Is it possible to express Gibbs free energy as following?

We know $$\Delta G =\Delta H -T\Delta S$$
$$\Delta S= \frac{Q\_{rev}}{T} $$
Hence $$\Delta G=\Delta H-Q\_{rev} $$




Yes, but only if the temperature T in the initial and final equilibrium states of the system is the same *and* you are able to identify a reversible path between the initial and final equilibrium states over which all the heat transfer takes place at the temperature T.


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