Chemistry - Thermodynamics - II Question with Solution | TestHub

Using formula

${\Delta }_{\mathrm{r}}{\mathrm{G}}^0=-{\mathrm{RTlnK}}_{\mathrm{p}}$

$25200=-2.3\times 8.3\times 400\log \left({\mathrm{K}}_{\mathrm{p}}\right)$

${\mathrm{K}}_{\mathrm{p}}={10}^{-3.3}={10}^{-3}\times 0.501$

$=5.01\times {10}^{-4}{\mathrm{Bar}}^{-1}$

$=5.01\times {10}^{-9}{ \mathrm{Pa}}^{-1}$

$=\frac{{\mathrm{K}}_{\mathrm{C}}}{8.3\times 400}$

${\mathrm{K}}_{\mathrm{C}}=1.66\times {10}^{-5}{ \mathrm{m}}^3/\mathrm{mole}$

$=1.66\times {10}^{-2} \mathrm{L}/\mathrm{mol}$

$=2$

Gibbs free energy can be visualized as the amount of useful energy present in a thermodynamic system that can be utilized to perform some useful work.