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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### Principles of Chemical Equilibrium

- Chemical equilibrium is a dynamic process, having equal rates of the forward and reverse reactions.
- The equilibrium constant describes the extent to which products are formed over reactants.
- Le Chatelier’s Principle states that when a system at equilibrium is disturbed, it will react to restore equilibrium.
- ICE tables can be used to calculate equilibrium concentrations or pressures.

### Principles of Chemical Equilibrium

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 Equilibrium Constant
- The Equilibrium Constant cont'd
- The Equilibrium Concentration and Constant for Solutions
- The Equilibrium Partial Pressure and Constant for Gases
- Relationship of Kc and Kp
- Heterogeneous Equilibria
- Manipulating K
- Manipulating K cont'd
- The Reaction Quotient Q
- Le Chatlier's Principle
- Problem-Solving with ICE Tables
- Problem-Solving with ICE Tables cont'd
- Problem-Solving with ICE Tables cont'd
- Summary
- Sample Problem 1: Calculate the Equilibrium Constant
- Sample Problem 2: Calculate The Equilibrium Concentration

- Intro 0:00
- Lesson Overview 0:08
- Introduction 1:02
- The Equilibrium Constant 3:08
- The Equilibrium Constant
- The Equilibrium Constant cont'd 5:50
- The Equilibrium Concentration and Constant for Solutions
- The Equilibrium Partial Pressure and Constant for Gases
- Relationship of Kc and Kp
- Heterogeneous Equilibria 8:23
- Heterogeneous Equilibria
- Manipulating K 9:57
- First Way of Manipulating K
- Second Way of Manipulating K
- Manipulating K cont'd 12:31
- Third Way of Manipulating K
- The Reaction Quotient Q 14:42
- The Reaction Quotient Q
- Q > K
- Q < K
- Q = K
- Le Chatlier's Principle 17:32
- Restoring Equilibrium When It is Disturbed
- Disturbing a Chemical System at Equilibrium
- Problem-Solving with ICE Tables 19:05
- Determining a Reaction's Equilibrium Constant With ICE Table
- Problem-Solving with ICE Tables cont'd 21:03
- Example 1: Calculate O₂(g) at Equilibrium
- Problem-Solving with ICE Tables cont'd 22:53
- Example 2: Calculate the Equilibrium Constant
- Summary 25:24
- Sample Problem 1: Calculate the Equilibrium Constant 27:59
- Sample Problem 2: Calculate The Equilibrium Concentration 30:30

### 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: Principles of Chemical Equilibrium

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

*Today's lecture from general chemistry is on the principles of chemical equilibrium.*0003

*Let's go ahead and take a look at the lesson overview.*0010

*We will first start off with a brief introduction and then get in right to*0012

*the core of everything which is basically what we call the equilibrium constant.*0017

*We are going to first define this equilibrium constant and then go into different*0021

*types of equilibria including what we call homogeneous equilibria and heterogeneous equilibria.*0026

*We will then go and see how we can change the value of K _{eq}.*0033

*We will also introduce something called the reaction quotient.*0039

*We then jump into a very fundamental principle in all of general chemistry.*0042

*That is called Le Chatelier's principle.*0046

*After that we will get into the quantitative part of the chapter which involves what we call ICE tables.*0050

*We will wrap everything up with a very brief summary followed by a pair of sample problems.*0056

*Chemical equilibrium, exactly what this is.*0065

*This refers to the simultaneous occurrence of a forward and a reverse reaction at the same rate.*0068

*We see that equilibrium is a dynamic process, not a static process.*0075

*For example, let's say you had two beakers here; these beakers are closed.*0079

*Let's say that this one beaker had H _{2}O in it.*0086

*Let's just say this is time zero.*0091

*At time zero, I have nothing but pure liquid water.*0094

*But we know this from everyday experience inside a water bottle.*0097

*If you let some time progress, we notice that the water level is going to drop a little.*0101

*That is because some of the liquid water has entered the gas phase.*0108

*But there is also going to be a point in time where this water level is not going to drop forever.*0116

*It is going to reach a minimum.*0125

*It reaches a minimum because as soon as the vaporization process occurs, the condensation process also occurs.*0129

*We say for this case that the rate of evaporation is equal to the rate of condensation.*0142

