the molarities, we can put them into our reaction quotient; the reaction quotient is going to be exactly the same thing.0913

Now, we have: Q is equal to the concentration, as we said, of HOCl squared, over the concentration of H2O, times the concentration of Cl2O, which is equal to (drop down a little bit here) 0.0833 squared, times 0.1867, times 0.0033; and we get 11.16.0922

Now, Q is 11.16; we said that K was equal to 0.0900; so, clearly, Q is much larger than K, which implies that the reaction will move to the right to reach equilibrium.0968

In other words, it is still moving--I'm sorry; not to the right--to the left!0992

The Q is bigger; yes, Q is bigger--it is going to move to the left.0998

Sorry about that; that means that there is too much product at this temperature, given the equilibrium, which is a fingerprint for that reaction; so, there is too much product; the product needs to decompose to form reactant.1004

It is moving to the left to reach equilibrium.1019

That is it; OK, now, again, notice that there are no units for that; there are no units for this particular equilibrium constant.1028

In fact, notice in the question: the question actually said, "Equilibrium constant equals"--it didn't say "K equals" or "KP equals."1040

If you are given Keq, K, or KP, it will specifically mean that we are talking about pressures, or we are talking about moles per liter; but in this particular case, because there is no unit, that actually means that the KP and the K are the same.1050

You remember the definition of the relationship between KP and K; well, the fact that there is no unit means that there is no...if you look at the equation, Δn=0, so that RT that we had is 1.1069

So, when it says the "equilibrium constant," but it doesn't specifically specify whether it is a K or a KP, well, it's the same thing--they are actually equal to each other, which is why we use the same number, .0900, with molarity and with the one that we just did, which was done in terms of pressure.1085

In both cases, we use the .0900; that comes from the fact that there is no unit that tells us that the K and the KP are the same.1103

These are the little things that you have to watch out for; in other circumstances, when you do have a unit, you have to watch out; you have to actually (if you are dealing with molarity and you are given the KP, you have to) either convert the KP to a K, or you have to convert the molarities to pressures, if you can, depending on what the problem is asking.1112

Again, these are the sort of things; there is a lot that is going to be going on--there is a lot that you have to watch out for.1132

It isn't just "plug and play"--you don't just put numbers into an equation and hope things will fall out.1139

You have to understand what is happening; this is real science, and real science means conversions, units, and strange things.1145

OK, so let's do another example.1155

Here is where we begin to actually explore some of the diversity of these equilibrium problems.1160

What we are going to do is: most of our learning is actually going to come through the problems themselves.1165

That is why we are going to do a fair number of these equilibrium problems; it's very, very important that you have a reasonably solid understanding of how to handle these things, because it's going to be the bread and butter of what you do for the rest of chemistry--certainly for the rest of the AP and the free response questions; the electrochemistry; acid-base; the thermodynamics.1170

It's precisely this kind of reasoning, and equilibrium is fundamental to it all.1191

That is why it comes before everything else does.1196

Equilibrium is chemistry; it is that simple.1198

OK, so let's do Example #2: let's do this one in red--how is that?1203

OK, Example 2: The question is going to be a bit long, but...let's see.1211

At a certain temperature, a 1.0-liter flask contains 0.298 mol of PCl3 and 8.70x10-3 mol of PCl5.1221

OK, now after the system comes to equilibrium (comes to eq), 2.00x10-3 mol of Cl2 gas was formed in the flask.1257

Now, PCl5 decomposes according to the following: PCl5 decomposes into PCl3 + Cl2 gas.1289

What we would like you to do is calculate the equilibrium concentrations of all species and the Keq.1310

OK, so we would like you to calculate the equilibrium concentrations of all the species (in other words, the PCl5, the PCl3, and the Cl2) and we would like you to tell us what the Keq is--what the K is.1337

OK, so let's read this again: At a certain temperature, a 1-liter flask contains .298 moles of PCl3, and 8.70x10-3 moles of PCl5.1349

After the system comes to equilibrium, 2.0x10-3 moles of Cl2 is formed in the flask.1362

Great! So, let's go ahead and start this off; so I'm going to go ahead and move to a new page so I can rewrite the equation.1368

We are going to do our little ICE chart here: Initial, Change, Equilibrium.1378

PCl5 decomposes into PCl3 plus Cl2; our initial concentration, our change, and our equilibrium concentration--which is what we actually want here.1383

It's telling me that PCl5 was 8.7x10-3 initially, right?1396

8.70x10 to the negative...oh, let's do...yes, that's fine; OK; I can just do this down below.1406

Notice: they give us (well, here; let me do this over to the side)--with the PCl5, they gave us 8.70x10-3 moles.1420

Now, that is not a concentration--that is moles, not moles per liter, but they said we have a 1.0-liter flask.1436

Again, this is one of the other things: we have to make sure to actually calculate the concentrations; so, in this case, it's going to be 8.70x10-3 moles, over 1.0 liters.1443

Well, because it's 1 liter--it's a 1-liter flask--the number of moles is equal to the molarity.1457

So, I can just go ahead and put these numbers here.1462

If this were not a 1-liter flask--if it were anything other than a 1-liter flask--I would have to actually calculate the initial concentration, and those are the values that I use in my ICE chart.1465

OK, so in my ICE chart, I'm working with concentrations, not moles.1476

OK, so we have: 8.70x10-3 molarity, and we said the PCl3 was 0.298, and there is no chlorine gas.1482

Well, they said that, at equilibrium, there is 2.0x10-3 moles per liter of chlorine gas.1496

So, chlorine gas showed up; so the change was: well, for every mole of chlorine gas that shows up, a mole of PCl3 shows up, and a mole of PCl5 decomposes, because the ratio is 1:1, 1:1, 1:1.1506

If 2.00x10-3 moles shows up, that means here, also, 2.00x10-3 moles shows up; here, it is -2.00x10-3 moles.1526

Again, we are using just basic intuition and what we know about the physical system to decide how the math works.1545

This is what chemistry is all about--this is the single biggest problem with chemistry.1553

There is nothing intuitively strange about any concept in chemistry--it's all very, very clear--it's all very, very basic as far as what is happening; there is nothing esoteric; there is nothing metaphysical going on.1557

It is just that...how does one change the physical situation into the math?1571

Well, this is how you do it; you have to know what is going on physically, and then the math should fall out; just trust your instincts.1576

One mole of this shows up; well, the equation says one mole of this shows up.1583

That means, if one mole of this shows up--that means one mole of this was used up; that is it.1587

So now, we do 8.7x10-3, minus 2.0x10-3, and we end up with 6.7x10-3, which I am going to express as a decimal; so, 0.067.1592

And then, we have this one; when we add it together, we end up with 0.300; and here, we have 2.00x10-3.1606

It is probably not a good idea to mix the scientific notation and decimals, but you know what--actually, it is not that big of a deal--it is what science is all about.1619

OK, so now, because we have the equilibrium concentrations, we have solved the first part of the problem.1627

At equilibrium, we are going to have .067 Molar of the PCl5, .300 Molar of the PCl3, and 2.0x10-3 Molar of the Cl2.1635

Well now, let's just go ahead and put it into our equilibrium expression, and find our Keq.1648

That is the easy part.1654

OK, so Keq is equal to...it is going to be...the Cl2 concentration, times the PCl3 concentration, divided by the PCl5 concentration.1656

That equals 2.00x10-3, times 0.300, divided by 0.067.1671

Yes...no; 6.7x10-3...oops, I think I have my numbers wrong here; this is supposed to be...yes, I should have just left it as scientific notation.1688

You know what, I'm just going to go ahead and leave it as scientific notation.1698

I shouldn't mess with things; 6.7x10-3, and we will write this as 6.7x10-3, also.1702

6.7x10-3: I ended up forgetting a 0; it was .0067; OK.1713

And then, when we solve this, we end up with...(let's see, what number did we get?) 8.96x10-2 Molar.1720

So, in this case, it does have a unit; this is a Keq--this is not a concentration.1732

That is why I am not a big fan of units when it comes to equilibrium constants; as far as equilibrium constants are concerned, I think that units should only be used to decide about conversions.1738

Other than that, I think they should be avoided; but you know what, we will just go ahead and leave it there.1753

This is the Keq; Keq equals 8.96x10-2.1756

That is actually a pretty small number; what does a small Keq mean?1761

That means most of this reaction is over here, on the left; there are not a lot of products.1765

Most of it is PCl5; that is what that means.1769

Don't let these numbers say that it is mostly this, because, remember: we started off with a certain amount of the PCl5; we started off with a whole bunch of the PCl3.1773

So, .298--it only went up to .3; that means it only went up .002; that is not very far.1786

So, don't let these equilibrium amounts fool you into thinking that the reaction went forward; it is this equilibrium constant which tells you the relationship between these three numbers under these conditions.1795

But, this is a fingerprint; this reaction at this temperature will not go very far forward.1811

Most of it is still PCl5; that is what is going on--don't let this .3 fool you; it doesn't mean that it has formed that much.1816

You already started at .298; you only formed .002 moles per liter of the PCl3--not very much at all.1824

It is confirmed by the Keq.1833

OK, let's do another example; that is what we are here to do.1835

Let's see what we have.1842

Yes, OK; let me write it out, and then...well, you know what, I am going to actually start a new page for this one, because I would like to see part of the problem while we are reading; OK.1845

This Example 3: Now, carbon monoxide reacts with steam (this is H2O gas) to produce carbon dioxide and hydrogen at 700 Kelvin (we don't need a comma there).1856

At 700 Kelvin, the eq constant (equilibrium constant) is 5.10.1901

Calculate the eq concentrations of all species if 1.000 mol of each component (each component means each species) is mixed in a 1.0-liter flask.1914

OK, our reaction is: they said: Carbon monoxide gas plus steam, which is H2O gas, forms carbon dioxide gas, plus hydrogen gas.1953

That is our reaction, and it is balanced; so it is 1:1:1:1--not a problem.1966

OK, so let's read this: Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen.1973

At 700 Kelvin, the equilibrium constant is 5.10.1978

So, we have our (let me do this in blue now) 5.10.1983

Calculate the equilibrium concentrations of all species if 1.0 mol of each of the components (that means 1, 1, 1, 1 mole of each) is mixed in a 1-liter flask.1987

OK, now the first thing we need to do is: we are going to (and again, this is sort of something we are always going to do--this is the procedure) write the equation, which we have.1999

We are going to write the equilibrium expression, and then we are going to check the reaction quotient to see where the reaction is at that moment, to see which direction it is actually going to be moving in.2010

Let's write the equilibrium expression for this, which is going to be the same as the reaction quotient.2019

It is the concentration of H2, times the concentration of CO2, divided by the concentration of CO, times the concentration of water.2026

OK, and that is also equal to the reaction quotient.2037

The first thing we want to do is calculate the reaction quotient to see which direction it is moving in.2041

Now, they say they put in 1.000 mol of each component.2048

Well, it is sitting in a 1.0-liter flask, so basically, I can work with moles: 1 mole, divided by 1 liter, is 1 mole per liter, so the concentration of each species is 1 mole per liter.2054

Well, now let's calculate the Q.2068

The Q equals...well, it is 1 Molar of the H2, 1 Molar of the CO2, divided by 1 Molar of the CO and 1 Molar of the H2O; the Q equals 1.2071

Now, the reaction quotient Q, which is equal to 1, is less than 5.10 (remember, they gave us the 5.10, which is equal to K).2088

When Q is less than K, that means the reaction wants to move forward to produce more product, in order to reach equilibrium.2096

That means it hasn't reached equilibrium yet; it is still moving forward--it is producing more product to reach equilibrium.2105

That means carbon monoxide and H2O gas are being depleted, and for each amount that these are depleted (because the ratio is 1:1), an equal amount of CO2 and H2 are being formed.2113

Now, we can do the actual equilibrium part of this problem.2124

So again, these "1 Molar"--that came from where we started at that moment.2128

At any given moment, I stick in 1 mole of each in a 1-liter flask, and let me find out what this value is.2133

It is 1; it is less than the equilibrium constant; that means it is going to move forward, to the right.2140

OK, so now let's do our equilibrium part, and we do that by doing our ICE chart, so let me rewrite CO + H2O (I tend to rewrite things a lot--sorry about that; I hope it's not a problem--I'm sure you have different ways of doing it yourself--as long as each of these is here...) goes to CO2 + H2.2146

OK, we have an initial concentration; we have the change; and we have our equilibrium concentrations, which is what we are looking for.2173

Our initial concentrations are 1.000, right?--that is how much we started with.2181

We stuck each of those in a flask.2186

Now, this reaction quotient tells us that the reaction is moving in that direction to reach equilibrium.2189

It is moving in that direction; that means CO is disappearing.2196

H2O is also disappearing by an amount x--that means a certain amount is decomposing--a certain amount of CO is being lost, is being converted.2202

Well, since it is moving to the right, and this is 1:1:1:1, that means this is +x; that means CO2 is forming for every 1 mole of this that is disappearing.2214

This is also +x; I hope that makes sense.2226

Our equilibrium concentration (at equilibrium, once everything has stopped, a certain amount of CO has been used up, so our equilibrium concentration) is going to be 1.00-x.2231

A certain amount of H2O has been used up; that is 1.00-x.2242

A certain amount of CO2 has formed, so it is going to be 1+x.2248

A certain amount of H2 has formed: 1+x.2252

These are our equilibrium (oh, wow, that is interesting; look at that--let's get these lines out of the way) expression, but notice: now, we have x, so we need to actually find x.2258

Fortunately, we can do that: we can plug it into the equilibrium expression, right?--because the equilibrium expression is a measure of these concentrations at equilibrium; that is what these values are.2274

We stick it in here; we know what the Keq is--it's 5.10; and we solve for x.2286

That is it; it is just an algebra problem.2291

So, let's go ahead and do that: so K, which is equal to 1.000+x, times 1.000+x (that is the CO2 and H2 concentrations), divided by the CO and H2O concentrations, which at equilibrium are 1.000-x, 1.000-x, and we know that that equals 5.10.2294

OK, so now let's just handle this algebraically.2329

This is (1.000+x) squared, over 1.000 (oops, too many 0's) minus x, squared, equals 5.10.2333

Now, I know that I said earlier (I think a lesson or two ago): when you are writing out the equilibrium expression, don't put the square--don't square it immediately--if two of the things are the same.2353

That is different than what I am doing now; I wrote the expression as each species separately, so that I can see that I actually have four species in my equilibrium expression.2363

Here, now, I am just dealing with the math.2373

Once you have actually written it out, then you can go ahead and write it like this to deal with the math.2375

Now, it's (1+x)2 over (1-x)2.2380

That is fine; but when you initially write the expression, don't cut corners; write down everything.2384

We want to know that the expression actually consists of four terms, not two terms, each squared.2389

OK, and now we just...well, we have a square here and a square here, so we'll just go ahead and take the square root of both sides.2395

We end up with 1+x over 1-x, equals 2.258, and then we multiply through to get 1.000+x=2.258-2.258 times x (I'm hoping that I am doing my math right here).2406

We end up with 3.258x equals 1.258, and x is equal to 0.386 Molar.2440

We found x; x is .386 Molar.2458

Now, if you go back to your ICE chart, it didn't ask for what x was; it asked for the final concentration.2464

Well, the final concentrations were the 1-x and the 1+x for those four species.2471

So, the CO concentration (carbon monoxide concentration), which also equals the H2O concentration, is equal to 1.000-x, 1.000-.386, equals 0.613 Molar.2476

The carbon monoxide and the water are at .613 molarity.2501

Now, the CO2 concentration, which also happens to equal the H2 concentration, is equal to (OK, let's see if we can clean this up a little bit; I'm not going to have these stray lines driving us crazy all day)...CO2 (I just really need to learn to write slower; I know that that is what it is) equals H2 concentration, equals 1.000+x, equals 1.000+0.386, =1.386 molarity.2505

OK, and my friends, we have done it; we have taken a standard equilibrium problem, and here is how we have approached it.2551

1) Write the equation; this is chemistry--chemistry always begins with some equation; don't just go into the math.2564

Look at the equation--the equation gives you all of the information that you need--in fact, it will tell you everything you want.2570

2) Write the equilibrium expression--write the K expression.2576

Write it out explicitly--don't count on the fact that you will know how to do it later on when you are ready to plug things in.2583

Write it out; don't cut corners; doing things quickly is not impressive--doing things correctly is impressive.2589

3) Find Q, the reaction quotient; find Q to decide which direction the reaction is going in--to decide the reaction direction, if any.2598

That is how we knew...in this problem that we just did, the reaction quotient was less than the equilibrium constant--so that means the reaction was actually moving forward.2614

Well, that is how we knew that the reactants get the -x and the products get the +x.2623

If it were the other way around and the reaction were moving to the left, that means the products would get the -x and the reactants would get the +x.2628

You have to do this; you have to do the reaction quotient--very, very important.2637

After the reaction quotient, well, you set up your ICE chart: Initial, Change, Equilibrium concentration.2642

Once you get a value for the equilibrium concentration, you put the eq concentration expressions (in other words, the concentrations that you calculated there--the equations/expressions) into the Keq expression.2652

Then, last but not least, you solve, depending on what they want.2685

If they want x, you stop there; if they want equilibrium concentrations, you take the x value; you plug it back into the equilibrium expressions; and you add and subtract until you get your equilibrium expressions.2689

If they want something else, they will tell you that they want something else.2701

Again, this process is what you are always going to be doing; this is the algorithm, the general, broad-strokes algorithm.2705

Write the equation; write the expression; find the Q; set up your ICE chart; put the equilibrium concentrations in; and then solve it.2714

Within this are the different variations that make up the different number of problems, which seem to be infinite (I understand completely).2723

That is all you are doing here.2731

OK, thank you for joining us here at Educator.com for our continuation of AP Chemistry in equilibrium.2734

In our next lesson, we are actually going to do more equilibrium problems, because again, this is a profoundly important concept; you have to be able to have a good, good, solid, intuitive understanding of what is going on.2740

So, we will do some more practice.2751

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to continue our detailed discussion of equilibrium.0005

We are just going to be doing examples.0009

We have talked about what equilibrium is, the expression; we have actually done a fair number of problems so far, and problems that are reasonably complicated.0012

But, I wanted to do some more that go a little bit more in-depth, as far as--they have some slight variations; they are a little more detailed and require a little bit more care--for the purposes of showing you where things can possibly go wrong, but mostly just to get you really, really accustomed to these equilibrium problems.0020

It has been my experience, from all of my students who have had a solid grounding in equilibrium--the rest of chemistry is a complete breeze for them.0039

They know exactly what to do; they know where to go; I virtually didn't have to teach the class after this.0046

We hammered the equilibrium so much that they knew exactly what was going on; that is why this is important.0051

If you can get your head around this, and be able to handle at least a good 70 or 80 percent of these problems, the AP test should be absolutely a breeze for you, I promise.0057

OK, so let's jump right on in.0066

Our first example is going to be a synthesis of hydrogen fluoride from hydrogen and fluorine gas.0070

Let's go: Example 1: So, in a synthesis of hydrogen fluoride gas...0077

Now notice, I didn't say hydrofluoric acid, because this is hydrogen fluoride gas--it's in gaseous form; remember, an acid is something that is when you take the thing and you drop it in the water, and it dissociates into a hydrogen ion and fluoride ion, or hydrogen ion and any other anion.0093

That is when it is an acid--the acid is the H+; when HF is together, it is not an acid--it won't do any damage.0111

Well, I won't say it won't do any damage, but it won't do any damage as an acid--let's put it that way.0119

In the synthesis of HF from H2 and F2 at a given temperature, 3.000 moles of H2 and 6.000 moles of F2 are mixed (it might be nice if I actually knew how to spell--yes, that would be good) in a 3.0-liter flask.0124

So again, we have moles, and we have a 3-liter flask; this time it is not 1-liter, so we are going to have to actually calculate the concentration.0168

OK, the eq constant at this temperature is 1.15x102.0177

What are the equilibrium concentrations of all species?0196

Or I should say, "What is the equilibrium concentration of each species?"0202

Being able to handle these ICE charts is the real key to the majority of the second half of chemistry.0218

The ICE chart itself is always going to be a ubiquitous feature of the problems that we solve.0225

The difference is what the ICE chart looks like: acid-base problems, equilibrium problems, thermodynamic problems, whatever it is...the ICE chart itself is going to change, depending on what the problem is asking.0229

Being able to handle that--that is the real deal; that is when you know you actually know what is going on--when you know how to arrange the ICE chart as necessary; the rest is just math, basic algebra.0242

OK, so in the synthesis of HF from H2 and F2 at a given temperature, 3 moles of hydrogen and 6 moles of fluorine are mixed in a 3-liter flask.0255

The equilibrium constant is 1.15x102, or 115.0267

What is the equilibrium concentration of each species?0272

OK, so let's write our equation; like we said, that is our process: H2 + F2 goes to 2 HF.0275

We want to make sure that the equation is balanced; we want to write our equilibrium expression--it is going to be the concentration of HF squared (stoichiometric coefficient), over the concentration of H2, times the concentration of F2.0283

OK, now let's go ahead and calculate the initial concentration of H2.0298

Well, it says we have 3.00 mol of H2 in a 3.000-liter flask; so we have 1.000 Molar H2.0304

That is what this little 0 here, down at the bottom is: H0; it just means the initial concentration.0319

If you want to put an i, that is fine, too.0324

F2, the initial concentration--it says we have 6 moles of fluorine gas, over (again, it's in the same flask, so it's) 3 liters, so our concentration is 2.00 Molar.0327

Well, Q--the next thing we want to do is, we want to calculate the Q in this case, the reaction quotient, to tell us what direction it is in.0343

There are a couple of ways to do this in this particular problem: we can just plug it in; the concentration of HF to begin with...well, there is no HF, so it's just 0...over 1, times 2, which is 0.0350

That is fine: you can do it that way, or you can just say, "Well, since there is no HF to begin with, I know the reaction is going to move in that direction; there is none of this yet--there is only this and this--so it has to move forward; there is nowhere else for it to go."0364

But, I think it is best to stick with the process, and just go ahead and put the numbers in, and do it that way.0382

OK, so this is 0, which implies that the reaction will move to the right.0388

Moving to the right means it will form product and it will deplete reactant.0397

H2 and F2 concentration will go down; HF concentration will rise.0402

Now, we can do our ICE chart: H2 + F2 (I'll give myself plenty of room here) goes to 2 HF; Initial, Change, Equilibrium.0407

We start off with 1 Molar of hydrogen; 2 Molar of fluorine; 0 Molar of HF.0421

This is going to deplete by x; this is going to deplete by a certain amount, x; this is going to increase by an amount, 2x, because of that 2.0429

That is the whole idea: 1, 1, 2.0440

We get equilibrium concentrations of 1.000-x, 2.000-x, and we get 2x.0443

Now, we plug in these equilibrium concentrations into this expression, and since we already know what that is, we solve the algebraic equation.0454

So, let's write: K is equal to...well, it's equal to the concentration of HF squared, so it's 2x squared, over 1.000-x, times 2.000-x.0472

That is our equation, and we know it is equal to 1.15x102.0494

They gave us the K; now, let's solve for x.0500

Well, this is just a quadratic equation; so you are just going to have to sort of do it.0503

You can...there are several ways you can do this: you can do it by hand, do the quadratic formula; you can go ahead and use your graphic utility, for those of you that have the TI-83s and 84s and 87 calculators.0515

You can handle these really, really easily; that is what I prefer; that is what I use; I have a TI-84--that is how I solve these.0527

Let's just go ahead and at least work out the algebra, and then we'll just write out the answer, presuming that we actually used a graphical utility or something like that to do it.0534

We end up with 4x2=1.15x102 (2.000-3.000x+x2), and we get 4x2=230-345x+115x2.0546

And then, we end up with 111x2 - 345 x + 230 = 0.0582

And again, when I put this into my graphical utility, and I solve, I end up with (again, this is a quadratic equation, so I have 2 values): the first x-value is 0.968 Molar, and the second root is going to be 2.14 Molar.0592

Now, we can't have both of these be true; there is only one equilibrium condition; one of these has to be true.0611

Here is how you decide: well, take a look at the original concentration--1 Molar of hydrogen gas, I think it was.0617

Well, 1 molarity - 2.14 molarity--that is going to give you a negative 1.14 molarity; you can't have a negative concentration, so that one drops out.0626

Let's do that in red; that one drops out--this is the x-value.0637

So, we have that .968 Molar is the x-value; we plug those back into the equilibrium concentrations in our ICE chart to get the following.0642

Our H2 concentration is 1.000-0.968=3.2x10-2 Molar at equilibrium.0653

Our F2 concentration is 2.000-0.968, is equal to (let's get a better-looking equals sign than that) 1.032 molarity.0669

And our final HF concentration is 2x, so it's 2 times 0.968, is equal to 1.936 molarity (you know what, these numbers are getting strange again...1.936 molarity).0687

And my friends, we have our final solution.0709

Notice, we started off with 1 Molar of H; we end up with 3.2x10-2; that means most of the H is used up.0714

That makes sense, because again, you are looking at a very, very large Keq; the Keq is 115--that means the reaction is very far to the right.0724

It favors the product formation, not reactants.0735

When it has come to equilibrium, virtually no reactants are left over.0740

The only reason that you have 1.32 moles per liter left over of the other reactant: because you started off with 2 moles per liter of that, and the stoichiometry of the H2 to the F2 is 1:1.0744

