For more information, please see full course syllabus of General Chemistry

For more information, please see full course syllabus of General Chemistry

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## Table of Contents

## Transcription

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### Energy & Chemical Reactions

- State functions are path-independent.
- The first law of thermodynamics states that energy is conserved during a chemical reaction.
- Energy can be transferred through heat and work.
- Calorimetry problems involve either constant-volume or constant-pressure scenarios.
- Enthalpy is the amount of heat transferred under constant pressure.

### Energy & Chemical Reactions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Lesson Overview
- Introduction
- The First Law of Thermodynamics
- The First Law of Thermodynamics Cont'd
- Calculating ∆U, Q, and W
- Calculating ∆U, Q, and W Cont'd
- Calculating ∆U, Q, and W Cont'd
- Calculating ∆U, Q, and W Cont'd
- Constant-Volume Calorimetry
- Bomb Calorimeter
- The Effect of Constant Volume On The Equation For Internal Energy
- Example 3: Calculate ∆U
- Constant-Pressure Conditions
- Calculating Enthalpy: Phase Changes
- Melting, Vaporization, and Sublimation
- Freezing, Condensation and Deposition
- Enthalpy Values For Phase Changes
- Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
- Calculating Enthalpy: Heats of Reaction
- Using Standard Enthalpies of Formation
- Using Standard Enthalpies of Formation
- Enthalpy From a Series of Reactions
- Coffee-Cup Calorimetry
- Summary
- Sample Problem 1
- Sample Problem 2

- Intro 0:00
- Lesson Overview 0:14
- Introduction 1:22
- Recall: Chemistry
- Energy Can Be Expressed In Different Units
- The First Law of Thermodynamics 2:43
- Internal Energy
- The First Law of Thermodynamics Cont'd 6:14
- Ways to Transfer Internal Energy
- Work Energy
- Heat Energy
- ∆U = q + w
- Calculating ∆U, Q, and W 8:58
- Changes In Both Volume and Temperature of a System
- Calculating ∆U, Q, and W Cont'd 11:01
- The Work Equation
- Example 1: Calculate ∆U For The Burning Fuel
- Calculating ∆U, Q, and W Cont'd 14:09
- The Heat Equation
- Calculating ∆U, Q, and W Cont'd 16:03
- Example 2: Calculate The Final Temperature
- Constant-Volume Calorimetry 18:05
- Bomb Calorimeter
- The Effect of Constant Volume On The Equation For Internal Energy
- Example 3: Calculate ∆U
- Constant-Pressure Conditions 26:05
- Constant-Pressure Conditions
- Calculating Enthalpy: Phase Changes 27:29
- Melting, Vaporization, and Sublimation
- Freezing, Condensation and Deposition
- Enthalpy Values For Phase Changes
- Example 4: How Much Energy In The Form of heat is Required to Melt 1.36 Grams of Ice?
- Calculating Enthalpy: Heats of Reaction 31:22
- Example 5: Calculate The Heat In kJ Associated With The Complete Reaction of 155 g NH₃
- Using Standard Enthalpies of Formation 33:53
- Standard Enthalpies of Formation
- Using Standard Enthalpies of Formation 36:12
- Example 6: Calculate The Standard Enthalpies of Formation For The Following Reaction
- Enthalpy From a Series of Reactions 39:58
- Hess's Law
- Coffee-Cup Calorimetry 42:43
- Coffee-Cup Calorimetry
- Example 7: Calculate ∆H° of Reaction
- Summary 47:12
- Sample Problem 1 48:58
- Sample Problem 2 51:24

### General Chemistry Online Course

I. Basic Concepts & Measurement of Chemistry | ||
---|---|---|

Basic Concepts of Chemistry | 16:26 | |

Tools in Quantitative Chemistry | 29:22 | |

II. Atoms, Molecules, and Ions | ||

Atoms, Molecules, and Ions | 52:18 | |

III. Chemical Reactions | ||

Chemical Reactions | 43:24 | |

Chemical Reactions II | 55:40 | |

IV. Stoichiometry | ||

Stoichiometry I | 42:10 | |

Stoichiometry II | 42:38 | |

V. Thermochemistry | ||

Energy & Chemical Reactions | 55:28 | |

VI. Quantum Theory of Atoms | ||

Structure of Atoms | 42:33 | |

VII. Electron Configurations and Periodicity | ||

Periodic Trends | 38:50 | |

VIII. Molecular Geometry & Bonding Theory | ||

Bonding & Molecular Structure | 52:39 | |

Advanced Bonding Theories | 1:11:41 | |

IX. Gases, Solids, & Liquids | ||

Gases | 35:06 | |

Intermolecular Forces & Liquids | 33:47 | |

The Chemistry of Solids | 25:13 | |

X. Solutions, Rates of Reaction, & Equilibrium | ||

Solutions & Their Behavior | 38:06 | |

Chemical Kinetics | 37:45 | |

Principles of Chemical Equilibrium | 34:09 | |

XI. Acids & Bases Chemistry | ||

Acid-Base Chemistry | 43:44 | |

Applications of Aqueous Equilibria | 55:26 | |

XII. Thermodynamics & Electrochemistry | ||

Entropy & Free Energy | 36:13 | |

Electrochemistry | 41:16 | |

XIII. Transition Elements & Coordination Compounds | ||

The Chemistry of The Transition Metals | 39:03 | |

XIV. Nuclear Chemistry | ||

Nuclear Chemistry | 16:39 |

### Transcription: Energy & Chemical Reactions

*Hi, welcome back to Educator.com.*0000

*Today's lesson from general chemistry is going to be our first discussion on thermodynamics.*0002

*We are going to be discussing energy and chemical reactions.*0010

*Let's go ahead and go over the lesson overview.*0015

*We are going to first start off with a brief introduction.*0018

*Because this is going to be totally different than all of the previous material we have been seeing.*0021

*We are then going to discuss the very important first law of thermodynamics.*0027

