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

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

### Spontaneity, Entropy, & Free Energy, Part IV

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- Intro 0:00
- Spontaneity, Entropy, Free Energy 0:29
- Standard Free Energy of Formation
- Example 1
- Reaction Under Non-standard Conditions
- Example 2
- ∆G = Negative
- ∆G = 0
- Diagram Example of ∆G

### AP Chemistry Online Prep Course

### Transcription: Spontaneity, Entropy, & Free Energy, Part IV

*Hello, and welcome back to Educator.com; welcome back to AP Chemistry.*0000

*Today, we are going to continue our discussion of free energy, and we are going to introduce the notion of reactions that take place under non-standard conditions.*0004

*How do we account for the free energy change that takes place for something that is not one atmosphere, or not one Molar concentration?*0014

*Well, this is what we are going to do today; and we are going to introduce a very, very, very important equation.*0022

*So, let us go ahead and get started.*0028

*In the last example of the last lesson, we used that ΔG=ΔH-TΔS to calculate the free energy of a given reaction.*0031

*Well, before we actually introduce this notion of non-standard conditions, let's just talk a little bit about the other way that we can actually get ΔG, instead of using the ΔH-TΔS.*0043

*Let's say--well, let's just start with a definition.*0058

*OK, definition: The standard free energy of formation, which is symbolized as ΔG with a little f down at the bottom, is the change in free energy that accompanies the formation of one mole of a substance from its constituent elements with all reactants and products in their standard states.*0063

*There you go: so, for example, the ΔG of formation of, let's say, water: H _{2} gas + O_{2} gas goes to 1 mole (right, 1 mole of a substance, so) of H_{2}O liquid.*0162

*If we want to calculate the standard entropy change for that, we would do whatever it is that chemists and physicists do, and this would actually end up being one-half here (right? no, the other way around--what am I doing?--this is H _{2}...oh, that's nice; I love that...plus 1/2 O_{2}, goes to H_{2}O).*0180

*There is some ΔG of formation associated with that, some number.*0206

*This number, the ΔG of formation--this is the number that you find at the end of thermodynamic tables.*0212

*This definition is actually exactly like the definition for enthalpy, if you remember, from way back then when we did thermochemistry.*0222

*There was something called the standard heat of formation (the standard enthalpy of formation), and it was the change in enthalpy that accompanies the formation of one mole of that particular substance (that we want the enthalpy of formation for), under standard conditions, with everything being in their standard states.*0229

*This definition is completely analogous: the idea here is that, instead of using the ΔH=ΔG-TΔS, we can actually use the equation as written, and just take the ΔG of the products, minus the ΔG of the reactants, and get the ΔG of the reaction.*0249

*We actually end up getting the same number; so we have tables, or we have the equation ΔH-TΔS.*0267

*So, let's just do a quick example here.*0274

*Example: What is the ΔG for 2 CH _{3}OH + 3 O_{2} (in other words, the combustion of methanol gas); this is gas; this is gas; 2 CO_{2} gas plus 4 H_{2}O.*0279

*We want to know what the standard free energy change is for this.*0310

*OK, now the entries...I'm going to actually write out the entries that you would see in a thermodynamic table, just so that you actually see it; I know that you can flip to the back of your book, but I would like you to see the numbers anyway.*0315

*The entry for...well, actually, I'm going to do the entry for all of them: so let's go to CH _{3}OH as a gas; O_{2} as a gas; CO_{2} as a gas, and (oops, here we go with the stray lines again; we don't want that) H_{2}O gas.*0325

*Now, we have entries that look like this: there is going to be something that says ΔH _{formation}, and it's going to say kilojoules per mole.*0358

*And then, there is going to be an entry that says ΔG of formation, and that is in kilojoules per mole.*0367

*And then, they are going to have S, which is going to be in Joules per mole-Kelvin.*0373

*You know what, I really, really, really need to make this a little bit more legible; I understand--my apologies.*0381

*S--and notice, there is no ΔS; this is just S--this is going to be in Joules per mole-Kelvin.*0388

