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

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

### Entropy & Free Energy

- Nature favors states of low energy and high entropy.
- Entropy is related to disorder or chaos, and is the focus of the second and third laws of thermodynamics.
- There are a series of equations that allow us to calculate both entropy andGibb’s free energy under a given set of conditions.

### Entropy & Free Energy

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
- Introduction to Entropy
- Entropy and Heat Flow
- Entropy and Heat Flow cont'd
- The Second Law of Thermodynamics
- Total ∆S = ∆S of System + ∆S of Surrounding
- Nature Favors Processes Where The Amount of Entropy Increases
- The Third Law of Thermodynamics
- Problem-Solving involving Entropy
- Problem-Solving cont'd
- Problem-Solving cont'd
- Problem-Solving cont'd
- Introduction to Gibb's Free Energy
- Gibb's Free Energy cont'd
- Standard Molar Free Energies of Formation
- The Free Energies of Formation are Zero for All Compounds in the Standard State
- Gibb's Free Energy cont'd
- ∆G° of the System = ∆H° of the System - T∆S° of the System
- Predicting Spontaneous Reaction Based on the Sign of ∆G° of the System
- Gibb's Free Energy cont'd
- Summary
- Sample Problem 1: Calculate ∆S° of Reaction
- Sample Problem 2: Calculate the Temperature at Which the Reaction Becomes Spontaneous
- Sample Problem 3: Calculate Kp

- Intro 0:00
- Lesson Overview 0:08
- Introduction 0:53
- Introduction to Entropy 1:37
- Introduction to Entropy
- Entropy and Heat Flow 6:31
- Recall Thermodynamics
- Entropy is a State Function
- ∆S and Heat Flow
- Entropy and Heat Flow cont'd 8:18
- Entropy and Heat Flow: Equations
- Endothermic Processes: ∆S > 0
- The Second Law of Thermodynamics 10:04
- Total ∆S = ∆S of System + ∆S of Surrounding
- Nature Favors Processes Where The Amount of Entropy Increases
- The Third Law of Thermodynamics 11:55
- The Third Law of Thermodynamics & Zero Entropy
- Problem-Solving involving Entropy 12:36
- Endothermic Process and ∆S
- Exothermic Process and ∆S
- Problem-Solving cont'd 13:46
- Change in Physical States: From Solid to Liquid to Gas
- Change in Physical States: All Gases
- Problem-Solving cont'd 15:56
- Calculating the ∆S for the System, Surrounding, and Total
- Example: Calculating the Total ∆S
- Problem-Solving cont'd 18:36
- Problems Involving Standard Molar Entropies of Formation
- Introduction to Gibb's Free Energy 20:09
- Definition of Free Energy ∆G
- Spontaneous Process and ∆G
- Gibb's Free Energy cont'd 22:28
- Standard Molar Free Energies of Formation
- The Free Energies of Formation are Zero for All Compounds in the Standard State
- Gibb's Free Energy cont'd 23:31
- ∆G° of the System = ∆H° of the System - T∆S° of the System
- Predicting Spontaneous Reaction Based on the Sign of ∆G° of the System
- Gibb's Free Energy cont'd 26:32
- Effect of reactant and Product Concentration on the Sign of Free Energy
- ∆G° of Reaction = -RT ln K
- Summary 28:12
- Sample Problem 1: Calculate ∆S° of Reaction 28:48
- Sample Problem 2: Calculate the Temperature at Which the Reaction Becomes Spontaneous 31:18
- Sample Problem 3: Calculate Kp 33:47

### 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: Entropy & Free Energy

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

*Today's lesson from general chemistry is on entropy and free energy.*0003

*This is basically going to be our second and final lecture on thermodynamics.*0009

*As usual we are going to start off with a brief introduction.*0013

*We are going to get into something very important in thermodynamics which is what we call entropy.*0017

*It is going to be related to spontaneous processes.*0023

*After we get into entropy, we are going to go through the second and third laws of thermodynamics followed by problem solving.*0027

*After entropy, we are going to then get into what is called Gibbs free energy followed by definition and problem solving.*0038

*Then we are going to go ahead and conclude with a summary and a pair of sample problems.*0047

*We have consistently said, you have heard me say over and over again that nature favors states of low energy.*0055

*In other words, reactions occur to obtain a lower energy state in the end.*0063

*This is also known as a downhill reaction.*0069

*In other words, nature tends to favor reactions that start in a high energy state and go to a low energy state.*0072

