Take R = 8.314 J mol^-1 K^-1 and T = 298 K. The standard Gibbs energies of formation are Delta_f G deg(NO2) = 51.3 and Delta_f G deg(N2O4) = 97.8 kJ mol^-1. For N2O4(g) 2NO2(g) (standard state 1 bar), a mixture has p_N2O4 = 0.50 bar and p_NO2 = 0.40 bar. The Gibbs energy change Delta G of the mixture and the direction of spontaneous change are

JEE Advanced levelSingle correct optionDifficulty 3 of 3ce157

Take $R = 8.314 J mo l^{- 1} K^{- 1}$ and $T = 298 K$ . The standard Gibbs energies of formation are $Δ_{f} G^{\circ} (NO X_{2}) = 51.3$ and $Δ_{f} G^{\circ} (N X_{2} O X_{4}) = 97.8 kJ mo l^{- 1}$ . For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ (standard state 1 bar), a mixture has $p_{N X_{2} O X_{4}} = 0.50$ bar and $p_{NO X_{2}} = 0.40$ bar. The Gibbs energy change $Δ G$ of the mixture and the direction of spontaneous change are

Δ G^{\circ} = 2 (51.3) - 97.8 = + 4.80 kJ mo l^{- 1}

, so

K = e^{- 4.80 \times 1000/ (8.314 \times 298)} = 0.1441

Q = (0.40)^{2} /0.50 = 0.32 > K

Δ G = Δ G^{\circ} + RT ln Q = 4.80 + (2.4777) (- 1.139) = + 1.98 kJ mo l^{- 1}

, which is positive, so the backward reaction is spontaneous.