You have used up half of it; that is the only reason that it looks like this number is so big, compared to this number.0757

The Keq tells you that it is mostly to the right; you started off with no hydrogen fluoride; you ended up virtually 2 moles per liter of hydrogen fluoride.0764

That is confirmed; so, the numbers match up; everything is good.0774

OK, let's see what is next.0780

Example 2: OK, this is going to be an example of an equilibrium problem where the Keq is actually very, very small.0785

And again, depending on the equation, you might run into some rather complicated things, like cubic, quartic, quintal equations, which you can certainly handle with your graphical utility--it's not a problem--that is the great thing about having graphical utilities.0799

But, the method we are going to show is going to be a slightly simplified version, if you just want to do the math quickly.0810

And then, we will give you a way of checking to see whether any simplifications you made were actually viable--whether you could actually get away with it.0816

There might be situations where you might simplify something, and you can't get away with it.0824

We will give you a rule of thumb for doing that.0827

OK, so Example 2: Gaseous NOCl decomposes to form gaseous NO and Cl2.0830

At 40 degrees Celsius, the eq constant is 1.4x10-5; it is very small--that means there is virtually no product at equilibrium.0858

OK, initially, 1.0 mol of NOCl is placed in a 2.0-liter flask.0876

Here we go again: what is the equilibrium concentration of each species?0901

OK, let's write our equation: 2 NOCl decomposes into 2 NO + Cl2.0923

Let's write our equilibrium expression--that is what we always do: it's going to be the concentration of NO squared, times the concentration of Cl2, divided by the concentration of NOCl squared (is that correct?--yes, that is correct).0933

OK, so now we do that, and let's see what else we have.0955

Initial concentrations--we have to do initial concentrations.0962

The NOCl equals 1.0 mol over...it looks like a 2.0-liter flask: is that correct?--yes.0967

Let me circle my numbers: 2 (oops, let me use blue) liters, 1 mole; that is my Keq; OK.0984

Let's go back to red; that is going to equal 0.50 moles per liter, and then I have my initial NO concentration, which is 0, and my initial Cl2 concentration, which is also 0.0997

Therefore, Q is equal to 0 squared times 0, over 0.5, equals 0, which is definitely smaller--the Keq is small, but it is still bigger than (the 1.4x10-5 is still bigger than) 0, which implies that the reaction moves to the right.1015

Moving to the right means we are forming product.1042

We are depleting reactant.1048

OK, so let's go ahead and set up our ICE chart.1053

We write our equation: 2 NOCl goes to 2 NO + Cl2; Initial, Change, Equilibrium.1056

We start off with 0.50 Molar and none of those; we are going to deplete this by 2x; we are going to form 2x, and we are going to form x.1069

Therefore, the equilibrium concentrations...just add them straight down: what you started with, what you lost, and what you end up with.1086

0.50 minus 2x; 2x; and x; now, let's go ahead and put it into our expression.1093

Our K is equal to 2x squared (that is the squared part--2x squared) times x to the 1 power, divided by that squared, (0.50-2x)2.1105

That equals 1.4x10-5; OK.1132

Well, all right, see now: it is getting a little complicated.1141

You are going to end up with 4x2 times x; it is going to be 4x3; I don't know...you are certainly welcome to go ahead and solve this, just because you have a graphical utility; it's not a problem--you can do that.1144

But, let me give you an alternate procedure, which actually makes things at least a little bit more tractable mathematically.1156

Because we notice that the Keq is very small, which means that the reaction is over here, mostly; that means not a lot of product; this is small, because that is small, meaning that there is not a lot of product--in fact, there is virtually no product at all...it is mostly NOCl still.1169

Well, because of that--because it is mostly NOCl still--that means that NOCl hasn't lost very much.1189

Very little of it has actually decomposed.1197

Well, very little of it has decomposed...well, we started with .5; that means x is probably really, really, really small compared to .5.1199

Because it is so small compared to .5, it is possible to take this term and just leave it out--ignore it.1212

Solve the problem, and then check to see whether it is actually valid or not that we did what we did.1224

So, this is how we do it: so again, to our first approximation, we are going to presume because this is small, it means that most of it is here; that means very little of this is going to decompose; in other words, very little of this is going to form.1233

Because this is small, we are saying it is so small compared to .5 that it is probable that the .5 minus the 2x is going to go unnoticed, so let me just ignore it and solve the easier problem.1246

OK, well, let's see what we can do.1256

We have: It's going to be 2x2 times x, over 0.50 squared, equals 1.4x10-5.1258

Well, what we end up with here is (what we are going to end up with, once we do all the multiplication): we are going to end up with: 4x3 is equal to 3.5x10-6; x3 is equal to 8.75x10-7; x is going to equal 9.6x10-3.1273

We found a value of x, 9.6x10-3; now, we want to check to see whether we were actually justified in ignoring it.1300

Well, if I take (yes, that is fine) 9.6x10-3, and if I divide by the 0.50, and I multiply by 100, I am trying to see what percentage of the .5 this 9.6x10-3 really is.1311

That is what I'm doing; I am trying to see how much of the .5 this is.1341

Well, as it turns out, it actually equals about 1.91%.1346

Now, we did, actually, 2x: so 2x would put us roughly at twice that; but, as you can see, the 1.91% is actually really, really small.1352

As it turns out, if you make an approximation of this nature, and the value that you get ends up being anywhere from about...anywhere less than 5 percent of your total value, it is actually (as a good rule of thumb) a valid approximation.1363

That means, virtually, the mathematics is not going to notice that you actually eliminated that.1379

So, because of the 1.91 percent, the x compared to the .5 (or, in this case, it's going to be 2x, but again, you are still going to be below 5%), you are actually pretty good.1384

And this allows us to sort of keep this value of x, as opposed to having to solve this entire equation, this cubic equation.1394

We just did it in a nice, simple way to get an answer, instead of having to use a graphical utility, and we got this value for x.1403

Given that value of x, we can go ahead and use it now to find our equilibrium concentrations.1411

Our NOCl concentration, final, is equal to 0.50, minus the 9.6...oops, this is going to be minus two times the 9.6x10-3.1418

Our final NO concentration (I will go ahead and let you finish off the arithmetic here) is going to equal 2 times 9.6x10-3.1442

Our final Cl2 concentration is going to equal, well, just x: which is 9.6x10-3.1461

The rest is just arithmetic, which I will leave to you--nice and straightforward.1473

Hopefully you are getting a sense of the general procedure and of the things that you have to sort of watch out for.1480

OK, so let's see: let's go ahead and close this off with an interesting type of problem; it should go pretty quickly, actually.1487

Let's do...yes, that is fine; we can start it on this page.1499

OK, Example 3: Calculate the value of the equilibrium constant for the reaction O2 gas + oxygen gas goes to ozone gas, given reaction 1 (this is going to be sort of a Hess's Law kind of problem): NO2 in equilibrium with NO + O, and K1 is equal to 6.8x10-49 (wow, that is really, really small) and that one, which is O3 + NO goes to NO2 + O2; K2 equals 5.8x10-34.1508

Let's see what they are asking: they are asking you to calculate the value of the equilibrium constant for this reaction (let's do this in blue), given these two reactions and their corresponding equilibrium constants.1593

Well, we know from Hess's Law that if we want to find a final reaction that involves some of the reactants that we have, we have to rearrange them by either flipping them or multiplying them by constants.1607

In doing so, if we add all of the equations together, and then get our final equation--well, when we did ΔHs, when we did enthalpies, we just added the enthalpy; with Ks, with equilibrium constants, it is actually different.1621

When we add equations to get a final equation, what we do to equilibrium constants is: we actually multiply them.1635

So, let's go ahead and do this one.1643

In order to actually come up with this, I am going to flip Equation 1.1646

I am going to flip Equation 1, and that will give me: NO + O goes to NO2.1654

Now, when I flip an equation, I take the reciprocal of the equilibrium constant, right?--because you are just flipping products and reactants.1664

That equilibrium constant, now, is 1.47x1048--huge!1678

I am also going to flip the second reaction; I am going to make the reactants the products and the products the reactants; when I do that, I end up with NO2 + O2 in equilibrium with NO + O3.1686

Well, this equilibrium constant--again, I flipped it, so I take the reciprocal of that, and I get 1.72x1033.1702

Now I add these two equations; NO2 cancels NO2; NO cancels NO; I am left with O + O2 goes to O3; this was the equation that we wanted.1714

Now, in order to get the final equilibrium constant, I have to...I don't add these; I multiply them; it becomes K1' times K2'.1735

It equals 1.47x1048, times 1.72x1033, and I end up with getting some huge number, if I am not mistaken.1749

2.53x1088: that is huge!--that means that any time oxygen gas and free oxygen atom come together, you will not find them separately!1766

This huge number tells me that the reaction is way to the right; there is no equilibrium here--not really.1782

Any time you have a bunch of equations, if you add them together to come up with a final equation, you multiply the equilibrium constants for all of the individual equations.1789

So, when we add equations to get a final net equation, our final K is equal to K1 times K2 times...all the way to Kn, if we had n equations; and that is it.1803

OK, so we have gone ahead and dealt with a fair number of problems in equilibrium.1837

Next time, we actually are going to continue our discussion of equilibrium; we are going to talk about Le Chatelier's Principle, and then we are going to do some more problems involving Le Chatelier's Principle.1843

And Le Chatelier's Principle, just to give you a little bit of a preamble, is basically just: If I have a system at equilibrium, if I stress that system out somehow--if I put pressure on it, meaning if I add this or heat it up or cool it down, what happens to that equilibrium.1854

So, I can shift the equilibrium; well, we know (well, I can tell you) that a system will always seek out equilibrium.1869

When I apply stress to a particular system, the system is going to respond by doing whatever is necessary to relieve that stress.1878

You know this from just being a human being: any time, any system that you apply stress to, the response of that system will be doing the things that relieve that stress.1885

Well, a chemical system behaves in exactly the same way, and it is a very deep, fundamental thing called Le Chatelier's Principle.1896

It is actually quite beautiful; so we look forward to seeing you next time.1903

Thank you for joining us at Educator.com and AP Chemistry.1907

We'll see you next time; goodbye.1910

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we are going to close out our discussion of equilibrium with a discussion of Le Chatelier's Principle in relation to equilibrium.0004

Now, that doesn't mean that we are just never going to talk about equilibrium again; we are--equilibrium is the central concept in chemistry and most sciences.0011

This was a general discussion about equilibrium--getting a little bit of practice in the techniques and how to subtract and add reactants to decide which way a reaction is going to go.0019

All of these things will show up when we discuss acid-base, solubility product, complex ion, equilibria...so, very, very important.0031

So now, we are going to discuss a qualitative method of deciding what happens when you actually (once a system is at equilibrium) place a stress on that system.0040

Your intuition is actually absolutely correct; when you place a stress on any system, the system is going to react in a way that it wants to offset that stress.0050

In chemistry, in an equilibrium situation, we call that Le Chatelier's Principle.0059

Let's jump in and get started.0064

Once again, it is a way of predicting the effect that a change in concentration or pressure or temperature has on the system that is already at equilibrium.0067

If we have a system at equilibrium, we can do a couple of things to it to stress it out: we can change the concentration of a particular reactant or product (in other words, add or subtract reactant or product); we can change the pressure of the system, either by directly changing the pressure or by changing the volume, or by adding an inert gas; or we can actually change the temperature (drop the temperature or increase the temperature).0077

When we do these things, the equilibrium position is going to change.0102

Now mind you, when we change concentration and when we change pressure, the equilibrium position might change, meaning there might be more or less of reactant or product; but the equilibrium constant does not change.0106

In other words, the equilibrium itself doesn't change; the position changes.0120

However, that is not the same with temperature, because, remember: the equilibrium constant is actually dependent on temperature.0125

So anyway, we will see more of that when we start doing some of the problems.0131

OK, so let's just go ahead and write out the principle as our fundamental thing to get started with.0135

The principle is: If a change is imposed on a system already at equilibrium (let me go ahead and not use that symbol; let me go ahead and just write out "at equilibrium"), the position of the equilibrium will shift in the direction that offsets the change.0146

Now, let me just describe what that means in terms of a chemical reaction.0200

So, let's just take aA + bB is in equilibrium with cC + dD.0205

Now, when we say it will shift in the direction that offsets the equilibrium, once the system actually reaches equilibrium, if I stress it out somehow, it will move in the direction to offset that change--the change that I have made.0213

What we mean is that it will either shift right or it will shift left; in other words, it will produce more product, or it will produce more reactant.0224

That is all that that means--shift equilibrium.0232

So remember, when we actually start a reaction, it moves in the forward direction; but eventually, when enough product develops, it starts to also move in the reverse direction.0235

At some point, the rate of the forward and reverse reactions is the same; that is what we call the equilibrium point--when the concentrations of this, this, this, and this are constant--when there is no more change.0244

That doesn't mean the reaction has stopped; it's a dynamic equilibrium--things are still going, except there is no net change.0255

So, a shift in equilibrium means will it move forward? Will it move back? And I can do things to push it forward and back.0260

That is essentially what chemical engineers do: they try to find ways to push reactions forward--the reactions that would not otherwise want to move forward.0267

How can we stress it out to make it go forward?0277

OK, the first...so we said that we can do a couple of things to it: we can affect, we can change (well, not change--we can--yes, let's go ahead and say change) the concentration by adding or subtracting; we can change the pressure of the system (that is another method by which we can affect the system), or the temperature.0280

Temperature, pressure, and concentration are the things that we can control, that we can actually do to a system at equilibrium.0314

When we effect these changes, what happens?0322

The first thing we are going to talk about is concentration.0325

Here we go: How does a system respond to a change in concentration?0333

OK, well, like this: If we add species X (and by species X it could mean any one of these--it could be A; it could be B; it could be C; it could be D)--I have this system here; it is in a reaction vessel; I can add or subtract any one of these species; here is how the system will react.0338

Well, if I add a species, X, well, the system is going to do what is necessary to offset the addition, which means (so, system offsets the addition) that it will consume X, which means it will deplete X, which means that the reaction will shift accordingly.0361

What "accordingly" means: it means away from X.0411

In other words (we will do an example, but I'll just do it here), if I have this system at equilibrium, and if I add B, that means I have pumped more B into here.0419

Well, the system wants to offset my addition of B, so it wants to reduce the concentration of B.0429

The only way to reduce the concentration of B is by actually using it up.0435

The only way it can use it up is by shifting the reaction forward to use up B.0440

Yes, it also ends up using up A, but the idea is that it is using up B.0445

It wants to offset what I have done to it: I have increased the concentration of B--it wants to push the concentration back down.0449

In order to push the concentration back down, it has to use it up.0455

In order to use it up, it moves the reaction forward; it produces more product, less reactant.0459

That is what it means; so in this case, we say that the equilibrium shifts to the right, towards the product.0465

Now, what happens when we subtract a species?0471

Let's say I have A, B, C, D in equilibrium, and let's say I siphon off...let's say O2 gas is one of the products that is formed--what happens if I siphon off O2 as is produced?0475

Well, the subtraction of X goes like this.0487

It is always the same: Subtract species X; well, the system will offset the subtraction...the system wants to offset the subtraction; that means it will...since I am subtracting X, it wants to do the opposite, which means it will produce X; that means that the reaction will shift in a direction which produces X.0491

It is that simple; you are just following the train of logic which produces X.0538

In the example of, let's say, aA + bB is in equilibrium with cC + dD, if I end up subtracting B (meaning if I pulled B away from a system already at equilibrium), well, the system is going to respond by wanting to increase the concentration of A.0545

In order to increase the concentration of A (in other words, produce A...I'm sorry, B), the reaction will shift in the direction that does that.0564

Well, the only way to produce B is to shift this way.0572

In other words, use up this and this so that more B is produced.0575

So, the reaction is going to shift to the left.0579

That is all you are doing here.0583

Let's do an example, a real-life example.0585

It's going to be Example 1: I have the equation 2 SO2 + O2 goes to 2 SO3.0592

All of these are gaseous: sulfur dioxide, oxygen gas, sulfur trioxide.0601

Now, we want you to predict the direction the reaction will shift when its equilibrium is disturbed (so this is a system already at equilibrium).0605

One disturbance we are going to do is: we are going to add SO2.0633

The second disturbance we are going to do: we are going to remove SO2.0640

The third disturbance we are going to do is: we are going to remove O2.0646

OK, well, so we have a reaction here; now, we want (the system is at equilibrium)--we add sulfur dioxide.0652

Well, if we add sulfur dioxide, the system wants to now reduce the sulfur dioxide--it wants to do the opposite of what we are doing.0663

We are adding; it wants to subtract; that means lower the sulfur dioxide concentration again.0670

In order to lower it, it has to use it up; in order to use it up, it is going to shift it to the right.0675

So, the equilibrium is going to shift to the right.0680

You can either write "shift to the right," or you can just do a right arrow; I tend to use arrows.0682

I like symbolics, when it works--when it is clear.0686

That is it; it is that simple.0690

What happens if I remove SO2?0691

Well, if I remove SO2 from an equilibrium mixture, the system is going to want to produce more SO2 to compensate for what I have taken away.0694

In order to produce more SO2, it has to shift the equilibrium that way.0702

That means product is going to decompose, O2 and SO2 are going to form, until the SO2 level comes back to such that the equilibrium constant is brought back to where it is.0706

So, in this particular case, it is going to shift to the left.0720

What happens if I actually remove...you know what, let's change this; I don't want to remove this; I want to remove SO3.0726

Now, what happens if I remove SO3?0733

Well, I go over here; SO3 is on the product side; if I remove SO3, if I pull it out of the flask, the system is going to want to compensate for that loss by producing SO3.0735

The only way to produce SO3 is by pulling the reaction forward, depleting this, depleting that, producing SO3.0748

The reaction is going to shift to the right.0757

That is all that is going on here.0760

Find the stress that you are putting on the system, and find out what the system has to do in order to do the opposite of that stress--it's that simple.0762

OK, so now, let's go ahead and talk about stressing out the system via pressure.0773

Now, instead of changing the concentration of a given species, let's see what happens when we increase the pressure or decrease the pressure on the system.0780

So, this is going to be a change in pressure.0790

This is the second of the effects that we can run on a system at equilibrium.0795

Well, as it turns out, there are three ways to change the pressure.0800

OK, we can add or remove reactant or product--gaseous; OK, I'm going to put "gaseous" in; so again, when we talk about pressure, we are talking about gases.0804

So, if I want to change the pressure, one of the ways that I can do it is by adding or subtracting actual gaseous product--reactant or product, if one of those species happens to be a gas.0823

If it's not a gas, then I can't do anything about the pressure.0835

Add or remove--well, by adding or removing a reactant or product, I handled that the same way that I did just a moment ago; I'm just adding or removing a reactant or a product.0837

I handle it--it's the same as changing the concentration.0849

That is what I am doing; when I am adding or subtracting a gas, yes, I am changing the pressure, but what I am doing is actually changing the concentration, because remember: pressure and concentration are actually the same thing.0852

They are just variations of the same thing.0862

So that one--I handle it just by addition or subtraction of reactant, like we did a moment ago.0866

The other way to change it is by adding an inert gas.0873

If I pump in an inert gas (in other words, a gas that doesn't react with anything, that just takes up space), I have changed the pressure of the system; how does that system react when I have actually added an inert gas?0877

We will see the answer in just a minute.0890

3: I can change the volume.0893

Change the volume: so, you know PV=nRT; if the volume goes up, the pressure goes down; if the volume goes down, the pressure increases, all else being equal.0896

The relationship between volume and pressure is an inverse relationship.0908

So, these three ways are the ways that I can actually change the pressure of the system.0911

Now, let's see how each one reacts.0916

Well, 1: as we said, handle the same way as the addition or the subtraction, changing the concentration.0918

This is handled the same as a change in concentration, because that is it; that is all you are doing.0929

"As a change in concentration..." so we already know this one.0938

2: If we add an inert gas, as it turns out, adding an inert gas--it changes the total pressure, but it doesn't change the equilibrium, because it's not reacting with any of the species.0943

So, adding an inert gas changes nothing--no change.0954

Yes, total pressure will increase in the system, but because the gas is not really reacting with anything, it doesn't really shift the equilibrium one way or the other.0964

The same concentrations of reactant and product are in there, and the Keq, which is the product of the products, over the product of the reactants.0972

Those concentrations don't change, so the Keq doesn't change.0988

Well, nothing is changing; so there is no change.0992

3: OK, this one--there definitely is a change.0996

If I do a volume decrease, that is the same as a pressure increase; these double arrows--they just mean "implies"--it's following a train of logic.1002

Volume decrease is the same as a pressure increase; the system now wants to offset the pressure increase; "wants to offset" means it will seek out a way to decrease the pressure back.1020

That means it will shift in the direction of fewer gas particles.1054

I won't do an example just yet; I want to actually talk about a volume increase, and then I will do the example, and it will make sense.1069

So, let's follow the train of logic here: if I do a volume decrease for a system at equilibrium, well, if I decrease the volume, the pressure is going to increase.1074

If the pressure increases, we know that the system is going to offset that increase, so it is going to want to seek out a way to decrease the pressure back.1084

Therefore, it is going to shift in the direction that has fewer gas particles, because fewer gas particles means less pressure.1091

That is what the system is going to do.1099

Now, what happens if we have a volume increase?1101

OK, so a volume increase: well, it's going to be just the opposite.1107

A volume increase means a pressure decrease, so the system wants to offset by a pressure increase.1110

How does it do that?--well, it shifts in the direction of more gas particles.1133

More gas particles means higher pressure; so once again, if I increase the volume of the system, I decrease the pressure.1149

Well, by decreasing the pressure, the system needs to offset that decrease that I just made, so it is going to want to increase the pressure.1158

The only way that a system has to increase the pressure is by shifting the reaction in a direction that produces more gas particles.1165

So, if the products happen to have more gas particles, it will shift in the direction of products, which means it will shift to the right.1173

So now, let's do an example.1180

Oh, I should give one last note here: I'll put N.B. for nota bene; if reactants and products have equal numbers of gas particles, there is no change.1184

I can increase the pressure, decrease it...do whatever I need to; there is no change, because now, the system has nowhere to go.1211

There are equal numbers of particles on the reactant side and on the product side, so there is nowhere for the system to go.1218

The total pressure will increase or decrease accordingly; the system will stay exactly the same--there will be no shift.1224

In other words, it won't move to the right to create more product; it won't move to the left to create more reactant.1230

OK, so let's do Example 2; let's do this in red.1236

Example 2: we have: CH4 gas + H2O gas is in equilibrium with carbon monoxide gas, plus hydrogen gas.1247

We have gaseous components everywhere--in reactants and products--we have 1, 2, 3, 4; OK.1261

Let's see: oh, I'm sorry, this is 3 H2; I was going to say, "There is something wrong here"; let's double-check: C, C, H2O gas; we have H2; that is 4; that is 6; that is 6; 1 oxygen; 1 oxygen; yes, OK.1270

We have 1 mole of CH4, 1 mole of H2O, 1 mole of carbon monoxide, and 3 moles of hydrogen gas.1288

OK, the question is: What happens if we (hmm, temperature--interesting; OK) reduce the volume?1295

This system--let's say, all of a sudden, we pump all of this from a 1-liter flask into a 250-milliliter flask; we reduced the volume; what happens to the equilibrium position?1319

Again, the equilibrium constant doesn't change; that is the whole idea behind the shift; the constant wants to stay the same, so it has to shift in a direction that will keep it the same.1333

OK, what happens when we reduce the volume?1344

OK, so I'm going to do some symbolism here: a down arrow is decrease; and up arrow is increase.1347

A volume decrease implies a pressure increase; a pressure increase means that a system wants a pressure decrease.1352

In order to get a pressure decrease, I need fewer gas particles.1368

The only way to get fewer gas particles is to shift the reaction to the left.1376

The reason is: I have (let me do this in blue) 1 mole, 1 mole; I have 1 mole...I have 3 moles here.1386

The products have 4 moles of gas particles; here, I only have 2 moles of gas particles.1396

If I increase the pressure, the pressure will want to decrease to offset; the only way that the system has in order to decrease it is by shifting the reaction that way, by diminishing the product and producing more reactant.1401

4 moles, 2 moles; there are fewer moles of gas particles if the reaction goes this way.1414

In that case, it will actually reduce the pressure as far as it can to reach equilibrium.1422

That is all that is going on here.1428

OK, let's say the opposite: What happens if we just directly decrease the pressure?1431

Well, if we decrease the pressure, the system is going to want to offset it by increasing the pressure.1445

The system wants to increase pressure; the only way to increase pressure is to move the reaction in a direction that produces more gas particles.1453

So, it will shift to the right.1465

That is it; it's all you are doing--follow the train of logic; everything will fall out.1468

OK, now let's talk about the third change that we can make to a system at equilibrium, and that is going to be changing the temperature (heating it up, cooling it down); what happens to the equilibrium?1474

Now, remember: we said that the equilibrium position changes when we change the pressure or change the concentration.1487

In other words, the actual values of the concentrations change, but the equilibrium constant does not change.1495

That is the whole idea behind a constant.1501

With temperature, it is different, because the equilibrium constant is actually dependent on the temperature.1507

For most practical purposes, it is not really a problem, but you should know that the Keq is temperature-dependent.1514

That is important; so, at a different temperature, you are going to have a different Keq.1525

An equilibrium constant--yes, it is a fingerprint for that reaction, but it is a fingerprint for that reaction at a specific temperature, which is why the problems always give you what the temperature is.1529