*That is going to be the natural tendency.*0079

*In this lecture, we are going to complete our discussion from thermodynamics and quantify the above statement.*0081

*In the end, we are going to see that nature not only favors*0089

*states of low energy but also states of what we call high entropy.*0091

*Consider the following closed reaction vessels that contain gas molecules.*0099

*Let me go ahead and draw a closed reaction vessel here.*0104

*We can make this one gas particle; the red one can be another gas particle.*0109

*The black one can be another gas particle; we are now going to...*0115

*Because these are gases, gases naturally expand; what can happen is the following.*0120

*What are some possible different configurations for these three gas particles if they naturally expand?*0128

*Situation number one is when all three gas particles are on the same side.*0135

*We can have a second situation when only the blue and the black are together and the red is different.*0140

*We can have another situation where the blue is by itself and the red and black are together.*0148

*You notice that these two situations here are actually the same*0162

*where we have two gas particles on one side versus one on the other.*0169

*It turns out that configuration number two actually has more what we call microstates.*0177

*Once again configuration number two is going to have what we call more microstates.*0196

*We can even do one more to show you how this is possible.*0204

*Now red and blue can be together with black by itself.*0209

*This is also considered a configuration number two.*0215

*Configuration number one, there is only one way to put three particles on the same side.*0220

*Configuration number one has one microstate.*0227

*Because gases expand, we are going to see that configuration number two is going to be the more probable configuration.*0237

*Why?--because there are a higher number of ways to do it.*0246

*There are a higher number of microstates to achieve, configuration number two, where one gas particle is on one side.*0250

*Two gas particles are on the other side.*0257

*We symbolize the term microstate with capital W.*0260

*For configuration number one, W is 1; for configuration number two, W is simply 3.*0264

*We can quantify that this is going to be more probable.*0273

*In other words, this is more likely to occur.*0281

*The equation we use to quantify this is following.*0286

*S is equal to k _{B} times the natural log of W where S is what we call entropy.*0289

*I want you to think of entropy as basically disorder or chaos.*0297

*We see that S is directly proportional to W.*0308

*In other words, as W goes up, so does S.*0313

*What does that mean though?*0318

*We already said that a high value of W means it is going to be more likely to occur*0321

*which says that very high entropy values are going to relate to more likely processes.*0326

*If W large, S is large; process more likely to occur.*0337

*In this equation, we see that there is a proportionality constant called k _{B}.*0351

*k _{B} is what we call the Boltzmann constant.*0356

*That is equal to 1.38 times 10 ^{-23} joules per kelvin.*0363

*Because natural log of W is going to be unitless, that means the units of entropy are also joules per kelvin.*0369

*This entire equation is known as the Boltzmann equation.*0377

*S is equal to k _{B} natural log of W.*0383

*The units of entropy are going to be joules per kelvin.*0386

*Let's go ahead and relate to entropy to what we have already discussed*0393

*in our first lecture from thermodynamics which involved heat flow and energy.*0397

*If you recall from our first thermodynamic lecture, if heat leaves the system, Q is going to be a negative value.*0402

*If heat enters the system, Q is going to be a positive value.*0410

*Entropy is going to be a state function and can be related to Q where ΔS is equal to Q over T.*0415

*Temperature, because you see capital T, has to be in kelvin here.*0426

*Again because this is a state function, what we are interested in is*0433

*a change in entropy just like we talked about for ΔE or ΔU.*0437

*If heat enters the system, Q is going to be positive which means ΔS is going to be positive.*0448

*Let's see if that makes sense.*0460

*Remember if heat enters the system, we are supplying thermal energy for molecules to absorb.*0462

*If they are going to absorb this thermal energy, that gives them more molecular motion such as vibrations.*0470

*That increases the number of states an electron can occupy.*0477

*Basically we are increasing the amount of disorder.*0483

*We are increasing the amount of chaos.*0485

*Yes, it does make sense that ΔS is a positive value.*0488

*ΔS is also related to heat flow during phase changes.*0494

*It is found that Q is equal to n times ΔH of the phase change.*0500

*This is from our first thermodynamics; from thermo I.*0506

*Therefore it follows that ΔS is equal to Q over T which equals to nΔH of the phase change over the temperature.*0514

*What does that mean then?*0525

*That means for endothermic processes such as melting and vaporization, endothermic means Q is going to be a positive value.*0527

