# For a given chemical reaction

Question:

For a given chemical reaction $\mathrm{A} \rightarrow \mathrm{B}$ at $300 \mathrm{~K}$ the free energy change is $-49.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the enthalpy of reaction is $51.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The entropy change of the reaction is $\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$.

Solution:

$\mathrm{A} \underset{\mathrm{T} 300 \mathrm{~K}}{\longrightarrow} \mathrm{B} \quad[\Delta \mathrm{G}]_{\mathrm{P}, \mathrm{T}}=-49.4 \mathrm{~kJ} / \mathrm{mol}$

$\Delta \mathrm{H}_{\mathrm{rxn}}=51.4 \mathrm{~kJ} / \mathrm{mol}$

$\Delta \mathrm{S}_{\mathrm{rxn}}=?$

$\Rightarrow$ From the relation $[\Delta \mathrm{G}]_{\mathrm{P}, \mathrm{T}}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$

$\Rightarrow \Delta \mathrm{S}_{\mathrm{rxn}}=\frac{\Delta \mathrm{H}_{\mathrm{rxn}}-[\Delta \mathrm{G}]_{\mathrm{P}, \mathrm{T}}}{\mathrm{T}}$

$=\frac{[51.4-(-49.4)] \times 1000}{300} \frac{\mathrm{J}}{\mathrm{mol} \mathrm{K}}$

$\Rightarrow \Delta \mathrm{S}_{\mathrm{rxn}}=336 \frac{\mathrm{J}}{\mathrm{molK}}$