• # question_answer 96)   $\Delta G$is net energy available to do useful work and is thus a measure of "free energy". Show mathematically that $\Delta G$ isa measure of free energy. Find the unit of$\Delta G$. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?

According to Gibbs Helmholtz equation: $\Delta G=\Delta H-T\Delta S$ $\Delta G=\Delta U+P\Delta V-T\Delta S\,\,\,\,\,\,\,\,\,\,.....(i)$ From second law of thermodynamics: $\Delta S=\frac{q}{T}$ (Reversible process)                    ... (ii) From first law of thermodynamics:                                 $\Delta U=q-{{w}_{rev}}$ ${{w}_{rev}}$ = Reversible work done by the system From (ii) and (iii) $\Delta S=\frac{\Delta U+{{w}_{rev}}}{T}$ or            $T\Delta S=\Delta U+{{w}_{rev}}$ Putting the value of $T\Delta S$ in (i) we get $\Delta G=\Delta U+P\Delta V-[\Delta U+{{w}_{rev}}]$ $=P\Delta V-{{w}_{rev}}$ $\mathbf{-\Delta G=}{{\mathbf{w}}_{\mathbf{rev}}}\mathbf{-P\Delta V}$ Thus, decrease in Gibbs free energy is measure of reversible work done by the system exclusive of work of volume expansion. From (i). If$\Delta H=+ve$, and $\Delta S=+ve,$ then process will be spontaneous when $T\Delta S>\Delta H$ ; in this case $\Delta G$ will be negative.

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