*Then the rest of the chapter is highly quantitative.*0034

*Once we have the introduced the different parameters that are at a focus*0037

*in thermodynamics, we are going to learn how to calculate each of them.*0042

*There is going to be different situations when we both have a volume and temperature change.*0045

*There is also going to be a situation when we have no volume change.*0051

*Then there is going to be a situation when we have no pressure change.*0057

*That is what pretty much this chapter is about, studying, introducing these very important physical parameters.*0061

*Then learning how to calculate them in each of their different situations.*0069

*We are going to finish up the chapter with a discussion of enthalpy.*0074

*Followed by our summary and then two sample problems.*0078

*Let's go ahead and get into the lesson now.*0083

*Remember we introduced the definition of chemistry way long time ago.*0088

*We said that chemistry is the study of matter and the changes that it undergoes.*0094

*We have already discussed the three states of matter.*0098

*We have already discussed the types of physical and chemical changes.*0101

*But there is always something that is associated with physical and chemical changes.*0105

*That is a change in energy.*0111

*When we talk about energy, energy can be measured and expressed in different units.*0115

*If you have a utility bill, especially the electricity, that is going to be usually expressed in terms of kilowatt hours.*0123

*If you are going to look at the side of a cereal box or anything*0131

*or any type of beverage, some nutritional information, nutritional information tends to use calories.*0136

*But in the physical sciences, we tend to use what is called the Joule.*0142

*The Joule is going to be abbreviated capital J.*0147

*The relationship between all three is the following.*0150

*One kilowatt hour is 3.6 times 10 ^{6} joules; one calorie is 4.184 joules*0153

*Now that we know the units of energy, let's get into some more theory.*0163

*What we are interested in in general chemistry is what is called internal energy.*0171

*You have probably taken physics by now.*0178

*You probably have already learned what is called kinetic energy and potential energy.*0181

*Where potential energy is just related to position and kinetic energy is the energy related to motion.*0187

*What internal energy is, internal energy is going to be the sum basically.*0197

*Internal energy is the sum of a system's potential energy and its kinetic energy.*0203

*Depending on what textbook you use, internal energy will be either symbolized with capital U or with capital E.*0209

*We are going to now introduce another term, what is called the state function.*0219

*Internal energy is an example of a state function; a state function is path independent.*0223

*In thermodynamics, it is pretty easy to always identify what a state function is.*0231

*A state function name is going to tend to be capitalized.*0235

*What do I mean by path independent?*0240

*Let's go ahead and look at the following reaction.*0243

*A solid plus B solid goes on to form AB solid.*0246

*This reaction is the direct combination of A in the solid state reacting with*0259

*B in the solid state to form the compound AB in the solid state.*0265

*Let's say that this reaction had associated with it a certain change in energy.*0269

*This is the most direct route; but I can do this in a stepwise motion.*0278

*For example, I can first go from A solid to A gas.*0283

*I can then go from B solid to B gas.*0290

*Once I have A and B in the gaseous state, I can combine the two.*0295

*Once in the gaseous state, that can then go on to form AB gas.*0300

*AB vapor can then condense back down to the solid.*0308

*What a state function means, what path independent means is the following.*0314

*If I take the direct route or if I do the stepwise route, in the end*0320

*the change in energy is going to be the same regardless of the path.*0330

*Δenergy same regardless of path.*0334

*Again state functions are path independent.*0345

*We really don't care about the step by step process that we take.*0348

*We only really care about the comparison between the initial and the final state.*0352

*Because of that, we are then often interested in not just internal energy but the change in internal energy.*0359

*What I am going to be using is the symbol ΔU to represent change in internal energy.*0367

*When we talk about internal energy, there are essentially two ways to transfer internal energy in and out of a system.*0377

*When I use the word system, I mean maybe a chemical reaction for example.*0386

*The two ways to transfer internal energy in and out of a system are through work and through heat.*0392

*Work is going to be symbolized lowercase w.*0401

*Heat is going to be symbolized with lowercase q.*0404

*Because they are lowercase, they are not state functions.*0407

*Remember from the previous slide that state functions tend to be capitalized.*0410

*What you want to pay attention to are the signs.*0418

*If a system performs work, then the work of the system is going to be negative.*0424

*If a system has work performed on it, we say that the sign is going to be positive.*0440

*Let's now look at the signs for q.*0457

*If heat enters a system, we say that the sign of q is going to be positive.*0460

*If heat leaves a system, we say that the sign of q for the system is going be negative.*0473

*You have to know these signs because when you are going to be given a*0485

*word problem, you have to be able to derive the sign from the sentences.*0489

*A lot of this many of you will recognize this comes immediately from physics.*0496

*We define work energy as the product of applied force over a certain distance.*0512

*Many of you will recognize the equation, work is equal to force times distance.*0513

*Heat energy basically, that is more literal.*0516

*Heat energy basically is going to be essentially thermal energy.*0519

*The change in internal energy equals the sum of work and heat energies.*0525

*The equation for the first law of thermodynamics is ΔU is equal to q plus w.*0533

*A very common application of the first law of thermodynamics is when the system experiences both a change in volume and in temperature.*0542

*Now we are going to learn to calculate the previous three parameters; ΔU, q, and w.*0553

*This is a very traditional type of problem.*0564

*This involves a piston and a cylinder like you have in your automobile engine.*0566

*The cylinder and piston are warmed by a flame.*0571

*The gases in the cylinder expand, doing work on the surroundings by pushing the piston outward.*0574

*If the system absorbs 559 joules of heat and does 488 joules of work, what is ΔU?*0579

*Basically we have a piston here inside a cylinder.*0587

*Here we have a lot of a hot gas inside.*0594

*This hot gas is going to push against the piston.*0599

*It is going to apply a force against it.*0604

*You are told that the system absorbs 559 joules of heat and it does 488 joules of work.*0608

