Question

The equilibrium constant for the reaction given below is $2.0\times {10}^{-7}$ at 300 K. Calculate the standard free energy change for the reaction;
$PC{l}_{5\left(g\right)}\rightleftharpoons PC{l}_{3\left(g\right)}+C{l}_{2\left(g\right)}$
Also, calculate the standard entropy change if $△H^o=28.40kJmo{l}^{-1}.$

• A. $△G^o=40kJmo{l}^{-1}$ $△S^o=30J{K}^{-1}mo{l}^{-1}$
• B. $△G^o=38.48kJmo{l}^{-1}$ $△S^o=-33.6J{K}^{-1}mo{l}^{-1}$
• C. $△G^o=45kJmo{l}^{-1}$ $△S^o=-35J{K}^{-1}mo{l}^{-1}$
• D. $△G^o=42kJmo{l}^{-1}$ $△S^o=40J{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is B $△G^o=38.48kJmo{l}^{-1}$ $△S^o=-33.6J{K}^{-1}mo{l}^{-1}$

The formula which relate Gibbs free energy $\Delta G^0$ and equilibrium constant $K_{eq}$ is $\Delta G^0=-2.303\times R\times T\times lo{g}_{10}\left[K_{eq}\right]$
$△G^o=-2.303\times 8.314\times 300lo{g}_{10}[2.0\times {10}^{-7}]$
$=+38479.8Jmo{l}^{-1}$
$=+38.48kJmo{l}^{-1}$
Also, $△G^o=△H^o-T△S^o$
$\therefore △S^o=\frac{△H^o-△G^o}{T}$
$△S^o=\frac{28.40-38.48}{300}$
$△S^o=-0.0336kJ{K}^{-1}mo{l}^{-1}=-33.6J{K}^{-1}mo{l}^{-1}$