Chemical Thermodynamics Part 3

Concept:Standard free energy change $\Delta_r G^\circ$ is the difference in Gibbs free energy between products and reactants under standard conditions.For the reaction $\mathrm{N_2O_4 \rightleftharpoons 2 NO_2}$, the graph plots $G$ against extent of reaction.Explanation:From the graph, $G_B^\circ$ (product state) is higher than $G_A^\circ$ (reactant state).Thus $\Delta_r G^\circ = G_B^\circ - G_A^\circ$ is positive, not $-5.40\ \mathrm{kJ\ mol^{-1}}$. Hence statement A is false.A positive $\Delta_r G^\circ$ means the forward reaction is non‑spontaneous under standard conditions, but some $\mathrm{N_2O_4}$ still dissociates to reach equilibrium. So statement B is false.At equilibrium (point E in graph), both $\mathrm{N_2O_4}$ and $\mathrm{NO_2}$ coexist. Reverse reaction does not go to completion. Hence statement C is false.When 1 mole of $\mathrm{N_2O_4}$ converts to equilibrium mixture, the graph shows $\Delta G^\circ = -0.84\ \mathrm{kJ\ mol^{-1}}$. So statement D is true.When 2 moles of $\mathrm{NO_2}$ convert to equilibrium mixture, the total $\Delta G^\circ$ is the sum of the free energy change from $\mathrm{NO_2}$ to pure $\mathrm{N_2O_4}$ ($-5.40\ \mathrm{kJ\ mol^{-1}}$) plus the change from pure $\mathrm{N_2O_4}$ to equilibrium mixture ($-0.84\ \mathrm{kJ\ mol^{-1}}$).Thus $\Delta G^\circ = -5.40 + (-0.84) = -6.24\ \mathrm{kJ\ mol^{-1}}$. So statement E is true.Answer:Correct statements are D and E only. Hence option B is correct.