It doesn't mean you have to use the temperature in a calculation; it just means "Know that it is at that temperature."1540

OK, so let's talk about what happens with temperature.1547

Temperature is actually really easy to deal with; let's do a quick review.1555

Exothermic means ΔH is (not equals) negative; ΔH is less than 0.1560

It also means that heat is released; heat is produced, in other words.1577

Endothermic is the opposite; that means the ΔH is positive (not equals, but is; the equals sign is used for numbers); in other words, heat is absorbed.1591

Or, a better way of saying it is "is required."1610

OK, so exothermic--now we will talk about how we deal with heat, as far as Le Chatelier's Principle is concerned.1617

Exothermic--heat is released; heat is produced; imagine that heat is a product--that is it.1627

Not only do you produce a chemical product, but you are also producing heat in the process.1634

Treat heat like a product; that is it.1639

Heat is a product, so you write it on the right side of the arrows.1643

Endothermic means heat is required; well, if heat is required, that means heat is a reactant--treat heat like a reactant--put it on the left side of the arrows.1650

Well, if I am writing heat as a product or a reactant, I can imagine it like a species.1666

If I increase the temperature of something, that means I am actually increasing the concentration of heat.1679

Now, I treat a change in temperature the same way that I treated an increase in concentration.1686

Well, let's write out what this looks like; let's do an example, actually.1693

Actually, you know what, let me just do a rough breakdown; so let me do aA + bB goes to cC + heat.1696

This is an exothermic reaction; I have written it; ΔH is negative.1706

I have written it as if heat is just another product; it is one of the things that is produced.1711

It produces a chemical, C, but it also produces heat.1716

Now, if I have this situation at equilibrium, what happens if I increase the temperature?1720

Well, if I increase the temperature, that means I am increasing the concentration of heat.1726

Well, if I increase the concentration of heat, the system wants to react by decreasing the concentration of heat; that means it wants to use up the heat--that means it is going to shift that way in order to use up the heat, to pull it away--just like the first part, when we did concentration.1732

If I do a temperature decrease on an exothermic reaction--well, with a temperature decrease, the temperature is going to want to offset by going up.1749

Well, the only way the temperature can go up is by producing more heat from the equation itself.1759

In order to produce more heat, the reaction has to shift to the right.1764

It is going to shift toward products, toward the production of heat--yes, and C, but of heat.1770

So again, when you are dealing with temperature, take a look to see which--whether your reaction is exothermic or endothermic.1776

If it's exothermic, write heat as one of the products; if it's endothermic, write it as one of the reactants.1783

And then, go back to treating it like it's just a reactant or a product that you are increasing or decreasing the concentration of.1788

So let's do an actual example.1796

We have: N2 + O2 is in equilibrium with 2 NO; so nitrogen gas and oxygen gas in equilibrium with nitrogen monoxide.1799

The ΔH for this process is 181 kilojoules.1813

I apologize; I tend to write my kilos with a capital K; I know it's probably bad practice, but...it's all right; it's not supposed to be, but...I'm going to be like that.1818

This is a positive 181 kilojoules; so this is an endothermic reaction.1826

OK, this is endothermic, so I'm going to write this reaction with heat as one of the reactants.1831

I am going to write heat, plus N2, plus O2, is in equilibrium with 2 NO.1836

So now, let's see what happens.1845

What can we do?...well, OK; if we increase (at part A) the temperature, what happens?1852

Well, if we increase the temperature, that means we are increasing the concentration of heat, because it is one of the reactants--it's an endothermic reaction.1861

It is going to want to decrease the concentration of heat to offset.1869

The only way to decrease it is to deplete it, which means it is going to push the reaction to the right.1872

So, the reaction will shift right.1878

In other words, shifting right means it will produce more nitrogen monoxide; in the process of producing more nitrogen monoxide, oxygen will deplete, N will deplete, and heat will deplete--it will use up the excess heat that I have put in there.1881

Well, if I do a temperature decrease--if I decrease the temperature--the system wants to increase the temperature back up.1895

The only way it has (the system has) to increase the temperature is by producing more heat.1903

The only way to produce more heat is by moving the reaction that way, to produce more heat, more N2, and more O2.1908

It is going to shift left.1915

That is all these shifts mean: is it going to move towards formation of more product or formation of more reactant?1919

Formation of more product means "shift right"; formation of more reactant means "shift left."1926

That is it; that is all that is going on here.1933

Let's do another example.1936

We have: 2 SO2 + O2 going to SO3 (well, it's 2 SO3; it's not balanced--and that is going to be 3 O2; there, that is a little bit better).1937

So, we have sulfur dioxide gas and oxygen gas, sulfur trioxide; I think I may have left the equation unbalanced earlier on; I apologize for that, if so.1956

So now, it says that the ΔH for this reaction is -198 kilojoules.1966

Well, a negative ΔH is exothermic; exothermic means it is releasing heat--it is producing heat--so I'm going to rewrite this as: 2 SO2 + 3 O2 is in equilibrium with 2 SO3 + heat.1973

Or, I can just write "energy"--and again, heat and energy are just the same thing.1990

So now, heat is a product; it is one of the things that is formed in this reaction.1993

Well, if I increase the temperature of this reaction at equilibrium--if I increase the temperature, the system is going to want to decrease that temperature back down.1998

The only way the system has to decrease the temperature is by shifting the reaction that way--using up heat and using up SO3 to produce more reactant; so it is going to shift left.2008

If I decrease the temperature, the system is going to want to offset by increasing the temperature.2022

The only way the system has to increase temperature is by producing more heat.2029

How does the system produce more heat?--it moves to the right to produce more sulfur trioxide and to produce more heat to compensate for the heat that I took away.2032

So, it ends up shifting right.2043

That is it.2049

So, we have a system at equilibrium, and there are a couple of things that we can do to it--well, three things that we can do to it.2052

We can add or subtract reactant or product; we can change the pressure of the system; or we can change the temperature of the system.2059

Each of those things causes the reaction to actually move in a direction to reestablish equilibrium.2069

That is the whole point: I have upset the equilibrium balance; the system needs to do what it can, if it can, to reestablish the balance.2076

That doesn't mean it can always do that; like we said, if we have 2 moles of gas on the reactant side and 2 moles of gas particles on the product side, and all of a sudden I either...let's say I change the pressure of the system by increasing it, well, there is no direction that the system can actually move in to reduce that pressure change, because both sides have 2 moles.2085

There isn't one side with a...so, basically, the system stays exactly where it is.2108

I have stressed out the system, but the system doesn't have a means of escaping, of alleviating that stress, so it stays there.2112

More often than not, it will--it will have an escape route; it will have an avenue by which to offset the stress.2120

But, it doesn't mean it will have; so you have to watch out for that.2126

OK, so let's do a final example here, and it will sort of be an all-encompassing example: What happens when we do all kinds of things to it?2130

So, let's have the equation example; let's do H2 (I really, really need to write a little bit more clearly, or at least slow down; I know that I tend to write quickly--I apologize) gas, plus I2 gas, is in equilibrium with hydrogen iodide gas.2140

Now notice: I didn't say hydroiodic acid; this is a gas--this is not aqueous.2164

When you take hydrogen iodide gas, and you bubble it in the water, that is when it dissociates into hydrogen ion and iodide ion.2169

That is when we call it hydroiodic acid.2177

This is just hydrogen iodide; it's not acidic--this won't hurt...well, I won't say it won't hurt you, but it won't hurt you as an acid.2179

OK, so now, what happens when H2 gas is added?2187

Well, if H2 gas is added, the system is going to want to move in a direction that actually depletes what I added.2195

It wants to offset it; in order to deplete H2, it has to move to the right, so it's going to shift right.2202

I'm going to use symbols here.2208

Part B: What happens when I2 is removed?2211

Well, if I remove I2, the system is going to want to offset it by producing more I2 that was there.2217

So, it is going to want to pull the reaction back this way, so it's going to shift to the left.2223

It is going to produce more reactant; yes, it will also produce more of this, but the idea is to produce more of this.2227

C: HI is removed.2235

Well, if I remove hydrogen iodide, the system is going to want to offset by creating more hydrogen iodide.2241

How do you create more iodide?--you shift the reaction to the right.2247

You don't do it; the system automatically shifts to the right (I'm sorry; I should be a little bit more precise).2252

You are putting a stress on the system; the system is reacting by moving to the right to offset the stress that you put on it.2257

Part D: Argon is added--argon gas.2266

Well, argon is a noble gas; it is inert; so we add an inert gas.2271

Well, if you are adding an inert gas, you are changing the total pressure of the system, but there is not really much that you can do; so, in this particular case, nothing is going to happen--no change.2276

OK, C, D, E: And...volume is doubled.2296

Well, if we double the volume, it is going to be sort of the same as the argon gas; even though it's inert...2308

Volume is doubled; that means the pressure is decreased, right?--volume is doubled; that means you are making the volume bigger, so you are decreasing the pressure.2317

But, you have 2 moles of particles on the right, and you have 2 moles of particles on the left; there is no place for the reaction to go, to offset the decrease in pressure.2330

It is just going to have a decrease in pressure; so there is no change.2341

OK, now let's do F: let's do a temperature increase.2347

The temperature is increased; OK...oh, I forgot to write...what is the ΔH here?--the ΔH of formation is 25.9 kilojoules.2354

So this is actually endothermic (because we need to know what the ΔH is, in order to decide what happens when we make a change to temperature).2366

So, if this is endothermic, that means that heat is one of the reactants--heat is one of the things required to actually move the reaction.2376

If temperature is increased (that means if I increase the concentration of heat), the system is going to want to offset by decreasing the concentration of heat that I put in; so, it wants to reduce this.2386

The only way to reduce this: the system must move in this direction; so it will shift to the right.2397

There you have it: Le Chatelier's Principle.2406

If you have a system at equilibrium, and you do something to mess with that equilibrium by either changing the concentration of a reactant or product, by changing the pressure of the system, or by changing the temperature of the system; the reaction is going to move in the direction that offsets that change.2410

It is that simple.2436

OK, thank you for joining us here at Educator.com, and thank you for joining us for equilibrium in chemistry.2438

This concludes our discussion of equilibrium.2445

Next time I see you, we will be talking about acids and bases.2448

Take good care; goodbye.2450

Welcome back to Educator.com; welcome back to AP Chemistry.0000

Last time we finished off our discussion of equilibrium, a very, very important topic, and now we are going to go on to probably the second central topic of chemistry, which is acids and bases.0005

They're very, very important in all aspects of chemistry, particularly in biochemistry, because much of the reaction that takes place in your body is, of course, acid-base reactions.0016

We are going to spend some time discussing acids and bases and writing reactions and talk about what is going on qualitatively, and then, we will also deal with it quantitatively, and we are going to introduce a new equilibrium constant, called the Ka or the Kb.0026

Now, fortunately, it isn't really anything new; it isn't new in the sense that it is a different way of writing the equilibrium constant.0043

It is just a different symbol for it; it is still exactly what you learned with equilibrium, and everything that we learned with equilibrium will again play the same part.0049

We will be doing ICE charts, so we will get plenty of practice in this.0058

If you feel like you don't have a solid grasp of equilibrium, acids and bases are a wonderful place to continue to practice that until you get a solid grasp of it.0062

Let's go ahead and get started with some definitions.0071

OK, what we are going to be using: there are a couple of models for acid-base behavior, and the primary model--at least, the one that is used most often--is something called the Bronsted-Lowry acid-base model.0075

Let's just talk about what that is; we'll just define it.0087

Bronsted-Lowry...the name itself doesn't really matter all that much--for historical reasons, I suppose it's nice to give credit to the people who deserve credit--but it is the model that is important: acid-base model.0093

Now, an acid is a hydrogen ion donor, and a base is a hydrogen ion acceptor--a very, very simple definition, and let's talk a little bit about what this means.0109

So, an acid is a compound that has an H+ that can come off--that can potentially come off.0133

I'll write "potentially come off"--in other words, it has a hydrogen ion to donate, if it needs to donate it.0161

More often than not, it will actually donate it--it will give up the hydrogen ion.0170

A base is something (I shouldn't say something--it's a compound--yeah, right, it's just "something")...is a compound that can potentially take that (not that; it will be that, but let's be as general as possible)...that can take an H+ ion.0176

That is it: so, a base is a compound that has a hydrogen attached to it, but the hydrogen can potentially separate from that compound, and just float around freely as hydrogen ion.0221

That is, in fact, the acid: when we speak about an acid, that is what an acid is: it is just free hydrogen ions floating around in solution--that is what does the damage.0233

A base is something that has the capacity to either take a hydrogen ion from something (an acid) that has it to give--in other words, rip it off--or, if there are free hydrogen ions floating around, a base can actually grab onto it.0241

That is it: so an acid is a hydrogen ion donor; a base is a hydrogen ion acceptor.0260

Acid-base chemistry--they come in pairs; when there is some acid, there is some base, usually; so we will see what that means in just a minute.0266

OK, there are two ways to write the reaction of an acid in water--or with water, I should say: to write the reaction of an acid with water.0277

And again, an acid is something that you are actually dropping in water, and once it is in water, it actually tends to come apart.0304

Here is what actually happens: we will just write H, and we will write A; so H is the H that actually comes off; this A can be anything--it's just a generic other part of the molecule.0312

...Plus H2O, goes to H3O, plus A-.0326

What has happened is that the H has gone from this HA over, and attached itself to the H2O; it has actually attached itself to the oxygen, and it is H3O+, now.0335

H2O is neutral; H+ is plus charge--that is why now, this has a plus charge; and because this was HA, it's a neutral ionic compound, but this H left its electron when it left (it left as H+), it has an A-.0346

There is another way to write this reaction, which is the way that I am certainly accustomed to doing it, and in certain circumstances, it helps to write it this way as opposed to this way.0365

It is why we actually have two ways of doing it, because it depends on the problem and how convenient it is to write one way or the other.0375

HA...it's the actual dissociation reaction: when you put this in water, it becomes free ion, H+ ion, and free A- ion.0382

This is a little bit more descriptive of what is really going on, in the sense that HA dissolves, like sodium chloride dissolves, and it breaks up into a hydrogen ion and the other ion (whatever the rest of the molecule is).0393

What is really happening is: there is an actual transfer.0409

This is the acid; it has the hydrogen to give up--it has the hydrogen to donate.0413

In this case, water is acting as the base; it has the capacity to take that hydrogen, or to accept that hydrogen.0418

So, when we write it this way, we actually talk about an acid conjugate base pair, which I will actually get to in just a minute; but I just want to sort of talk about these two equations, and how to represent them.0426

When we talk about an acid dissociating, or an acid reacting with water, we can either write it this way, or we can write it this way.0443

This H3O+ and this H+ are the same thing.0452

This H+ is just an H+; when it is attached to an H2O, we write it as H3O+, but the chemistry is the same.0456

Now, let's go ahead and write the equilibrium constant for this.0466

The Keq--well, we said the Keq is the concentration of product, concentration of product, divided by concentration of reactant, concentration of reactant.0470

This is liquid water; it doesn't show up--remember, liquids and solids don't show up in the equilibrium constant.0483

This is going to be: the concentration of H3O+ in moles per liter, raised to its coefficient (which is 1), the concentration of A-, raised to its coefficient (which is 1), divided by the concentration of HA; this is aqueous.0488

OK, water doesn't show up; when we have an acid written like this, or (let me actually write this version of it: H+, A-, over HA) when we are dealing with an acid, we don't call it Keq; we call it Ka.0507

This is the acid dissociation constant; it is the equilibrium constant for the reaction of an acid with water, or for the reaction of an acid dissolved in water.0530

When you dissolve it in water, we think about it as just a dissociation, like any ionic compound.0543

But what is really happening--it is actually reacting with water.0547

The acid is giving up the H; the water is acting as the base, accepting the H.0550

In some sense, when you look at this, this equation, this equilibrium--what it expresses, what it represents...it represents a competition between water and the conjugate base for this H.0555

In other words, does the H stay with the A, or does it go with the water?0570

If it goes with the water, that means the reaction is over on this side: H3O+ is formed.0574

If the H doesn't want to leave the A--if it wants to stay with the A--that means most of the reaction is on this side.0580

That is what the Ka measures.0586

So, the equilibrium constant is called the acid dissociation constant.0588

It is the same as any other equilibrium, except it is representing what is happening when the acid reacts with water, when the acid is dropped into water.0595

That is it; so, from now on, when we are discussing acid-base chemistry, it is going to be Ka.0605

Later on, when we discuss bases, we are going to have an analogous equation, and it is going to be Kb.0610

But it is just an equilibrium constant.0615

And what does an equilibrium constant represent? It represents the extent to which a reaction has moved forward.0619

Once it reaches equilibrium, how much of this, how much of this, how much of this?--it's a measure of the extent of reaction.0626

How far forward has it gone? How far has it not gone?--that is all it is a measure of.0634

The higher the Ka, that means the dissociation; that means it's to the right, because the numerator is a bigger number.0639

OK, now, let's talk about polyprotic acids really quickly.0648

For acid dissociation, 1 H+ leaves at a time.0656

For example, if I have the acid H3PO4, which we know of as phosphoric acid, it doesn't do this: H3PO4--it doesn't just dissociate into 3 H+s plus PO43-.0683

It doesn't do that; that is not how acids behave.0699

This doesn't happen; what does happen is that 1 H+ leaves at a time, so the dissociation of phosphoric acid is three steps, and it looks like this.0704

H3PO4 + H2O is going to be in equilibrium with H3O+ + H2PO4-.0722

That is the first dissociation; there is some Ka associated with that first equilibrium.0733

And then, H2PO4 has two more Hs to give up; so, H2PO4 (actually, you know what, I think I'm going to write it as just a regular dissociation--I'm sorry; I think it will be a little more clear) now gives up another one.0739

It gives up 1 at a time: H+, and it becomes HPO42-; there is an equilibrium associated with that.0762

OK, and then we have HPO4; it has another H+ to give up, and now it's PO43-; there is a third acid dissociation constant.0773

And each one of these is actually different; these phosphoric acid dissociates to differing degrees--these are not the same number.0787

In other words, this equilibrium might be really far forward; this equilibrium may be really far this way; it just depends.0796

The key idea is: for a polyprotic acid (polyprotic just means having more than one hydrogen to give up), it gives them up one at a time.0805

Very, very important: for acids, it releases one at a time; it doesn't happen all at once.0813

So, H2PO4: it doesn't just release both hydrogens--it releases one completely, and the second one comes off very little, in fact--so, just something that you should know.0819

OK, so let's talk about acid strength.0830

Acid strength: Acid strength is defined by the position of the equilibrium constant--or by the position of equilibrium; in other words, which is measured by the equilibrium constant.0837

It's measured, it's defined, by the position of the equilibrium; and the equilibrium, we said, was some acid plus water, going to hydronium ion...or just H3O+...0856

We call it hydronium, by the way--let me just write that name down: H3O+ is called a hydronium.0874

Anything with a positive charge tends to have an -onium ending; it's the same as H+--you can treat them the same.0882

So, the acid strength is defined by the position of the equilibrium; if the equilibrium lies really far to the right, that means, if this acid has completely given up all of its Hs over to water, it is a strong acid.0891

So, the farther to the right the equilibrium is, that means it is a stronger acid.0903

We talk about a strong acid being something that actually does not want to keep its H's; it wants to give them up completely.0914

It wants to be A-, and it wants its H to be H+.0922

It does not want to be together.0928

OK, now, let's talk about...let's see; so, actually, let me write that again.0930

Let me say: farther to the right, stronger acid--so, when we talk about acid strength, we are talking about relative strength; it's always...we are comparing it to something.0942

Stronger acid--well, that means a higher Ka, right?--because the Ka is products over reactants; well, farther to the right means there is more product and very little reactant--a really big numerator, a really small denominator, and a huge number--stronger acid.0958

OK, now let's go back to blue here.0976

A strong acid implies a weak conjugate base.0983

Acids and bases come in conjugate pairs--"conjugate base"; we get the conjugate base by just pulling off the hydrogen from the acid.0988

So, let's write our equation again: HA + H2O is in equilibrium with H3O+ + A-.0999

This is the acid; this is its conjugate base.1010

I pulled off the H; that leaves my conjugate base.1021

If I take a base (in this particular case, again, this is the acid--this has the H to give up; water is acting as the base--it is going to take that H, or accept it, if it wants to), when it accepts the H and it becomes H3O+, this is called the conjugate acid.1026

That is it; if you have an acid, you pull of the H; you are left with a conjugate base.1048

If you have a base, and if you add a hydrogen ion to it, you have its conjugate acid.1053

That is it--nothing more than that; no more, no less: don't read any more into that.1060

OK, so a strong acid means a weak conjugate base, which is exactly what you would expect.1064

A strong acid does not want to hold onto its H's; it will give them up completely; that means it has a weak conjugate base.1070

A weak conjugate base means that it doesn't want this H; that is the whole idea.1077

A strong base wants the H--will take the H; it's a competition.1084

A weak acid has a strong conjugate base.1091

A weak acid is one that does not want to give up its H's; and the reason it doesn't want to give up its H's is because its conjugate base is so strong that it literally holds onto its H's without letting go.1104

That is the idea; so, weak acid means conjugate base.1116

When you are thinking about this, you have to pick a perspective; you have to pick a point of reference; that is the whole idea.1121

We decide to take the reference point as the acid itself; we call it strong; we call its conjugate base weak.1127

That means the conjugate base doesn't want the H; it gives it up.1134

A weak acid is one that does not want to give up its H; that means that its conjugate base is very, very strong--it will not release its H.1139

So, it's really a question of perspective; you need a point of reference; and the point of reference that we have taken is the acid.1149

We talked about a strong acid and a weak acid: a strong acid has a weak conjugate base; a weak acid has a strong conjugate base.1157

OK, for strong acids, no Ka exists...no Ka is listed, we should say.1172

So, for example, hydrochloric acid is a very strong acid; its dissociation (it's not even in equilibrium)--it completely breaks up into free H+ and free Cl-.1184

You won't find any of this at all.1195

Well, the Ka for this is H+(Cl-)/HCl.1198

Well, since there is a whole bunch of this and a whole bunch of that, and virtually none of this, this denominator is really tiny; the numerator is really big.1207

So, you have a Ka which is huge; in fact, the Ka is so huge that we can't even measure this.1216

Because we can't measure the Cl concentration accurately, we don't even list a Ka for it; it's just off the charts.1222

Strong acids--full dissociation; that is the idea.1228

Once again, strong acid means full dissociation.1233

That means this H and Cl completely come apart; in solution, you will never find any Cl attached to an H.1240

It will free Cl- and free H+.1249

That is what makes it a strong acid; because there is so much H+ floating around freely, that is what does the damage.1252

So, when we say "strong acid," we are talking about a chemical property.1259

Strong acids do damage because there is so much H+; there is so much H+ because the H and Cl do not want to stay together--they want to be separate.1263

Strong acid--full dissociation: that's very important.1274

OK, let's do an example to get a sense, or a feel for what is going on, for some numbers here.1279

So, Example 1: HF, which is hydrofluoric acid, has a Ka of 7.2x10-4.1290

It's a small number; it's a weak acid--it's not very far forward.1301

Most of it is in this form; OK, so let me write the dissociation: HF goes to H+ + F-.1305

When I measure this, this, and this, and I put these on top of that--the Ka--I get this number.1314

It's pretty tiny; a tiny Ka means that most of this equilibrium will be found on this side, the reactant side.1320

In other words, there is very little dissociation; most of it just stays HF.1327

Hydrocyanic acid, HCn--it has a Ka of 6.2x10-10--wow, really, really small.1333

So, when I see HCn--when I drop some of that in water (that is Cn-), this would be the dissociation.1343

This is so tiny that this is telling me that most of this reaction is over on this side.1351

I won't find a lot of free H+ and Cn-; I'll find some--I was able to measure it--but most of it is this.1356

That is what the small Ka means; a small Ka means it hasn't dissociated very much.1362

A big Ka means it has dissociated a lot more.1368

Between these two, this is bigger; this is smaller; this is a stronger acid, because its Ka is bigger--it has dissociated more.1371

It is still a weak acid, but compared to hydrocyanic, it's stronger than hydrocyanic.1383

That is the difference; when we speak about the strength of acids, we are often going to be comparing; it's relative.1390

We are going to be talking about 2 or 3 or 4 species.1396

Now, the question is: which one has the stronger conjugate base?--that is the question.1402

Which acid has the stronger conjugate base?1409

Well, for HF, the conjugate base is--just take off the H; this is the conjugate base.1421

For HCn, it is the acid minus the H; that is the conjugate base.1427

Which has the stronger conjugate base? Well, the stronger conjugate base comes from the weaker acid, right?1433

Weak acid=strong conjugate base; OK, between these two, the weaker acid is the hydrocyanic acid.1445

In other words, it's a weak acid because it doesn't dissociate very much; that is confirmed with this really, really small Ka.1456

It doesn't dissociate very much; it's a weaker acid; it has a stronger conjugate base.1464

In other words, the base is so strong that it doesn't want to release its hydrogens; it wants to hold on to it.1470

That is what a strong base does: a strong base grabs onto its hydrogens and holds onto it.1476

A strong acid wants to give up its hydrogens; that is why we have strong acid, weak base--yes, stronger acid, weaker base; weaker acid, stronger base.1482

So, in this case, F- is a weaker conjugate base than Cn-.1495

That is it; we are using Ka values to decide, depending on what species we are talking about.1509

