Chemistry - Thermochemistry Question with Solution | TestHub

Gibbs Helmholtz Equation $\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T}\Delta \mathrm{S}$ ....(1) differentiate this equation w.r.t. temperature at constant pressure $\begin{array}{l}{\left(\frac{\partial \Delta \mathrm{G}}{\partial \mathrm{T}}\right)}_{\mathrm{P}}={\left(\frac{\partial {\mathrm{G}}_{\mathrm{y}}}{\partial \mathrm{T}}\right)}_{\mathrm{P}}-{\left(\frac{\partial {\mathrm{G}}_{\mathrm{x}}}{\partial \mathrm{T}}\right)}_{\mathrm{P}}...(2)\\ =-{\mathrm{S}}_{\mathrm{y}}-\left(-{\mathrm{S}}_{\mathrm{x}}\right)\\ =-\left({\mathrm{S}}_{\mathrm{y}}-{\mathrm{S}}_{\mathrm{x}}\right)=-\Delta \mathrm{S}...(3)\end{array}$ where $\Delta \mathrm{S}$ change in entropy on combining equation (1) \& (3) we get $\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}{\left(\frac{\partial (\Delta \mathrm{G})}{\partial \mathrm{T}}\right)}_{\mathrm{P}}$ ...(4) Equation (4) is on alternative form of Gibbs Helmholtz equation Dividing equ. (4) by ${\mathrm{T}}^2$, we get $\frac{\Delta \mathrm{G}}{{\mathrm{T}}^2}=\frac{\Delta \mathrm{H}}{{\mathrm{T}}^2}+\frac{1}{\mathrm{T}}{\left[\frac{\partial (\Delta \mathrm{G})}{\partial \mathrm{T}}\right]}_{\mathrm{P}}$ on rearrangement, we get $\begin{array}{l}{\left[\frac{\partial (\Delta \mathrm{G})}{\partial \mathrm{T}}\right]}_{\mathrm{P}}=-\frac{\Delta \mathrm{H}}{{\mathrm{T}}^2}\\ \therefore \Delta \mathrm{H}=-{\mathrm{T}}^2{\left[\frac{\partial (\Delta \mathrm{G})}{\partial \mathrm{T}}\right]}_{\mathrm{P}}\end{array}$