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We know that ,

$ \Delta G = \Delta H - T \Delta S$

We know that $ \Delta H = q = T \Delta S$

so,

$\Delta G = 0$

Where is my mistake here?

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  • $\begingroup$ the formulas are for the change in Gibbs free energy yet the title talks about the value of the Gibbs free energy - which do you mean? $\endgroup$ – user245141 May 11 at 11:04
  • $\begingroup$ yeah made the question more accurate $\endgroup$ – DDD4C4U May 11 at 11:20
  • $\begingroup$ Essentially answered here: physics.stackexchange.com/questions/218068/… $\endgroup$ – ratsalad May 12 at 14:21
  • $\begingroup$ It is close but the thing is that in that answer he explains why it doesn't work in that case. I am trying to find the reason why it would work in the general case compared to that case. $\endgroup$ – DDD4C4U May 12 at 19:33
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The change in Gibbs free energy is zero for a reversible process (as long as no non-PV work is being done). The equivalence that you provide, $\Delta H = q = T \Delta S$, is only valid for reversible processes at constant $T$ and $P$. If the process is irreversible then $T \Delta S \neq q$, and thus $\Delta G \neq 0$.

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