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### Course: Physical Chemistry (Essentials) - Class 11 > Unit 6

Lesson 13: Gibb's free energy# A look at a seductive but wrong Gibbs spontaneity proof

A look at why the "proof" of the relation between changes in Gibbs Free Energy and Spontaneity is wrong in many textbooks. Created by Sal Khan.

## Want to join the conversation?

- Just pointing this out. The reason why delta S environment can be approximated as Qrev,environment/T (which is = -Qrev,sys/T) is that any changes to the environment (which is MUCH larger than the system) can be assumed to be reversible. Since the environment is so big, the changes to it can be assumed to take infinitesimal steps, which allows the assumption of reversibility. (Sorry I had to put this in the questions as the comments don't allow so many characters!)(13 votes)
- i still dont understand free energy...(6 votes)
- I am confused now. because Sal had used the Q/T formula for the spontaneous irreversible process in one of his previous videos.

http://www.khanacademy.org/science/physics/thermodynamics/v/entropy-intuition?qa_expand_key=ag5zfmtoYW4tYWNhZGVteXJpCxIIVXNlckRhdGEiTHVzZXJfaWRfa2V5X2h0dHA6Ly9ub3VzZXJpZC5raGFuYWNhZGVteS5vcmcvMDA4YjRiOGU0ZDdlOGY1NmI0ZTlmNjk2NTAwYWE2YTgMCxIIRmVlZGJhY2sY6QcM

the answer seems to indicate that the use of Q_ir / T for delta S_environment is not so problematic after all?(3 votes)- The equation isn't wrong, it just makes an assumption about the process, that it is an irreversible system. If you are "proving" the equation is true, you cannot start with such an assumption.(7 votes)

- is the equation G=H-TS applicable to all processes?(3 votes)
- If he wants to be really rigorous, then the assumption of constant pressure is false, and he can't derive the term \DeltaH(2 votes)
- Hi Sal, I hope you read this, please let me know what you think. I have to say I agree with most of the textbooks on this,

First I think the Qrev and Q irrev are the same in your example at least, but even if they were not, it would not be a problem. They are the same since when you have friction, system has to do more PV work due to friction, so it loses more energy compared to frictionless situation.. Now this extra heat loss is exactly equal to heat produced by friction, so when it absorbs friction heat, it brings it down to the level of reversible one,so now it can absorb the same amount of heat as the reversible process from surrounding.

This might look wrong at the surface , you might say that q is not a state function, so it cannot be the same for rev and irrev processes. Yes , but your example is only making friction the difference between the two processes which is not the general case when we talk about irreversible processes . In other words, you are assuming that the only energy used extra to reversible one is for friction. but this is not a complete definition of irreversible processes. Irreversible processe are also those not done infinitesimally slowly, so extra work is done for other things other than friction such as accelerating the piston for example. in such cases, the two Qs will not be equal.

This brings me down to the second point I made above. Even if they are not equal, it would not affect the proof. You are making a statement which I think is the gist of your argument and that is at2:20: “we can only use delta S envi = Q abs/T for reversible processes”. however that is not the case. I am not sure where you have seen this requirement. The correct requirement is this: delta S sys = Qrev/T. however the change in the entropy of the enviroment is only due to the heat that system is giving to it by our process, may it be reversible or irreversible.

One thing to pay attention to is that the heat provided by the friction is an internal exchange in the system and we are not dealing with that. This heat is produced by the system and used by the system. We do not concern ourselves with such cases. We are only concerned about that part of heat that is exchanged. An example I can give you of this is that when we have a chemical reaction is closed system which produced gas, this extra production of gas is not making the system hotter. However if we inject some gas from outside to the system, this will make the system hotter. But why?

This is why: in both cases some work is being done: when we inject gas, we are pushing new gas by some force to the system, and this does work. When the system makes new gas it has to do the same work. However in the second case, the work the system is doing is given back to it by the energy produced , and they cancel each other out. So as you see we have both work done by system (to push new gas into old gas) and heat produced in the system.non of these will be taken into account in our calculations since these are internal interactions of the system. Howeve the work we do to push new gas to the system is to be taken into account, which then tells us that some heat will be transferred out of the system due to this.(4 votes)

- … But isn't there a spontaneous reaction that is reversible to which this applies; namely potassium nitrate (or other soluble salts) dissolving in water? KNO3 dissolves spontaneously in water and can be brought to equilibrium as a saturated solution at a defined temperature where both solid and solution are present. Wouldn't this be an example of a spontaneous system in equilibrium (i.e. reversible) allowing for such a derivation ultimately leading to Delta G = -RTlnKsp and the Gibbs-Helmholz equation?(2 votes)
- In the previous video, sal substitutes delta S for the environment as -Q(IR)/T which is equal to the heat absorbed by the environment (Q(abs)/T). In this video, how can he say delta S (env) is not equal to Q(abs)/T(2 votes)
- But in the previous video, Sal used dS_env=Q_irr/T (1).

Even by taking into account the fact that Q_irr<Q_rev, how is the use of (1) justified since the environment DOES NOT exchange heat reversibly? Thanks.(2 votes) - Isn't there is a marked difference between Thermodynamic Reversibility and the Chemial Reversibility(the one we study under concepts such as equilibrium)?