*You see immediately how important it is to pay attention to the sentences because you have to derive the signs.*0615

*ΔU is equal to q plus w; let's go ahead and look at q.*0621

*You are told that the system absorbs 559 joules.*0628

*That means heat is entering the system; that is going to be +559 joules.*0632

*In addition, w, you are told that the system is doing, it is performing 488 joules of work.*0638

*That is going to be a negative sign.*0646

*When all is said and done, we get an answer of 71 joules for ΔU.*0650

*It is quite straightforward to use this equation that represents the first law of thermodynamics.*0657

*Let's now introduce the equations for work and for heat because each of them have their own.*0666

*Let's first introduce the work equation.*0673

*Work is equal to -P _{external} times ΔV.*0677

*In this case, P _{external}, this is going to be a constant external pressure.*0682

*That is the key; the pressure has to be constant or else you cannot you this equation.*0690

*Of course ΔV, we immediately recognize as change in volume or V _{final} minus V_{initial}.*0697

*When fuel was burned in the cylinder equipped with a piston, the volume expands*0708

*from 0.255 liters to 1.45 liters against an external pressure of 1.02 atm.*0712

*During this process, 875 joules of heat is given off.*0719

*Calculate ΔU for the burning of the fuel.*0723

*Work is equal to ?P _{external} times ΔV.*0730

*You are told in the problem that the external pressure is 1.02 atm; times ΔV.*0737

*That is going to be V _{final} minus V_{initial} or 1.45 liters minus 0.255 liters.*0749

*When all is said, we are going to get an answer of -1.22.*0757

*Our units are going to be atmospheres times liters together.*0762

*We always want to try to get to joules.*0768

*The conversion factor between liter atmosphere and Joules is the following.*0772

*One liter atmosphere equals 101.3 Joules.*0777

*We are just going to use dimensional analysis.*0782

*-1.22 atmosphere times liter times 101.3 joules over 1 liter atmosphere.*0785

*That is going to be equal to -123 joules.*0793

*That is going to be our value for w.*0797

*Let's see if this sign makes sense.*0800

*Does it make sense that the sign of work here is going to be negative?*0803

*We have a piston.*0808

*The piston is exerting a downward pressure of 1.02 atm.*0811

*If my system is the...*0818

*If gas is trapped inside the piston, they have to do work to push against this downward pressure.*0822

*You are told that the piston is going to expand from 0.255 to 1.45.*0828

*In order for that increase in volume to occur, the system will be performing work to go against the downward pressure.*0834

*Yes, the negative sign of w does make sense.*0843

*The system is performing work; very good.*0847

*Now that we have introduced the work equation, let's now tackle the heat equation.*0852

*The heat equation for all you pre-meds is mCaΔT.*0858

*You will recognize that as MCAT; you will hear q being the MCAT equation.*0863

*M is just your mass, usually in grams.*0869

*ΔT is your change in temperature; T _{final} minus T_{initial}.*0872

*C is something new; C is what we call the specific heat capacity.*0878

*The specific heat capacity is in units of joules per gram degree Celsius.*0882

*What specific heat capacity is, it is formally defined as the following.*0887

*It is the energy required to raise the temperature of 1 gram of a compound by exactly 1 degree Celsius.*0891

*What does the specific heat capacity physically mean?*0900

*If you think of a metal, metals tend to get very hot very very quickly.*0905

*They are excellent conductors of thermal energy.*0910

*It doesn't take much to heat a metal.*0914

*We say that metals are going to have relatively smaller heat capacities.*0918

*They don't retain the heat as much.*0921

*The opposite of metals, remember insulators.*0925

*Insulators such as Styrofoam are going to have smaller heat capacities.*0928

*It take a lot of energy to heat up Styrofoam.*0933

*That is what we know and have experienced.*0937

*One very important compound of course is going to be liquid water.*0943

*You should just know its specific heat capacity to be 4.18 joules per gram degree Celsius.*0948

*That is relatively large, especially compared to metals.*0953

*If you think about it, yes it does require lots of energy to heat up water.*0956

*So q is equal to mcΔT.*0962

*Now let's go ahead and look through an example.*0966

*You have a 0.055 kilogram sample of water and 27.5 degree Celsius.*0969

*You are told that it absorbs 725 joules of heat.*0975

*Calculate its final temperature.*0978

*q is equal to mcΔT.*0982

*Let's plug in what we know and see if we can solve for what is remaining.*0988

*You are told that the water absorbs 725 joules of heat.*0994

*That means heat is entering the system, which means it is going to be a positive sign.*0998

*q is equal to +725 joules; that is equal to the mass of the water.*1003

*Remember we want to use mass in units of grams.*1011

*That is going to be 55 grams.*1014

*Times the specific heat capacity of water which is going to be 4.18 joules per gram degree Celsius.*1018

*Finally times ΔT where ΔT is equal to T _{2} minus T_{1}.*1029

*T _{2} is what we are trying to solve for.*1034

*The initial temperature is 27.5 degrees Celsius.*1038

*We are in good shape; it looks like we only have one unknown.*1044

*That is the final temperature which is what we are trying to solve for anyways.*1049

*T _{2} is going to be equal to 30.7 degrees Celsius.*1053

*Always double check; does it make sense that T _{2} is greater than T_{1}?*1058

*Yes, absolutely because you are told that the sample of water is absorbing heat.*1064

*That heat is going into the system, into the water to warm it up.*1068

*Yes, our final temperature had better be greater than our initial temperature.*1073

*Again for thermodynamics, it is always a good habit to double check your work and ask yourself, does it physically make sense?*1077

*There is different ways of calculating the amount of energy that is involved in a process.*1088

*One way is using a constant volume process.*1096

*Now we introduce the word calorimetry.*1101

*All calorimetry is, it is the determination of the amount of energy transferred in a chemical process.*1103

