Answer
$1.5\times10^{7}$
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[2\Delta_{f}G^{\circ}(NOCl,g)]-[2\Delta_{f}G^{\circ}(NO,g)+\Delta_{f}G^{\circ}(Cl_{2},g)]$
$=[2(66.08\,kJ/mol)]-[2(86.55\,kJ/mol)+(0\,kJ/mol)]$
$=-40.94\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K_{p}$
$\implies K_{p}=e^{-\frac{\Delta _{r}G^{\circ}}{RT}}=e^{-\frac{-40.94\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$
$=1.5\times10^{7}$