*We have: -201, 0 (remember, 0--elements, the enthalpy is 0), and we have -394, -242; the ΔG is -163, and the ΔG for elements is also 0; this happens to also be 394, which is really interesting, in and of itself; -229.*0398

*And then, we have 240, 205, 214, and 189.*0426

*The entropy of an element is not 0: oxygen gas is a disordered system; there are a lot of particles of gas: 205--that is a lot of entropy.*0435

*It's not 0; it's not the same as these.*0443

*When I run this--when I take 4 times the ΔG of H _{2}O (so I'm working in this column right here--I just wanted you to see what the other entries are), plus 2 times the ΔG of CO_{2}, minus 2 times the ΔG of CH_{3}OH gas, minus 3 times the ΔG of O_{2}...when I actually run the reaction, I end up with (I'll run the calculation) a ΔG.*0447

*Notice, the f is gone; those were ΔG's of formation of the individual elements; that is what is in the table; the final outcome is just my ΔG, my free energy.*0477

*And it's not in moles; it's not in kilojoules per mole; it's just in kilojoules.*0488

*And the reason is because I have accounted for the moles from the balanced reaction: I am taking 2 times something, 3 times something--I am using the stoichiometric coefficients.*0492

*I end up with -1378 kilojoules--137,800 Joules.*0502

*That is a lot of energy; that reaction--the burning of methanol--is highly spontaneous under standard conditions.*0515

*So, that is what this means.*0526

*Let's see: OK, so now, we have a good sense of dealing with free energy; we treat it the same way we treat ΔH, or H, or S--we just take products minus reactants.*0530

*We also have the equation ΔH-TΔS, to calculate it that way.*0542

*Now, let's think about what this actually means.*0547

*This number here tells me that it's spontaneous as written: that means it will happen (eventually) if I don't do anything.*0551

*That doesn't mean it's going to happen quickly; it just means that it will happen, and I don't have to do anything about it.*0559

*Thermodynamically, it's spontaneous.*0565

*Spontaneity is a thermodynamic statement, not a kinetic statement; it just says that the energy (from your perspective) of the CO _{2} and the water is lower than the energy of the methanol and the oxygen gas.*0567

*So, it's a downhill thing.*0587

*We just needed to get it over that hump; that is the kinetics part.*0590

*OK, so now let's think about what this means: this ΔG is a measure of the tendency, the potential, for a reaction to move forward.*0592

*Well, let's say it starts moving forward: well, as it starts to move forward, more CO _{2} and water are going to form; methanol is going to be used up; oxygen is going to be used up.*0606

*Let's say that half the reaction has gone forward: does that mean, at that point, that the free energy is the same?*0619

*As it turns out, no: the idea is that free energy wants to get down to 0--or, in this case, wants to get up to 0.*0626

*0 free energy means the system is at equilibrium; so, you can have a negative free energy; that means that the reaction, as written, wants to go to the right.*0635

*It's spontaneous to the right.*0648

*If it's positive, that means the reaction wants to go to the left.*0650

*The idea is: the reaction wants to get to a point where the ΔG is 0.*0655

*So, as a reaction proceeds forward, the ΔG actually changes until it gets to a point where it's 0; that is equilibrium--that is what all reactions do--they move toward equilibrium.*0660

*So, let me write this down: As a reaction proceeds toward equilibrium, the ΔG changes until it hits equilibrium--until it reaches 0, which is the equilibrium point.*0675

*OK, now, we have been calculating ΔG with that little 0 on top--standard ΔG's under standard conditions (1 atmosphere pressure; 25 degrees Celsius; 1 Molar concentration for aqueous species--standard conditions).*0723

*Well, what happens when we run a reaction not under standard conditions--what if it's at 200 degrees Celsius; what if it's 3 atmospheres of one gas and 17 atmospheres of another gas?*0746

*What happens then--how does it affect the ΔG?*0761

*Is it affected at all?*0764

*As a matter of fact, it is; so, when we change the conditions, when we change the pressures, the temperatures, the ΔG of the reaction as written changes.*0766