*ΔS is positive.*0540

*This agrees with the previous slide where when we put energy in, everything gets a little more chaotic and a little more disorderly.*0542

*For example, as you go for melting, you go from a very orderly state, a solid, to a less orderly state, to a liquid.*0553

*When go from a liquid to a gas, a liquid is less orderly.*0563

*A gas is going to be the most disordered state.*0567

*Everything is a lot more chaotic there.*0572

*ΔS is going to be positive when heat enters the system.*0576

*Here we have the phase change backing up this point.*0580

*In other words then, we can summarize this very nicely.*0586

*If a process is likely to occur, then ΔS is going to be a positive value.*0590

*We say that the process is expected to be spontaneous if ΔS is positive.*0596

*This brings us into our second law of thermodynamics.*0605

*The second law of thermodynamics states that a process will be spontaneous if ΔS total is greater*0609

*than 0 where ΔS total is equal to ΔS of the system plus ΔS of the surroundings.*0616

*In other words, nature is going to favor processes that contribute to the overall entropy of the entire universe.*0623

*In other words, nature favors processes where the amount of entropy or disorder increases as a result of the process.*0631

*We see this occurring in nature quite a bit.*0640

*If you think of a virus, what does a virus naturally do?*0643

*A virus is going to naturally expand and go from cell to cell consuming the resources there and moving on and on.*0648

*If viruses expand, they get more disorderly, more chaotic.*0658

*We see this in herd migration.*0662

*Herd migration, flocks of animals tend to naturally expand from area to area.*0667

*They tend not to stay just in a single location.*0678

*Of course the universe itself, the universe theoretically started off as a single point.*0682

*It is just constantly expanding and expanding.*0689

*Again we see that entropy is seen in several naturally occurring examples.*0694

*Then you may be wondering, if ΔS positive is going to be for a spontaneous process, what does it mean when entropy is zero?*0704

*That brings us into the third law of thermodynamics.*0715

*Zero entropy really does not exist because the conditions to obtain zero entropy do not exist experimentally.*0718

*The third law of thermodynamics states that entropy is zero only for a perfect flawless crystal at absolute zero.*0728

*But absolute zero has not been obtained experimentally.*0736

*Only in the most extreme conditions where we have zero thermo motion, that is molecules cease to move, entropy is completely zero.*0741

*In other words, entropy is always going to be a nonzero quantity in practice.*0750

*Let's now focus on problem solving involving entropy.*0758

*The following are typical types of questions involving entropy.*0762

*Sometimes a question can ask you simply to predict the sign of ΔS given a chemical reaction.*0765

*This is going to be an example of a qualitative process, of a qualitative problem.*0770

*Basically when the process is a physical change, ΔS is going to be positive for*0777

*any endothermic process just like we talked about, including melting, vaporization, and sublimation.*0781

*When you go from a solid to a liquid, from a liquid to a gas, or from a*0788

*solid to a gas, ΔS of these processes are going to be expected to be positive.*0793

*They are opposite processes, are all exothermic which means ΔS is expected to be negative.*0801

*Liquid to solid, gas to liquid, and gas to solid.*0807

*Freezing, condensation, and deposition; again ΔS is expected to be negative.*0812

*Let's go on to another type of problem.*0822

*Another type of problem involves a quantitative problem.*0828

*What you want to do here is pay close attention to the physical states.*0834

*From a solid to a liquid to a gas, entropy increases.*0838

*We have already seen that.*0841

*What you want to keep track of is the change in physical states from reactant to product side.*0843

*Let's go ahead and look at an example.*0850

*Fe solid plus O _{2} gas goes on to form Fe_{2}O_{3} solid.*0856

*We are going to need two of these, three of these, and two of these.*0864

*Excuse me... we will need four irons.*0869

*You notice that we start off with a solid and a gas.*0873

*But on the product side, we have no gases here; no gases on product side.*0878

*If we are eliminating the highest disorderly state which is a gas, we are starting off with it.*0887

*We are not winding with it; we are actually getting less disorderly.*0894

*ΔS for this process expected to be a negative value.*0898

*Let's go on to another type of problem.*0903

*If the reaction involves all gases, what you want to keep track of*0906

*is the net change in gas molecules from reactant to product side.*0909

*For example, N _{2} gas plus 3H_{2} gas goes on to form 2NH_{3} gas.*0913

*On the left side, we start off with four gas molecules.*0923

*On the right side, we wind up with only two gas molecules.*0930

*For all intents and purposes, gas molecule A is going to be the same as gas molecule B for entropy counting purposes.*0935