*If we were to write this out in a chemical reaction, this would be H _{2}O liquid.*0158

*Now we introduce a new type of reaction arrow which is this.*0164

*That goes to H _{2}O gas; this is our equilibrium arrow.*0169

*It basically shows that the forward direction is happening simultaneously with the reverse direction.*0180

*Let's now get into how we can represent equilibrium numerically.*0190

*Consider the following reaction.*0198

*Small a big A plus small b big B, equilibrium sign and then small c big C small d big D.*0199

*In this reaction, let the lowercase letters represent the stoichiometric coefficients.*0207

*You know the moles after we balance the chemical equation.*0222

*It turns out that after a chemical reaction has reached equilibrium, it is experimentally determined that the ratio of*0226

*product to reactant concentration raised to the stoichiometric coefficients is actually constant at any given temperature.*0235

*Basically the concentration of C raised to the c power times the*0243

*concentration of D raised to the d power over the concentration of A*0249

*raised to the a power times the concentration of B raised to the b power.*0256

*This whole ratio of products to reactants raised to the coefficients is equal to some constant that we call K.*0262

*Sometimes you are going to see this called K _{eq}.*0272

*This is formally what we call the equilibrium constant.*0275

*The only thing that can change the value of K _{eq} is temperature.*0285

*The equilibrium constant is temperature dependent.*0290

*We can typically represent K in two different ways.*0302

*K _{c} is when molarities are used; K_{p} is when partial pressures are used.*0307

*Every textbook is a little different.*0326

*But for the partial pressures, the typical units are going to be atmosphere or bar.*0328

*Once again K _{c} and K_{p} are just the equilibrium expressions when*0338

*molarities and partial pressures are used for K _{c} and K_{p} respectively.*0343

*For solutions, the equilibrium constant can be expressed in units of molarity just like we discussed.*0353

*However it turns out that the equilibrium constant is unitless.*0361

*K _{c}, how do we go ahead and do that?*0364

*Mole over liter raised to some power divided by mole over liter raised to some power.*0368

*It turns out that the equilibrium expression is always referenced to 1 molarity.*0378

*This is really moles over liter per 1 molar.*0385

*This is moles over liter per one molar raised to the y power and raised to the x power.*0394

*As you can see that after this is done, we see that all units cancel.*0402

*K _{eq} is actually one of the few unitless values in all of your general chemistry studies.*0411

*That is K _{c}.*0420

*It turns out that if we choose to do our problem with K _{p}, maybe this is going to be in atmospheres.*0422

*This is also relative to 1 atm, raised to some power divided by atm raised to some power.*0429

*We see it again that the units cancel; K _{eq} again is unitless.*0438

*What is the relationship between K _{c} and K_{p}?*0451

*K _{c} and K_{p} are directly proportional to each other.*0455

*Depending on the reaction, sometimes they are nearly identical.*0461

*But the exact relationship between the two is the following where*0465

*K _{p} is equal to K_{c} times RT over Δn where Δn is*0469

*equal to the moles of gaseous product minus the moles of gaseous reactant.*0479

*Once again K _{p} and K_{c} can be very similar.*0489

*However they are not quite the same thing.*0495

*You should really check with your instructor to see what he or she prefers.*0499

*For chemical equilibria, we are usually dealing with very small concentrations.*0507

*Because of this, we assume that pure solids and pure liquids are going to remain relatively unchanged.*0512

*If you have the following reaction, A solid plus B aqueous goes on to form C aqueous plus D solid,*0520

*it is only the aqueous species that are going to affect the equilibrium, affect K _{eq}.*0532

*Hence pure solids and pure liquids are going to be assigned an arbitrary value of 1 when incorporated into the expression for K.*0546

*That is they have no effect.*0555

*K _{c} for this reaction here would be simply the concentration of C times 1 divided by the*0557

*concentration of B times 1 which is just equal to the concentration of C over concentration of B.*0566

*Once again pure liquids and pure solids do not appear in the expression for K, in the K _{eq} expression.*0573

*Again this is going to be specifically for heterogeneous equilibria.*0593

*What are some ways where we can manipulate K?*0599

*The first way of manipulating K is by multiplying an entire chemical equation by a factor.*0602

