QJEEM21Ch52d5: JEE Main 2021 Chemistry - Class 11 JEEM21Ch52d5 Solution: For the reaction, 2NO 2 (g) N 2 O 4 (g), when S = 176.0 JK 1…

Given, $\Delta$ H = $-$ 57.8 kJ mol ^{$-$ 1}

$\Delta$ S = $-$ 176 JK ^{$-$ 1} mol ^{$-$ 1}

T = 298 K

Using Gibb's free energy relation

$\Delta$ G = $\Delta$ H $-$ T $\Delta$ S

where, $\Delta$ G = change in Gibb's free energy

$\Delta$ H = change in enthalpy

T = temperature

$\Delta$ S = change in entropy

$\Delta$ G = 57.8 kJ/mol $-$ [298 K $\times$ ( $-$ 176 Jk ^{$-$ 1} mol ^{$-$ 1})]

= 57.8 kJ/mol $-$ $\left( {298 \times {{ - 176} \over {1000}}kJ} \right)$ [ $\therefore$ 1 kJ = 1000 J]

= $-$ 5.352 kJ/mol

| $\Delta$ G | = 5.352

Hence, answer is 5.