This Ka means HF is a stronger acid than HCn.1515

It means its conjugate base is weaker than Cn-; that is what is going on here.1520

OK, so now, let's discuss water as an acid and a base.1526

Let's define something called amphoteric: an amphoteric substance is one that can behave as an acid or a base.1538

It should be "and a base," because it actually does both; "as an acid or a base"--in other words, something that can go both ways.1561

Well, let's look at the dissociation of water.1571

I'm going to write it slightly differently.1575

I'm going to write the first water: so now, we are talking about water as a species dissolved in water.1577

It's plain old water; I know it's a little weird, but think about it this way.1583

I'm going to write it as HOH + H2O.1587

These are both water, but I wanted to emphasize what is happening.1591

This HOH, believe it or not, actually dissociates just a little bit; it breaks up into H+ and OH-.1596

Well, this H goes over here to become H3O+ + OH-.1602

Another way of writing this is HOH, as free dissociation, without involving the water species, breaks up into H+ + OH-1611

Well, let me rewrite it: HOH is in equilibrium with a little bit of H+ and a little bit of OH-.1622

The Keq for this is equal to...well, this is liquid water; liquid doesn't show up in the equilibrium concentration; this is aqueous; this is aqueous; this is floating around, in other words, and this is floating around in the water, to a certain degree.1634

It is equal to the H+ concentration times the OH- concentration--the products.1650

Well, as it turns out, for water, we actually call it Kw.1656

We have done the experiment; and, as it turns out experimentally, we have actually measured, at 25 degrees Celsius...experimentally, at 25 degrees Celsius, the hydrogen ion concentration is equal to the hydroxide ion concentration; that is what this equation says.1662

For each H that is produced, one OH is produced.1685

They have the same concentration; it equals 1.0x10-7 moles per liter.1688

The concentration of H+ and OH- in standard, neutral, run-of-the-mill water, is 1.0x10-7.1696

Well, what does that mean?--Kw is H+ times OH-, so Kw equals 1.0x10-7 squared, equals 1.0x10-14.1706

Water has a Ka of 1.0x10-14.1723

We give it a special name; we call it Kw for water--it's very, very important that you know this.1728

Kw is 1.0x10-14 at 25 degrees Celsius--always.1734

OK, what does this mean?--it does have a meaning.1743

This means (and here is where it gets really, really important): in any, in any aqueous solution, at 25 degrees Celsius, no matter what is in it--no matter what--no matter what is in it (in the solution), the product of H+ and OH- must always equal 1.0x10-14.1753

OK, we need to stop and think about this, because this is really important.1811

Any aqueous solution, no matter what is in it--sodium chloride, potassium phosphate, the complex ion, magnesium hydroxide, phosphoric acid, permangenic acid--anything--if I were to measure the hydrogen ion concentration and the hydroxide ion concentration, they have to--the products have to equal--have to equal 1.0x10-14.1816

That is what this says; at any given moment, an aqueous solution (in other words, water--something that is in water)--the product of the hydrogen ion and the OH- ion concentration always 1x10-14.1842

That doesn't mean that each one is 1x10-7; that is just for neutral water.1858

It just says that, if...so this is equal to the H+ concentration times the OH- concentration--that means, if the H+ concentration rises, the OH- concentration has to drop in order to retain the constant value of 1.0x10-14.1863

So again, they don't have to equal each other; their product has to be a constant.1883

If one goes up, the other goes down; that is the relationship--that is the mathematical relationship.1888

OK, their product has to be 10-14; and it is going to be very, very important in just a minute.1894

So, let's just...if the concentrations do equal each other, we call it neutral.1899

If the hydrogen ion concentration is bigger than the OH- concentration, we call it acidic.1910

That is where we get the idea of "acidic"; so, when you say that orange juice is acidic, that means that, if you take some orange juice, and if you measure the concentration of H+ versus the concentration of OH-, the H+ concentration is going to be higher.1919

But, the product of the two concentrations is still going to be 10-14.1934

If the H+ concentration is less than the OH- concentration, we call that solution basic.1940

That's it; OK, so let's do an example.1948

Nice, simple math: Calculate H+ and OH- (I'm really, really sorry about my handwriting; I know) concentration for the following solutions.1956

OK, let's see: Oh, "at 25 degrees Celsius."1982

A: We have: 1.5x10-5 molarity OH-.1991

Well, we know what the concentration of OH- is; it's right there--it's 1.5x10-5.2000

Well, we know that the H+ concentration, times the OH- concentration, at 25 degrees Celsius, equals 1.0x10-14, so we just plug it in to find the H+ concentration.2010

H+ concentration equals 1.0x10-14, divided by the OH- concentration; I have just done a simple division.2026

1.0x10-14 (ten to the negative fourteen, not fifteen), divided by OH-, which is 1.5x10-5, and I get a concentration of 6.7x10-10.2038

Which one is a higher concentration? OH- is 10-5; H+ is 10-10; the OH- is a bigger concentration--the solution is basic.2058

It will still hurt you, so don't...it doesn't mean that a base won't hurt you; an acidic solution will hurt you--a basic solution will also hurt you.2072

OK, the idea is: we want a neutral solution.2080

By the way, blood pH is just about...we will get to that in a minute; sorry about that.2083

OK, so let's do another one; let's do a 2 Molar solution--2.0 Molar solution of H+.2091

Well, again, we have the H+ concentration: in this case, a 2.0 Molar solution, a solution that contains 2 molarity H+; that is the hydrogen ion concentration.2101

Well, we know that the H+ concentration, times the OH- concentration, equals 1.0x10-14; therefore, the OH- concentration equals 1.0x10-14, divided by 2.0, which was that.2112

And we end up with an OH- concentration of 5.0x10-15--a very, very small number; in this case, the hydrogen ion concentration hugely, hugely out-values the OH- concentration.2132

This is acidic--I would say highly acidic; that is it; very, very dangerous.2149

OK, let's do another one here.2158

Example 3 (or, actually, I'm not sure if it's Example 3, but...I'll just do an example): At 100 degrees Celsius, the Kw equals 5.13x10-13.2163

So remember: we said that, at different temperatures, the equilibrium constant is different.2183

At 25 degrees Celsius, the Kw is 1.0x10-14; but if I change the temperature to 100 degrees Celsius, now it has gone up to 5.13x10-13.2187

It has gone up.2199

The question to you is: Is this reaction, a dissociation of water into H+ plus OH---is it endo- or is it exo-thermic?2202

OK, well, let's see what happens here: at 100 degrees Celsius, they are telling me that the Kw equals 5.13x10-13.2227

They want to know if this reaction is endo- or exothermic.2240

Well, let's see what we did: we increased the temperature, and by increasing the temperature, we actually pushed the reaction forward, because that is what this Kw is telling us.2243

This Kw of 5.13x10-13 is bigger than the 1.0x10-14, right?--negative 13 is bigger than negative 14.2256

That means that the equilibrium has actually shifted to the right; that means more product has been produced.2267

That means that the reaction shifted to the right when I increased the temperature.2273

Well, if it increases to the right when I increase the temperature, that means that temperature, or heat, must have been one of the reactants, because, if it's one of the reactants, and if I increase this temperature, the system is going to want to offset what I do to it (which is an increase) by decreasing the temperature.2283

Well, in order to decrease the temperature, it has to use up this extra heat that I have pumped into it.2311

In order to use up the heat, it has to shift the reaction that way; it has to produce more H and more OH-, which is exactly what it did, because now the equilibrium constant has gone up.2317

When an equilibrium constant goes up, that means the reaction has moved to the right.2329

The numerator of the equilibrium constant has increased; so, in this case, because heat is on the left side of this equilibrium arrow, this is an endothermic reaction.2333

So notice, we used our discussion of equilibrium and Le Chatelier's Principle to actually find out, qualitatively, that this is an endothermic reaction.2346

In order to push this reaction forward--in order to produce more H+ and more OH-, I actually have to increase the temperature, which means that it is an endothermic reaction.2356

I don't know what the ΔH is, but I do know that it is going to be positive for this reaction.2368

OK, B: what is the H+ and OH- concentration in a neutral solution at 100 degrees Celsius?2375

Well, we just said that Kw is equal to the concentration of OH- times H+ at 100 degrees Celsius.2398

We gave it already; it's 5.13x10-13.2407

Well, in a neutral solution--go back a slide or two--neutral means that the OH- concentration is equal to the H+ concentration.2412

So, if I call this X and call that X, I end up with: X squared is equal to 5.23x10-13.2425

I get: X is equal to 7.16x10-7 molarity.2435

At 100 degrees Celsius, there is 7.16x10-7 moles of H+ per liter of water, and 7.16x10-7 moles of OH- per liter of water.2444

That is all that is going on here; molarity is moles per liter.2461

OK, let's see what else we can do.2466

Now, we are going to close this discussion off by introducing the pH scale.2474

Now, I have to (I probably shouldn't say this, but I'm going to go ahead and do it, because I tell all my students this)...working with concentrations is perfectly valid--concentration is something we always work with; concentration of H+, concentration of OH-, concentration of Cl-...2479

With acids and bases, because these concentrations tend to be really, really small, as you have noticed (you know, 10-7, 10-6...), chemists have come up with a way to have numbers that actually are a little more easy to deal with--normal numbers like 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.2499

So, they have created this thing called a pH scale.2516

Here is what it means: the p of anything is equal to negative log of the concentration of that anything.2519

So, in other words, if I talk about pH, that is equal to the negative log of the hydrogen ion concentration.2532

If I talk about pOH, that is equal to the negative log of the hydroxide ion concentration.2540

If I say pKa (so this p function means whatever it is...p of whatever--just take the negative log of that whatever), pKa is the negative log of Ka.2546

It is a way of taking numbers that are really odd, like 6.6x10-3, and turning them into actual numbers that you can deal with: 3.16--something like that.2562

Personally, I don't care for it myself; I like dealing strictly with what we are dealing with.2571

We are dealing with concentrations; the equilibrium constant is expressed in concentrations; so, I think it is perfectly valid to deal with concentrations.2578

But again, this is so deeply entrenched in chemistry that literally nobody talks about concentration anymore when they speak about acid-base.2588

They speak about pH and pOH and pKa.2596

That is the only reason; but they are actually the same thing.2599

It is not like we are introducing a new concept; we are just introducing a way of changing the numbers of concentrations that tend to be really small, and turning them into numbers that make more sense, that look better--not make more sense, just look better, aesthetically.2605

OK, so one more example: if I said, "What is the pCl-?" well, that is going to be the -log of the Cl- concentration, if we happen to be talking about Cl-.2618

It is just a p function, and it is the negative log of the concentration we are talking about.2630

OK, let's do an example.2636

What is the pH and pOH of a solution which is 1.4x10-3 Molar hydroxide?2642

OK, well, they want the pH, and they want the pOH.2668

Well, I know the pH is the negative log of the hydrogen ion concentration; pOH is the negative log of the hydroxide ion concentration.2672

Well, they give me the hydroxide ion concentration, so why don't I just deal with that one first.2681

pOH is equal to negative log of the OH- concentration, which is negative log of 1.4x10-3.2685

I just stick in this in the calculator; I take the logarithm of it; I change the sign; and I end up with 2.85.2699

So notice, 2.85 is a much more attractive number than 1.4x10-3; at least, that is what many people think.2709

I, personally, don't think so, but that is fine; I'm a chemist; we deal with pHs; we'll deal with pHs.2716

But, don't think it is something different; it is not.2722

If you want to, you are more than welcome to deal strictly in concentrations, and if you ever have to find a pH, just at the last minute take the negative log of it: no harm, no foul.2724

Now, we want the pH; well, the pH--we know that the H concentration, times the OH- concentration, is equal to 1.0x10-14.2735

The H+ concentration equals 1.00x10-14 (well, let me just stick with 1.0; let's just do 2 significant figures here), divided by 1.4x10-3.2748

That is going to equal 7.2x10-12, and then, when I take the pH of that (that is going to be the negative log of the H concentration), -log of 7.2x10-12 equals 11.14.2767

11.14 pH; 2.85 pOH: notice something--what is 11.14 plus 2.85? Yes, you guessed it--it's about 14.2791

Now, let's explain why.2806

We said that Kw is equal to the hydrogen ion concentration, times the hydroxide ion concentration.2812

Let's take the negative log of both sides.2822

-log of Kw is equal to -log of this whole thing: H+, OH-...2824

Well, -log of Kw is pKw; that is it--equals -log of H+, right?--plus the negative log of OH-.2836

You end up with pH plus pOH; well, pKw is the negative log of 1.0x10-14, equals pH plus pOH.2860

Because Kw is 1.0x10-14, the negative log of that is 14; that is our final relationship.2884

So, for any aqueous solution at 25 degrees Celsius, pH plus the pOH of the solution equals 14.2895

This is just a restatement of the fact that the H+ concentration times the OH- concentration equals 1.0x10-14.2914

These are the same thing, except one deals with these numbers (these scientific notation numbers, the small ones); one deals with numbers that are a little bit more tractable.2928

In any aqueous solution, the product--the product--of the hydrogen ion and hydroxide concentrations is 10-14.2937

The pH plus the pOH of that solution is equal to 14.2948

OK, so we have introduced acid and base; actually, we have introduced acid mostly; we have talked a little bit about base, mostly in the context of conjugate base.2955

We have talked about the dissociation of acids and what happens in water.2964

We have introduced the idea of an equilibrium constant for that acid dissociation.2970

We have introduced the idea of water acting as both acid and base, an amphoteric substance.2976

And we have introduced the very, very important property that the product of the hydrogen ion and the hydroxide ion concentrations is 10-14; or, if we want to use the p scale, we can talk about the pH and the pOH equaling 14 for any aqueous solution at 25 degrees Celsius.2981

Next time, we will continue our discussion of acids and bases, and we will actually get into some of the more equilibrium-type problems that are involved.3000

Thank you for joining us here at Educator.com and AP Chemistry.3008

We'll see you next time; goodbye.3010

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

In our last discussion, we introduced the basic notions of acids and bases; actually, we didn't talk about bases so much--we will actually talk about those mostly in the next lesson.0004

We introduced the notion of an acid, and we said an acid is something that has hydrogen ions that it can give up.0014

Often, what happens is: you will take a particular acid, like hydrogen chloride, drop it in water, and that will dissociate, and it will separate into hydrogen ions and chloride ions.0020

Or, it may not separate at all, which is part of the topic we are going to be talking about today: weak acids.0030

We talked a little bit about the pH scales, the relationship between the hydrogen ion concentration and the hydroxide ion concentration in water.0035

We said that each one of them is 10-7, to come up with a total of 10-14.0045

Right now, in a minute, I am actually going to write down some of the basic things we discussed, and then we are just going to launch right into a series of problems on how to handle the pHs of weak base solutions.0050

A weak base--I'm sorry, a weak acid--is something that doesn't dissociate completely; so let's just jump in and get started and see what we can work out.0064

OK, so, last time, we said that, remember, we have water, which acts as both an acid and a base.0073

H2O--it dissociates into H+ plus OH-.0080

We said that the equilibrium constant for this--because this is liquid water, this is aqueous, this is aqueous--so our Ka for this, which we ended up calling Kw for water, was just the hydrogen ion concentration, times the hydroxide ion concentration.0087

It always equals 1.0x10-14 (let me make that 14 a little bit more clear), at 25 degrees Celsius.0106

So, the Keq, the Ka...these things change at temperature, but normally, we are talking about 25 degrees Celsius.0119

This means that any aqueous solution, no matter what is in it, the hydrogen ion concentration times the hydroxide ion concentration is always going to equal 1.0x10-14.0125

That is very, very convenient, because, no matter what one of these is, we can always find the other--and that is the whole idea.0137

Today, we are going to talk about weak acids; and weak acids are ones that actually don't dissociate completely.0145

Let's talk about a strong acid first: hydrogen chloride--when we drop this is water, the dissociation reaction for this is H+ + Cl-; it's a strong acid--strong means it completely dissociates in water, just like sodium chloride (salt) or anything else that is just completely soluble.0153

Well, as it turns out, a weak acid doesn't do that; a weak acid behaves like this: like we said last time, HA--it will react a little bit with water, and it will actually establish a little bit of an equilibrium.0174

What you will end up getting is something like H3O+, plus the conjugate base of this weak acid.0187

Another way of writing this, which I personally prefer, is just the standard dissociation, without mentioning the water: H+ + A-.0195

Now, we said there is a Ka associated with this; and the Ka is equal to the H+ concentration.0205

And again, sometimes I will use the square brackets; sometimes I won't, simply to save myself a little bit of time.0210

But, these are all concentrations in moles per liter.0216

A- over HA: now, most of the time, for the weak acids, these Kas are going to be very, very small.0219

They are going to be things on the order of -4, -5, -8, -10, even -12 and -15.0227

What that means: because a Ka is small, a small equilibrium constant means that the reaction has not moved very far forward.0235

That means most of it is here; so, if I took, let's say, some hydrofluoric acid (which is not really hydrofluoric acid--it's hydrogen fluoride, and I dropped it in water to create a solution of hydrofluoric acid), it doesn't actually dissociate too much.0244

Yes, it does produce a little bit of this and this; it does come apart a little bit; but it is a weak acid, which means that most of the hydrofluoric acid will be floating around as HA, as the entire species, in aqueous solution.0260

You still can't see the Ha in there; it is still a homogeneous solution; but it is solvated, so it isn't a hydrogen...0274

This is why, when we talk about acids, acids are always that particular compound, found in water.0281

We just automatically presume that we are talking about an aqueous solution.0289

As it turns out, hydrogen fluoride, when you make it, is not an acid--it's just hydrogen fluoride.0293

Things become acids when you drop them in water; but, because we always find them in water, because we are always talking about aqueous chemistry, we just often refer to them as acids.0298

A weak acid is one that doesn't dissociate very much.0307

Now, we need to find a way to actually find the pH of these solutions; and that is pretty much what we are going to be doing today, in this lesson; we are just going to be doing several problems.0311

Some of them are exactly like the ones before--maybe slightly different; but the idea is to develop a sense of turning what we have discussed (as far as the basic definitions of pH acid)--now we want to quantify it--now we want to use some of the math behind it.0319

What you are going to discover is: these problems that we are going to do are exactly like the equilibrium problems that we did in the previous chapter.0336

This whole ICE chart--the Initial, the Change, the Equilibrium concentration--that is going to happen over and over and over and over again.0343

You are going to be doing a ton of these; and I am specifically concentrating on doing a number of these problems, because it is in the process of doing these problems that we actually get a sense of what is going on with these acid-base chemistries.0350

There is going to be a lot of commentary, a lot of talk.0362

This is where most of the deep chemistry is happening; and, if you understand this stuff, the AP exam will be an absolute piece of cake, I promise you.0365

This is where most of the problem takes place, and we want to make sure we really understand it, because it's beautiful chemistry; it's applicable chemistry; and...well, let's just do it.0375

OK, so we are going to start off, actually, with a strong acid, just to get a sense.0385

The first example is going to be: Calculate the pH of a 0.15 Molar nitric acid solution (nitric acid is HNO3), and the pH of a 1.5x10-10 Molar HCl solution.0389

OK, so let's do the first part: let's calculate the pH of a .15 Molar nitric acid solution.0420

All right, when we do these acid-base problems (strong acid, weak acid...later on when we work with weak base, strong base; and later on when we work with buffers), it is always the same thing.0426

We want to develop a nice, systematic approach, because, as the species in the solution get more complicated, it's all that much more important to keep track of what is going on.0437

We are going to use these problems to develop a sense of the chemistry; we are going to let the chemistry decide how the math works.0446

You need to understand the chemistry; if you don't understand the chemistry, the math is irrelevant.0451

It will mean nothing to you; but, if you understand the chemistry, the math is a piece of cake.0456

In fact, it is pure instinct; there is nothing here, mathematically, that you don't know or can't follow--there is nothing mysterious happening.0460

We just want to make sure you understand what is happening.0467

OK, so first of all, we notice that it is a strong acid; a strong acid means full dissociation--that means this is going to happen.0470

That means HNO3 is going to completely dissociate into H+ ion plus NO3-; that means H+ is going to be floating around freely, NO3 is going to be floating around freely, and there is going to be nothing that is left.0479

That is why there is an arrow pointing one way--this is not an equilibrium.0491

Strong acid--full dissociation.0494

So now, what we ask ourselves--always the first question we ask ourselves: What is the major species--list the major species in the solution.0497

Well, the major species in solution are...this will decide what the chemistry is, because there is always going to be something that is going to dominate the chemistry in the solution.0504

Well, major species: we have...of course, we have H2O--that is our solvent; we have H+ floating around; and we have NO3- floating around.0517

That is it; well, they are asking for the pH--what is pH?--remember, we said that pH is just a different way of dealing with the hydrogen ion concentration.0527

It is the negative log of the hydrogen ion concentration, in moles per liter.0537

Well, look at the major species we have in solution: we have H2O; we have H+ and NO3.0543

There are two sources of hydrogen ion: one of the sources is the dissociation of the nitric acid itself; well, the other source is the dissociation of water (remember, water also comes apart and produces a little bit of this).0548

But, because this is such a strong acid that all of these hydrogen ions...actually, what it is going to end up doing is: you know that this is (under normal conditions) just 10-7 molarity--the hydrogen ion concentration.0564

It is 7 orders of magnitude smaller than this; it is virtually...you could completely neglect it; as a matter of fact, because this dissociates (remember Le Chatelier's Principle?), all of the hydrogen ion that is floating around from the dissociation of this strong acid actually pushes this equilibrium this way.0579

So, there is even less hydrogen ion coming from water itself.0598

For all practical purposes, we can ignore that; the dominant species here is this; our problem is already done.0602

It says we have .15 Molar HNO3; well, strong acid...0.15 moles per liter HNO3 produces .15 Molar H+ and .15 Molar NO3-, because it's 1:1, 1:1; that is the ratio.0611

This is easy; the pH is just minus the logarithm of 0.15; that is it, because the concentration of H+ in a .15 Molar HNO3 solution is .15 moles per liter of H+.0626

That is the free thing floating around; and we get 0.82.0642

That is it--nice and simple: strong acid--all you have to do is use the molarity of that acid.0648

Just take the negative log of it; you automatically have the pH, because you have full dissociation.0655

Now, let's do the second part, B: Now, we have a 1.5x10 to the negative (11 or 10?--what did we say here)...1.5x10-10 Molar HCl solution.0660

OK, again, let's see what the major species are in solution.0676

Well, again, we have HCl; it's a strong acid; it breaks up into H+ + Cl-; well, 1.5x10-10 Molar HCl is full dissociation.0680

That implies that it is 1.5x10-10 Molar H+.0696

Well, think about this for a minute (and again, this is why you don't just want to jump into the math--you are going to get the answer wrong if you just all of a sudden take the negative log of this, like we did the first one): look at this number.0701

1.5x10-10: that is a really tiny number; compared to the auto-ionization of water, water is 1.0x10-7; 1.0x10-7 is three orders of magnitude bigger than 1.5x10-10.0712

So, virtually, for all practical purposes, this hydrogen ion concentration that comes from this hydrochloric acid, which is a strong acid, is completely negligible.0730

The reason it is negligible is because the concentration of the H is so little.0741

Yes, it's a strong acid, so our instinct tells us, "Well, if it's a strong acid, then it must be the dominant force in the solution"; it's not.0746

There is so little of it that it doesn't have any effect; so for all practical purposes, the H+ concentration here is what it would be in water at 25 degrees Celsius--which is 10-7.0753

Well, the negative log of 10-7 is 7.0768

So, our pH is 7 in this particular case.0773

We took a look at the chemistry; we took a look at the numbers involved in the chemistry--the concentrations; that helps us decide what is going to happen next--what math to use.0777

Here, .15 is a pretty high concentration; 1.5x10-10: that is three orders of magnitude less than the normal concentration of just standard water--the hydrogen ion concentration in standard water.0787

We can ignore it, despite the fact that it is a strong acid.0801

Don't let names fool you; let the chemistry tell you what is happening--pull back from the problem and think about what is going on.0804

Science and chemistry--well, all of science, not just chemistry...there is a lot happening; you have to be able to keep track of multiple things.0811

It is not just about solving the problem; solving the problem is easy if you know what is happening.0820

If you don't know what is happening, solving the problem will always be a nightmare; you will always be a victim--you will always be sort of a slave to algorithmic procedures.0826

And once something--once a problem deviates from that algorithmic procedure, you are going to be lost.0835

You don't want to be lost; you want to be able to think--always think; that is the idea.0840

OK, so now, let's go on to weak bases; and this is where the mathematics and the situation actually gets interesting.0844

We are going to present--we are going to do the same thing: we are going to present a nice, systematic approach for all of these problems.0852

Weak acid, weak base; strong acid, strong base; buffer solution; the same steps are the habit we want to get into.0857

OK, so let's do Example 2: Calculate the pH of a 1.5 Molar hydrofluoric acid solution.0865

So again, we could say hydrogen fluoride solution; it's the same thing--it just means you have taken hydrogen fluoride, and you have dropped it into some water, and you have created a solution.0888

Hydrogen fluoride is your solute; water is your solvent; now, it becomes an acid.0897

Even though we speak about HF being hydrofluoric acid, it's only an acid when it has been dropped in water; it's very important to remember that.0902

OK, the Ka for hydrofluoric acid (because it is a weak acid, it has a Ka; remember, strong acids don't have Ka--the equilibrium lies far to the right, so the denominator is virtually 0) is 7.2x10-4.0911