*For example, let's take A aqueous plus 2B aqueous goes on to form 3C aqueous.*0608

*In this case, K as you see is equal to the concentration of C cubed*0621

*divided by the concentration of A times the concentration of B squared.*0627

*Let's go ahead and take this chemical reaction and multiply everything through by 2.*0633

*2A aqueous plus 4B aqueous goes on to form 6C aqueous.*0638

*It turns out therefore that this K _{new} is going to be equal to concentration of C to the sixth*0647

*divided by the concentration of A squared times the concentration of B raised to the fourth power.*0654

*We see very nicely that K _{new} is simply equal to the original K_{c} squared.*0661

*The rule of thumb is the following.*0670

*That when multiplying a reaction by a factor, K is going to be raised to that factor.*0672

*K is raised to this factor.*0695

*That is the first way of algebraically manipulating K.*0699

*Once again this is by multiplying through a chemical reaction by a factor.*0702

*The second way is by taking the reverse reaction.*0709

*For example, let's take not A plus 2B going to 3C*0712

*but 3C aqueous goes on to form A aqueous plus 2B aqueous.*0717

*In this case, K is equal to concentration of A times the concentration of B squared divided by the concentration of C cubed.*0726

*We see that this is going to be 1 over k _{forward}.*0739

*k _{reverse} is simply the reciprocal of k_{forward}.*0745

*The third way of changing K is by adding a series of individual chemical reactions together to form a net balanced chemical equation.*0753

*Let's go ahead and see.*0760

*For example, A aqueous plus B aqueous goes on to form C aqueous.*0765

*C aqueous plus D aqueous goes on to form E aqueous.*0777

*Let's go ahead and add these two together.*0788

*When we add these two together, we are going to get A aqueous plus B aqueous plus D aqueous goes on to form E aqueous.*0790

*You see that C is going to be cancelled out because it is going to be formed and consumed simultaneously.*0803

*What we want to look at now are the expressions for k for each of these.*0815

*Here k, I will call this k _{1} for reaction one, is equal to the*0819

*concentration of C divided by the concentration of A times the concentration of B.*0823

*Here k _{2} is equal to the concentration of E divided by the concentration of C times the concentration of D.*0830

*Here I will call this third one k _{net}.*0841

*That is equal to the concentration of E divided by the concentration of A times the concentration of B times the concentration of D.*0845

*We see that when we add individual reactions together to get a net reaction, the equilibrium constant*0855

*of the net reaction is simply equal to the product of each individual equilibrium expression constant.*0864

*That is the third way of algebraically manipulating K _{eq}.*0874

*Let's now go on to another aspect of the equilibrium constant.*0884

*It is what we call the reaction quotient.*0887

*The equilibrium constant is good when we actually have the values at equilibrium.*0890

*But what happens if we use values not at equilibrium?*0894

*When concentrations or pressures are inserted into the expression for K that are not at equilibrium,*0899

*the ratio of products to reactants is now what we call the reaction quotient symbolized Q.*0906

*Q is going to be equal to products raised to some power divided by reactants raised to*0911

*some power except that these molarities and partial pressures are not at equilibrium.*0916

*The significance is the following.*0942

*Q can be used to predict which way a reaction will shift to reobtain a state of equilibrium.*0944

*Basically we are going to have the following general rules.*0976

*If Q is greater than K, that means we have too much product, not enough reactant.*0980

*We are going to shift left.*0987

*If Q is less than K, we have too much reactant relative to product.*0992

*We are going to shift right.*1001

*Finally if Q is identical to K, we are at a equilibrium state.*1005

*Shifting left is the same as making more reactant.*1015

*Shifting right is the same as making more product.*1028

*At equilibrium, there is no net change; neither direction is favored over the other.*1036

*Once again Q can be very useful for determining what direction a reaction will shift if any.*1044

*Now we have come into one of the most fundamental principles from all of general chemistry.*1055

*This is called Le Chatelier's principle.*1060

*Le Chatelier's principle tells us that when a system at equilibrium is disturbed,*1063

*it will react to counteract the disturbance in an attempt to restore equilibrium.*1068

*One of the best examples we can think of is from us.*1078

*When we bleed, when we lose blood, what is the natural thing that our body does in order to counteract this?*1082

