Gibbs free energy in chemical reaction

I know this kind of question has been asked many times but I think this is a bit different.


We have $G = H - TS$. Here I am talking about chemical reactions at constant pressure and temperature. We can write $\Delta G = Q - Q\_\mathrm{rev} = \Delta H - T \Delta S$.


For an irreversible chemical reaction, for example an exothermic one, if $q = -10~\mathrm{kJ~mol^{-1}}$ (exothermic), then $Q\_\mathrm{rev}$ could be $-9$, $-8$, $-7$, etc. The difference $(Q – Q\_\mathrm{rev})$ would be negative ($-1$, $-2$, $-3$, etc.). This represents the portion of $Q$ that can be converted into non-PV work. The remainder of the heat is dispersed into the surrounding and is wasted. Right?


So here $Q$ as absolute value is more than $Q\_\mathrm{rev}$, so we have more heat generated than a reversible chemical reaction that can be used as work ($Q = 8~\mathrm{kJ~mol^{-1}}$ and $Q\_\mathrm{rev} = 2~\mathrm{kJ~mol^{-1}}$). So why is it said that a reversible reaction produces more useful heat? Again, I am talking about chemical reaction and not mechanical process.


