Question

Starting with the thermodynamic relationship G = H + TS, derive the following relationship $\Delta G$ = -$T△S_{Total}$

Answer 1

$G=H-TS$

on differentiating

$\Delta G=\Delta H-T\left(\Delta S\right)-\left(\Delta T\right)S$

the $r\times n$ occurs at constant temperature, $\Delta T=0$

$\Delta G=\Delta H-T\left(\Delta S\right)⟶\left(1\right)$

We also know that entropy change of surroundings.

${\Delta S}_{surrounding}=\frac{-\Delta H}{T}$

$\Delta H=-T{\left(\Delta S\right)}_{surr}⟶\left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$

$\Delta G=-T.{\Delta S}_{surr}-T\Delta S$

$\Delta G=-T({\Delta S}_{surr}+\Delta S)$

we know, $({\Delta S}_{surr}+\Delta S)={\Delta S}_{total}$

$\Rightarrow \Delta G=-T{\left(\Delta S\right)}_{total}$

Thus, the relationship $\Delta G=-T{\left(\Delta S\right)}_{total}$ is derived.