Making sense of ∆G and ∆G°, when it comes to equilibrium

January 31, 2012
Categories: AP | AP TOPIC 13 | Big Idea 5

Consider the two** equations that deal with Delta G (∆G).

**Since this post was originally written in January 2012, the AP exam has changed. One of the changes was to remove equation #2 below from the equations & constants sheet. As such, I think that knowledge of it, and the consequences associated with it, are unlikely to be tested quantitatively on the exam in the future, but nevertheless, I still feel that understanding that conditions other than standard ones will cause ∆G to take on new values is a useful reference point.

Equation 1:

∆G° = – RT ln K

Since K is the equilibrium constant, we are at equilibrium, the amounts of products and reactants in the mixture are fixed, and the sign of ∆G° can be thought of as a guide to the ratio of the amount of products to the amount of reactants at equilibrium and therefore the thermodynamic favorability of the reaction.

If it so happens that products and reactants are equally favored at equilibrium, then ∆G° is zero, BUT ∆G° is not *necessarily* ZERO at equilibrium.

Equation 2:

∆G = ∆G° + RT ln Q

Since Q is NOT the K, and we are NOT necessarily at the equilibrium position, the sign of ∆G can be thought of as a predictor about which way the reaction (that has reactants and products defined by Q), will go.

If ∆G° is negative at equilibrium, then we will have lots of products at equilibrium, meaning Q needs to be bigger (greater than 1) to approach K. As Q gets larger (i.e., as we get more products), the term ‘RT ln Q’ gets increasingly positive, and eventually adding that term to a negative ∆G°, will make ∆G = 0, equilibrium will be established and no further change occurs.

It is possible that Q could already be too large and therefore ∆G is positive. IF so, then the reaction will need to from more reactants, reduce the value of Q, and allow ∆G to reach zero, i.e., allow equilibrium to be established.

If ∆G° is positive at equilibrium, then we will have lots of reactants at equilibrium, meaning Q needs to be smaller (less than 1) to approach K. As Q gets smaller (i.e., as we get more reactants), the term ‘RT ln Q’ gets increasingly negative, and eventually adding that term to a positive ∆G°, will make ∆G = 0, equilibrium will be established and no further change occurs.

It is possible that Q could already be too small and therefore ∆G is negative, IF so, then the reaction will need more products, increase the value of Q, and allow ∆G to reach zero, i.e., allow equilibrium to be established.

In short, it is ∆G (NOT ∆G°) that will be zero at equilibrium and the sign of it (generated by the combination of ∆G° and RT ln Q in Equation #2), will define which way the reaction proceeds.

2 Comments

  1. Jerome Smith

    But delta G naught is the delta G at standard condition. 25 C, 1 atm. So delta G naught is constant for a given reaction. It is related to K at the equilibrium temp since then delta G is 0.

    Reply
    • Daniel

      The deviation of delta G from delta G0 is given by: delta G = delta G0 + RTlnQ, where Q = product/reactants expression.

      When Equilibrium is obtained, delta G = 0, So delta G0 = -RTln K, ie. Q is now K as Q was for non equilibrium.

      So delta G0 is not necessarily 0 be it at 25 deg C and 1 atmosphere or otherwise. Delta G = 0 at equilibrium, not delta G0.

      Reply

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