Chemistry - Thermodynamics - II Question with Solution | TestHub

We can use the relationship between the equilibrium constant ($\mathrm{K}$) and the standard Gibbs free energy change ($\mathrm{\Delta G}°$) to calculate the value of $\log \mathrm{K}$ at $298 \mathrm{K}$

Given,

$$\begin{gathered} ∆{\mathrm{H}}^{\mathrm{o}} = – 54.07 \mathrm{kJ} {\mathrm{mol}}^{–1} \\ ∆{\mathrm{S}}^{\mathrm{o}} = 10 {\mathrm{JK}}^{–1} {\mathrm{mol}}^{–1} \end{gathered}$$

We know, 

$∆\mathrm{G}°=∆\mathrm{H}°-\mathrm{T}∆\mathrm{S}°$

$=-54.07-\frac{298\left(10\right)}{1000}$

$=-57.05 \mathrm{kJ}/\mathrm{mole}$

$∆\mathrm{G}°=-2.303\mathrm{RT} {\mathrm{logK}}_{\text{eq }}$

$-57.05\times 1000=-2.303\times 8.314\times 298 {\mathrm{logK}}_{\text{eq }}$

$-57.05\times 1000=-5705 {\log }_{\mathrm{eq}}$

$10={\mathrm{logK}}_{\text{eq }}$