Answer
$7.2\times10^{24}$
Work Step by Step
We find:
$\Delta G^{\circ}=-RT\ln K_{p}$
$\implies \ln K_{p}=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{-141.8\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}=57.23345$
$K_{p}=e^{57.23345}=7.2\times10^{24}$
