Answer
(a) $K_{p}=6.2\times10^{-7}$
(b) $\Delta G=44.4\,kJ/mol$
Work Step by Step
(a) $\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[\Delta G_{f}^{\circ}(PCl_{3},g)+\Delta G_{f}^{\circ}(Cl_{2},g)]-[\Delta G_{f}^{\circ}(PCl_{5},g)]$
$=[(-269.6\,kJ/mol)+(0)]-(-305.0\,kJ/mol)=35.4\,kJ/mol$
$\Delta G^{\circ}=-RT\ln K_{p}\implies \ln K_{p}=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{35.4\times1000\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273.15)K}$
$=-14.3$
$K_{p}=e^{-14.3}=6.2\times10^{-7}$
(b) $Q=\frac{P_{PCl_{3}}P_{Cl_{2}}}{P_{PCl_{5}}}=\frac{(0.27)(0.40)}{0.0029}=37.24$
$\Delta G=\Delta G^{\circ}+RT\ln Q$
$=35.4\,kJ/mol+(8.314\,Jmol^{-1}K^{-1})(298.15\,K)(\ln 37.24)$
$=35.4\,kJ/mol+(8.97\times10^{3}\,J/mol)$
$=35.4\,kJ/mol+8.97\,kJ/mol$
$=44.4\,kJ/mol$