That is a kind of a small number: 7.2x10-4; well, what does a small Ka mean?0927

A small Ka--it means that there is very little dissociation--that is the whole idea.0935

You need to know what these things mean; they are not just numbers and parentheses and concentrations; it means something, physically.0941

It means there is very little dissociation.0948

Very little dissociation--that means that HF...the dissociation of HF into H+ + F-...that means mostly, if we looked at this, it's going to be mostly hydrogen fluoride, not hydrogen ion and fluoride ion.0952

That is what the small Ka means.0966

OK, in other words, it is mostly here; the equilibrium is mostly here; in solution, it's still just HF floating around, not H+ and F- floating around freely; so, we will say mostly here.0969

Well, the next step is...let's list our major species to see what is going on.0981

Let me actually list the major species here: our major species in this solution are going to be...again, because there is very little dissociation, it's a weak acid, you are going to find mostly HF, and you are going to find H2O.0986

Both of these are weak acids: HF, as we said, has a Ka of 7.2x10-4; well, H2O has a Ka, which we call Kw; it is 1.0x10-14.1001

Well, between 10-4 and 10-14, this is 10 orders of magnitude bigger; it's huge--this is virtually nonexistent compared to this.1018

Because this is so much bigger, we can ignore any contribution that H2O does in its dissociation.1027

Its contribution of hydrogen ion is completely negligible--not even measurable, to be honest with you.1035

This is going to dominate the equilibrium in the solution.1040

So now, we want to see: at equilibrium, when everything has stopped and settled down, what is the pH of that solution?1043

Well, here is how we do it: We write down the equilibrium expression, HF goes to H+ + F-; we have I, we have C, and we have E (our initial concentration, change in concentration, and equilibrium concentration).1050

Our initial concentration was 1.5 Molar; there is, before the system has come to equilibrium, 0 and 0; so again, imagine, when you are doing these problems--you are taking something; you are dropping it in water; and it is before anything happens.1065

Before it starts to come apart, that is the initial conditions.1084

Now, HF is going to dissociate a little bit, so that means a little bit of it is going to disappear; and for every amount that disappears, that is how much H and F show up, right?1088

1:1, 1:1...when 1 molecule of this breaks up, it produces 1 ion of H+ and 1 ion of F-.1100

That is why we have -x, +x, +x, when they put +'s here.1108

And now, our equilibrium concentration is just adding these: 1.5-x; when everything has come to a stop, that is how much we have of the HF; that is how much we have of the H; and that is how much we have of the F-.1113

Now, we can write our expression.1126

Ka, which is equal to 7.2x10-4, is equal to...well, it's equal to the H+ concentration, times the F- concentration, over the HF concentration, at equilibrium.1128

That equals x, times x, divided by 1.5-x.1145

Now, let me rewrite this over here; so let's...part of one equation: 7.2x10-4; now, what we want to do...again, we want the pH.1155

We want to take hydrogen ion concentration and take the negative log of it; so we are looking for x.1166

We want to solve this equation for x.1172

Now notice, we are going to end up with a quadratic equation here; there are a couple of ways that we can handle this.1175

We can go ahead and (because you have graphing calculators, presumably--most of you) you can go ahead and use your graphing calculator to find the roots of this quadratic equation.1180

Or, you could solve the quadratic equation--turn this into a quadratic--multiply through and turn it into a quadratic--use the -b, plus or minus radical b2 minus 4ac over 2a to solve for x.1189

Or, we can actually simplify the procedure to make this a little better for us.1201

And again, the simplification is based on standing back and taking a look at the chemistry.1205

This is a weak acid; there is going to be very little dissociation here.1210

Even though there is an equilibrium, there is going to be very little, actually, of the H+ concentration--so little, in fact, that compared to 1.5 Molar--compared to this, this is going to be so tiny that the 1.5 probably won't even notice that it is gone.1214

So, we can simplify this expression by writing it as: 7.2x10-4 equals x2 over 1.5.1230

Now, when we find the value of x, we need to check to see if our assumption, if our approximation, is valid; and we will show you how to do that in just a minute.1245

Let's go ahead and just solve this first, and then we'll check to see if it's valid.1253

When we multiply through, we get x2 = 1.08x10-3, and we get x=0.033 moles per liter.1257

Well, this is the hydrogen ion concentration, because that was x in our ICE chart.1270

Now, we need to check if our approximation--our eliminating x from the denominator here--was valid.1275

Here is how you do it: most Kas that are listed in tables have an error of about plus or minus 5%.1280

Now, this value, .033, 0.033--if it is 5% or less than the original concentration that you subtracted from...1.5 times 100; you want to calculate a percentage; if this is 5% or less of this number, then our approximation is valid, meaning the difference is not really going to be noticeable--that your numbers are actually pretty good.1289

If it is above about 5 or 6 percent, that means you can't make this approximation--you have to solve the quadratic equation.1322

So again, we are taking a look at the chemistry, and we are letting the Ka tell us that we have a weak acid--very little dissociation.1329

So, this x, compared to 1.5, is probably going to be very, very tiny--so much so that we are probably not even going to notice a difference: 1.5, 1.49--it's going to be so small.1337

Well, if it falls into the 5% rule, then it's valid; if it doesn't, then it's not valid; so the first thing to do is go ahead and simplify, and check to see if this is the case.1349

In this particular case, we get 2.2%; well, 2.2% is definitely less than 5%, which means our approximation is valid, and we can use this value of x; it's perfectly good.1359

So, now that we have that, we have our H+ concentration of .033, our pH equals -log of 0.033, is equal to 1.48; there you go.1372

Now, I personally (just a little aside) don't care for pH; I think once you have found the hydrogen ion concentration, you are talking about concentration.1387

Taking a number and fiddling with it so that it has a more attractive number, like 1.48--I personally don't think 1.48 is more attractive or easier to deal with than .033, or 3.3x10-2.1396

It's a perfectly valid number; here, we know we're talking about concentration; pH--it's a little weird, because again, the lower the pH, the more acidic.1409

If you remember, you have a pH scale that runs from 0 to 14; 7 is neutral--that is the pH of plain water.1419

Below 7, we are talking about an acidic solution; above 7, you are talking about a basic solution.1427

The lower the number (6, 5, 4, 3, 2, 1)--that is more acidic; what that means is that there is a greater concentration of hydrogen ion.1434

It's a little strange: a lower number means more powerful; but again, this has just become part and parcel of the standard chemical practice in the industry, across the board.1444

But again, as long as you understand what is happening, that is what is important.1456

OK, so let's do another one; this time, let's calculate...let's see; this is going to be Example 3.1461

Let's calculate the pH of a 0.15 (this time, 0.15) Molar solution of HClO, hydrogen hypochloride--otherwise known as hypochlorous acid.1472

The Ka for this equals 3.5x10-8.1501

The first you want to notice is: look at the Ka; this is a very, very small number--that means we are talking about a very weak acid, very little dissociation.1507

When you take hydrogen hypochloride and drop it into water, it's going to stay hydrogen hypochloride; it's not going to come apart into hydrogen ion and hypochloride ion too much.1515

A little bit, it will; but not too much.1526

The fact that we have a number that we can measure means there is some dissociation; but again, this says that it is a weak dissociation.1529

OK, so now let's see which equilibrium is actually going to dominate in this.1536

We have two sources of hydrogen ion: we have the HClO, which breaks up into H+ + ClO-, potentially; and the Ka for this, as we just said, is 3.5x10-8.1542

Then, we have H2O; that is the major species in the water; that is H+ + OH-, but again, the Kw for this is 1.0x10 to the (oops, not 14) negative 14.1558

Well, 10 to the -8--even though it is a small number, it is a lot bigger than 10 to the -14; 6 orders of magnitude bigger.1577

The dominant reaction here, the dominant equilibrium, is this one.1584

This is virtually not even noticeable.1589

So, let's go ahead and do the problem.1596

Since the HClO equilibrium is going to dominate, we are going to write HClO, and I just am one of those people that likes to write everything--so, H+ + ClO- (I know that it's a little...I'm just writing it over here, but I just like to make sure that everything is clear).1598

This is initial, change, equilibrium (and let's erase some of these stray lines here); so, our initial concentration of HClO, before the system has come to equilibrium--when we just drop it in before it has come to equilibrium--this is 0.15 Molar.1619

There is no hydrogen ion, and there is no hypochloride ion.1635

Well, a certain amount of it dissociates; and it produces, for every mole that (we're not going to have that; we want this to be clean; +x) dissociates, it produces a mole of H+; it produces a mole of ClO-.1639

Equilibrium concentration is 0.15-x; this is +x; this is +x; and now, we can go ahead and write our equilibrium.1658

Our Ka is our hydrogen ion concentration, times our hypochloride ion concentration, over our hydrogen hypochloride.1671

When we do this, in terms of numbers, we get: 3.5x10-8, which is our Ka; it is equal to x times x, over 0.15-x.1681

And again, we want to simplify, so this is going to be approximately equal to x squared, over 0.15.1697

Then, we'll check to see if that approximation is actually valid.1705

When we solve this, we get x2 is equal to 5.25x10-9, and we get that the x, which is equal to the hydrogen ion concentration, is equal to 7.24x10-5.1709

Now, let's check the validity; so I'll write "check the validity of our approximation"--in other words, what we did here.1728

We'll take 7.24x10-5, and we'll divide it by the initial concentration, 0.15; so, when we do that and multiply by 100, it equals (let me drop it down a little bit here) 0.048%.1735

Wow, look at that; that means only 0.048% is actually dissociated; that is a very, very, very weak acid--virtually none of it has come apart.1753

This is clearly below 5%; our approximation is good; so we can use this number, and when we take the negative log of 7.24x10-5, we end up with (let's see what we ended up with here) 4.14.1765

4.14--and this our pH, and this is our hydrogen ion concentration.1788

Again, my preference is for the actual concentration; pH is just a number.1794

OK, that is good; so hopefully, you are getting a sense of what it is that we are doing; we are looking at Ka; we're choosing the major species; we're deciding which of those major species is going to dominate in the solution.1800

We know what chemistry is going to dominate; usually there is one species, one equilibrium, that dominates--everything else can be ignored.1812

And then, we just write out our ICE chart: Initial concentration, the Change equilibrium, and then we set it equal to the equilibrium constant, and we just deal with the math at that point; it's a simple algebra problem.1820

OK, now let's do a problem where we have a mixture of weak acids.1831

Same exact thing: it's going to be more species in water, but one of them is going to dominate; so let's do this.1834

Example 4: Calculate the pH of a solution that contains 1.2 moles per liter of hydrogen cyanide, or hydrocyanic acid, whose Ka is 6.2x10-10--very, very weak--and 4.0 Molar HNO2 (hydrogen nitrite or nitrous acid), whose Ka is equal to 4.0x10-4.1843

Also, calculate the concentration of the cyanide ion at equilibrium.1902

OK, so we want to find the pH; so we have a mixture--we have this certain amount of water; we drop in some 4-Molar hydrogen nitrite, or nitrous acid; we drop in 1.2-Molar hydrocyanic acid; we want to know what the pH of the solution is, and we want to know what the concentration of the free cyanide ion is in that solution.1925

OK, let's list our major species with their particular equilibriums.1949

Well, let me go back to blue here; so, major species in water.1955

That is what we are doing; we always want to do the major species to decide what chemistry is going to dominate.1963

Well, HCn is a weak acid; that means it is mostly HCn--it hasn't dissociated.1967

A strong acid dissociates--weak acids, not very much.1973

We also have HNO2 floating around in that, and we have H2O.1977

Well, the HCn equilibrium (H+ + Cn-)--we said that the Ka is equal to 6.2x10-10.1983

The (oops, wow, look at that; that is a crazy line; OK) HNO2 equilibrium (NO2-)--the Ka for this one is equal to 4.0x10-4.1996

And of course, we have the (oh, here we go again; wow)...last but not least, we have the H2O equilibrium, which is also another source of hydrogen ion, plus OH-; the Kw equals 1.0x10-14.2022

So now, let's compare: 10-10; 10-4; 10-14.2042

We can virtually ignore the 10-14--it's too tiny.2047

Between 10-4 and 10-10, this one--the HNO2; this is going to dominate the chemistry in this solution.2050

Because that is going to dominate, we can ignore any contribution of hydrogen ion from hydrogen cyanide, and we can ignore any contribution of H+ from water.2058

And again, we are doing this because there are three sources of hydrogen ion: some can come from the HCn; some can come from HNO2; some can come from H2O.2069

But, the Ka of the HCn and the H2O are so tiny that they are negligible, so this is what controls the chemistry of the solution; that is the chemistry we concentrate on.2078

Therefore, let's go with HNO2: H+ + NO2-; now, you are going to do the same thing that you did before.2089

You are going to make a little ICE chart: Initial, Change, Equilibrium; you are going to do your simplification; you are going to check it; everything is going to be fine; you are going to end up with a pH of...2103

I hope you actually run through and do this, based on the previous two examples.2113

So, for this one, you are going to end up with a pH of 1.40.2118

Now, the question is (the second part): How do we find the cyanide ion concentration?2126

OK, let's write down the cyanide equilibrium; we need to find the cyanide ion concentration, so we need to work with the cyanide equilibrium.2132

HCn goes to H+ + Cn-.2141

Ka equals 6.2x10-10.2147

Now, let's stop and think about what this means.2152

This is saying that, in a given solution, where you have cyanide ion, hydrogen ion, and HCn (hydrogen cyanide) floating around in solution--these three species floating around in some concentration each--the equilibrium concentration (in other words, the concentration of this, times the concentration of that, divided by the concentration of this) equals 6.2x10-10.2156

Remember, that is what an equilibrium expression is: equilibrium positions can change, but the constant stays the same--the relationship among these three--that is what equals this.2184

It doesn't matter where these H+ come from; this is talking about: at any given moment, if you have cyanide, hydrogen ion, and hydrocyanic acid in solution, the concentration of this times that, divided by the concentration of this, always equals that.2196

This gives us a way to find it; it doesn't matter where these H+ come from--they can come from HCn, or they can come from any other source, any other source.2215

In this case, the primary source of the hydrogen ion is the nitrous acid.2226

It is the dominant species in the water; it is the one that is going to give up its hydrogen ion and suppress the others.2232

So, at equilibrium, there is a certain concentration of hydrogen ion, and that concentration of hydrogen ion was the 1.4--well, the pH was 1.4; the 1.4 came from the negative log of the hydrogen ion concentration.2239

Now, let's go ahead and work this out.2257

This says...so we know that the Ka is equal to H+ concentration, times Cn- concentration, over HCn concentration.2263

That equals 6.2x10-8; this is the equilibrium expression.2278

Well, let's see: let's go ahead...so at equilibrium, we have...not 6.2x10-8; we have 6.2x10-10; yes, that is 10.2285

That means 6.2x10-10; now, we go back on our ICE chart, and we read off this, this, and this.2300

We ended up with: our equilibrium concentration is 0.04.2310

That was the hydrogen ion concentration that gave us the pH of 1.4.2321

We have the Cn- concentration, and, at equilibrium, the HCn concentration was 1.2-x.2325

OK, again, we are just dealing with a basic math problem, but as it turns out, we have a little bit of a problem; we have this x here...smaller than normal...suppresses the HCn dissociation...let's see here.2335

Let me see, how do I want to go ahead and explain this to make it most reasonable?2358

We have that; we have that; we have the cyanide concentration; OK.2364

Let me erase something here.2369

OK, so let me rewrite this here: HCn, H+ + Cn-.2378

Now, at equilibrium, we started off with 1.2 Molar of the HCn, right?2389

This was 1.2 Molar.2404

Now, we have the H+ concentration, which is 0.04; this is the equilibrium concentration.2407

This is what we want--this Cn- concentration.2412

Now, it's true that, of the HCn, some of it will dissociate; so the fact of the matter is, the equilibrium concentration is going to be 1.2-x, but again, remember what we said: the dominant equilibrium here was the nitrous acid.2417

It is so dominant that it is actually going to suppress this dissociation.2437

Anything that HCn would actually produce, any H+, is virtually nonexistent, because not only was it small to begin with, but because there is so much of the H+ from the nitrous acid, it's actually going to push this even more that way, so there is going to be even less.2442

So, for all practical purposes, we don't even have to worry about any HCn dissociating.2458

Therefore, our final equilibrium will end up actually being: 6.2x10-10 is equal to 0.04 (which was the hydrogen ion concentration we found from the HNO2 dissociation), times the Cn- concentration, over the HCn concentration (which was 1.2 Molar).2464

And then, when we solve for this, we end up getting a Cn- concentration (and this time, I will use the square brackets): 1.86x10-8 Molar.2490

As you can see, 1.86x10-8 Molar is so tiny--that means virtually...there is virtually no cyanide ion floating around.2504

That is based on this equilibrium.2515

So again, if you have a mixture of acids, one of those acids will dominate the equilibrium.2518

If you are asked for the concentration of a species from the other acid, the one that is not dominant, you can still use...you just write out its equilibrium, and it just means that--in this particular case, we wanted cyanide; well, the equilibrium expression for HCn says that cyanide concentration, times H+, divided by this, equals this.2524

Well, we have had this from what we just calculated from the dominant species; we have this, which is the original concentration.2546

And then, all we have to do is solve for this, because we already have this.2555

1, 2, 3, 4; we have 1, 2, 3 of them; we solve for the fourth.2559

There you go; I hope that made sense.2564

OK, let's see: we have taken a look at some weak acid problems; we have listed the major species; we take a look at the major species to decide which one is dominant.2569

And usually, in this particular case (for weak acids), we just see which one has the higher Ka value.2582

Whichever one has the higher Ka value, that equilibrium, that acid, will dominate the chemistry of the solution.2588

We write its equation; we list its initial concentration, the change in the concentration, and the equilibrium concentration; and then, we put the equilibrium concentrations, in terms of x, into the equilibrium expression, and we solve for x.2595

More often than not, they are going to ask for calculating the hydrogen ion concentration or, actually, the pH.2610

More often than not, they will just straight-out ask for the pH.2616

So, thank you for joining us here at Educator.com and AP Chemistry for weak acids.2619

Our next topic we are going to be discussing is going to be percent dissociation; we are going to spend a fair amount of time on weak bases.2624

Take care; goodbye.2631

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to continue our discussion of acids and bases--a profoundly important topic.0004

We are going to be talking a little bit about percent dissociation, sort of a continuation of our weak acid topic from last lesson.0009

But mostly, we are going to be talking about strong and weak bases; and, as you are going to see, how you handle the calculations is exactly like we did for weak acids.0017

Anyway, let's just jump in and talk about percent dissociation, and move forward and see what we can do.0027

So again, the idea is to do a fair number of problems to get you comfortable with what is going on--comfortable with the chemistry--how to handle this intuitively and turn that intuition into mathematics.0033

OK, so weak acids: remember, we said that a weak acid is some species, like for example, let's say, hydrogen fluoride (which was an example that we did in the last lesson): that dissociates a little bit into H+ and F-.0045

Now, not too much; again, that is the whole idea behind a weak acid.0062

Well, there is this notion called percent dissociation; we want to know how much of the original HF actually came apart--how much of it dissociated.0065

Well, a percent is always the same thing; a percent is always the part over the whole, times 100.0075

So, by definition, our percent dissociation is equal to the amount dissociated (and the amount dissociated is always this amount, or this amount, because again, if .2 moles dissociates of this, well, .2 moles is produced of that and .2 moles is produced of that...so it's always the amount dissociated), divided by the initial amount (or the initial concentration--either way--the initial concentration); it's the part over the whole, times 100.0080

That is it; a percent dissociation--you are just measuring: "To what extent has this weak acid actually come apart? 10%? 5%? 3%? 2%?"0121

Strong acids--100% dissociation; percent dissociation for an acid, a strong acid, is 100%.0132

You are going to find, in general, for weak acids, anywhere from about .5% up to maybe about 5 or 6% for a weak acid.0138

Like, for example, your vinegar solution: you are talking about acetic acid, which is maybe 2% dissociated; it's very little.0146

It is actually kind of interesting--with such little dissociation, and yet, it has that really, really strong, strong acidic quality to it.0156

OK, as a quick example, let's think about the last example that we did when we talked about the hydrogen fluoride.0164

Remember, we found that the hydrogen ion concentration for the fluoride was 0.033 Molar, in the last lesson.0173

Well, the initial hydrogen fluoride concentration that we started with (so initial is a little 0 down at the bottom)--it was 1.5 Molar.0183

Therefore, our percent dissociation is equal to 0.033 over 1.5, times 100%; you ended up with 2.2 percent.0193

That means, when we stuck that 1.5 Molar hydrogen fluoride solution...well, when we have that 1.5 Molar hydrogen fluoride solution...only 2.2% of the hydrogen fluoride has dissociated into H+ and F-.0207

That means the other 97.8% is still hydrogen fluoride, floating around in solution--pure hydrogen fluoride, not hydrofluoric acid.0225

Acid is when it is this way; but again, we just sort of have become accustomed to saying "hydrofluoric acid," but it's important to distinguish.0234

When it is together like this, yes, it is an acid, because it is in water--but it is not dissociated; it is actually still together.0242

Only 2.2% of the original amount has dissociated; that is it--it's all this is.0249

OK, let's list a general trend, actually--a general trend with percent dissociation.0256

The more dilute the solution, the greater the percent dissociation.0269

It is a nice little thing just to sort of keep in the back of your mind: in other words, if I have, let's say, a 1.0 Molar solution of HF, it's going to be some percentage.0287

Well, if I have a .5 Molar solution of HF, which is more dilute (right?--lower the concentration), the percent dissociation is actually going to be higher.0299

If I have a 0.1 Molar HF solution, the percent dissociation is going to be even higher.0309

That doesn't mean that the pH is going to be lower; that doesn't mean that the concentration of the hydrogen ion is going to be higher.0317

As you have lower and lower concentrations, you are actually going to end up with less concentration of H+; it just means that there is a greater dissociation, that more of the original acid has actually completely come apart into free H+ and free conjugate base ion.0324

That is what that means; be very, very certain to distinguish between the two.0341

Lower concentration, more dilute solution, gives a greater percent dissociation.0346

The percent dissociation is not a measure of acidity; pH is a measure of acidity (in other words, how much hydrogen ion is actually floating around freely).0356

Remember, we said: when we talk about an acid, we are talking about free H+ floating around in solution; more H+, more acidic, more damage.0366

That is it; so let's use this idea of percent dissociation to actually calculate a Ka (an equilibrium constant).0375

Our first example is going to be: In a 0.100 Molar lactic acid solution, the lactic acid is 3.8% dissociated.0384

OK, and for those of you who are biology majors and biochemists, lactic acid is the product that is produced by your cells under anaerobic conditions (under anaerobic respiration).0423

When you start to get really, really fatigued, that is the lactic acid building up in your muscles, until your body can get enough oxygen going to actually start burning the sugar aerobically (meaning with oxygen), as opposed to anaerobically (without oxygen).0435

So, dissociated...Calculate the Ka for this acid.0449

Calculate the equilibrium constant for this acid.0456

OK, so let's go ahead and write down an equilibrium expression.0466

I'm just going to write HL for lactic acid; actually, you know what--I think I'll write HLac, is in equilibrium with H+ + lactate ion.0471

Well, they tell me that the initial concentration is 0.100 Molar, and we have none of that; the change was -x, so it becomes x and x.0484

That means the equilibrium concentration is .100-x, x, and x.0496

Well, the Ka for this is equal to...well, sure enough, the same equilibrium expression: the hydrogen ion concentration, times the lactate ion concentration, times the concentration of lactic acid (or hydrogen lactate).0502

Now, we are looking, this time, for Ka; that is what we want to find, which means that we need this number, this number, and this number in order to multiply and divide--which means we need x.0516

Once we have x, we have this and this, and we can subtract from .1 to get this; we multiply and divide, and we get our Ka.0527

How can we find x?--well, they tell us that the percent dissociation is 3.8 (is it 3.8?)...yes, 3.8%.0535

What does that mean?--that means that this number, the amount that is dissociated (which is the same as this number, x), divided by the original .100, times 100, equals 3.8.0545

So, we have a way to find x; and now, we just go ahead and solve it.0561

We get: x is equal to 0.0038; there you go--nice and simple!0565

Now, we have our values; now, the hydrogen ion concentration (which is this one) is going to be 0.0038; the lactate ion concentration is 0.0038 (that is what this is, right?--they are the same; for every mole of this broken up, it produces a mole of this and produces a mole of that); and our HLac concentration is going to equal the original .100, minus 0.0038.0574

When we put all of these values in, we get: Ka is equal to 0.0038, times 0.0038, divided by 0.0962 (that is this one....962...).0613

And, when we run this number, we get 1.5x10-4.0639

So again, if you understand the chemistry, you should be able to do the math.0645

We have this; we set up the ICE chart; we knew that we needed to find x.0651

But, they gave us x already--they gave us a way to find x; they told us the percent dissociation.0656

That means the amount that was dissociated; well, the amount dissociated is the amount produced of the H+, the amount produced of the lactate, divided by the initial amount, which was .1.0661

So, this over that or this over that, times 100, is 3.8; we have x; we plug it in; and we get our Ka.0671

This is a common problem; you may actually run across it in your AP exam.0679

OK, now let's go on to discuss bases.0684

All right, now, the base is the opposite of an acid.0688

An acid is something that has hydrogens that it wants to give away; a strong acid--it gives up all of its Hs (in other words, it comes apart completely); a weak acid is one that sort of doesn't really give up its Hs too easily--it actually holds onto them.0693