*We can do this using a constant volume process.*1111

*This is what we call bond calorimetry.*1115

*The energy released from a combustion reaction can be determined experimentally in a bomb calorimeter.*1118

*Let me go ahead and provide a rough sketch of what a bomb calorimeter is.*1125

*Basically you have a small cup; inside the cup is going to be your reactants.*1131

*Here is the key; this cup is going to tend to be made of stainless steel.*1142

*We know stainless steel to be very very hard and very very rigid.*1154

*Because it is quite rigid, its volume is not going to change, at least on the macroscopic scale.*1158

*That is how we guarantee a zero change in volume.*1167

*Stainless steel maintains volume, maintains a fixed volume during the reaction.*1172

*This little cup of stainless steel, that contains your reactants.*1183

*This is essentially hooked up to a fuse.*1187

*This is why we call it a bomb calorimeter.*1190

*Basically you are going to ignite the fuse.*1192

*The fuse is going to provide a spark, a source of energy.*1195

*You are going to have basically a combustion reaction occur inside the stainless steel apparatus.*1199

*On the outside of this stainless steel apparatus is a vessel.*1206

*This is going to be some type of insulative material.*1213

*This vessel then is going to have a thermometer embedded in it.*1221

*The thermometer is going to allow us to determine what the temperature change is during the reaction.*1227

*Of course the temperature is going to go up because it is a combustion reaction.*1234

*Sometimes the bomb calorimeter is going to have water too.*1239

*This is going to be with or without water.*1244

*Basically the water is going to provide a very nice heat sink for us*1247

*so that we can dissipate the thermal energy coming off of the combustion reaction.*1251

*What we are assuming is the following.*1261

*That all of the heat energy that is given off by this combustion reaction is going to be essentially completely transferred to the surroundings.*1263

*Basically the surroundings of this experiment are going to be the bomb calorimeter itself.*1274

*If there is the water, it is also the water.*1282

*Anything that is capable of absorbing the energy.*1285

*All of this is equal to zero.*1292

*q of reaction plus q of bomb plus q of water is equal to zero.*1294

*There is no heat lost to the outside environment.*1298

*All of it is going from the reaction inside the bomb calorimeter to the bomb calorimeter itself.*1301

*If there is water, to the water itself.*1308

*The equation for the heat of the bomb calorimeter is simply equal to the heat capacity of the calorimeter times ΔT.*1312

*Please make a note that this is going to be independent of its mass; mass independent.*1321

*Something very interesting happens.*1333

*What we want to look at is now the effect of constant volume on our equation for internal energy.*1335

*We know that ΔU is equal to q plus w, which equals to q minus PΔV.*1341

*If the volume is not changing, ΔV and this whole term essentially goes to zero.*1350

*We very nicely have the relationship of ΔU is strictly equal to q.*1356

*That is, in a constant volume process, all of the energy that is*1362

*transferred is in the form of heat, not in the form of work.*1368

*All energy transferred as heat.*1374

*Again constant volume process, ΔU equals to q.*1383

*Let's go ahead and now look at a typical problem.*1393

*You have 1.550 grams of C _{6}H_{14}; it is undergoing combustion in a bomb calorimeter.*1400

*Right away, bomb calorimeter, immediately I would underline that because that tells me that ΔV is zero.*1407

*So ΔU is q; q of the reaction of course.*1416

*You are told that the temperature changes from 25.87 to 28.13.*1422

*Remember that is what is expected; the temperature is going to increase.*1426

*Remember this is a combustion reaction after all.*1429

*They are always going to be giving off energy in the form of heat.*1431

*Calculate ΔU in kilojoules per mole.*1436

*You are told that the specific heat capacity of the calorimeter is 5.73 kilojoules per degree Celsius.*1439

*Let's go ahead and solve the problem; you are asked to calculate ΔU.*1446

*But because this is bomb calorimetry, ΔU is equal to q of the reaction.*1450

*In addition we know that q of the reaction plus q of the bomb calorimeter*1457

*plus q of water is equal to zero from the previous slide.*1464

*But do you see water being mentioned in this example?*1468

*There is none; this term is not applicable.*1472

*We have the relationship that q of the reaction is equal to ?q of the bomb.*1479

*This is equal to ?C of the bomb times ΔT.*1486

*Let's go ahead and plug in everything we know.*1494

*That is going to be equal to -12.95 kilojoules.*1496

*The question asks us for ΔU in units of kilojoules per mole.*1508

*You see here that we have the mass of the reactant.*1513

*All we have to do is get that to moles; then divide our kilojoules over the moles.*1517

*That is going to be 1.550 grams times 1 mole over its molar mass which is approximately 86 grams.*1523

*That is going to give us x moles of the reactant, C _{6}H_{14}.*1536

*When all is said and done and we take the kilojoules over the moles, we are going to get -718.5 kJ per mole.*1544

*When you do these problems, always pay attention to the units that are requested for ΔU.*1554

*That is how we do a typical bomb calorimetry problem.*1561

*In addition to having a calorimetry performed under a constant volume process, we can also do a constant pressure process.*1567

*This is going to be a slight mathematical derivation here.*1576

*We know that ΔU is equal to q plus w, which is equal to q minus PΔV.*1580

*From thermodynamics, there is going to be a new state function that is going to introduced right now.*1586

*This state function is what we call enthalpy; enthalpy is going to be symbolized ΔH.*1592

*By definition from thermodynamics, ΔH is equal to q plus PΔV.*1597

*We will call that equation two; let's combine equations one and two right now.*1602

*ΔH is equal to U plus ΔV, which is equal to q minus PΔV plus PΔV, which is equal to q.*1608

*In other words, at constant pressure conditions, enthalpy and heat are identical.*1619

*They are the same thing.*1626

*In addition, if the volume change is negligible, you see from the equation that ΔH and U are approximately the same value.*1629

*This is how we define enthalpy.*1640

*Enthalpy is essentially heat energy under the constraint of constant pressure.*1642