*Because we start off with four gas molecules and we only wind up with two,*0945

*we are actually decreasing the amount of disorder in this process.*0949

*ΔS is expected to be negative for this example.*0953

*Sometimes the problem is going to ask you to calculate ΔS*0958

*for what is called the system, surroundings, and total where*0960

*ΔS total is equal to ΔS system plus ΔS of the surroundings.*0963

*We have talked about system and surroundings in thermodynamics I.*0967

*If you recall, Q of the system equals to ?Q of the surroundings.*0972

*This is going to come into play.*0978

*If a refrigerator coolant absorbs heat from stored food items, the heat then vaporizes the coolant which boils at -27 degrees Celsius.*0979

*ΔH of vaporization is equal to -22.0 kilojoules per mole.*0988

*Calculate ΔS total when 1.471 moles of the coolant vaporizes exchanging heat with the food items that are stored at 4 degrees Celsius.*0993

*What we have to do is the following.*1002

*We know we are using this equation for sure.*1005

*ΔS total is equal to ΔS of the system plus ΔS of the surroundings.*1007

*Let's go ahead and define the system; it doesn't matter which one as the coolant.*1018

*Let's go ahead and define the surroundings as the food items inside the refrigerator.*1024

*This is then going to be equal to...*1031

*We know that ΔS is equal to Q of the system over T.*1035

*The ΔS of the surroundings is going to be equal to Q of the surroundings over T.*1043

*But we also know that this relationship here where Q of surroundings is equal to ?Q of the system.*1050

*We can then go ahead and plug everything in.*1060

*That is going to be equal to nΔH of vaporization over the temperature of the coolant.*1062

*That is going to be equal to ?nΔH of vaporization but this time over the temperature of the food.*1072

*Really the only difference is the temperature that the system and surroundings are stored at*1080

*if we assume complete transfer of heat from the system to the surroundings.*1085

*When all is said and done, we are going to get -0.248... don't forget the units.*1092

*Here this is going to be kilojoules per kelvin or -248 joules per kelvin.*1098

*Again you must be careful on how you define your system and surroundings.*1108

*Be consistent throughout the entire equation.*1113

*Finally another type of problem involves standard molar entropies of formations which is the*1118

*change in entropy when a compound is formed from its constituent elements under standard conditions.*1123

*Recall that we covered standard molar enthalpies of formation and used the summation equation.*1131

*Similarly ΔS of the reaction is going to be equal to summation nS of the products minus summation nS of the reactants.*1137

*These values here again just like for the enthalpy, you are going to look up in the appendix.*1148

*Don't forget that n is going to be the stoichiometric coefficient.*1157

*Unlike standard molar enthalpies of formation, known compound is going to have a standard molar enthalpy of formation.*1169

*Why?--because the zero is only obtained under the conditions of the third law of thermodynamics.*1175

*Let's go ahead and examine our... excuse me... this is very important.*1182

*Again you never can assume that the molar entropy of formation is a zero value ever.*1191

*It is only true for ΔH and, we are going to learn, for one more--what we call ΔG later.*1200

*Let's examine our final state function.*1207

*This is an introduction to Gibbs free energy; free energy is ΔG.*1210

*It is defined as the amount of energy available to do work.*1215

*We said that mother nature favors high entropy already and low energy.*1220

*It is that energy that we are really talking about.*1225

*We will see that the spontaneous processes will all have ΔG less than zero.*1227

*To go ahead and do this, we can do ΔG is equal to ΔH minus TΔS.*1241

*This is the equation that relates all three thermodynamic properties together.*1258

*We already said that a positive ΔS is going to be a spontaneous process.*1266

*We want ΔG to be negative for a spontaneous process.*1279

*What conditions give you ΔG negative?*1287

*ΔG negative is going to be spontaneous under the following conditions--for very large ΔS values and at high temperatures.*1290

*As you can see, if ΔS is large and temperature is large,*1312

*this term is going to be larger than this term giving us*1318

*an overall negative ΔG value; very large ΔS at high temperatures.*1322

*Basically ΔH is going to be a relatively negative value; ΔH, small as possible.*1329

*Again this is what we call Gibbs free energy.*1345

*Just like our previous state functions, we have standard molar free energies of formation*1350

*where ΔG of the reaction is equal to summation nΔG of the products minus summation nΔG of the reactants.*1355

*Like enthalpy, the free energies of formation are zero for all compounds in the standard state.*1363

