Answer
$-106\,kJ/mol$
Work Step by Step
$\Delta _{r}G^{\circ}=-RT\ln K=-RT\ln K_{p}$
$=-(8.314\,Jmol^{-1}K^{-1})(298\,K)(\ln 4.4\times10^{18})$
$=-106\times10^{3}\,J/mol=-106\,kJ/mol$
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