*The remainder of this lecture deals with calculating enthalpy under various conditions.*1652

*There is numerous ones; there is about four to five.*1662

*Which just shows you how important enthalpy is to this chapter.*1664

*The first way of calculating enthalpy involves phase changes.*1669

*Every phase change which we know to be physical changes of course is going to be associated with a change of energy.*1674

*Melting, vaporization, and sublimation; think about it.*1682

*If you are going to melt something, you have to put energy into it.*1686

*They all require energy.*1688

*Those processes that require energy are what we call endothermic processes.*1692

*Endothermic processes tells us the following.*1697

*That the sign of ΔH is going to be a positive value.*1700

*The opposite directions, freezing, condensation, and deposition, all give off energy and are exothermic processes.*1706

*For exothermic processes, the sign of ΔH is going to be a negative value.*1714

*Enthalpy values for phase changes are commonly given in units of kilojoules per mole under standard conditions.*1721

*What do we mean by standard conditions?*1733

*Basically standard conditions are provided for us to level the playing field*1735

*because there is so many different parameters we can perform experiments at.*1739

*It is very for us to come up with a universal accepted condition.*1744

*That is going to make everything a lot easier.*1755

*Standard conditions are defined as the following.*1757

*If we are dealing with a gas, it is going to be 1 atmosphere of pressure.*1760

*If we are dealing with a solution, it is going to be a 1 molar concentration.*1765

*The temperature is going to be 25 degrees Celsius.*1772

*Again 1 atm, 1 molar, and 25 degrees Celsius.*1776

*Let's go ahead and look at the following question.*1784

*How much energy in the form of heat is required to melt 1.36 grams of ice?*1787

*You are given the ΔH of these phase changes to be 6.00 kilojoules per mole.*1792

*Anytime you see the letters f-u-s, that means a fusion.*1798

*That is going to be for the melting process; melting process.*1802

*Anytime you are dealing with a phase change, it is quite simple.*1809

*q is equal to n times ΔH of the change.*1812

*Very straightforward equation where n is going to be moles of your compound.*1821

*ΔH of the phase change is going to be like we said in kilojoules per mole.*1827

*It is pretty straightforward.*1833

*We have to get the 1.36 grams of the ice to moles.*1834

*1.36 grams times 1 mole over 18.016 grams.*1837

*Then we want to multiply this by kilojoules per mole.*1845

*That is going to give us 0.453 kJ for our final answer.*1851

*Does it make sense that the sign of ΔH here is positive?*1857

*Remember this is a melting process so absolutely this is going to be an endothermic process.*1861

*Basically you want to add ΔH as a new conversion factor, kilojoules per mole; very good.*1867

*That is the first way of calculating enthalpy we are going to talk about via a phase change.*1877

*The next way of calculating enthalpy is what we call the heat of a reaction.*1884

*Remember that we said that every chemical process is going to have a change in energy associated with it.*1888

*This change in energy is what we call ΔH of the reaction.*1895

*ΔH of the reaction is going to be symbolized ΔH _{r}.*1898

*It is usually given to the right of a chemical reaction.*1903

*Basically this is telling me that if I am going to perform this reaction exactly one time,*1908

*and I got the perfect amounts, mole to mole ratios of reactants,*1916

*it is going to release exactly 906 kilojoules of energy in the form of heat.*1919

*Calculate the heat in kiloJoules associated with the complete reaction of 155 grams of ammonia.*1927

*Anytime you are given a ΔH of a reaction problem, all you are going to do is*1934

*you are going to using ΔH of reaction as a conversion factor; again new conversion factor.*1938

*Basically the conversion factors are the following.*1947

*You can have -906 kilojoules for every 4 moles of ammonia.*1950

*You can have -906 kilojoules for every 5 moles of O _{2}, etc.*1959

*We have seen this before where we are using the stoichiometric coefficients directly into a conversion factor.*1966

*Again I want to remind you, this is all the same material.*1973

*We are just using a slight twist.*1976

*But it is completely within your realm of what you know how to do already.*1979

*Let's go ahead and solve the problem; 155 grams of ammonia.*1985

*Again this is stoichiometry because we are asked to relate any two items in the chemical reaction.*1990

*One of those items is energy.*1995

*Times... we have to get everything into moles.*1998

*Remember that is always our first step.*2000

*Divided by roughly 17 grams of ammonia.*2002

*Then I am going to use my conversion factor.*2005

*That is going to be -906 kilojoules divided by 4 moles of ammonia.*2009

*When all is said and done, I am left with -2065 kilojoules as our final answer.*2016

*Again this is how you deal with what is called heats of reaction.*2025

*They are typically given to the far right of a chemical reaction.*2027

*The next scenario with calculating enthalpy is what we call standard enthalpies of formation.*2035

*Standard enthalpies of formation is going to be symbolized as ΔH _{f}.*2042

*The formal definition is the energy change associated with the formation of a compound*2047

*from its constituent elements in their standard states at 1 atm in 25 degrees Celsius.*2053

*Let's take apart this definition; let's talk about water vapor.*2060

*The elements that are involved are of course hydrogen and oxygen.*2067

*We have already discussed these.*2072

*Hydrogen exists as a diatomic gas under standard conditions.*2074

*Oxygen exists as a diatomic gas under standard conditions.*2078

*The reaction representing the formation of water vapor under standard conditions is H _{2} gas plus 1/2O_{2} gas goes on to form water vapor.*2084

*This reaction is -241.8 kilojoules for the heat of formation.*2097

*Every textbook always has an appendix.*2104

*You are going to find in the appendix a very comprehensive table of heats of formation for many many compounds.*2108

*You probably don't have to memorize them.*2118

*You will either be given the information or be able to look it up in the appendix when you do problems.*2120

*There are two notes that you want to be aware of now.*2127

*ΔH of formation and all ΔH values, ΔH reaction, ΔH formation, etc, they are all affected by the reaction, the coefficients.*2132