A base is the opposite; a base is something that actually...strong base: let's actually define a strong base.0707

A strong base--it fully dissociates to produce OH---so, for example, sodium hydroxide.0716

Sodium hydroxide: when you take solid sodium hydroxide (it's kind of like hard rocks) and drop it in water, it dissolves; it dissociates.0731

It produces free sodium ion and free hydroxide ion.0741

It is a basic solution; it is a base, because when it dissociates, it produces OH-.0745

Remember, water is also a base; water breaks up into hydrogen ion and OH-; it's amphoteric--it's an acid and a base.0751

This is just a base.0758

Potassium hydroxide, another strong base: notice, I wrote the arrow in one direction--full dissociation.0760

K+, OH-: if you had a solution of sodium or potassium hydroxide, you wouldn't find any NaOH, any KOH, in solution; it would all be this, this, this, this--free ions in solution.0766

Let's do a quick example.0783

Handle it the exact same way, except now, in reverse.0787

Calculate the pH of a 4.0x10-2 Molar NaOH solution.0791

Well, let's see: NaOH is a strong base; any of your alkali metals (sodium, potassium, all of those in the first group)--with a hydroxide, they are all strong bases; they all fully dissociate.0808

"Strong base" means full dissociation.0822

OK, so the major species in water: again, nothing new--we handle it the same way.0829

Take a look at what it is: it's a strong base; there is going to be full dissociation; now, let's see what is floating around in water to decide what is going to dominate the equilibrium.0834

The major species in water floating around: you have sodium ion; you have hydroxide ion; and you have H2O.0844

Well, we know that H2O is also a source of hydroxide ion, right?--because H2O dissociates into H+ + OH-.0853

But, this is 10-7; let's write the Ka for this--the Kw is 10-14 (1x10-14; I just ignored the 1).0863

Well, this is a strong base, and this is 4.0x10-2 Molar; that means 4.0x10-2 Molar sodium hydroxide has fully dissociated, and has produced 4.0x10-2 moles of hydroxide.0876

Well, 10-2; 10-14; a huge difference--that is 12 orders of magnitude--so this is virtually...you can ignore it.0894

The species that is going to dominate the chemistry is this; the sodium is not going to do anything at all--it just sits there like a spectator ion.0905

So, our pOH (or actually, let me)...our OH- concentration is equal to 4.0x10-2, because we have full dissociation.0913

In other words, NaOH goes to Na+ + OH-, in case you are not sure what I am talking about.0930

It starts off with 4.0x10-2 Molar--full dissociation.0941

This is 00 minus 4.0x10-2; all of it dissociates--this ends up being 0.0946

This ends up being 4.0x10-2, 4.0x10-2.0954

That is what we mean by the OH- concentration, is that; because all of it is gone away--there is no more of that left in solution.0961

It is that; now, let's calculate the pOH; remember, p is just a function--it means the negative log of something.0970

Negative log of 4.0x10-2; we end up with...actually, you know what, I did this a little differently.0976

I did it this way: OH- equals that; well, what do we know about the hydroxide ion concentration and the hydrogen ion concentration in any aqueous solution?0988

They multiply to 10-14.1000

So, OH- times H+ equals 10-14.1002

We are looking for pH, not pOH; so the H+ concentration is 10-14, divided by 4.0x10-2, equals 2.5x10-13.1013

And then, we will take the pH, equals the negative log of that number (2.5x10-13), and we get 12.6--basic solution.1030

Remember what we said: if the pH is above 7 and below 14, or just above, you get a basic solution.1042

This confirms the fact that this is a basic solution--the pH.1049

pH is the standard by which we decide.1053

OK, now let's talk about weak bases.1059

Weak base--all right, the general reaction for a weak acid, we said, was this: we said, if we had an acid, plus water, we'll go to hydronium ion (which is the same as H+), plus the conjugate base of the weak acid.1065

Now, the weak base general reaction is this (the general reaction; I'll just use B for base): it is: the base, plus water, goes to BH+ + OH-.1095

In other words, let me rewrite this another way: B, plus (let me write H2O as HOH)--what is happening is that the base actually takes a hydrogen ion away from water.1120

But, it doesn't take it as a hydrogen atom; it takes just the hydrogen ion.1135

Hydrogen leaves its electron with the hydroxide, which is why you end up with BH+ + OH-.1139

I want to show you why this happens.1148

Most bases, weak bases--they have a lone pair of electrons, and we'll do an example in a minute.1151

H is here; this is usually not something that you are going to do until Organic Chemistry, but I want you to see what is happening, because I want it to make sense to you.1157

These electrons--they reach out and they actually grab the H; they rip it away from the water molecule.1165

When these electrons move in, these electrons move out, OK?1173

H cannot be attached to two simultaneously (well, it can for hydrogen bonding; but for our purposes); this base wants this, so when it takes this, it goes this way, and it kicks these electrons; they move onto the hydroxide, and what you end up with is this.1180

Because this is only coming as a proton, minus its electron--it's coming as an H+--now this whole species has a plus charge, and you have an OH-.1201

This general reaction is the reaction of a weak base with water; this is very important reaction--it always happens like this.1211

We said that a base is something that produces hydroxide ion; well, here you go.1221

A base doesn't necessarily have to have hydroxide ion in it; so, for example, you know that sodium hydroxide produces hydroxide by dissociation.1226

Well, an example like ammonia--well, let's actually use a specific example.1236

NH3, which has a lone pair, plus HOH (I'm going to write it as HOH, water) goes to NH4+ + OH-.1245

OH- is still produced in solution, but the OH- doesn't come from the base itself; it is the base that reacts with water.1259

Water is now acting as the acid; it is giving up its H; this ammonia is acting as the base--it pulls the H off.1268

Now, it's NH4+ and it's OH-; so it's producing OH- in a roundabout way--not by dissociation--it's doing it by breaking up the water.1274

This is...and there is, of course, a Kb associated with this.1286

It is an equilibrium constant expression for the reaction of a base (or something, some species) that will extract a hydrogen ion from water to produce the conjugate acid of this base and hydroxide ion.1294

The Kb is exactly what you think it is; this is liquid; this is aq; this is aq; and this is aq; it is equal to NH4+ times OH-, over NH3.1311

That is it; this is the generic expression for a weak base; it is the base, plus the water, plus the BH+, plus OH-.1332

It is taking the hydrogen from water, leaving the hydroxide ion.1342

In the process, it produces the hydroxide ion.1346

It produces the hydroxide ion; the solution becomes basic; that is why it is called a base.1351

It is also called a base because it actually takes the hydrogen ion from something (in this case, water).1356

It is acting like the base; water is acting like the acid.1363

In the weak acid equilibrium, this is the acid; this is the base; now, these are the lone pair of electrons that are taking this hydrogen and producing that as a conjugate.1367

Back and forth: it's just a competition between two things for the hydrogen ion.1379

Here, the base...it just depends on what the equilibrium is; that is all acid-base chemistry is--it's a competition between two different species for the hydrogen ion in between.1385

It is a tennis game between the two: which one is stronger?--the stronger one will take the hydrogen ion.1393

OK, so again, a base--whether weak or strong--produces OH-.1401

This is the take-home lesson: a weak base, strong base...they all produce OH-.1418

A strong base does it by dissociation; a weak base does it by hydrogen abstraction from water.1423

OK, so now, let's do a couple of examples here.1430

Oh, let me just give a couple of versions: Strong base--we already mentioned one, potassium hydroxide; it dissociates into potassium ion plus OH-; here is your OH-.1436

A weak base--I'm going to use something called pyridine, and it is a molecule that looks like this, and it reacts with water.1450

Let me actually write it as HOH; you will find it very, very convenient to write water as HOH--I certainly prefer to do so, although a lot of people censure me for it--I don't know why.1462

It becomes N + OH-; again, here is your OH-; it happens in a roundabout way.1475

Here it comes apart; here it takes hydrogen from the water to produce the hydroxide.1488

Let's do an example.1494

All of it will make sense as we sort of do more examples: OK.1498

Let's see: what can we do?1503

Oh, calculate the pH of a 13.0 Molar NH3 solution.1508

The Kb is equal to 1.8x10-5.1526

OK, so here, because we are dealing with a weak base, we are producing hydroxide ion concentration.1533

The x-value that we found is actually going to be hydroxide ion concentration; we are going to have to use the relationship of hydroxide, times hydrogen ion concentration, equals 10-14 to find the hydrogen ion concentration, and then take the negative log of that.1540

If we just take the negative log of the hydroxide ion concentration we find for a weak base problem, we are not going to get the pH; we are going to get the pOH.1555

Now, you can do that; that is fine, as long as you take the pOH and subtract it from 14, and that will give you the pH; because again, we have chosen hydrogen ion concentration as the standard--pH as the standard.1563

Well, we can calculate the p of anything.1576

OK, so let's see what we have: well, let's take a look at the major species.1580

Again, same procedure: major species: we have NH3 (weak base--it's a weak base, which means that it is not dissociated--it stays mostly NH3), and its Ka is 1.8x10-5.1586

The other species is water, because water is also a source of hydroxide.1607

But, its Ka (which is Kw) is equal to 1.0x10-14; -5; -14; I think I'll go with the -5.1612

We can ignore this one; this is going to dominate the equilibrium--this is going to dominate the solution.1622

The hydroxide ion in this solution is going to come mostly from the chemistry of ammonia.1629

So, let's write it out: NH3 + HOH (same equation over and over again: base plus water goes to conjugate acid plus hydroxide; "conjugate acid": conjugate acid just means stick an H on top of it--attach an H to it and put a plus sign on it) + OH-; the Initial; the Change; the Equilibrium.1635

I know you are going to get sick of these ICE charts.1658

Let me see where are we (where are we, where are we, where are we); oh, there we are: 13 Molar; this is before anything happens--initial means before the system comes to equilibrium.1662

Before this takes place, water doesn't matter; there is no ammonium, and there is no hydroxide yet.1672

A certain amount of ammonia disappears; OK, now this is kind of interesting.1680

NH3 is not coming apart like an acid is, but we put a -x because NH3, as ammonia, is disappearing.1686

What is forming is ammonium; does that make sense?1696

Before, when we had H, like HF, and it dissociated, we knew it was -x; and it produced x amount of H and x amount of F-.1700

But here, it is still -x, even though this NH3 is actually taking something from the water and becoming more.1715

It is becoming NH3 to NH4+.1724

But what is happening is that NH3 is disappearing as a species; NH4+ is showing up; OH- is showing up.1726

That is the idea; you have to get your mind around the physical reality of it.1735

This doesn't matter; x is showing up; hydroxide is showing up; our equilibrium concentration is going to be 13...actually, this is 13.0, if I am not mistaken.1740

Yes, I think it's three significant figures; sorry about that: 13.0, 13.0-x; that doesn't matter; let's make sure this is clear; that doesn't matter; that doesn't matter; that doesn't matter; this is +x, +x; handle it the same exact way.1753

So, we have: Kb is equal to the ammonium ion concentration, times the hydroxide ion concentration, divided by the ammonia concentration.1773

Let's put some numbers on this: 1.8x10-5 is equal to x, times x, divided by 13.0-x.1788

Well, again, because we are talking about a weak base (1.8x10-5), x is probably going to be pretty small compared to the 13.0.1800

Let's just ignore it for the time being, and we will check the validity of our approximation in a minute; and just leave it as 13.0.1812

x squared is equal to 2.34x10-4, and x is equal to 0.0153 Molar, which equals the hydroxide ion concentration.1820

Well, the pOH equals the negative log of the hydroxide ion concentration, equals negative log of 0.0153, is equal to 1.82; and pH + pOH is equal to 14; pH plus 1.82 is equal to 14, so our big fat pH is 12.18 (bigger than 7--a lot bigger than 7).1838

This is a basic solution; it's confirmed.1876

There you go; that is it--handle it the exact same way as an acid, except it's different; it's a base--it's a weak base.1882

Instead of a Ka, we have a Kb; nothing is different; everything is exactly the same.1891

Initial, Change, Equilibrium; decide what the major species are; decide which species is going to dominate the solution--which is going to produce, in this particular case, the most hydroxide ion.1896

It is the weak base; it's weak, but it is still stronger than the water.1907

OK, now let's do the percent; I'll put in quotes "dissociation."1913

Well, percent dissociation works for acid, because an acid is going from HA; it is actually dissociating into H+ + A-.1920

A base, like NH3, is actually becoming BH+ + OH-...so we say "percent dissociation," but what is more appropriate for a base is percent association (n other words, percent of the association of B with the H to produce H+).1929

They still call it percent dissociation; but as long as you know that now, you are talking about a base; it is a reverse process.1953

So, percent "dissociation"--I actually call it percent association, because I like things to make sense--I don't like things to just drop out of the sky.1959

OK, so we said that it is the OH-; it's the amount of thing produced--over the initial concentration of what we had.1969

We have 0.0153 Molar, over 13.0 Molar, times 100, equals 0.12%.1983

That means, of the NH3, only .12% of the ammonia actually ripped off a hydrogen from the water to become ammonium ion and produce that much hydroxide.1998

That is it; that is all this means.2016

OK, so let's see what we have here: let's try one more, and I think we'll wrap it up for weak base discussion.2020

Example 3: Calculate the pH of a 0.10 Molar (this time) methylamine, which is CH3NH2, solution; so a methylamine is just like an ammonia, except it has, instead...I've taken out one of the hydrogens and put a CH3 on there.2031

And the Ka for methylamine is 4.38x10-4.2070

I would like you to see what this actually looks like, because I'm a big fan of structures.2078

NH3 looks like this, as you know, and it has a lone pair of electrons; well, methylamine--it has its 2 hydrogens (not a problem), but now, it has a CH3 attached to it; so, it's the same thing; it's like an ammonia; it has a lone pair there; but instead of an H, it just has that.2082

It behaves exactly the same way; it's a base; it pulls a hydrogen off the water to produce hydroxide.2104

So, let's write our major species: well, we have the CH3NH2, and we have our H2O.2110

Well, 4.38x10-4 versus 1.0x10-14: yes, I think we can ignore the 1.0x10-14 as a source of hydroxide.2123

Most of the hydroxide in this solution is going to come from this--a weak base, but still a stronger base than that.2134

So, let's write our CH3NH2 + HOH (or you can write H2O, not a problem--same equation--plus water) goes to CH3NH3+ (just add a hydrogen and stick a plus charge on it) + OH-.2140

We have an initial; we have a change; we have an equilibrium.2159

0.10; nothing; 00 (before anything happens); as the system comes to equilibrium, this species disappears; this species appears; and this species appears.2163

At equilibrium, we are left with .100-x; this doesn't matter; that doesn't matter; this is +x; this is +x; now, we have: 4.38x10-4 is equal to this times that, divided by that.2176

x squared over 0.10-x approximately equals x squared over 0.10.2196

Now, when we do this, we end up with the following (when we do this approximation): we end up with x equal to 6.6x10-3.2206

Let's check the validity of this; let's see if our approximation here, from going from here to here, is valid.2219

Well, 6.6x10-3 over 0.10, times 100: guess what--it actually equals 6.6%.2224

6.6% is too high; it's close to the 5, but it really is too high.2237

That means that you have a choice; you can either...you have to go back; you can't use this approximation, in other words.2243

You have to actually solve this whole equation as it is.2249

You can't eliminate the x from here; you have to solve this equation; you have to do it--either solve it as a quadratic equation (which is not a problem--it's easy enough to do; it's just numbers; you have a calculator--you can do it), or I'm going to show you a method called the method of successive approximations, which is a really, really great technique, if you don't want to use the quadratic.2255

It is an older technique; it still works--there are a lot of computer programs that are actually based on this method of successive approximations.2276

Here is how it works: OK, so now let's go back, and we said that we have 6.6x10-3, right?2283

So, when we did this approximation of 4.38x10-4 equals x2 over 0.10-x, approximately equal to x2 over 0.10; we got a value of...our first value...we got 6.6x10-3 for x.2295

Let me write x=this.2325

x equals 6.6x10-3; OK, we checked this 6.6x10-3; we divided by the .10, and we got 6.6%; that is too high.2329

Instead of going back and solving this, here is what you do: you take this first value, and you put it back in for x, and you solve this equation again.2342

In other words, you take .10; you subtract 6.6x10-3; and you solve the equation x2 over the number that you get here, when you subtract this from this.2351

You solve this, and you get a second value for x.2365

Now notice, I didn't put it here and here; that doesn't make sense.2369

What I am doing is: I'm going to use this first value to get closer with a second value.2374

I'm going to use the second value to get closer with a third value.2380

When any two successive values actually match each other, that means I have hit my point.2384

For those of you that are familiar with something called the Newton-Raphson method of solving for the roots of an equation, this is somewhat similar to that.2389

You are basically just sort of converging on a value, and when two successive values are equal, that means you have hit your point; you are not going to go any further.2397

So here, we were slightly off; I use this to put it back into this one, leaving this alone; I solve for x, and a second value I get is x=6.39x10-3.2407

This is a pretty significant difference; this is fairly significant here.2422

And now, I check this one; well, I actually don't really--I just stick it back into this again; I take .10-6.39x10-3 in the denominator; I leave the x2 on top; and I solve this equation again.2427

My third value that I get when I solve for x: I get x=6.40x10-3.2443

I stick this back in there; I do it again; I get a fourth value, x=6.40x10-3.2453

Two values, one after the other, match; I can stop there.2464

x=6.40x10-3.2468

x happens to be my hydroxide ion concentration of 6.4x10-3.2473

Now, let me see: how else did I handle that?2481

And the pOH, which is the negative log of the 6.4x10-3, gives me 2.19, and then pH is equal to 14-pOH; I end up with 11.8.2483

That is my basic solution.2502

So again, when you are presented with a value, and you try to approximate it by leaving this x off, and it turns out that that x value is higher than 5% of the original value that you took it from; well, you can take the value that you got and stick it in there.2507

Just subtract it from that value in the denominator, run this calculation again for x, and just keep doing that over and over again.2523

Whatever value you get, put it back in; when you get two values that are the same, that is when you stop--that is your value; you have converged on it.2531

You have gone from...if this is the real value, let's say you started over here; you are going bounce here; you are going to bounce here; you are going to bounce here; you are going to converge on it.2538

It really is nothing more than a Newton-Raphson method.2547

Or, you can just solve the quadratic equation.2552

It's up to you; personal taste--I think it's sort of nice to do whatever you feel comfortable with, because it certainly makes the act of problem-solving much more relaxing and much more enjoyable whenever you are doing something that you like to do.2554

OK, so I'll go ahead, and we will stop there for our discussion of weak bases.2567

Next time, when we meet, we are going to talk about polyprotic acids and the acidic and the acid-base properties of regular salts.2573

So, thank you for joining us here at Educator.com, and we'll see you next time; goodbye.2581

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we are going to continue our discussion of acid-base equilibria, and we are going to talk about polyprotic acids.0004

If you remember, in the first lesson that we discussed acids in, we said that polyprotic acids are simply a fancy name for acids that have more than one hydrogen that they could possibly give up.0010

So, for example, sulfuric acid (H2SO4), carbonic acid (H2CO3)...probably the most important of the polyprotic acids is H3PO4, phosphoric acid.0021

It is how the inside of a cell actually maintains its pH; it is the buffer solution inside of the cell.0033

And the carbonic acid, the H2CO3, is how the blood maintains its pH; it is a buffer solution outside of the cell.0040

We are going to discuss polyprotic acids today and talk about they behave; and it is not altogether different than anything that you have seen already, as far as weak acids and strong acids.0048

We'll just jump right in, do a quick couple of definitions, and we'll start with the examples.0058

OK, so once again, the really important thing about all acids (polyprotic or otherwise--well, polyprotic especially) is that acids dissociate one hydrogen ion at a time.0065

So, in the case of, say, carbonic acid (let me actually write that out first: so acids dissociate one hydrogen ion at a time), which is H2CO3, it's in equilibrium, so it dissociates into H+ + HCO3-, and there is a Ka associated with this, an acid dissociation constant, 4.3x10-7.0076

And notice, we have a little 1 here, so Ka1, because it gives up one ion at a time.0112

Now, this HCO3- has another H that it can give up, and it also behaves like an acid; so it now shows up on this side, and it is in equilibrium with the proton that it gave up: CO32-.0118

And this is Ka2; this is the second acid dissociation constant, which is equal to 5.6x10-11.0135

One thing you want to notice here, which is a general trend for all polyprotic acids, is that the first Ka is going to be a certain number; the second Ka is one that is very, very small: 10-11...very tiny--certainly much smaller than this.0143

So basically, when it comes to polyprotic acids, it's really the first dissociation that counts the most.0160

The only exception to this case (which we will see in a little bit when we discuss sulfuric) is sulfuric acid, and the reason being that sulfuric acid is strong in its first dissociation--it behaves just like hydrochloric acid or hydrobromic acid--but in the second dissociation, it's actually a weak acid, so it's slightly different depending on the particular molarity of the solution that we are trying to find the pH of.0168

But, we will get to that in a minute.0191

Another example would be phosphoric acid, like we mentioned, and it has three dissociations, because it has three protons to give up.0193

H3PO4 is in equilibrium with H+ + H2PO4-, and the first acid dissociation constant is 7.5x10-3.0200

And then, this H2PO4- dissociates into H+ + HPO42-; so make sure you keep track of the charges.0214

It's very, very important, especially with these polyprotic acids; they all sort of look alike--the H2PO4, the HPO4, the PO4, H3PO4; it gets very, very confusing, so definitely work very carefully, very, very slowly.0228

Chemistry is very symbol-heavy, and it's easy to get lost in the symbols and make silly mistakes, simply for that--as opposed to conceptual mistakes.0241

The second dissociation constant, the Ka2, is 6.2x10-8.0251

And then, this species, HPO4, the hydrogen phosphate (which is 2-), becomes H+ + PO43-.0258

Now, the phosphate is finally alone, and the third dissociation constant is equal to 4.8x10-13; so clearly, 10-3; 10-8; 10-13: you see the pattern.0270

At is dissociates more, it becomes harder and harder to actually pull off that second and third hydrogen.0285

Let's just go ahead and do an example, and I think everything will make sense.0291

Example 1: We want to calculate the pH of a 4.0 Molar H3PO4 solution, as well as the equilibrium concentrations of H3PO4, H2PO4-, HPO42-, and PO43-.0297

So, not only do we want the hydrogen ion concentration (in other words, the pH, when we actually take the negative log of it), but we want to find the concentration of all of the species in solution, once you take that phosphoric acid and drop it into water to create a 4 Molar solution.0349

What is the concentration of each of those species, in addition to the pH?0365

OK, well, let's just go ahead and start.0368

The first thing we do is the same thing we always do: we check to see the major species in the water before anything happens.0371

That is the whole idea; we want to see what is in there before the situation comes to equilibrium, before the system comes to equilibrium.0377

We want to check out the chemistry; we want to make some decisions about what is going to dominate--what chemistry is going to dominate.0384

There is always going to be one that dominates.0390

OK, so major species: well, phosphoric acid is a weak acid; you notice, Ka1 is 7.5x10-3; that is kind of small, so it's a weak acid.0392

A weak acid means that it's not going to dissociate very much.0405

So, most of it is going to be in this form: H3PO4.0409

That H3PO4 is just going to be floating around in solution, not very dissociated.0413

The major species in the solution: well, you have an H3PO4, and your other major species is just the water.0418

Well, both of these contribute hydrogen ions; well, the Ka of, or the Kw of, water (which happens to be the Ka of water) is 1x10-14.0425

Well, the Ka of this one is 7.5x10-3, 11 orders of magnitude bigger; so we can ignore water's contribution to the hydrogen ion concentration; this is the dominant species.0435

Because that is the dominant species, that is the equilibrium that we are going to work with.0447

The same thing we have always done: major species; decide which is dominant, which is going to control the chemistry; that is the one you concentrate on--ignore everything else.0451

So, let's go ahead and write our equilibrium expression and do our ICE chart.0459

Let's see...yes, that is fine.0465

OK, so we have: H3PO4 that dissociates into H+ + H2PO4-.0470

We have our initial concentration, our change, and our equilibrium.0482

Our initial concentration is 4.00 molarity, and this is 00, so this is before anything has had a chance to come to equilibrium.0486

A certain amount of this H3PO4 is going to dissociate, and for each amount that dissociates, 1:1, 1:1 ratio--that is the amount that shows up in solution of the other species.0495

Therefore, our equilibrium concentration is 4.00-x; this is x; this is x; and now that we have our equilibrium concentrations in terms of x, and we know what the Ka is, we set it equal to each other.0505

7.5x10-3 is the Ka; it is the x, the hydrogen ion concentration, times x, the dihydrogen phosphate concentration, divided by the phosphoric acid concentration, 4.00-x.0521

We do our normal approximation, which we can check the validity of; so this is equal to x2 over 4.00, just to make our math a little bit easier.0540

When we do this, we get x2 (actually, you know what, I'm just going to skip this line altogether; and I'm just going to go ahead)...and you multiply by the 4; you take the square root; you end up with x is equal to 0.173.0548

That is equal to the hydrogen ion concentration; therefore, the pH is the negative log of the hydrogen ion concentration--you get a pH of 0.76.0563

That is our first part: 0.76 is our pH; let's check the validity of what we have; by checking the validity, that means...0574