*Your body is going to try to make more blood.*1091

*When we lose blood, our bodies try to make more in order to counteract the lost.*1092

*This is a nice example of Le Chatelier's principle.*1112

*But now we are going to apply this to chemical reactions.*1114

*In Le Chatelier's principle, you notice that there is the word disturb.*1119

*There are several ways of disturbing a chemical system that is at equilibrium already.*1124

*Number one is a change in reactant or product concentration.*1127

*Number two if we are dealing with gases, it is going to be a change in reactant or product partial pressure.*1132

*The third one is the change in reaction volume.*1137

*The fourth one is going to be a change in temperature.*1140

*Let's now study each of these.*1144

*Basically if the concentration of reactant goes up, then we are going to shift away from it.*1150

*We are going to shift right.*1163

*If the concentration of product goes down, we don't have enough of it.*1165

*We are going to shift right; that is one situation.*1170

*If the partial pressure of a reactant goes up, partial pressure is just the same as concentration.*1178

*We know from ideal gas law that pressure is proportional to amount.*1189

*If partial pressure of the reactant goes up, we are going to shift right.*1194

*If the partial pressure of the product goes down, we are going to shift right.*1204

*For temperature, I am going to do this on the last slide because I am running out of room right now.*1215

*I will come back to changes in temperature and also to change in vessel volume.*1228

*But now let's get into some problem solving with some ICE tables.*1235

*We will now approach chemical equilibrium from a quantitative view.*1238

*An ICE table allows for one to study a chemical system at three points in time.*1241

*Basically what are all reactant and product initial amounts?*1246

*That is what the I stands for.*1250

*During the reaction on the way to equilibrium, what are their changes in concentration/pressure?*1252

*That is what the C stands for.*1256

*Finally what are their final volumes once equilibrium has been achieved?*1258

*That is what the E stands for.*1262

*We are going to look at this right now.*1263

*Consider 2 water gas goes to 2H _{2} gas plus O_{2} gas.*1266

*K is equal to 2.4 times 10 ^{-5} at some temperature.*1271

*At equilibrium, the concentration of H _{2}O gas is 0.11 molar.*1275

*The concentration of H _{2} gas is 0.019 molar; calculate O_{2} at equilibrium.*1279

*The very first thing we want to do is set up our ICE table.*1285

*2H _{2}O gas goes on to form 2H_{2} gas plus O_{2} gas.*1289

*What I like to do, I just like to set up the letters I-C-E right underneath it.*1296

*We just fill in this table right now.*1300

*You are told that at equilibrium, the water is 0.11 molar and that the H _{2} is 0.019 molar.*1303

*We don't know what O _{2} is.*1313

*We can just call that the concentration of O _{2} at equilibrium.*1315

*We know the expression for K is equal to 2.4 times 10 ^{-5}.*1322

*That is going to be equal to the concentration of H _{2} squared times*1328

*the concentration of O _{2} divided by the concentration of H_{2}O squared.*1332

*That is going to be equal to 0.019 squared times the concentration of O _{2} at equilibrium divided by 0.11 squared.*1338

*When all is said and done, the concentration of O _{2} at equilibrium is going to be 8.0 times 10^{-4} molar.*1349

*Don't forget the units.*1359

*You notice that we didn't have to fill in the rest of the table*1363

*because we were already at halfway there to the equilibrium values.*1366

*This is a nice usage of the ICE table.*1371

*Let's go ahead and do another example though.*1374

*A 1 liter flask was filled with so much of SO _{2} and so much NO_{2} at some temperature.*1377

*After equilibrium was reached, 1.3 moles of NO was present.*1382

*The reaction is the following; calculate the value of K _{c} at this temperature.*1385

*The problem wants us to calculate K _{c} at this temperature.*1390

*We are going to go ahead and set up the ICE table--I, C, and E.*1396

*SO _{2} here is going to be 2.00 molar initially.*1406

*O _{2} is going to be 2.00 molar also.*1411

*After equilibrium was reached, 1.30 moles of NO gas was present.*1415

*That is what goes right here, 1.30 molar.*1419

*Let's fill in the rest of the table.*1426

*If no initial values are mentioned of SO _{3} and NO, it is safe to assume that they are zero.*1427

