ΔG is zero at equilibrium but ΔG° = -RT ln K is not, why? Also, where's the third law?

7 answers & replies

• [+][–]Ishita GoyalDropperSample· 29d ago

  $\Delta G$ is for the actual mixture at a given moment, $\Delta G^\circ$ is for the standard state (everything at standard conditions, 1 bar). They are connected by

  $\Delta G = \Delta G^\circ + RT\ln Q$

  At equilibrium $Q = K$ and $\Delta G = 0$, so $0 = \Delta G^\circ + RT\ln K$, which gives $\Delta G^\circ = -RT\ln K$. Not a contradiction, just two different quantities. $\Delta G$ tells you which way the reaction goes from where you are. $\Delta G^\circ$ tells you where equilibrium lies.

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  • [+][–]Rahul DubeyDropperSample· 29d agoAuthor

    Clear. ΔG tells direction, ΔG° tells where equilibrium sits. Writing this as a headline in my notebook.

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• [+][–]Tushar PawarDropperSample· 29d ago

  I think ΔG° is also zero at equilibrium, which makes K = 1, so at equilibrium reactants and products are in equal amounts. It's the same condition really.

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• [+][–]Siddharth RaoDropperSample· 29d ago

  Tushar, ΔG° = 0 only when K = 1. That's one special case. For most reactions K is far from 1. Example at 298 K: $\Delta G^\circ = -2.303RT\log K$ and $2.303RT \approx 5.7$ kJ/mol, so for $K = 10^5$ you get $\Delta G^\circ \approx -28.5$ kJ/mol, clearly not zero, yet that reaction sits at equilibrium with ΔG = 0 when the mixture reaches Q = K. Large negative ΔG° means K is large. Positive ΔG° means K < 1.

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  • [+][–]Tushar PawarDropperSample· 29d ago

    ah ok, ΔG° = 0 only if K = 1. I mixed ΔG and ΔG° again. thanks!

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• [+][–]Kavya NaikDropperSample· 29d ago

  Your numerical, in the right units. At the switchover temperature $\Delta G = 0$:

  $T = \frac{\Delta H}{\Delta S} = \frac{30000\ \text{J/mol}}{100\ \text{J/(K mol)}} = 300\ \text{K}$

  So above 300 K, $-T\Delta S$ wins and the reaction is spontaneous. Always convert kJ to J before dividing, that was my mistake too, amma kept scolding me.

  Summary of signs: $\Delta H -$, $\Delta S +$ always spontaneous. $\Delta H +$, $\Delta S -$ never. Same signs, depends on temperature (both $+$: high T favours, both $-$: low T favours).

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• [+][–]Lakshmi TripathiDropperSample· 29d ago

  Third law matters because it lets us have absolute entropies. At 0 K a perfect crystal has only one possible arrangement (one microstate), so $S = 0$. From there you measure heat capacities up in temperature and get an absolute value of $S$ at 298 K. That's why tables list standard molar entropies $S^\circ$ as actual positive numbers. Enthalpy has no such zero point, so tables give $\Delta H$ of formation only, never absolute $H$. That's the reason you can calculate $\Delta S^\circ_{reaction}$ = sum of $S^\circ$ products − sum of $S^\circ$ reactants.

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