*Something like O _{2} gas, H_{2} gas, N_{2} gas, Cl_{2} gas, Br_{2} liquid, and I_{2} solid just to name a couple of examples.*1369

*These are all going to have a ΔG of formation which is equal to ΔH of formation equal to zero.*1385

*But again S, the molar entropy of formation, is not going to be zero whatsoever.*1397

*Let's now study the temperature dependence of ΔG.*1409

*It is found that we already said that ΔG of the system is equal to ΔH of the system minus TΔS of the system.*1414

*The reason why this is important is because of the following--this term right here.*1424

*ΔS of the system is in this equation.*1429

*But the second law tells us that ΔS of the universe has to be a positive value.*1433

*If we go strictly by entropy, we need to know the ΔS of the total, not just of the system or the surroundings.*1446

*However this equation here tells us that all I have to know is ΔG of the system; that is all.*1454

*Let's go ahead and see what conditions are going to be right for this.*1466

*We can go ahead and make a chart here; ΔH, T, and ΔS.*1474

*Then this is going to be sign of ΔG.*1481

*Let's go ahead and delete temperature here--sign of ΔG, and then ΔH, ΔS.*1488

*Let's say ΔH is a negative value--if this is less than zero and ΔS is a negative value.*1495

*If ΔH is negative and ΔS is negative, it will be very difficult to get ΔG to be negative.*1505

*The only time ΔG can be negative is going to be at a low temperature here.*1516

*What if ΔH is negative and ΔS is positive?*1523

*If ΔH is negative and ΔS is positive, then this sign of ΔG is going to be negative at high temperature.*1531

*What if ΔH is positive and ΔS is positive?*1547

*If ΔH is positive and ΔS is positive, ΔG is going to be negative at high temperatures.*1551

*Finally what if ΔH is positive and ΔS is negative?*1560

*If ΔH is positive and ΔS is negative, ΔG will never be negative according to the equation at any temperature.*1565

*This is going to be greater than zero at all temperatures.*1574

*Again as you can see, ΔG of the system is greatly influenced by*1579

*not only the signs of ΔH and ΔS but also the kelvin temperature.*1585

*Finally let's go ahead and examine the effect of reactant and product concentration on the sign of free energy.*1595

*It is found that ΔG of the reaction is equal to ΔG ^{0} plus RT natural log of Q.*1602

*Remember what Q is?--Q is what we call the reaction quotient.*1609

*This equation relates reactant and product concentration to the free energy not at*1618

*standard conditions which means the concentrations are not going to be 1 molar.*1623

*Or the pressures are not going to be 1 atm; or partial pressures not equal to 1 atm.*1630

*Finally it can also be shown at equilibrium that ΔG not of the reaction is equal to ?RT natural log of K*1639

*where this is our equilibrium constant which we have discussed so much in the previous sections.*1647

*This is any equilibrium constant.*1654

*This can be K _{a}, K_{b}, K_{p}, K_{sp}, K_{f}, etc; any of them.*1656

*Basically this equation tells us that if K is large, the reaction is product favored.*1667

*ΔG of the reaction is a negative value; that makes sense.*1672

*If K is very large, that means the reaction is highly product favored.*1677

*In other words, the reaction as written is going to be likely to occur.*1682

*If the reaction is likely to occur, ΔG of that reaction should be negative.*1687

*Let's go ahead and summarize our thermodynamics lecture before getting into our problems.*1694

*We see that nature not only favors states of low energy but also now high entropy.*1701

*Entropy is basically disorder or chaos.*1707

*It is the focus of the second and third laws of thermodynamics.*1710

*Basically what we have seen in this lecture is that there is a series of equations that govern entropy and free energy.*1713

*They allow us to calculate both of them under a given set of conditions.*1722

*Let's go ahead and do sample problem number one.*1729

*Calculate ΔS of the reaction under standard conditions for the following.*1732

*We have aluminum oxide reacting with 3H _{2} going on to form two aluminums plus three waters.*1738

*Anytime you see a balanced chemical equation and it asks you simply to calculate*1746

*ΔS not of the reaction, you are going to use the summation equation.*1751

*Use ΔS of the reaction is equal to summation n S molality of all products minus summation n S molality of all reactants.*1755

*I am just going to setup the problem for you.*1772

*Again you are going to look up these values in the appendix of your textbook.*1775

