Answer
(b) $7\times10^{5}$
Work Step by Step
We find:
$\Delta G^{\circ}=-RT\ln K_{p}$
$\implies \ln K_{p}=-\frac{\Delta G^{\circ}}{RT}=-\frac{-33.3\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}$
$=13.44$
$K_{p}=e^{13.44}=7\times10^{5}$