So let's do it over here; let's say "check validity of our approximation"--I mean, we know it's good, but it's good to just sort of check it.0583

0.173, which is x, divided by 4.0, times 100; you end up with 4.3%.0592

It's close to 5, but it's still under the 5% rule, so we're still actually pretty good; it's not a problem.0600

A perfectly good, valid approximation: we don't need to do the quadratic equation; this is a good concentration; this is the pH.0606

Now let's see where we stand: we found the pH, but we also wanted to find the concentrations of all the other species in solution.0613

Let's see what we have; we have the H+ concentration; that is equal to x, which is 0.173 Molar.0620

Well, the H concentration also happens to be the H2PO4- concentration.0629

The H2PO4- concentration is the same thing, 0.173 Molar.0634

Well, since we have x, we also have the phosphoric acid concentration (H3PO4), which is equal to 4.00-0173, and I'll just go ahead and...no, we'll just do 3.8 Molar (I'll just go ahead and round it up a little bit; sorry about that).0641

So, that is going to be 3.8 Molar; so now, the only thing that we are actually missing is: we are missing the HPO4- concentration and the PO43- (this is 2-; see, again, I am making the same mistakes; I have symbols and charges all over the place).0667

So now, we need the concentration of the hydrogen phosphate, and we need the concentration of the phosphate ion.0688

Well, like we said, an equilibrium is established; it doesn't matter where these species come from (the phosphate, the hydrogen...); the Ka that we have for a given species, for a given equilibrium; if it involves the species that we want, that is good enough.0693

The equilibrium that we do have, that involves the HPO42-, is the following.0713

We have the H2PO4 concentration; well, that is in equilibrium with H+, plus the HPO42-, which is the species that we are looking for.0721

Well, we have this concentration, and we have this concentration; that is this and this; and we also have the second Ka, the second dissociation constant.0734

That is what this is here: this is the second dissociation of phosphoric acid, and the Ka for this one, Ka2, is equal to 6.2x10-8.0742

So now, we can just basically rearrange the equilibrium expression and solve for HPO42-.0757

And then, for the PO43-, we do the same thing with Ka3, the third dissociation constant.0764

Let's go ahead and write everything out, so that we see it.0769

Ka2, based on the equation that was just written (the dissociation of the H2PO4-), becomes the concentration of H+, times the concentration of HPO42-, over the concentration of H2PO4- (oh, this is crazy--all kinds of charges and symbols going on).0774

Therefore, I rearrange this to solve for this species, because I have this, I have this, and I have this; it's a simple algebra problem.0797

We have HPO42-, the concentration--moles per liter--is equal to the Ka2 times the H2PO4- concentration, divided by the H+ concentration.0806

We get: that is equal to (I'm going to move this over here) 6.2x10-8, which is our Ka, times 0.173, which was our concentration of H2PO4; it also happens to be the concentration of our H+.0825

So, these cancel, of course, leaving us with 6.2x10-8 molarity for...that is the HPO42- concentration--that is one of the last things that we needed.0849

Notice, it's very, very, very small.0866

Well, now we want to know the PO43- concentration; that is the final concentration--the final species whose concentration we want.0870

OK, well, now that we have found the HPO42- concentration (it's this), we have an equilibrium.0880

The third equilibrium is (the third dissociation of phosphoric acid--excuse me): HPO42- (oh, wow, you see what I mean--all of these charges, all of these symbols floating around; you're bound to make mistakes--you have to go very carefully) is in equilibrium--it dissociates into H+ + PO43-.0886

We can write a Ka for this: the third dissociation constant is equal to the H+ concentration, times the PO43- concentration, over the HPO42- concentration (which we just found).0912

We rearrange this, and you get: the PO43- concentration is equal to the Ka3, times the HPO42- concentration, divided by the H+ concentration.0929

So, the Ka3 is going to be (let me see: what is the Ka3?) 4.8x10-13, times the HPO4 concentration, which we just found, which is 6.2x10-8, over the hydrogen ion concentration, which we found in the first step: .173.0953

Well, we get a very, very tiny number: we get 1.7x10-19 molarity; that equals the PO43- concentration.0976

There you have it.0990

So, a polyprotic acid: you handle it in the exact same way--most polyprotic acids are going to be weak acids.0993

The only exception is sulfuric; sulfuric is strong in its first dissociation, weak in its second dissociation; we are going to handle a problem in just a minute.0998

But aside from that, you handle it in exactly the same way.1007

And then, if you happen to need the concentrations of the other species--a further dissociation (so, phosphoric acid, dihydrogen phosphate, hydrogen phosphate, phosphate), you use the concentrations that you found, and then you use the next dissociation, and the next dissociation, with the appropriate Ka, to find the concentrations of all the species that you want.1010

About the only difficult problem here, as you saw, was just keeping track of the symbols and not losing your way in the symbology.1035

That is going to be the hardest problem, which, as far as I am concerned, that is the problem you always want--you don't want conceptual problems; you want mechanical problems; those are easy to deal with.1042

OK, so let's do another example; this time, we are going to deal with sulfuric acid, and we are going to do a couple of variations of it.1050

So, let's just start Example 2: We want to calculate the pH of a 1.2 Molar H2SO4.1057

OK, well, so sulfuric acid: let's write down its dissociations.1074

Its first dissociation is: H2SO4 dissociates into H+ + HSO4- (the hydrogen sulfate ion, also called the bisulfate ion).1079

This Ka is large: we said it's a strong acid in its first dissociation; it's large--it doesn't even have a number.1091

You remember, strong acids don't have Kas; they are huge.1097

The second dissociation is: HSO4- dissociates into H+...1101

Actually, I'm not even going to give the equilibrium sign here; this is going to be a one-sided arrow pointing in one direction--full dissociation; there is no H2SO4 left--very, very important to remember that for strong acids.1107

For strong acids, there is an arrow pointing to the right, and that is it; it's not in equilibrium.1119

We have HSO4 going to H+ + SO42-, and the Ka (it's small, but it's actually still pretty large, relatively speaking): 1.2x10-2.1125

So, a little larger than the other weak acids, but it still behaves as a weak acid; there is an equilibrium here; it is still less than 1.1138

OK, we want to calculate the pH; well, let's do what we always do--let's see what the major species are.1145

Be careful here: the major species in solution--you have taken sulfuric acid; you have dropped it into water; sulfuric acid is strong in its first dissociation.1152

Therefore, a solution of sulfuric acid is entirely composed of free hydrogen ion, free hydrogen sulfate ion, and (the other species in water is) H2O.1163

There you go; now, again, we are calculating a pH here; once again, strong dissociation; it's a strong acid in its first dissociation; that means in species, before any equilibrium is reached, it's this and this and this.1179

"Strong acid" means full dissociation--it breaks up completely: there is none of this left.1197

It is all that in solution: this floating around, this floating around, this floating around; what is the hydrogen ion concentration?1202

Well, the question is: Since it's a strong acid in its first dissociation, can we treat it just like any other strong acid and just take the molarity and just take the negative log of it?1211

Can we just do -log of 1.2?1220

Because 1.2 moles per liter produced 1.2 moles per liter of H and 1.2 moles per liter of that, and this is actually pretty weak, it's probably not going to contribute too much.1223

However, we need to make sure; so let's ask our question: Can we just treat this like any other strong acid problem?1235

The answer is maybe; so, we have to check.1263

The reason we have to check is, we need to see if this actually does contribute anything.1269

It is small, but it's not so small (like on the order of 10-5 or 10-6, 10-7; it's 10-2.1274

When you get to the 10-2 range, you're going to have to be careful; and we'll see in a minute what a general rule of thumb is.1284

A general rule of thumb is: when you have second-dissociation constants, such as 10-2, if you are below about 1 molarity, you can't really ignore it; you have to include it.1291

So, not only do you have to take the 1.2 Molar, but some of this HSO4, because it also dissociates, is going to produce a little bit more H+; so you might have to include that; that is what we are going to check.1302

We are going to check to see if the second dissociation is significant--produces enough H+ so that we have to include it into the 1.2.1317

Let's write: first of all, maybe, so we have to check; so now, mind you, our final H+ concentration that we are going to take the negative log of to find the pH is going to equal, in this case, the 1.2+x.1327

This x is going to be any hydrogen ion that comes from this dissociation.1345

It's strong in this first dissociation, so I know that, before anything happens, I know that at least 1.2 moles per liter of hydrogen ion is floating around in solution.1352

However, I need to know if the second dissociation also produces enough hydrogen ion; so I have to add it to this.1360

That is the thing; we are going to compare x and 1.2 to see what the real difference is; if it's small, we can ignore it, and just treat it as a regular strong acid problem; if it's not small, we have to include it before we take the final negative log of that final hydrogen ion concentration.1369

I hope that makes sense.1384

The ICE chart is going to look slightly different; OK.1386

Well, let's go ahead and do our ICE chart, then: so again, we know that we have the 1.2of that, so we don't have to worry about this.1389

We have to worry about this equilibrium: how much H+ is this HSO4 going to produce under these circumstances?1397

So, the equilibrium that we want to look at is the HSO4-: H+ + SO42-.1405

I hope that makes sense: major species, ICE charts...you do ICE charts with weak acids; you don't do them with strong acids.1414

I have at least 1.2 moles per liter of H+ floating around, because it's a strong acid in its first dissociation.1421

I need to check the second dissociation to see if it produces a significant amount of H+ to add to the 1.2.1428

So, Initial, Change, Equilibrium: our initial concentration of HSO4 is 1.2 Molar, because it is produced in the same way that this is produced.1436

When H2SO4 dissociates, it produces 1.2 Molar of this, 1.2 Molar of that.1448

Now, what is our initial H+ concentration?1453

Here is where we have to be careful, and where the ICE chart is different: our initial H+ concentration is also 1.2, because we have free H+ floating around and free HSO4 floating around; that is what these two are.1455

But, there is no SO42- yet, before anything else happens.1471

In solution, before the system comes to equilibrium, this and this are the same from the first dissociation.1475

Now, the change is going to -x here; this is going to be +x here; +x here; we add vertically down; we get 1.2-x for the HSO4; we get 1.2+x...like I said, 1.2+x is going to be our final hydrogen...and we get +x.1482

Now, we form our equilibrium; let's see, our Ka is (oops, here we go with the stray lines again; they show up in the most interesting places; OK--let me just do it off to the side here) 1.2x10-2, equals 1.2+x, times x, over 1.2-x.1504

So now, we have to solve this equation.1542

OK, here is what we are going to do: we are going to presume that x is small in order to do our approximation; then, we are going to check our approximation to see if it's valid.1544

So, when we take x small, watch what happens to this approximation.1556

OK, it's approximately equal to...that means this x disappears--not this x, this x.1561

So, it becomes 1.2 times x, over 1.2.1568

Well, these cancel, and I am left with: x is equal to 1.2x10-2, which is the same as 0.012.1579

I hope you'll forgive me; I actually prefer to work in decimals, as opposed to scientific notation.1590

I like scientific notation, but I just like the way that regular numbers look.1594

OK, now, we said that x is .012; now, we have to go up here: our final hydrogen ion concentration is going to be 1.2 (the original amount floating around), plus the x, plus the 0.012 that came from the second dissociation.1599

That equals 1.2.1627

It equals 1.212, but it's 1.2 to the correct number of significant figures.1632

So again, significant figures become a little bit of an issue; I personally don't care about significant figures all that much; a lot of people sort of criticize me for that, but you know what, it doesn't really matter; I'm more interested in concepts than I am in actual significant figures.1640

I think they are fine, and I think it's good to sort of use them, and in this context, yes, it sort of helps, because you notice: 1.2 plus .012--this is actually pretty small compared to this.1652

Since, working with significant figures, you still end up with 1.2, this says that you are actually justified in ignoring the second dissociation.1665

In other words, it doesn't really contribute all that much, in terms of hydrogen ion; you can just take the negative log of the 1.2.1674

So, here we have the 1.2; so the pH is equal to -log of 1.2, and you get a (let me see what...yes) -0.079 for a pH.1682

Yes, a pH can be negative; that means it's a really, really strong acid.1699

Now, if you ignore the significant figures, like I personally tend to do, I'll just go ahead and put that number down, too.1703

If you ignore the significant figures, and take the final hydrogen ion concentration to equal 1.212--if I include that .012, there is a little bit of extra hydrogen ion in there--well, you know what--your pH is going to be slightly different.1711

This is .084; 0.079, 0.084...you know, it's a small difference; if you are doing analytical work, it's important--for normal work, it's not altogether that important.1737

But you are going to end up with a pH of .08; that is really what it comes down to.1749

So, as it turns out here, the rule of thumb ends up being the following.1754

We notice that, for a solution which is about 1.2 Molar, roughly 1 Molar, the contribution of the hydrogen sulfate dissociation can be ignored.1765

You can treat it like a strong acid problem.1776

Below about 1.0 Molar, 1 Molar, .9 Molar...you can't ignore it; the second dissociation does produce a significant amount of hydrogen ion that has to be included in the concentration that came from the first dissociation.1779

You have to add those two; so, the first dissociation and then a little bit of the second dissociation...that is when you have your final hydrogen ion concentration.1796

For solutions of H2SO4 more dilute than approximately 1 Molar, the full expression for the equilibrium constant must be used.1805

In other words, we can't approximate; we have to do the 1+x, times x, over 1-x; we have to solve the quadratic, which means the quadratic expression must be solved.1842

OK, so now, we'll do a quick example; I'm actually just going to set it up for you (I'm not actually going to solve it), just so you see what it looks like.1864

The Example 3 is: Calculate the pH of a 0.010 Molar H2SO4.1872

This is pretty dilute: .01 Molar--a lot less than 1.1889

So, we can't ignore the second dissociation; OK.1896

Let's take a look at what we have; we have our major species, and again, we have the H+ from the first dissociation; we have the HSO4 from the first dissociation; and we have water.1900

These two are going to dominate the equilibrium; we know what this is already--this is just .01 Molar, because it's fully dissociated (strong).1918

The final hydrogen ion concentration is going to be 0.010+x, and x is the hydrogen ion concentration we get from the dissociation of the HSO4-.1927

That is the equilibrium we want to look at, HSO4- goes to H+ + SO42-; we have Initial; we have Change; we have Equilibrium.1943

The initial concentration is 0.010, right?--and 0.010, strong, dissociates into this and this (first dissociation).1955

This is 0; the change is -x, +x, x; we get 0.010-x over here; we get 0.010+x over here; we get x over here, and we write Ka2, which is 1.2x10-2, is equal to 0.010+x, times x (oops, let me just take out this parentheses here), and then, that is going to be over 0.010-x.1965

That is it; we have to solve this equation.2003

Multiply through; get your x2; get your x term; get your...it's going to be ax2+bx+c=0.2007

b and c and a--they can be negative, if they need to be.2019

And then, you just plug it into the quadratic equation, or use your graphing utility (your TI-83, 84 calculator, your graphing calculator) to find the roots of this equation.2022

Now, you are going to get two roots; check to see what makes sense--one of the roots is completely not going to make sense.2031

The other root will make sense, and that is the whole idea.2037

So, your final hydrogen ion concentration (oops, let's make this a little bit...): when you get x, you are going to get x equal to some number; we'll just put a little box there.2041

Your final hydrogen ion concentration is going to equal the 0.010, plus this boxed number, and your pH is going to equal the negative log of your final hydrogen ion concentration.2054

And that is it: so, for polyprotic acids, treat them like any other equilibrium; usually, most polyprotic acids are weak acids; it is going to be the first dissociation that dominates.2071

You can pretty much ignore the second, third, fourth...I don't think...I don't even know if there is a...well, there is, but...2081

Weak acid--you can pretty much ignore the second and third dissociations.2086

For sulfuric acid, maybe; maybe not.2092

If you are talking about 1 Molar or above, you can ignore the second dissociation; if you are talking about more dilute than 1 Molar (.8, .7, .6...below that), the second dissociation is going to contribute a significant amount of hydrogen ion.2094

You have to take the amount that comes from the original concentration, the first dissociation; find the amount that comes from the second; add them; and then take the negative log of that.2110

I hope that helps; again, you see: it's pretty much all the same--slightly different when you are dealing with multiple dissociations in the case of sulfuric acid.2119

Thank you for joining us here at Educator.com.2128

We will see you next time to discuss the acid-base properties of salts.2130

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

Today, we are going to talk about salts and their acid-base properties.0004

Let's just jump on in, start with a definition, and start doing problems, because I think that is probably the best approach.0008

A salt is a generic term for an ionic compound.0015

Sodium phosphate; silver chloride; lead iodide; sodium chloride; you name it--potassium permanganate--these are all salts, because they are ionic compounds.0036

It is just a positive ion and a negative ion; we just call them all salts.0050

Salt, sodium chloride, is specific--just happens to take that name; but when we talk about salts, we are talking about any ionic compound.0054

Now, as you know (or as you should know), when you take an ionic compound and you drop it in water, it is either going to dissolve, or it's not.0062

If it dissolves, it breaks up into its free ions, and those ions are just floating around freely in the water; so let's write that down, actually.0069

When salts dissolve, their ions float around as free ions.0081

OK: so, for example, if we had something like sodium chloride (which is a solid, as you know--it's salt), if we drop it in water (an arrow with a little water on it--that means you have dissolved it in water), you end up creating Na+ + Cl-.0107

Now, I am not usually going to write the aq, the subscripts, but that is what that means--it just means "dissolved in water"--aq.0119

You have sodium ions floating around, and you have chloride ions floating around.0125

Well, often, when this happens, the anion of the salt that you drop into water and dissolve--it is actually going to be the conjugate base of a weak acid.0129

Let's say that again: Often (let's write it down), the anion--the negatively-charged ion, this one--will be the conjugate base of a weak acid.0141

Example: sodium fluoride--if we take sodium fluoride, and if we dissolve it in water, we produce sodium ion, and we produce fluoride ion; so there is F- floating around, and there is Na+ floating around.0166

Now, you have seen this F- before; it is the conjugate base of the weak acid hydrofluoric acid, which means--a conjugate base just means you have taken an acid and you have taken off the H.0182

The thing that you are left with--that is the conjugate base.0195

So, over here, I am going to write; you have seen it as (oops, let me do this in red to separate) this: HF is in equilibrium with H+ + F-.0197

There is a certain Ka associated with this; in fact, it's 7.2x10-4.0215

The Ka equals 7.2x10-4.0221

So notice, this time, it doesn't show up as the acid; you didn't take the acid and drop it in solution--you dropped a salt whose negative ion, the anion, happens to be the conjugate base of a weak acid.0226

When this happens--when the anion of a salt that you dissolve in water happens to be the conjugate base of a weak acid--it actually reacts with water as a base.0240

Because remember--notice--as an acid, this is a weak acid; a weak acid means the equilibrium is on the left; that means it wants to be this way--it doesn't want to be this way.0249

That means, if you put some free F- anywhere near some H+, or anywhere near a source of H+, it is going to take that H+, and it is going to move over in this direction until this equilibrium is established.0261

So, weak acid--strong conjugate base; a strong base means that it has a tendency, a very high affinity, for hydrogen ions--for protons--for hydrogen ions; it wants to take them.0275

When you take a salt and dissolve it in water, and all of a sudden create this free base, which happens to be the conjugate base of a weak acid (hydrofluoric acid), here is what it does: it actually behaves as a base now.0286

It does this reaction: it reacts with the water that is floating around in solution, and it actually takes the hydrogen ion from the water; it is acting as a base--it takes hydrogen ion.0301

This time, water is acting as the base; it becomes HF + OH-.0313

Notice what we have done: in this process, this base takes the H, creates hydrofluoric acid, and in the process it creates hydroxide ion.0321

When this happens, you create a basic solution; so if you took a neutral, normal water, which is pH 7, and if you drop in some sodium fluoride, well, the fluoride ion will pull hydrogens off of the water, creating hydroxide ion.0330

The pH of that solution is going to go up; it is going to become a basic solution--that is what a basic solution is: it's when the concentration of hydroxide ion is higher than usual.0345

So, any time you have a salt where the anion happens to be the conjugate base of a strong acid...conjugate base of a weak acid; forgive me--where it happens to be the conjugate base of a weak acid, it is going to react as a base in this reaction.0356

People call this a hydrolysis reaction; I don't really like that name--just know that now, this conjugate base is going to act as a base.0374

You have seen this reaction before; this is the reaction of a base with water.0382

There is a Kb associated with this.0387

There is a Kb associated with this: it is HF, OH-, over F- (because water is liquid water).0390

Well, how do we know what the Kb is?0403

Here is the best part about it: the Kb, the relationship--we know the Ka already, but what is the relationship between the Ka and the Kb, since now the F- is actually behaving as a base, and not this way, as part of the weak acid equilibrium?0407

Now, it is involved in a basic equilibrium, where it is actually pulling off a hydrogen ion.0422

Here is the relationship--all Kas and Kbs are related by this equation: Ka, times the Kb, is equal to Kw.0429

So in this case, if you want to do a problem with this, and you need to treat this as a Kb problem, base problem, because this reaction is that of a base with water, all you do is: you take the Kb is equal to Kw over Ka for that species.0438

This is 10-14; in this particular case, it's 7.2x10-4, and you find the Kb, and you run this problem as a base problem.0459

We have already done it: weak acids, weak bases--that is all that is going on here.0470

So, let's actually do a problem, and it will make sense.0474

Let's see...OK; so, Example 1: Calculate the pH of a 0.35 Molar sodium fluoride solution.0479

The Ka is equal to (sorry, I had better say which Ka of what)...the Ka of hydrofluoric acid (the actual acid, the weak acid, where the base is coming from) is 7.2x10-4.0507

OK, what do we always do first?--we check the major species in solution to decide what chemistry is going to dominate.0529

Major species: well, you have sodium fluoride: sodium fluoride is completely soluble--remember the solubility chart from earlier in the year?0535

If you haven't memorized it, not a problem--just check it out; sodium fluoride, alkali metal, halogen--completely soluble.0545

That means what you have floating around in solution is sodium ion; you have fluoride ion; and you have H2O.0552

Well, we notice sodium ion doesn't do anything; it doesn't affect anything.0560

However, the anion happens to be the conjugate base of a strong acid, HF.0565

I keep saying "strong acid"; what is wrong with me?--weak acid, weak acid, weak acid.0571

So, F- is the conjugate base of a weak acid, hydrofluoric acid.0577

Well, it's the conjugate base of a weak acid; so what it is going to do--it is actually going to react as a base with the water, the following reaction.0582

It is going to be: F- + HOH goes to HF + OH-.0591

That is the reaction; this reaction is the reaction of a base with water.0608

There is a Kb associated with it.0614

We want to find the pH of this; well, now that we have our reaction, that we know what reaction is going to take place, we do our ICE chart.0616

Well, what is the initial concentration of the F-?--well, since we have full dissociation, it's 0.35.0626

Water doesn't matter; there is no HF formed yet--this is before anything happens, and there is no OH- before anything happens.0634

A certain amount of F- is going to disappear; as a species, it's going to react with H to become HF.0642

It is going to disappear; water doesn't matter; that means HF is going to show up, and OH- is going to show up.0649

There is absolutely nothing under the sun when it comes to these; we have done these several times--we know how it works, but now, because this is a base reaction, we need the Kb.0659

Kb is equal to Kw over Ka, equals 1.0x10-14, divided by 7.2x10-4, and we get a Kb of 1.4x10-11.0670

So now, we do 1.4x10-11 is equal to x, times x, divided by 0.35-x.0696

Well, look how small this is; you know what, x is going to be pretty small, so chances are, we can do the approximation.0709

It equals x2 over 0.35.0716

When we solve for x, we get x=2.2x10-6; but notice, this was not hydrogen ion concentration; we created a base in this reaction: x is equal to the OH- concentration, which implies that the pOH is equal to the negative log of this.0722

The pOH ends up being 3.2 (was that correct?--yes), and (wait, is that...yes, it is) then the pH is equal to 14 minus the pOH, which equals 10.8.0742

Sure enough, a pH of 10.8 means that it is a basic solution.0773

We had a salt; the anion was the conjugate base of a weak acid; therefore, it is going to react as a base with the water in a standard base reaction.0778

That is what bases do: bases take hydrogen from water to create hydroxide ion.0794

The equilibrium expression for that: we said it's a Kb, not a Ka; it's not an acid dissociation--it is a base association, if you will; it's a base constant.0799

Well, we have the Ka; normally, hydrofluoric acid is listed as a Ka, because it behaves mostly as an acid.0810

Therefore, the Ka is listed; but the relationship between Ka and Kb is Ka times Kb equals 10-14, which is Kw.0818

We solve for the Kb; we treat it like any other weak base problems; and we solve it; we get the pOH, in this case, because it's a base--because we want the pH, we take 14 minus that, 10.8.0826

I hope that makes sense.0839

OK, let's do Example 2: oh, actually, before that, let me actually...so let's stop there, and now let me go back to blue ink...0841

Now, if you have a salt (now we are going to talk about the cation; we mentioned the anion; now we are going to talk about the cation) where the cation is the conjugate acid of a weak base, then this cation will act as an acid and create hydrogen ion--create an acidic solution.0858

So, you have to watch the salt; you take a look at the salt, dissolve it; you take a look now--you not only look at the anion--you look at the cation as well.0919

For the first one, we said if the anion happens to be the conjugate base of a weak acid; now, if the cation happens to be the conjugate acid of a weak base.0928

It is going to act as an acid, and it is going to produce an acidic solution.0944

