### For the reaction $2NO_2 (g) \rightleftharpoons N_2O_4 (g)$ ....

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