*Let's go ahead and do the change.*1434

*The SO _{2} is going to go down by some amount x.*1438

*O _{2} is going to go down by some amount x.*1441

*SO _{3} goes up by some amount x.*1443

*O is going to go up by some amount x.*1452

*We notice because these are all 1:1 mole ratio.*1457

*At equilibrium, it is just going to be the addition of the two rows added together; sum.*1465

*That is going to be 2.00 minus x, 2.00 minus x, x, and x.*1472

*But guess what? x is 1.30 molar because we were told that in the beginning; that is very nice.*1479

*Therefore this is 1.30 molar; this is 0.70 molar; this is 0.70 molar.*1485

*We know enough to calculate the value of K _{c}.*1494

*K _{c} is going to be 1.30 squared divided by 0.70 squared.*1498

*That is going to give us an answer of 3.5 for our answer.*1504

*This is how we use the ICE tables to help us come up with the equilibrium constant.*1517

*I want to quickly summarize this and then jump into sample problems.*1526

*Chemical equilibrium is a dynamic process where the rate of forward and reverse reactions is equal.*1531

*Equilibrium constant K quantifies how reactant or product favored a reaction is once it has reached equilibrium.*1536

*If K is between 0 and 1, we say that it is reactant favored.*1543

*If K is greater than 1, we say that it is product favored.*1552

*Le Chatelier's principle states that a system at equilibrium when disturbed will react to counteract the disturbance.*1559

*We already saw the concentration and partial pressures.*1565

*But let's go ahead and see what I left out before; this is now temperature.*1573

*For an exothermic reaction, if the temperature increases, it turns out we are going to shift left.*1579

*If the temperature decreases, we are going to shift right.*1593

*Basically you treat the word heat as a molecule.*1599

*Again if you treat the word heat as a molecule on the product side, it becomes*1609

*a lot more intuitive as you will see what happens when we result in the change.*1616

*Another thing we can disturb an equilibrium with is with the volume.*1622

*If the volume of the reaction vessel goes up, that means the pressure is lower.*1629

*That means we are going to shift to make more gases.*1636

*Shift toward side of reaction with more gas molecules.*1640

*Once again if the volume goes up, we are going to shift toward the side of the reaction with more gas molecules.*1654

*Those are the ways we can disrupt a system that is at equilibrium and see how it is going to counteract the stress.*1662

*Finally ICE tables allow for determination of the reactant and product concentrations/pressures given the equilibrium constant K.*1668

*That is our summary.*1676

*Now let's go ahead and jump into a pair of sample problems.*1678

*Consider 2SO _{3} gas goes to 2SO_{2} plus O_{2}.*1681

*Initially 12 moles of SO _{3} is placed into a 3 liter flask at some temperature.*1685

*At equilibrium, 3 moles of SO _{2} is present.*1689

*Calculate the value of K _{c} at this temperature.*1692

*Let's go ahead and rewrite this.*1696

*2SO _{3} gas goes on to form 2SO_{2} gas plus O_{2} gas.*1698

*Let's go ahead and see what we can fill in, I-C-E.*1707

*Initially 12 moles is placed in a 3 liter flask of SO _{3}.*1712

*That is going to be 4.0 molar.*1716

*At equilibrium, SO _{2} is present, 3 moles; that is going to be 1.0 molar.*1721

*Usually you can fill in two-thirds of the table with just one or two sentences.*1728

*SO _{2} initial is not mentioned; that is zero.*1733

*O _{2} initial is not mentioned; that is zero also.*1735

*We have to pay attention to the coefficents.*1740

*The coefficients tells us the relative amount of reactant and product that is going to be involved.*1742

*Really this is going to be -2x; SO _{2} is going to be +2x.*1747

*O _{2} is just going to be +x.*1752

*At equilibrium, we get 4.0 minus 2x, 1 molar, and x.*1756

*The nice thing about this is that because we know the 2x and the 1 molar, we can conclude that x is 0.5 molar.*1763

*When we do that, we are going to get here 0.5 molar.*1776

*Here for SO _{3}, we are going to get 3.0 molar.*1780

*Great, we now have all of our molarities at equilibrium.*1786

*K _{c} is equal to the concentration of SO_{2} squared times the concentration*1791

*of O _{2} divided by concentration of SO_{3} squared, all equilibrium values.*1797

