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 0 answersPost by bob singh on April 13, 2015Professor,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 answersLast reply by: Professor HovasapianSun Apr 21, 2013 8:33 PMPost by Antie Chen on April 21, 2013Excuse 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 answerLast reply by: Professor HovasapianWed Sep 26, 2012 11:00 PMPost by nguyen yen on September 26, 2012Dear Dr Hovasapian,I'm confused with the value of R when you talked about RTlnQ and the pressure is expressed in atmWe use 8.31 and in other case it's 0.082 Please explain.

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

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 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

### 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

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: H2 gas + O2 gas goes to 1 mole (right, 1 mole of a substance, so) of H2O 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 H2...oh, that's nice; I love that...plus 1/2 O2, goes to H2O).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 CH3OH + 3 O2 (in other words, the combustion of methanol gas); this is gas; this is gas; 2 CO2 gas plus 4 H2O.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 CH3OH as a gas; O2 as a gas; CO2 as a gas, and (oops, here we go with the stray lines again; we don't want that) H2O gas.0325

Now, we have entries that look like this: there is going to be something that says ΔHformation, 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 H2O (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 CO2, minus 2 times the ΔG of CH3OH gas, minus 3 times the ΔG of O2...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 CO2 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 CO2 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 H2 gas, going to CH3OH 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 H2 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 CO2, times the partial pressure of H2, 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

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 H2 goes to CH3OH 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 H2; 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 H2 gas, and here I have the CH3OH, 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 H2 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 H2, and the CH3OH, 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 H2 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