Answer
a) $1.48\times10^{90}$
b) $2.09\times10^{-26}$
Work Step by Step
a) $\Delta G^{\circ}=\Sigma n_{p}\Delta G_{f}^{\circ}(products)-\Sigma n_{r}\Delta G_{f}^{\circ}(reactants)$
$=[2\Delta G_{f}^{\circ}(CO_{2}, g)]-[2\Delta G_{f}^{\circ}(CO,g)+\Delta G_{f}^{\circ}(O_{2},g)]$
$=[2(-394.4\, kJ)]-[2(-137.2\, kJ/mol) +(0)]$
$=-514.4\, kJ/mol$
$\Delta G^{\circ}= -RT\ln K\implies \ln K=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{-514.4\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}$
$=207.6226$
$K=e^{207.6226}=1.48\times10^{90}$
b) $\Delta G^{\circ}= [2\Delta G_{f}^{\circ}(H_{2}, g)+\Delta G_{f}^{\circ}(S_{2},g)]-[2\Delta G_{f}^{\circ}(H_{2}S,g)]$
$=[2(0\, kJ/mol) +(79.7\, kJ/mol)]-[2(-33.4\, kJ/mol)]$
$=146.5\, kJ/mol$
$\ln K=- \frac{146.5\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}$
$=-59.13$
$K= e^{-59.13}=2.09\times10^{-26}$
