For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g) $, when $\Delta S = -176.0 JK^{-1} $ and $\Delta H = -57.8 kJ mol^{-1} $, the magnitude of $\Delta G $ at 298 K for the reaction is _ _ _ _ $kJ mol^{-1} $. (Nearest integer)

Solution

We have, $\Delta G = \Delta H – T \Delta S $

$\Delta S = -176.0 JK^{-1} = -0.176 kJ K^{-1} $

$\Delta G =  -57.8 – 298 \times (-0.176) \approx -57.8+52.5 = -5.3 $

$\therefore |\Delta G | = 5.3  kJ mol^{-1} $

Ans: 5