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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!)
• i still dont understand free energy...
• 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?
• 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.
• is the equation G=H-TS applicable to all processes?
• If he wants to be really rigorous, then the assumption of constant pressure is false, and he can't derive the term \DeltaH
• 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 at : “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.
• … 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?
• 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
• 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.