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I'm confused about two equations and how they relate to each other. These are $\Delta S = \Delta Q/T$ and $\Delta G = \Delta H - T\Delta S$.

To derive the Gibbs' free energy change for the universe you assume a transfer of heat $\Delta Q$ from system to surroundings. The entropy change in the surroundings is then $+\Delta Q/T_{surr}$ and in the system it is $-\Delta Q/T_{syst}$.

Then by stating: $\Delta S_{univ} = \Delta S_{surr} + \Delta S_{syst}$ we get $\Delta S_{univ} = \Delta Q/T_{surr} + \Delta S_{syst}$.

Then assuming $\Delta Q_{syst} = -\Delta H_{syst}$ we get get $T_{surr}\Delta S_{univ} = -\Delta H_{syst} + T_{surr}\Delta S_{syst}$. $-T_{surr}\Delta S_{univ} = \Delta H_{syst} - T_{surr}\Delta S_{syst}$ so that $\Delta G_{univ} = \Delta H_{syst} - T_{surr}\Delta S_{syst}$.

My confusion stems from what happens to G in the system and surroundings when this heat transfer happens. If $\Delta G_{syst} = \Delta H_{syst} - T_{syst}\Delta S_{syst}$, wouldn't $\Delta G_{syst}$ have to equal zero, so that $-\Delta Q = T_{syst}\Delta S_{syst}$, i.e. so that $\Delta H_{syst} = T_{syst}\Delta S_{syst}$. Same for the surroundings.

But then if that is the case how can $\Delta G_{univ}$ change if both $\Delta G_{surr}$ and $\Delta G_{syst}$ are both zero? What am I missing here?

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  • $\begingroup$ there is no $\Delta Q$, just $Q$, containing $\Delta$ implicitly. // Rather $\mathrm{d}S \ge \dfrac{\delta Q}{T}$ $\endgroup$
    – Poutnik
    Commented Apr 9 at 21:16

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