- Gibbs free energy and spontaneity
- Gibbs free energy and spontaneity
- Gibbs free energy example
- More rigorous Gibbs free energy / spontaneity relationship
- A look at a seductive but wrong Gibbs spontaneity proof
- Changes in free energy and the reaction quotient
- Standard change in free energy and the equilibrium constant
- 2015 AP Chemistry free response 2c
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.
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- At3:38he says "equal to zero", but he writes "> 0". Which one is right?(11 votes)
- Can someone explain how this proof is different than the one he gives? Is there a distinction between Q_s and Q_ir?(7 votes)
- Sal's example substitutes the above-mentioned equation (Qenvironment/T) into the equation for the reversible reaction. Not the irreversible one. For the irreversible one he just acknowledges that the environment puts less Q in, because the irreversible reaction generates some of its own heat through friction.
To use the above-mentioned Q_environment with irreversible processes, we first have to define the Q_environment/T term as being less than it appears to be due to friction. When we define it as being less we end up subtracting this smaller number from Delta-S_irreversible system term, and getting an answer > 0. (as the previous proof demonstrated)
So to answer your second question, there isn't really a distinction between Q_s and Q_ir, the distinction happens when we look at the magnitude of the Q_environment. In Sal's proof, he takes friction into account by saying that Q_environment will be lower than Q_s. The above example doesn't do this.(11 votes)
- 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!)(12 votes)
- I am confused now. because Sal had used the Q/T formula for the spontaneous irreversible process in one of his previous videos.
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.(6 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.(3 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?(1 vote)
- 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(1 vote)
- 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.(1 vote)
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.