Answer
$K_{p}=5\times10^{17}$
When $\Delta_{r}G^{\circ}$ is negative, $K_{p}\gt1$
Work Step by Step
$\Delta _{r}G^{\circ}=-RT\ln K_{p}$
$\implies K_{p}=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{-100.97\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$
$=5\times10^{17}$
When $\Delta_{r}G^{\circ}$ is negative, $K_{p}\gt1$