Answer
$K=5\times10^{-31}$
Work Step by Step
$\Delta_{r}G^{\circ}=\Sigma n_{p}\Delta_{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[2\Delta_{f}G^{\circ}(NO,g)]-[\Delta_{f}G^{\circ}(N_{2},g)+\Delta_{f}G^{\circ}(O_{2},g)]$
$=[2(86.55\,kJ/mol)]-[(0)+(0)]$
$=173.1\,kJ/mol$
$\Delta _{r}G^{\circ}=-RT\ln K$
$\implies K=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{173.1\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298.15\,K)}}$
$=5\times10^{-31}$