*ΔH is proportional to the coefficients.*2145

*If I take the same reaction and I double it, I multiply it through by 2,*2148

*the ΔH value is going to get increased proportionally.*2154

*If I take the reverse reaction though, the sign is also going to flip.*2160

*For the reverse reaction, it is the same magnitude but opposite sign.*2166

*How do we go ahead and use ΔH of formation to solve ΔH of the reaction?*2174

*A typical problem is represented as the following.*2183

*Calculate ΔH of the reaction for the following given the heats of formation.*2185

*Again if you are not given it, you need to be able to look it up in the appendix.*2190

*This is going to be a very easy equation to remember.*2196

*ΔH of the reaction equals the following.*2202

*Summation n, ΔH of formation of all products, minus summation n ΔH of formation of all reactants,*2206

*where n is going to be the stoichiometric coefficient from the balanced chemical equation.*2212

*That is why again it is so important to make sure your chemical equation is balanced.*2223

*Or else this problem is going to be trouble.*2227

*We are going to take the products first.*2232

*You notice that only Al _{2}O_{3} and Fe_{2}O_{3} are given to you.*2235

*What is not given is iron solid and aluminum solid.*2241

*The reason why is because they are the elements in their standard state--aluminum solid monatomic and iron solid monatomic.*2244

*What you want to remember is the following.*2254

*That for elements in their standard states, ΔH of formation is formally going to be equal to zero.*2256

*That is what you will always want to remember.*2273

*That requires you to know those elements in their standard state.*2277

*Again most of them are monatomic, except the following.*2281

*H _{2} gas, N_{2} gas, F_{2} gas, Cl_{2} gas, Br_{2} liquid, I_{2} solid.*2286

*Also of course remember mercury is going to be a liquid under standard conditions.*2301

*Again these are the ones that you want to remember.*2309

*These are going to be zero ΔH of formation.*2312

*The rest are typically going to be monatomic solids.*2316

*Let's go ahead and do this; we are going to look at the products.*2324

*That is going to be aluminum oxide.*2327

*That is going to be -1675.7 kilojoules per mole.*2329

*My coefficient for aluminum oxide here is just 1.*2336

*The other product, iron, is just 0.*2342

*Remember summation tells me I am going to add everything together.*2347

*That takes care of my products.*2350

*I am now going to subtract this with the same equation for the reactants.*2351

*This is going to be for iron oxide.*2360

*Minus 1 mole of iron oxide times -824.2 kilojoules per mole.*2362

*Plus 0 for aluminum solid; summation products minus summation reactants.*2371

*We are going to get -852 kilojoules for the ΔH of the reaction.*2381

*Again it is a very easy equation to know; it really sticks out.*2387

*This is a summation equation again for ΔH of formation values under standard conditions.*2392

*Yet another way to calculate enthalpy is from a series of reactions.*2400

*We introduce now a very important law from thermodynamics is what we call Hess's law.*2406

*Hess's law tells us basically the following.*2412

*That the whole is equal to the sum of the parts.*2414

*That is, the change in enthalpy for an overall reaction equals to the sum of the enthalpy changes for each individual step.*2417

*Again the whole is equal to the sum of the parts.*2425

*Imagine we have a reaction through a three step process.*2432

*If I am to combine all these reactions, all we are going to do is combine reactions.*2437

*We are going to add up everything.*2442

*We are going to sum all reactants on one side.*2444

*We are going to sum all products on one side.*2449

*We are basically adding chemical reactions together.*2453

*We go ahead and do that; let's see what we get.*2457

*2NO gas plus O _{2} gas plus N_{2} gas plus O_{2} gas plus 2N_{2}O gas goes on to form all of the products.*2460

*2NO _{2} gas plus 2NO gas plus 2N_{2} gas plus O_{2} gas.*2477

*You want to treat reactions basically as any algebraic equation.*2490

*Now we can go ahead and cancel like terms on opposite sides of the equation.*2497

*Let's see what we can get rid of.*2505

*One O _{2} here is gone; that means one O_{2} here is gone.*2510

*This N _{2} is gone which means we only have one N_{2} remaining right here.*2514

*2NO on the left is going to cancel with the 2NOs on the right.*2519

*We have a net balanced combined reaction of 2N _{2}O gas plus O_{2} gas going on to form 2NO_{2} gas plus N_{2} gas.*2523

*This is our overall net equation.*2539

*Basically what Hess's law is telling me is that when I combine these three reactions,*2543

*ΔH of the overall reaction is simply equal to the sum of the individual ΔHs.*2549

*ΔH _{1} plus ΔH_{2} plus ΔH_{3}; again this is Hess's law; very good.*2556

*The final way to calculate enthalpy is what we call a coffee cup calorimeter.*2566

*It is called coffee cup calorimetry because the calorimeter is good old Styrofoam.*2578

*Styrofoam, we know to be a very inexpensive readily available insulator.*2584

*This type of calorimetry is going to more or less guarantee constant pressure conditions*2590

*because it is going to be whatever the atmospheric pressure is at the time.*2596

*q of the reaction is going to be equal to ?q of the solution.*2603

*Basically for coffee cup calorimetry, you have a coffee cup with a aqueous solution usually with your reactant in here.*2608

*All it is is the thermometer inserted and usually some type of lid is here used.*2619

*You see that this reaction is pretty much going to always be at whatever the atmospheric pressure is going to be.*2627

*Coffee cup calorimetry is not used for combustion reactions but instead for aqueous reactions.*2632

*That is a big difference right there; aqueous reactions, not combustion.*2641

*The system is equal to q of the reaction.*2647

*If the reaction is giving off energy, where is it going to go?*2650

*It is basically going to go to the aqueous environment, to q of the solution.*2654

*We are basically assuming that the Styrofoam material, because it is such a good insulator,*2661

*it is not involved in the transfer of energy; assuming Styrofoam not involved.*2667