*Rather than going through a sort of discussion of a derivation of this equation that I'm going to write down, I'm just going to write down the equation, because we just want you to be able to use it and understand that this ΔG is under standard conditions, but we're not always running reactions under standard conditions.*0777

*We often...most of the time, we're not doing it under standard conditions.*0799

*So, for reactions run under non-standard conditions, the ΔG of the reaction equals the standard ΔG, plus a certain term--a correction factor, if you will, for the non-standard part.*0803

*That is the whole point; at standard conditions, it's this; but if it's non-standard, then we're going to have to add a little term--either plus or minus, one way or the other--to adjust for the fact that it's non-standard.*0838

*That is RT ln(Q), and Q is exactly what you think it is: it's the reaction quotient.*0849

*If I have (let me do this one in red...I thought I said red...let's see) aA + bB going to cC + dD, the reaction quotient is equal to the concentration of C, or the pressure of C (depending on if it's aqueous or gas), raised to the power of c, times the concentration of D raised to the power of d, over A raised to the power of a, B raised to the power of b.*0856

*At any given moment, under those circumstances, we use Q for at that moment; we use K for equilibrium concentrations.*0896

*So, that is what this is; this is a very, very important equation.*0905

*It is telling me that, if I am going to run a reaction under non-standard conditions (let's say higher pressure--and we will deal with pressure mostly with our discussions), it is equal to...well, if I want to know what the free energy change of a reaction is, at a non-standard...I go ahead and I calculate the free energy change under standard conditions, and then I add to it R (which is the gas constant--I'll write that down: R=8.3145--that is Joules per mole-Kelvin), times the temperature (which is the absolute temperature in Kelvin), times the logarithm of the reaction quotient.*0909

*The reaction quotient is this thing, as stated in the particular problem.*0950

*Now, let's just go ahead and do an example.*0959

*Oh, and by the way, notice: ΔG and R--R is Joules; ΔG...when you calculate ΔG, more often than not, you will calculate it in kilojoules, depending on if you are going to use the thermodynamic data in the back or the equation.*0962

*Make sure your units match, OK?--so Joules, Joules or kilojoules, kilojoules: so just make sure that the units match.*0977

*OK, so let's go ahead and do our example.*0986

*Example: Calculate (and notice, there is no 0--there is no little degree sign on top, because it is not standard anymore; this is the ΔG of a reaction) ΔG of the reaction at 25 degrees Celsius (so, in this case, at least the temperature will be) for the reaction CO (carbon monoxide gas), plus 2 H _{2} gas, going to CH_{3}OH liquid.*0990

*We want to calculate the ΔG of the reaction, at 25 degrees Celsius, for the formation of methanol from carbon monoxide gas and hydrogen gas, when the partial pressure of the CO gas is equal to 6 atmospheres (so notice, it's not 1 atmosphere anymore), and the partial pressure of the H _{2} gas equals 2.0 atmospheres (not one atmosphere anymore).*1035

*So now, the pressures are higher.*1059

*This is a gas; this is a gas; this is a liquid; so--this is a liquid--it doesn't show up in the expression for the reaction quotient.*1062

*Let's go ahead and write this out (let me actually write the equation again; let me do it in blue).*1071

*We have: carbon monoxide gas, plus two moles of hydrogen gas, is going to form one mole of methanol liquid.*1079

*OK, so we have that the ΔG of the reaction is equal to the standard free energy change, plus RT ln(Q).*1091

*Well, let's see what Q is, first: Q is equal to the concentration of the products, divided by the concentration of the reactants.*1103

*In this case, these are gases, so we are going to use pressures; this is a liquid, so it doesn't show up in the numerator (it's just 1 up there).*1111

*The partial pressure of CO _{2}, times the partial pressure of H_{2}, squared (the stoichiometric coefficient: you know this already--that is what this is--just the reaction quotient under gaseous conditions).*1121

*OK, so now, let's go ahead and calculate: we have taken care of this; we'll plug it in in just a minute.*1134

*We know what T is: it's going to be 298; we know what R is--that is a constant; logarithm is just a mathematical operation; we need to calculate what ΔG is.*1142