Let's do an example, and I think it will make sense.0949

And again, it is the chemistry that you want to understand: you want to take a look at the species and decide how it is acting.0954

That is really what is going on; in this previous problem, we saw that, when we dissolved the sodium fluoride, we have F- floating around freely.0962

Well, what is F- going to do?--F- happens to be the conjugate base of a weak acid, so it is going to actually behave as a base; it is going start taking Hs away from water.0969

If you write down the reaction, everything should fall out.0980

OK, calculate the pH of a 0.10 Molar ammonium chloride solution; the Kb of NH3 equals 1.8x10-5.0985

OK, our major species: well, we know that anything that involves ammonium and chloride is going to be fully soluble; therefore, what is floating around in solution is ammonium ion, chloride ion, and H2O.1013

Chloride--let's look at the anion first--chloride is the conjugate base of a strong acid, HCl.1031

So, Cl is not going to take any H from anything; it is just going to float around freely--we can ignore it.1040

However, NH4+, ammonium, is the conjugate acid of a weak base.1047

How do we find the conjugate acid?--just stick a hydrogen ion onto it...of the weak base ammonia.1056

Ammonia is the weak base; its conjugate acid is that, right?--because it comes from ammonia.1062

When you put ammonia in water, it takes a hydrogen from water; it becomes ammonium, and it creates hydroxide.1074

However, in this case, we didn't just drop ammonia into solution; we dropped the actual ammonium ion into solution as a salt.1084

We dropped it as a salt; now, there is just free ammonium ion floating around--all ammonium ion.1093

Well, ammonium ion is the conjugate acid of a weak base, which is ammonia.1099

Therefore, it is going to now behave as an acid.1104

Its equilibrium is going to be this: it is going to actually dissociate into H+ + Cl-...no, plus NH3.1107

There is a Ka associated with this, because this is an acid dissociation reaction: it is something that has a hydrogen ion to give up.1122

It gives it up, and it creates its other side; so now, this Ka is equal to Kw over the Kb.1130

Kb is the Kb that we have for ammonia, the conjugate base of that.1138

So now, let's solve the problem.1144

NH4+ is the dominant species; it is the conjugate acid of a weak base--it is going to behave as an acid, so we write it: behaving as an acid.1147

We are going to create H+; we are going to create an acidic solution, because we dropped it in water that was neutral.1164

We have Initial, we have Change, and we have Equilibrium; well, the initial concentration of NH4 is 0.10.1171

There is no H to start with; there is no ammonia to start with; this is -x; this is x; this is x; +x; this is 0.10-x; this is +x; this is +x.1180

Now, it's behaving as an acid, so we need a Ka.1194

Well, a Ka is equal to Kw over Kb.1199

That is equal to 1.0x10-14, divided by 1.8x10-5; the Ka for this reaction is equal to 5.6x10-10.1204

That tells me, this 5.6x10-10...it's a small number; it is a weak acid.1222

But, it is still an acid--it is going to produce some H+...it's weak, but it still behaving as a weak acid--it's going to give up its H+ into floating around freely in the water; that is the acidic solution.1228

The pH of the solution is going to drop.1239

So now, let's take 5.6x10-10 is equal to x, times x, over 0.10-x, which is approximately equal to x squared over 0.10; we end up with x=7.4x10-6, which does equal the hydrogen ion concentration, because here, we are directly forming hydrogen ion, and therefore, the pH of this is the negative log of that, which ends up being 5.1.1241

Sure enough, 5.1 is an acidic solution.1279

So there you have it: recap: if you have a salt floating around in solution (if you have a salt and it dissolves), if the anion is the conjugate base of a weak acid, it will create a basic solution.1284

You treat it as a base equilibrium that will react with water, in other words--pull off the hydrogen ion.1301

If the cation is the conjugate acid of a weak base, it will behave as an acid--you write the acid equilibrium and use the Ka to solve it as a weak acid problem, like you have done before.1307

It will create an acidic solution.1318

A second type of cation--a second type of species (you know, it might be nice if I actually wrote my words properly here--what do you think?) that creates an acidic solution--is a highly-charged metal ion.1321

A good example is aluminum 3+; 3+ is a pretty high charge; you are also going to find solutions like, for example, chromium 6+, manganese 5+, 4+...things that have a really high charge after about 1 or 2.1362

Here is what is going on: if you take a salt like aluminum chloride, I'm going to actually write out the chemistry of what happens, because I want you to see what happens--how the species forms in solution, and then how we treat that species in solution, just like any other weak acid.1376

If you take aluminum chloride, and you dissolve it in water--well, you know that aluminum chloride is completely soluble, so you are going to end up with aluminum 3+, plus 3 chloride ions--it completely dissociates into its free ions.1395

Chloride doesn't do anything; it's the conjugate base of a strong acid, so it doesn't behave as a base--it just floats around.1408

However, aluminum (because it is so highly charged, and because it is in water)--the water molecules, the actual molecules themselves, surround the aluminum; and in the case of aluminum, you get something that looks like this.1415

I am actually going to draw the structure, so you see it; I'm going to write it out first.1431

Aluminum (so I'm going to bring it over here)--the aluminum 3+ actually associates with 6 water molecules (and I'm going to draw the lone pairs on them) to create this species called a complex ion.1435

Al(H2O)6, and the whole charge on the species is 3+; this is usually how we write it.1456

Any time you have a metal that is surrounded by, in this case, water (we will just deal with water for the moment), and this actually looks like something--you have an aluminum ion, and you have an H2O, an H2O...I'm going to draw this 3-dimensional structure.1465

I really shouldn't, because you really don't have to know this for the time being; we may talk about it a little bit later, towards the end of the course, but I think it's important; it's nice--there is nothing here that you shouldn't be able to sort of visualize.1487

There are these 6 water molecules: OH2, OH2, OH2, OH2...and it is actually the oxygen, believe it or not, that is attached to the aluminum here.1499

In a normal bond, it is just surrounding it; so you have these 6 water molecules, and they are arranged; four of them are arranged in a plane; so aluminum is in the middle; one here, here; one is here; and back here.1512

These dashed lines mean it is going back; it's facing away from you.1525

These wedges mean it's coming towards you; these straight lines mean there is one on top and one on bottom.1528

What you have is the following: what you have is aluminum, in the center; you have a water molecule here, a water molecule here, a water molecule back here, a water molecule back here; one up here, and one down here.1535

There are six of them around it; and this whole species is carrying a 3+ charge; well, of course it's carrying a 3+ charge--aluminum is 3+, and water is a neutral molecule, so when they aggregate (when they sort of grab onto aluminum, if you will--6 of them), you create this thing called a complex ion.1548

Well, here is what happens with this complex ion: now I'm going to take this complex ion; as it turns out, it's so highly charged that it actually ends up pulling a lot of the electrons towards itself, and actually creates an acidic hydrogen.1569

One of these hydrogens, believe it or not, actually comes off.1584

And here is the chemistry of it (and this is what is important: the structure you don't have to understand; the chemistry is what is going on): in solution, this is produced.1588

When you drop aluminum chloride into water, aluminum ion is formed; water is attracted to the aluminum ion, and six water molecules arrange themselves in a pattern around the aluminum to create this complex ion.1598

This complex ion has an acidic hydrogen; it gives it up.1612

Here is how it gives it up: Al(H2O)63+ (I'm going to leave off the brackets; I think it's just extra symbolism that is not necessary)--it dissociates into H+ + Al; now there is just a hydroxide attached to it, and there is 5 waters.1615

One of the waters has lost its hydrogen; now it's 2+.1639

Notice, this is just a standard dissociation of an acid.1644

You have this species that has this H that it can give up; it gives it up; it's right there.1649

The rest of it--it doesn't even matter what it is; all that matters is this.1656

You have a species that has a hydrogen ion; it gives up that hydrogen ion to create some conjugate base; this is what is important.1660

You treat this like any other weak acid; there is a Ka associated with this--we measured it, and the Ka of this happens to be 1.4x10-5.1667

That is it; this is just HA dissociating into H+ + A-; this A---yes, it happens to be a very, very complex-looking thing (we call it a complex ion), but you treat it the exact same way.1679

Don't let the makeup of the thing that you are discussing confuse you; it's the chemistry that matters--the chemistry is just: some species gives up a hydrogen ion, and then ends up as something else.1694

The hydrogen ion is what is important; the equilibrium is handled the exact same way.1708

So now, let's do a problem.1713

Example 3: This is what is important in science--you need to understand what is going on underneath.1717

The individual identities of the species--they don't really matter, as far as what is going on; they matter for the individual case that you happen to be dealing with, as a researcher, as a lab scientist, as a doctor, whatever it is, but the chemistry is all the same.1726

It is still just some species, some acid, that has a hydrogen to give up, and it gives it up.1742

The mathematics is handled exactly the same way; the species, the identity, is entirely irrelevant.1747

It is entirely irrelevant; this is what we want you to do.1752

Our ideal is to get you to think abstractly, to think big-picture; if you can handle the big picture, you will know what is going on with the little picture; the little details are just incidental--they change from problem to problem.1756

But, the big picture doesn't change; that is what is important.1769

In science, what you want to concentrate on is what doesn't change.1772

The things that change...well, they are incidental.1777

That is when you know something is important--if something is not changing, that is what you want to concentrate on.1780

It is the chemistry that is important; OK.1785

Sorry about that lecture.1788

So, Example 3--we have: Calculate the pH of a 0.010 Molar AlCl3 solution.1790

OK, so the first thing we do: major species. 1808

Well, here is what we know: when we have some aluminum salt that is dropped in water, the aluminum is going to float around freely as ion; that is the first thing that is going to happen--the aluminum chloride is going to dissociate.1813

Aluminum is...the water molecules are going to aggregate around aluminum, 6 of them are, and they are going to form the species Al(H2O)63+.1827

Now, it isn't important that you know the name of it, but this is called hexaaqua-aluminum (3).1841

You will actually do the naming towards the end of the course, when you talk about coordination compounds, but this is the species that actually forms in solution.1846

If we took a picture of the solution, that is the species that we find.1854

We find every aluminum ion surrounded by six water molecules, and that whole thing is carrying a 3+ charge.1858

The only other species in there is water.1864

Well, it is true--water does contribute some hydrogen ion--but it is Ka of 10-14; this one has a Ka of 1.4x10-5.1867

10 to the negative 5 is a lot bigger than 10 to the negative 14, so water can be ignored.1881

This is the dominant species in the water that will control the pH of the solution.1885

We know what this does: it behaves as an acid.1893

Al(H2O)63+ dissociates into H+ + AlOH(H2O)52+--just an acid dissociation; that is it.1896

This is being created; let's do Initial; let's do Change; let's do Equilibrium.1919

What is the initial concentration?--well, all of the aluminum is dissolved; that means all of this is formed; 0.010--there is nothing formed yet; there is nothing formed yet.1924

A certain amount is going to dissociate--that is how much is going to show up of the other species.1935

0.010-x, +x, +x; we have the Ka; we have the equilibrium expression; so we just put it in.1940

1.4x10-5 (I hope you're not getting sick of these problems; I know it's just over and over--it's the same thing; that is nice--we like patterns) equals x times x, divided by 0.010-x.1951

We can approximate this with x squared over 0.010; now, you might think to yourself, "Well, wait a minute; 0.010 is pretty small, and x...we are talking about 10 to the negative 5 here...maybe."1972

As it turns out, when you check the validity, the error ends up being about 3.7%; we are still below 5, so we are good; this is a perfectly good approximation.1987

So, x is equal to 3.7x10-4, which equals the hydrogen ion concentration; that implies that the pH is equal to 3.43.1998

3.43--OK, I think I want to write this a little bit slower, so that all these wacky lines don't show up--equals 3.43.2013

How is that?2026

So notice, it's handled exactly the same way; the identity of the species is irrelevant--it's behaving as an acid; it's giving up a hydrogen ion.2027

It's a weak acid, 1.4x10-5; there is an equilibrium; we have to use an ICE chart.2036

We are done; that is nice; OK.2043

Now, our final little situation here: what if we have this situation?2048

What if we have the situation: NH4F--what if we have ammonium fluoride?--it is a perfectly valid salt.2052

Ammonium chloride...we can form ammonium fluoride.2070

Salt--you drop it into water; what happens?--well, it's going to dissolve, if it's completely soluble; it's going to dissolve into NH4+ and F-.2074

Well now, I have a little bit of a problem: we have an anion which is the conjugate base of a weak acid, hydrofluoric acid; so it is going to behave as a base, and it is going to pull hydrogen off of water to produce hydroxide ion.2083

So, it is going to create a basic solution; but, we have ammonium ion also floating around in solution.2099

It is the conjugate acid of a weak base, ammonia, and it is going to behave as an acid, as a weak acid, itself.2106

It is going to give up its hydrogen ion to create an acidic solution.2115

The F- is going to go ahead and create basic solution; this is going to create an acidic solution; well, what is the final solution going to be--acidic or basic? How do we decide?2118

Well, when you have a situation where both of the ions are species that react, and one produces base; one produces water, it gets very, very complicated--that is the short answer.2128

The equilibrium gets complicated, and you actually will be dealing with stuff like this, if you go on to study analytical chemistry.2141

If you are a chemistry major, usually in your third year, you will take an analytical chemistry course, and there are ways to handle this mathematically.2147

Pretty complex; for our purposes, we just want to be able to sort of give a qualitative answer.2154

We want to be able to say is the solution acidic or basic, without specifying what the pH is.2159

That is actually very, very easy to do.2164

It comes down to this: you do a quick test--if (I'm actually going to write this a little further to the left--excuse me) the Ka for the acidic ion (meaning this one) is bigger than the Kb for the basic ion, which is this one, well, the solution is acidic.2168

In other words, if the Ka for this is bigger than this, that means that the equilibrium for this is farther to the right; it produces more hydrogen ion than this produces hydroxide ion.2207

Therefore, there will be more hydrogen ion in solution; therefore, the solution will be acidic.2219

That is what this is saying: so, you need to find the Ka of this; you find the Kb of this; remember, Kb is 10 to the 14 over Ka of the acid, and this Kb is 10 to the 14 over Ka.2223

I'm sorry; Ka is 10-14/Kb for the conjugate base, what we did earlier.2238

You compare the two; the bigger one--that will dominate.2245

If this is bigger, it will be acidic; if this is bigger, it will be basic.2249

So, if the Ka is less than the Kb (the Ka for the acidic ion, less than the Kb for the basic ion), then your solution is going to be basic.2254

And of course, the last possibility (always three possibilities when it comes to ordering: less than, greater than, or equal to): If the Ka is equal to the Kb, well, you know exactly what that is; that is going to get a neutral solution.2265

And now, let's do our final example: Example 4: Will a solution of aluminum sulfate (AlSO4, Al2(SO4)3) be acidic or basic?2283

Well, Al2(SO4)3 dissociates into Al2, aluminum 3+, plus 3 SO42-, so yes; we have an anion--negative ion--which is the conjugate base of a weak acid.2315

The weak acid, in this case--just add one H.2339

OK, it's HSO4-; not H2SO4; it's HSO4---add one H.2343

And aluminum happens to be that thing that forms that species, Al, the hexaaqua-aluminum (3); so this is going to create a basic solution; this is going to create an acidic solution; we need to compare the two.2350

So, we want to know the Ka of this; the Ka--we already know that one; that is going to be 1.4x10-5.2369

And we want to know the Kb of this; the Kb of this is 10 to the -14, over the Ka of this, which is 1.2x10-6.2383

I hope you guys saw what I did; I found out the species that are floating in solution; this--the conjugate acid of that is the HSO4.2401

This actually just forms this species; we have the Ka of this; we find the Kb; we don't take the H2SO4; we add one H to it, OK?--one H at a time.2411

It dissociates one at a time; it associates one at a time.2422

We get a Kb, is equal to 1.3x10-13; well, this is hugely bigger than this, which means that our solution will be acidic.2427

This aluminum, this aluminum species, will dominate the acidity of the solution; we will get an acidic solution; and that is how you handle it.2444

So, thank you for joining us here at Educator.com to discuss the acid-base properties of salts.2454

In our next lesson, I'm going to close off with just a brief discussion of some oxides, and then we will go ahead and move on to further aspects of acid-base equilibria.2459

We will talk about buffer solutions and titration curves.2470

Take care; see you next time; goodbye.2472

Hello, and welcome back to Educator.com; welcome back to AP Chemistry.0000

Today, we are going to close out with a couple of comments--close out the acid-base discussion with a couple of comments on some oxides--the covalent oxides and the non-covalent oxides.0004

We are just going to briefly touch on them--not too much, because I really want to get into the common ion effect and buffer solutions.0017

It is going to be a very, very, very important application of acid-base chemistry, probably the most important application of acid-base chemistry.0023

But, I didn't want to leave out this tail-end of sort of straight, normal, run-of-the-mill, routine acid-base stuff, and I wanted to talk about some of the oxides, because the chemistry does come up, and it's a little strange how certain oxides produce acidic and basic solutions.0032

If you remember, last time we talked about salts; and when they dissolve (the ones that are soluble), either the anion happens to be the conjugate base of a weak acid, or the cation happens to be the conjugate acid of a weak base.0050

Well, these things react with water, and they actually produce acidic or basic solutions.0064

Well, kind of the same thing happens with oxides; so let's take a quick look, and we'll just write down what they are.0069

We won't get too much into the chemistry of it, but I am going to demonstrate, actually, how it happens, because I want you to be able to see that this doesn't just fall out of the sky--that it is just a movement of electrons and things actually do make sense.0077

It isn't just about memorizing reactions; it's about understanding why things react the way that they react.0090

OK, so let's write down...covalent oxides (and you will see what we mean in just a minute) will produce acidic solutions in water.0095

In other words, if I take a covalent oxide, if I drop it in water, it is actually going to form an acidic solution.0120

Here are some examples: SO3: SO3 is a covalent oxide--"covalent" because you are talking about a nonmetal-nonmetal bond.0125

Sulfur and oxygen are both nonmetals; it's a covalent bond.0134

Plus H2O: it ends up forming H2SO4, and when H2SO4 forms, it's a strong acid in its first dissociation, so it dissociates into H+ + HSO4-.0138

That is why it is an acidic solution: you drop SO3 into water; it forms H2SO4; H2SO4 dissociates, producing hydrogen ion; therefore, you have an acidic solution.0153

SO2: if you bubble that into water, you end up with H2SO3, which is sulfurous acid, which is a weak acid.0164

That is why we have a double arrow here, showing equilibrium; but again, the dissociation is the same: HSO3-.0175

CO2: if you bubble CO2 into water at high pressure, you end up with carbonic acid.0182

Carbonic acid is a weak acid, and there is an equilibrium with that and the hydrogen carbonate ion.0193

So, in each case, you see that H+ is being produced down the line, downstream of the actual reaction that is taking place.0199

And then, let's just do one final one; let's do NO2 + H2O: it produces (it's actually going to be 2 NO2) nitric acid, and it also produces nitrous acid.0207

You will find both species in solution; nitric acid is strong, so it dissociates into H+ + nitrate; and here, we are going to have an equilibrium with H+ + nitrite.0221

There you go: covalent oxide--"covalent," meaning that the thing bonded to the oxygen--they are both nonmetals, basically; the thing bonded to oxygen is a nonmetal, so sulfur...any nonmetal.0235

These oxides, when dissolved in water, react with water to form acidic solutions.0249

Now, ionic oxides, on the other hand: ionic oxides produce basic solutions in water.0255

And you know, ionic means metal-nonmetal; so ionic oxide is basically a metal with oxygen; so, an example would be calcium oxide.0275

With calcium oxide, when it reacts with water, what you end up producing is calcium hydroxide.0284

Potassium oxide (K2O) plus water: you end up producing potassium hydroxide.0294

Potassium hydroxide is a strong hydroxide, and it dissociates into 2 hydroxides, plus 2 potassium ions.0304

Calcium hydroxide is moderately soluble; actually, it's not very soluble at all, but it is slightly soluble; therefore, it dissociates into 2 OH-, plus calcium 2+.0313

We write it as an equilibrium, because in fact, most of it is actually over here.0326

This is mostly a solid; this is potassium hydroxide; this is an aqueous solution; so this just means that it is not very soluble, but enough of it does dissolve to produce a hydroxide ion, and hydroxide ion--that is why we have a basic solution.0330

So again, an oxide--a covalent oxide--produces acidic solutions when reacting with water; an ionic oxide (basically, oxygen plus a metal, so any ionic bond between oxygen and a metal)--when those react with water, they produce basic solutions.0349

OK, now I'm going to quickly discuss how these are actually formed--just going to give some quick examples.0370

This is definitely not something that you have to know; however, I want you to see it, because I think it is important to see it; it is important to get a sense of the chemistry and to feel comfortable with it.0375

Those of you who go on to study organic chemistry: this is going to be a huge part of what you do--this sort of what we call arrow-pushing...movement of electrons.0385

Let's do the CO2 example: the CO2, plus H2O, goes to H2CO3, which dissociates into H+ + HCO3-.0394

Here is what actually happens: CO2 is a linear molecule; each oxygen is double-bonded to the carbon.0409

Water, as you know, or as you should know, maybe (and if you don't know, that is not a problem--we are actually going to get to it, because you remember: in this particular AP Chemistry course, we actually skipped over the bonding--I wanted to get to this stuff--equilibrium and acids and base--and I will actually return to the bonding)--a water molecule consists of oxygen single-bonded to hydrogens, and it has a couple of lone pairs.0415

Well, as it turns out, these lone pairs--what they do is: they attack the carbon, and they push the electrons onto one of the oxygens--one pair of electrons that are part of the double bond here.0443

What you end up getting is something that looks like this: O (now that oxygen has an extra electron), and then you have the oxygen, this and this, and 1 lone pair.0457

Well, because you have three things bonded to an oxygen, that is actually an extra bond; so now, this is carrying a positive charge.0469

You remember, charge has to balance on both sides of an arrow; here, this is neutral--everything is neutral here--net.0476

Here, this is minus; this has to be plus; that is how this works.0483

You will learn more about that later; I am just trying to get you to see what, exactly, goes on.0487

What happens next is: there is a transfer of a proton, and I am going to represent it like this: these negative charges actually take this hydrogen, and they push the electrons onto this oxygen, and what you get is OH, OH; there, you have your H2CO3.0492

Now, this ends up going into equilibrium with...that is how it happens.0514

Water attacks the carbon dioxide; there is a shift; there is a transfer of a proton over to oxygen; and what you end up with is carbonic acid.0522

Carbonic acid dissociates to produce H+.0529

I just wanted you to see that things don't just fall out of the sky; it actually makes sense what happens--this is reasonable.0533

OK, now let's do a calcium oxide example.0541

We have calcium oxide, plus H2O, going to calcium hydroxide.0545

Here is how it happens: calcium...this is an ionic bond--this is 2+; this is O2-; well, O2-...here is what happens: H, again, with the oxygen; these electrons over here...this is fully...there are 8 electrons around here.0554

They actually take one of these; they push the electrons onto here; what you end up with is, now, calcium 2+ plus an OH- plus another OH-.0570

This O2- takes an H+ and becomes OH-, right?--2 minus, 1 plus, is minus 1; it's an OH.0583

That leaves an OH, so that is the other OH-.0589

Now, these form your calcium hydroxide, which is mostly insoluble--it's actually kind of a solid--it's pretty solid; but it is moderately (you know what, I'm not going to have these stray lines running around all over here, so let me do it this way--let me go slow)...0594

Calcium hydroxide is in equilibrium with calcium 2+, plus 2 hydroxide ion, and there is your hydroxide ion to form a basic solution.0615

I just wanted you to see it: covalent oxides form acidic solutions; ionic oxides form basic solutions--standard chemistry.0627

It will show up on the AP Chemistry exam, in terms of actual reactions; so they might say something like "aluminum oxide," "iron oxide--does it form an acidic or a basic solution?"0639

Well, iron oxide is an ionic oxide; therefore, it forms a basic solution.0649

If you have sulfur trioxide, does that form an acidic or a basic solution when dropped in water?0655

Sulfur trioxide is a covalent oxide: when you put it into water, it forms an acidic solution.0661

Qualitatively, it does show up on the AP exam; but I wanted you to see why--what actually happens chemically.0667

I want you to get a sense that these things--it's not magic; it's just atoms that are slamming into each other, and electrons are moving around; that is all that is happening--very intuitive stuff, really.0674

OK, now we are going to move on to a profoundly, profoundly important concept called the common ion effect.0685

The common ion effect, in its application, is used to talk about buffer solutions; so let's talk about the common ion effect--let's get a little bit of the math and chemistry under our belts.0693

And then, we will talk at length about buffer solutions; there are going to be, actually, two or three lessons strictly devoted to buffer solutions, because they are profoundly important.0705

Your blood is a buffer solution; it is buffered by the carbonic acid buffer system, the carbonate buffer system.0715

It is used to maintain the pH of the blood somewhere between about 7.2 and 7.4.0719

The phosphate buffer system inside the cell--that maintains the pH at a certain value inside the cell.0727

If pH goes up or down, terrible things start to happen; there is actually a very, very narrow range, as far as healthy body function, when it comes to blood pH and intracellular pH.0735

OK, so let's start with a definition.0748

Definition: When an equilibrium already exists in a solution, the shift that occurs when you add an ion (a little metathesis here) already involved in the equilibrium is called the common ion effect.0752

It is just an application of Le Chatelier's Principle,