*That is why you don't see any term for the Styrofoam material.*2675

*Basically all of the energy that is given off by the reaction or absorbed by the reaction*2681

*is going to come from or go to the aqueous solution that the reaction is placed into.*2686

*q of reaction is equal to ?q of solution.*2693

*In other words, q of the system equals ?q of the surroundings.*2696

*We know the equation for q; we have introduced it before.*2703

*It is the MCAT equation; that is equal to ?mcΔT.*2706

*Let's go ahead and look at the following reaction.*2711

*Silver nitrate is reacting with hydrochloric acid to form silver chloride and nitric acid.*2714

*You have 50 milliliters of 0.1 molar silver nitrate and 50 milliliters of 0.1 molar HCl.*2720

*You mix it in a coffee cup calorimeter.*2727

*The temperature rises from 23.4 to 24.21; calculate ΔH of the reaction.*2730

*You are told that the density of the solution is 1 gram per milliliter.*2738

*You are given the heat capacity of the solution.*2742

*We are assuming it to be that of water.*2745

*We are combining 50 milliliters of one reactant with 50 milliliters of another.*2748

*Both of them are aqueous.*2753

*We are assuming that the density of solution is roughly 1.*2755

*We can get the mass of the solution.*2757

*1.00 gram per milliliter times the 100 milliliters of the solution is going to give me 100 grams of the solution.*2761

*We know that q of the reaction is equal ΔH of the reaction*2773

*because this is constant pressure after all; coffee cup calorimetry.*2781

*That is going to be equal to ?q of the solution.*2785

*That is equal to -100 grams times the heat capacity which is given.*2790

*Times the ΔT which is going to be 24.21 minus 23.40 degrees Celsius.*2796

*We are going to get an answer of -339 kilojoules.*2804

*Once again ask yourself, does it make sense that my sign of ΔH here is negative?*2809

*You are told that the temperature increases, which means it is warm to the touch.*2815

*Which means it is going to be giving off energy; this is an exothermic process.*2819

*Yes, our sign of ΔH had better be negative; this answer does absolutely make sense.*2823

*That is how we do coffee cup calorimetry.*2830

*Let's now take a moment and summarize the key points from this opening introductory chapter on thermodynamics.*2834

*We learn the first law of thermodynamics is basically a restatement for the law of conservation of energy.*2841

*That energy can neither be created nor destroyed.*2846

*Everything you have is not going to be lost at all; it is conserved.*2849

*However we introduced the mathematical form of the first law, where ΔU is equal to q plus w.*2857

*This is going to be true for any closed isolated system.*2865

*Basically after our brief introduction, we spent a lot of time doing calculations, calculating these different parameters of thermodynamics.*2869

*To calculate these parameters, you must consider the conditions.*2877

*Do we have volume constant?--do we have pressure constant?*2880

*We saw that the equations are different for each of the situations.*2885

*We learned a very important experimental technique called calorimetry.*2890

*Calorimetry is used to determine the energy change of a chemical reaction under either constant volume or constant pressure conditions.*2894

*Constant volume conditions, we did this using a bomb calorimeter.*2905

*Constant pressure conditions, we did this using a coffee cup calorimeter.*2915

*Finally we spent a great deal of time on enthalpy.*2920

*We learned many many ways of calculating enthalpy.*2924

*Remember enthalpy is essentially heat transferred under the constraint of constant pressure.*2927

*Let's now spend some time on a pair of sample problems.*2935

*Here we are given a chemical reaction; you are given a ΔH of reaction.*2941

*Right away I know that this is going to be a type of a stoichiometry problem.*2946

*I am going to be incorporating the ΔH of the reaction as a conversion factor.*2951

*Again we are going to be incorporating this as a conversion factor with of course the coefficients.*2961

*What mass of C _{4}H_{10} is required to produce 1500 joules of heat?*2972

*Produce, what does produce mean?*2980

*Produce means it is going to be giving off heat.*2982

*What is my sign?--is it positive or negative?*2987

*It is going to be negative; -1500 joules.*2990

*We are going to be using ΔH of the reaction as a conversion factor.*2997

*The reactant we are interested in is a C _{4}H_{10}; this equation is already balanced.*3000

*1 mole of C _{4}H_{10} over -26 of 58000 joules.*3006

*I am then going to go ahead and multiply this by the molar mass of the compound, 58 grams roughly over 1 mole.*3018

*When we get this, we get 32.7 grams of C _{4}H_{10} required.*3029

*Once again this was a nice stoichiometry problem where we used ΔH of the reaction as a conversion factor with the stoichiometric coefficients.*3039

*Pay attention to this.*3051

*If we had forgotten the negative sign, we would get then a negative mass, which physically doesn't exist.*3054

*Right away again always check if your signs are correct.*3067

*Again if we had not placed this as -1500 joules, we would have gotten a negative mass, which is not going to make sense.*3072

*Let's now move on to the final sample problem two.*3086

*Consider the following series of chemical reactions.*3090

*Right away you see three reactions; ΔH 1, 2, and 3.*3092

*You pretty much know this is a Hess's law type of problem.*3097

*The question asks us to calculate ΔH of the reaction for N _{2}O gas plus NO_{2} gas going to 3NO gas.*3100

*If I try to add up these reactions right away, I do not get the balanced overall reaction.*3113

*This means that I must manipulate my previous three reactions.*3118

*Must manipulate; must change at least one reaction.*3123

*Remember we went over the two ways to manipulate a chemical reaction.*3134

*We can either multiply through by some type of factor; multiply through.*3139

*Or we can reverse the reaction.*3148

*Remember if you multiply through all of the coefficients by a factor, ΔH gets multiplied by the same factor.*3154

*If you reverse a reaction, remember the sign of ΔH also changes.*3170

*Those are the two warnings that you must remember whenever manipulating a chemical reaction.*3179