*Let's go ahead and do the products first.*1779

*This is going to be 2 moles of aluminum times the standard molar entropy of formation of aluminum solid.*1781

*Because it is summation, I am going to add... plus.*1799

*Let me put that in parentheses.*1803

*Plus 3 moles of the water vapor times its standard molar entropy.*1804

*This is all in brackets; that is going to be subtracted from the reactant part.*1813

*Here this is going to be 1 mole of Al _{2}O_{3} solid times*1819

*its molar entropy plus 3 moles of H _{2} gas times its molar entropy.*1830

*Again you must pay attention to the physical states because the molar entropy of formation for say water vapor*1848

*is going to be different than from liquid water which is going to be different than from solid water.*1856

*Once again pay attention to the physical states.*1862

*When this is all said and done, you are going to get your answer*1865

*in of course units of joules per kelvin or kilojoules per kelvin.*1868

*It depends what the question specifies of course.*1874

*Let's go ahead and do sample problem number two.*1880

*The reaction N _{2}O_{4} going to 2NO_{2} is not spontaneous under standard conditions.*1883

*Calculate the temperature at which the reaction becomes spontaneous.*1889

*What that translates to is, in other words, at what temperature does ΔG become negative?*1893

*Anytime you are asked to calculate a temperature at which something becomes spontaneous,*1902

*pretty much you always use this equation: ΔG is equal to ΔH minus TΔS.*1908

*They are asking you when does this change sign?*1916

*In other words, when does ΔG go from positive to negative?*1920

*What you simply is you set this whole thing equal to zero.*1926

*Zero is equal to ΔH minus TΔS; you solve for temperature.*1931

*Temperature is going to be equal to ΔH over ΔS.*1936

*This is all going to be standard conditions of course.*1941

*To get ΔH, you are going to use the summation equation.*1946

*To get ΔS, you are going to use the summation equation*1950

*which means you have to look up the formation values in the appendix.*1954

*Look up formation values in appendix; appendices.*1959

*When you do this, you are going to get a temperature value.*1969

*At temperatures lower than T, ΔG is going to be positive.*1976

*At temperatures higher than what you calculate for T, ΔG is going to be a negative value.*1987

*Once again anytime you are asked to calculate the temperature at which something becomes spontaneous,*2006

*almost always you are going to use this equation: ΔG is equal to ΔH minus TΔS.*2015

*Set ΔG equal to zero and solve for the temperature.*2021

*Finally our third sample problem, calculate ΔG of the reaction at 25 degrees Celsius*2029

*for the formation of 1 mole of C _{2}H_{5}OH gas from C_{2}H_{4} gas and H_{2}O gas.*2036

*The first thing we probably want to do is translate this into a balanced chemical equation.*2047

*The reactants are C _{2}H_{4} gas; this is going to react with H_{2}O gas.*2054

*That is going to form 1 mole of C _{2}H_{5}OH gas.*2063

*Let's see if everything is balanced.*2069

*Two carbons, six hydrogens, one oxygen; yes, this is nicely balanced.*2071

*You are asked to calculate ΔG of the reaction at 25 degrees Celsius.*2078

*Again ΔG of the reaction can simply be calculated from ΔH of the reaction minus TΔS of the reaction.*2083

*Again to get the ΔH, you are going to use the summation equation.*2094

*To get the ΔS, you are going to use the summation equation.*2099

*Again the kelvin temperature is already specified for you right there.*2102

*That gives you ΔG of the reaction; they then ask you to calculate K _{p}.*2108

*The only equation in this lecture that involves the equilibrium constant is ΔG is equal to ?RT natural log of K.*2114

*We can then insert this ΔG of a reaction in there.*2126

*K therefore is equal to 10 raised to the -ΔG ^{0} over RT*2131

*where of course R is equal to 8.314 joules per k mole.*2140

*Always double check your work.*2151

*Make sure that K is a positive value because again*2152

*we saw in previous lectures that K can never ever be negative.*2158

*That is our lecture from general chemistry on thermodynamics.*2165

*I want to thank you for your time and attention.*2169

*I will see you next time on Educator.com.*2171

0 answers

Post by Maria Mazos on January 24 at 08:20:56 PM

18:24-can y please explain why -0.248?

1 answer

Last reply by: Maria Mazos

Tue Jan 24, 2017 8:18 PM

Post by Okwudili Ezeh on October 30, 2014

Please could you clearly state how you got the answer to the problem you solved at 18:24. You did not plug in any numbers and so how are we expected to understand what you did?