*This is going to be 0.5 squared times 0.5 divided by 3.0.*1805

*When all is said and done, this is going to give us a K _{c} of 0.056 for this reaction.*1815

*That is another usage of the ICE table.*1827

*Finally the last sample problem is sample problem number two.*1831

*At a particular temperature, K is equal to 3.75.*1834

*If all four gases had initial concentrations of 0.800, calculate their equilibrium concentrations.*1837

*Let's go ahead and set it up--I, C, and E.*1843

*0.800 molar initial, 0.800 molar, 0.800 molar, and 0.800 molar.*1848

*Because we have nonzero amounts of all reactants and products, we don't know which way we are going to shift.*1858

*This is a twist on things; now we have to solve for Q.*1868

*Q is going to tell us which way we are going to shift.*1872

*This is going to be equal to concentration of SO _{3} initial times the concentration of*1875

*NO initial divided by concentration of SO _{2} initial times the concentration of NO_{2} initial.*1882

*That is just going to be equal to 1, 1.00 which is going to be less than K.*1891

*Because Q is less than K, that means we are going to shift to the right.*1900

*Therefore the sign that goes here is going to be ?x.*1907

*This is ?x; this is +x; this is +x.*1911

*We are going to get 0.800 minus x; 0.800 minus x.*1916

*This is going to be 0.800 plus x.*1923

*This is going to be 0.800 plus x; K is equal to 3.75.*1926

*That is equal to 0.800 plus x squared divided by 0.800 minus x squared.*1934

*We can use what we call a perfect square to solve for this.*1946

*The square root of both sides is what we are going to do next.*1951

*When we go ahead and do that, we are going to get 1.93 is equal to 0.800 plus x divided by 0.800 minus x.*1956

*When all is said and done, x is going to be equal to 0.25 molar.*1966

*All we do next is simply plug it back into all of the expressions at equilibrium.*1972

*The concentration of SO _{3} at equilibrium is equal to the concentration of NO at equilibrium.*1978

*That is going to be 0.800 plus 0.25.*1986

*I will let you guys do the sum; that is going to be 1.050 molar.*1992

*The concentration of SO _{2} at equilibrium is equal to the concentration of NO_{2} at equilibrium.*2004

*That is going to be equal to 0.800 minus 0.25.*2012

*We are going to get 0.55 molar for these guys.*2020

*That is use of the ICE table when we have all four initial values present.*2026

*Once again you have to find which way it is going to shift.*2034

*You do that by evaluating Q.*2036

*That is our lecture from general chemistry on the principles of chemical equilibria.*2039

*I want to thank you for your time.*2045

*I will see you next time on Educator.com.*2046

1 answer

Last reply by: Professor Franklin Ow

Thu May 28, 2015 12:35 PM

Post by Smriti Sharan on May 20, 2015

How do you know what it is not at equilibrium? And how do you know when it is?

1 answer

Last reply by: Professor Franklin Ow

Fri Apr 3, 2015 11:02 PM

Post by Sachin Ambulkar on April 3, 2015

If you are given 2SO2(g) + O2(g) (equilibrium arrow) 2SO3(g) + heat, and the system is at equilibrium, increasing the temperature increases the number of moles of which substance(s) when equilibrium is reestablished?

1 answer

Last reply by: Professor Franklin Ow

Fri Apr 3, 2015 11:01 PM

Post by Saadman Elman on January 19, 2015

I agree with Jack Miars, In sample problem 1, the concentration of SO2 at equilibrium is 1 which is already given by the way. You said, it's 0.5. (Typo). Please clarify it. Thanks.

1 answer

Last reply by: Professor Franklin Ow

Sat Aug 16, 2014 4:20 AM

Post by Neil Choudhry on August 14, 2014

How did you get 4.0M for SO3 in Sample problem 1? It says initially it had 12.0 .

1 answer

Last reply by: Professor Franklin Ow

Fri Feb 7, 2014 10:26 AM

Post by Jack Miars on February 5, 2014

Also, on sample problem 1, x=.5 not the concentration of SO2. It should be 1

1 answer

Last reply by: Professor Franklin Ow

Fri Feb 7, 2014 10:25 AM

Post by Jack Miars on February 5, 2014

You didn't come back to how temperature and volume affect Q and K