*ΔG, standard free energy change--when I look at a thermodynamic table, I'm going to end up with...I'm going to do this one, minus 166 kilojoules (I'm going to...actually, let me do it this way); it's 1, times 1, times -166, minus 1 times -137, plus 2 times 0 (right?--we use the standard--something that we have done all along).*1152

*You end up with a ΔG of -29 kilojoules, which is equal to 29,000 Joules.*1193

*Good; so now, we have: our ΔG of reaction is equal to -29,000, plus 8.3145 Joules per Kelvin, times 298 Kelvin, times the logarithm of 1, over...well, what was the partial pressure of CO?--it was 6.0 atmospheres; the partial pressure of the hydrogen was 2.0 atmospheres squared.*1203

*When we do all of the math, we end up with -29,000, plus a -7,874; I wanted you to see it in both of its forms.*1245

*Under standard conditions, we just calculated that ΔG is 29,000; now, to that, based on the fact that the pressures are actually higher, this equation tells me that it's even more spontaneous, because this number is negative.*1263

*The total is going to be -36,874 Joules; so the reaction as written is even more spontaneous under conditions of higher pressure.*1278

*That is all this is: we use this equation to find the free energy change of a reaction under conditions that are non-standard.*1294

*That is it; that is all that this equation allows us to do.*1304

*Based on the reaction quotient, that means as written, I just throw into a flask CO gas at 6 atmospheres and hydrogen gas at 2 atmospheres; the free energy change for this reaction is going to be -36,874.*1307

*This is highly spontaneous: it has a tendency--it's going to start to move toward equilibrium; that is the whole idea.*1322

*OK, so let's see what else: we want to notice a couple of things about this.*1330

*Notice: ΔG is more negative, so it's more spontaneous, under conditions of higher pressure (or, I should say, under conditions of pressures higher than standard--pressures higher than 1 atmosphere).*1337

*But we could have predicted this from Le Chatelier's Principle: watch.*1379

*CO (now again, prediction is one thing: we are actually able to get a number, so--qualitative: we could predict it qualitatively, but we want to be able to do the quantitative; we want a number for it) + 2 H _{2} goes to CH_{3}OH liquid.*1383

*Here, we have three particles of gas; here, we have no particles of gas.*1405

*Well, under 1 atmosphere of pressure, it's at ΔG; now, I have 6 atmospheres of CO; I have 2 atmospheres of the H _{2}; I have increased the pressure of the system.*1412

*If I increase the pressure of the system, well, the system is going to want to offset that pressure, right?*1424

*That is the whole idea: it's Le Chatelier's Principle--it's going to do the opposite of--it's going to try to get itself back to where it was, to offset the effect.*1431

*Increasing the pressure, the system is going to want to decrease the pressure; how does it decrease pressure?--well, it decreases pressure by lowering the number of particles that are bouncing around against the walls to create that pressure, which means it's going to move in that direction.*1439

*It is going to be more spontaneous; it is going to push the reaction forward.*1455

*You see, all of these things are coming together; that is the whole idea behind chemistry--we want to get you to see the big picture: spontaneity, equilibrium, Le Chatelier's Principle, quantitative aspects, free energy, entropy--all of this is tied together.*1459

*OK, now, the last thing that we are going to talk about is the following.*1476

*ΔG=0: we said that, when the ΔG of a reaction (either at standard or non-standard conditions)--when it equals 0, that means equilibrium.*1482

*It does not mean completion; and here is what we mean by that.*1502

*If I start with some CO gas and some H _{2} gas, and here I have the CH_{3}OH, energetically, the free energy change is negative, right?*1509

*So, the ΔG is less than 0; it's going from a higher energy to a lower energy--that is what makes this spontaneous--but here is what happens.*1524

*As the CO and the H _{2} are used up (let me do this in red), they are going to diminish; OK, there is going to be less of this, more of this.*1533

*I'm going to come up; it's going to come to a point where there is going to be a mixture of the CO, the H _{2}, and the CH_{3}OH, when there is an equality.*1547

