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.