*I see that NO _{2} here is on the reactant side.*3188

*But the only place where NO _{2} appears is on the product side here.*3191

*For the first reaction, I am going to go ahead and I am going to reverse it.*3195

*2NO _{2} goes to 2NO plus O_{2}.*3199

*In addition, I only see one NO _{2} in the whole problem, but I have two here.*3206

*Which means I am going to have to multiply through by 1-1/2 to get rid of that 2, to reduce it to 1.*3214

*That means I have -1-1/2 ΔH _{1}.*3221

*ΔH _{2} is just N_{2} plus O_{2} going to 2NO.*3230

*That is going to remain exactly the same.*3236

*Finally the third reaction is 2N _{2}O going to N_{2} plus O_{2}.*3238

*That looks good right now because you see that N _{2}O is on the reactant side in the overall reaction.*3249

*Here it is on the reactant side in this reaction.*3254

*But what is the difference?--it is the factor of 2 again.*3257

*Now we are going to multiply through by 1-1/2 again to get rid of the 2.*3260

*That going to be 1-1/2 ΔH _{3}.*3271

*When all is said and done, when I add up all three of these equations that we have changed,*3278

*that we have made changes to some, we get the overall reaction, N _{2}O plus NO_{2} going to 3NO.*3286

*Hess's law tells me that my ΔH of the reaction is simply going to be equal to the sum of the individual ΔHs.*3295

*-1-1/2 ΔH _{1} plus ΔH_{2} plus 1-1/2 ΔH_{3}.*3304

*This problem is a great illustration of Hess's law.*3314

*That was our first lecture on thermodynamics.*3319

*Thank you for using Educator.com; I will see you all next time.*3324

0 answers

Post by Abdullah Alqahtani on June 24, 2017

YOU are very bad lecturer

1 answer

Last reply by: Professor Franklin Ow

Wed Jul 20, 2016 11:02 PM

Post by Magic Fu on July 20, 2016

In 15:44. Why Styrofoam has smaller heat capacities, and still need lot of heat? Shouldn't it be the opposite instead?

0 answers

Post by Peter Ke on September 16, 2015

At 45:00, why did you multiply 1.00 g/mL x 100mL because 1 Liter = 1000 mL so I was thinking that you should multiply it by 1000 and get 1000 grams of solution?

Also, when you got -339 kJ shouldn't it be 339 J without the kilo? Because if you were to set up the equation -100g(4.18 J/gC)(24.21 C - 23.40 C), don't the grams and celsius cancel each other out and you would be left with J?

Please correct me if I'm wrong.

0 answers

Post by Peter Ke on September 15, 2015

At 23:12. I understand how you got Qreaction = - Qbomb, because you subtract Qbomb on both side but I don't understand how you got -Cbomb(DeltaT). Please explain. Thx!

0 answers

Post by Peter Ke on September 15, 2015

At 16:04, can you show me how you got 30.7, cause the answer I got was like 495.1 which totally made no sense. So, please show me how you got 30.7. Thx

0 answers

Post by Peter Ke on September 15, 2015

At 8:59, why 488 is negative? Please explain it in a easier way. Thx

1 answer

Last reply by: Professor Franklin Ow

Mon Jun 1, 2015 5:41 PM

Post by Ivan de La Grange on May 31, 2015

For Sample problem 1, shouldn't it be .0327 grams?

2 answers

Last reply by: Denise Bermudez

Sun Mar 8, 2015 10:04 AM

Post by Denise Bermudez on March 2, 2015

hi professor

can you please explain or re-do how you got the answer of 30.7 degrees celsius. I set the equation correctly but i cant figure out how you got your answer with it. The part that confused me was the *4.18 J/ g celsius.

thanks!

0 answers

Post by Saadman Elman on January 31, 2015

Please look at 13:00 - 14:00 min, You did first part of the problem but you didn't do the second part of the problem. The question was asked Calculate delta U for burning of fuel if the system gives off 875 joule. When all is said and done, the answer that you will end up is -998.6 Joule which is the answer for delta you. You skipped that part. Please verify my answer.

0 answers

Post by Saadman Elman on January 9, 2015

Correct me if i am wrong. You said Insulator such as Styrofoam has small heat capacities. Since they retain the heat so i think they have higher heat capacities.

1 answer

Last reply by: Ryan Lozon

Sun Dec 21, 2014 11:34 AM

Post by Ryan Lozon on December 21, 2014

You mean insulators such as styrofoam having larger heat capacities than metals, right? Because they require more energy to raise one gram by 1 Celsius?

3 answers

Last reply by: Professor Franklin Ow

Mon Aug 4, 2014 11:31 PM

Post by brandon joyner on July 1, 2014

Why does 725 have a negative sign?

2 answers

Last reply by: Peidong, He

Sun Mar 16, 2014 11:53 AM

Post by Peidong, He on March 15, 2014

should insulators, such as Styrofoam, have a higher specific heat capacity?

0 answers

Post by KyungYeop Kim on July 9, 2013

Regarding a specific situation in which a nonspontaneous reaction under standard conditions becomes spontaneous at lower temperature, how can I describe this phenomenon in relation to enthalpy, entropy, and free energy? and how can I explain it in terms of temperature change? I've succeeded so far in determining that Î”G>0 since it's nonspontaneous under standard conditions, but what about Î”H and Î”S?

Given the equation = Î”G = Î”H*TÎ”S; I think the fact that temperature(T) can go either from positive to negative or negative to negative seems to confuse me. Are we assuming, in saying lowering temperature, that the T goes from negative(-) to negative(-)?

I know it's a complex problem, and I apologize if I'm asking too much, but I would like to know what the answer is and why. Thank you always!

0 answers

Post by KyungYeop Kim on July 9, 2013

I have a question about effective nuclear charge. As you go down a group in the periodic table, why is it that the effective nuclear charge decreases? from what I know, is it true that as the attraction decreases down the group, it somehow counterbalances the increase in nuclear charge? I'm confused.