# Spontaneity of a reaction with ∆G=0

## Homework Statement

Note: the following question is from a recent biochemistry exam, for which the grades have been posted and the exam questions and answers have been made available to students. I asked my professor about the answer to this question, and when I explained the issue, we resolved that we would investigate the question further to determine the correct answer.

I have a question regarding spontaneity:

1. Please select the TRUE statement regarding thermodynamics in reversible reactions:
a. When ∆G for a reaction is zero, the reverse and forward reactions will proceed spontaneously equally.
b. When ∆G for a reaction is more than zero, the reaction will proceed spontaneously in the reverse direction.
c. Both a and b are true.
d. Neither a nor b is true.

I think that the answer should be "b," but the key says that the answer is "c."

Answer "b" is clearly correct, but what about choice "a"? The question I am exploring is whether a reaction at equilibrium is (or can be) spontaneous. I thought that spontaneity is a process by which a system moves to a lower, more thermodynamically favorable energy state, and as such is controlled by the sign of ∆G.

Part of the issue is the clarification as to whether spontaneity describes the micro- or macro- state of the system. It is clear that chemical equilibrium is a dynamic process, and that at equilibrium the forward and reverse reaction rates are equal. (But "rates" takes us away from thermodynamics.)

So, when we say that ∆G<0 and a reaction is spontaneous in the forward direction, it is not spontaneous in the reverse direction, even though some molecules on a micro-state level may be converting in the reverse direction, the overall macro-state trend is that the system moves to a lower, more thermodynamically favorable energy state.

I am having a hard time finding links that address this question directly. Most say that ∆G determines spontaneity, and give the usual "if ∆G<0, spontaneous...∆G>0, non-spontaneous... ∆G=0, equilibrium." I want to know when ∆G=0, can we even say that such a reaction is spontaneous?

Please post what you think is the answer to the above multiple choice question is and any relevant support (such as textbook pages/reputable web pages) would be greatly appreciated.

∆G=H-T∆S

## The Attempt at a Solution

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It all hinges on how you define "spontaneous". The usual view would be that at delta G = 0 the reaction would still be spontaneous because the forward reaction can still proceed without any energy barrier stopping it. But that is only one view. Product formation would no longer be spontaneous because the forward reaction would be exactly matched by the reverse reaction. And arguing that the forward reaction can still proceed means "spontaneous" does not work, because even with delta G positive the forward reaction will still proceed; it is just that the reverse reaction proceeds faster.

I think that your answer is the better one; that "spontaneous" really has to mean making definite progress towards the products of a reaction. But there are good reasons in thermodynamics for regarding a process as spontaneous when delta G is actually equal to zero: they have to do with the concept of a "reversible" reaction as an ideal, where a system must be taken from 100% reactants to 100% products with delta G = 0 throughout.

I see the question as flawed anyway: Any answer other than d) should be marked as correct, since any of a), b), or c) are at least arguably true statements.

In reading through our textbook ("Biochemistry," 6th ed. by Berg et. al.) for our next exam, I found a review (on page 208) of spontaneity that was not listed in the index, and addresses the question I posted above.

Mainly, they emphasize that "A reaction can occur spontaneously only if ∆G is negative" and "A system is at equilibrium and no net change can take place if ∆G is zero."

Attached is a pdf of that page.

Hopefully this helps any other students with this question.

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