*It doesn't mean ("spontaneous")--this negative 36,000 that we got doesn't say that, if I put this in a flask and somehow the reaction takes place, that all of a sudden all of the carbon monoxide and the H _{2} are going to vanish, and the only thing left in the flask is going to be the methanol.*1563

*That is not what this is saying.*1582

*"Spontaneous" is a measure of the tendency of a system to reach equilibrium, not completion.*1584

*When we say "completion," that means there are no more reactants left.*1591

*When we say "equilibrium," that means it has reached a point where there is a little bit of everything left: that is what ΔG means.*1595

*So, pictorially, it looks like this.*1603

*In some sense (again, I'm not sure if this is the best way to represent it, but it's not a bad way of representing it): ΔG=0, equilibrium--it represents a low point in energy.*1612

*That means...over here, the ΔG is negative; it's spontaneous in this direction; it is going to seek out this--in other words, it's going to form more.*1634

*But, if it went to completion...as it turns out, at that point, as written, it is going to end up with a positive ΔG.*1646

*Well, a positive ΔG means it's spontaneous in the reverse direction; so now, the reaction is going to go this way.*1658

*It is going to go this way and this way until it reaches this point, where the ΔG of the reaction is equal to 0.*1664

*It is going to seek out equilibrium; it's not going to seek out completion.*1673

*Let me say that again: Reactions...ΔG going to 0...reactions will seek equilibrium; they will not seek out completion.*1677

*That doesn't mean that there are going to be no reactants left, and that it's going to be all products.*1687

*That is not what "spontaneous" talks about; "spontaneous" is a measure of the tendency of a reaction to seek out equilibrium.*1691

*It wants to get to a point where ΔG equals 0; it's like a ball rolling down a hill--it will roll down a hill, it will roll up the other side, and it will roll back down, roll up this side, until it finds a point of lowest energy.*1699

*Reactions move toward the valley of lowest free energy (zero).*1714

*I shouldn't say "lowest free energy"--zero, because negative free energy implies that it is actually moving toward that 0.*1720

*This is what you want to remember: Completion is not the same as equilibrium.*1727

*Yes, there are some reactions...like, for example, if you put hydrogen and oxygen in a flask and you ignite it--yes, it is going to be all water; there is going to be no hydrogen and no oxygen in there--at least, none that is measurable.*1733

*But, believe it or not...there is none that is practically measurable, but believe it or not, it actually is in equilibrium.*1744

*It has gone so far--there is so little hydrogen and oxygen gas left--that, for all practical purposes, yes, that reaction has come to completion.*1753

*In that case, completion and equilibrium are almost the same.*1759

*But, that doesn't mean that they are; the system is actually at equilibrium--there is still just a little bit of hydrogen and oxygen left, because the free energy for that reaction is here.*1764

*It is not going to end up going to form all water or all something else.*1775

*Think about that for a little bit.*1780

*OK, next time we get together, we are actually going to do a series of problems, because I know we have done only a handful of problems for the thermodynamic section.*1782

*We spent a lot of time discussing it, but we definitely need to wrap our minds around it by doing more practice.*1790

*So, the next time I see you, we are going to introduce one more concept of equilibrium, and then we are going to spend our time on some problems.*1797

*Thank you for joining us here at Educator.com.*1803

*See you next time; goodbye.*1806

0 answers

Post by bob singh on April 13, 2015

Professor,

For some reason, around 8:29, I am not getting -1378 kJ for my delta G. My calculations are:

(2(-394)+4(-229))-(-163). I get -1704 kJ.

3 answers

Last reply by: Professor Hovasapian

Sun Apr 21, 2013 8:33 PM

Post by Antie Chen on April 21, 2013

Excuse me, what's the whole name of Q? and in the calculating of Q, liquid and solid won't involve because they don't have pressure?

1 answer

Last reply by: Professor Hovasapian

Wed Sep 26, 2012 11:00 PM

Post by nguyen yen on September 26, 2012

Dear Dr Hovasapian,

I'm confused with the value of R when you talked about RTlnQ and the pressure is expressed in atm

We use 8.31 and in other case it's 0.082

Please explain.