I have read/been taught that thermodynamic reversibility is just a process carried out in infinitesimally small steps in infinite time, such that all individual steps are in equilibrium, like reducing the pressure over a gas by, say, removing one grain of sand's worth pressure.(as opposed to the irreversible change where the entire sand bag is removed)

Hence, thermodynamic reversibility should not be connected with a spontaneity of the process, right?(2 votes) - So how do you actually calculate the change in entropy of the environment for an irreversible process? Is it Qirrev / T?

But Sal says this is applicable only when the process is reversible. But he himself is found to use this equation in the previous video while proving that the total entropy change is greater than zero?(2 votes)

## Video transcript

In the video that I just did,
where I try to more rigorously prove the Gibbs free energy
relation, and that if this relation is less than zero
then this is spontaneous. I took great pains to make sure
that we use the proper definition of entropy. That every time that we said,
OK, a change in entropy from here to here is the heat
absorbed by a reversible process divided by the
temperature at which it was absorbed. And the change in entropy of
the environment is the opposite of that and, of course,
that is equal to zero. And I was very careful to
use this definition. And so you might have been
asking, hey Sal, there's a much simpler definition or
proof in my textbook. And I don't if it's in your
textbook, but it's in some of the ones that I've seen
and in some of the web pages I've looked at. Where they use a much simpler
argument that gets us eventually to this Gibbs
free energy relation. And I thought I would go over
it because as far as I can tell, it's incorrect. And what the argument tends to
go, is it says, look, the second law of thermodynamics
tells us, that for any spontaneous process, that delta
S is greater than zero. I agree with that completely
right now. And in order for delta S, and
that's delta S of the universe, is greater
than zero. And that means that delta S of
the system plus dealt S of the environment is going to
be greater than zero. and then this is the step that
you'll often see in a lot of textbooks and a lot of websites
that I disagree with. They'll say delta S of the
environment is equal to the heat or, let me say, the heat
absorbed by the environment, divided by the temperature
of the environment. And let's just say for
simplicity that everything here it's in some type of
temperature equilibrium. And it tends to be when we're
dealing with stuff in our chemistry sets in our
labs, whatever else. But the the reason why I
disagree with this step right here, that you see in a lot of
textbooks, is that this is not saying anything about the
reversibility of the reaction. You can only use this
thermodynamic definition of entropy if you know this heat
transfer is reversible. And when we're doing it in
general terms, we don't know whether it's reversal. In fact, if we're saying to
begin with that the reaction is spontaneous, that means
by definition that it's irreversible. So this is actually an
irreversible transfer of heat, which is not the definition
of entropy. The thermodynamic definition
of entropy is a very delicate one. You have to make sure that it
is a reversible reaction. Obviously, in a lot of first
year chemistry classes this doesn't matter. You're going to get the
question right. In fact, the question might be
dependent upon you making this incorrect assumption. So I don't want to confuse
you too much. But I want to show you that
this is not a right assumption. Because if you're assuming
something is spontaneous, and you're saying, OK, the change in
entropy in the environment is equal to the amount of heat
the environment absorbs, divided by T-- this is wrong
because this is not an irreversible reaction. But let's just see how this
argument tends to proceed. So they'll say, OK, this is
equal to delta S of our system plus the change in heat of our
environment, divided by the temperature of our
environment. They'll call this for the
environment, and that of course has to be
equal to zero. And they'll say, look, the
heat absorbed by the environment is equal to the
minus of the heat absorbed by the system. Right? It's either the system is
giving energy to the environment or the environment
is giving energy or heat to the system. So they are just going to be
the minus of each other. So the argument would go, well
this thing I can rewrite. This equation is the change in
entropy of the system, instead of writing a plus Q of the
environment here, I could write a minus Q of the system
over T is greater than zero. And then they multiply both
sides of this equation by T and you get T delta S of the
system minus the heat absorbed by the system is greater
than zero. You multiply both sides of this
by negative one and you get the heat absorbed by the
system minus the temperature times the change in entropy of
the system is greater than zero-- I'm sorry, is less than
zero when you multiply both sides by a negative, you
switch the signs. And then if you assume constant
pressure, this is the change in enthalpy
of the system. So you get the change in
enthalpy minus the temperature times delta S of the system
is less than zero. And they say, see this shows
that if you have a negative Gibbs free energy or change in
Gibbs free energy, then you're spontaneous. But all of that was predicated
on the idea that this could be rewritten like this. But it can't be rewritten like
that, because this is not a reversible process. We're starting from the
assumption that this is a spontaneous irreversible
process. And so you can't make this
substitution here. And that's why in the earlier
video I was very careful not to make that substitution. I was very careful to say, oh,
you know, the change in entropy of an irreversible
system that goes from here to here is the same as the change
in entropy as an irreversible system that goes from
there to there. Or let me say this
differently. The change in entropy of a
reversible system from there to there is the same as
an irreversible system from there to there. Although you don't know what
goes on in between for the irreversible. And so that's why I made
this comparison. This thing, and this thing are
the same, but then we compared the heat absorbed by an
irreversible system, and we showed that it's less than the
heat absorbed by a reversible system because it's generating
its own friction. And from that, we got this
relation, which we were able to then go and get the Gibbs
free energy relation. So, anyway, I don't want to make
a video that's too geeky or too particular, or
kind of trying to really pick at the details. But I think it's an important
point to make because so much of what we talk about,
especially in thermodynamics, is our definition of entropy. And it's very important we use
the correct one, and we don't take what I would argue are
incorrect shortcuts, because this is not the definition
of entropy right there.