Answer
(b) 1.53 kJ/mol
Work Step by Step
We find:
$Q=\frac{[C]}{[A][B]}=\frac{0.405}{(0.315)(0.315)}=4.08$
$T=(25+273)K=298\,K$
$R=8.314\,Jmol^{-1}K^{-1}$
$\Delta G^{\circ}=-1.95\,kJ/mol$
$\Delta G=\Delta G^{\circ}+RT\ln Q$
$=-1.95\,kJ/mol+(8.314\,Jmol^{-1}K^{-1})(298\,K)(\ln 4.08)$
$=-1.95\,kJ/mol+3480\,J/mol=-1.95\,kJ/mol+3.48\,kJ/mol$
$=1.53\,kJ/mol$
