For the reaction, N_2O_4 rightleftharpoons 2NO_2$N_{2}O_{4} \rightleftharpoons 2NO_{2}$, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is -5.40text kJ mol^-1$-5.40\text{ kJ mol}^{-1}$. B. As Delta G^ominus$\Delta G^{\ominus}$ in graph is positive, N_2O_4$N_{2}O_{4}$ will not dissociate into NO_2$NO_{2}$ at all. C. Reverse reaction will go to completion. D. When 1 mole of N_2O_4$N_{2}O_{4}$ changes into equilibrium mixture, value of Delta G = -0.84text kJ mol^-1$\Delta G = -0.84\text{ kJ mol}^{-1}$. E. When 2 mole of NO_2$NO_{2}$ changes into equilibrium mixture, Delta G$\Delta G$ for equilibrium mixture is -6.24text kJ mol^-1$-6.24\text{ kJ mol}^{-1}$.

Choose the correct answer from the options given below :

### Core Logic Let's analyze the statements based on the given Gibbs free energy (G) vs. extent of reaction plot.

A. Standard free energy change (Delta_r G^circ$\Delta_r G^{\circ}$) is the difference between standard free energies of pure products (point B) and pure reactants (point A): Delta_r G^circ = G_B^circ - G_A^circ$\Delta_r G^{\circ} = G_B^{\circ} - G_A^{\circ}$. Since B is higher than A, Delta_r G^circ$\Delta_r G^{\circ}$ is positive, not -5.40text kJ mol^-1$-5.40\text{ kJ mol}^{-1}$. Statement A is false. B. Even if Delta_r G^circ$\Delta_r G^{\circ}$ is positive, the minimum of the curve (equilibrium state E) lies between the pure reactant and product states. Therefore, partial dissociation occurs to reach equilibrium. It is false to say it will not dissociate at all. Statement B is false. C. The minimum E is not at fraction = 0 or 1, meaning an equilibrium mixture exists. Reverse reaction does not go to completion. Statement C is false. D. From 1 mole of pure reactant (point A) to the equilibrium mixture (point E), the drop in Gibbs energy is 0.84text kJ mol^-1$0.84\text{ kJ mol}^{-1}$. Thus Delta G = -0.84text kJ mol^-1$\Delta G = -0.84\text{ kJ mol}^{-1}$ is correct. Statement D is true. E. The difference from pure products (point B, equivalent to 2 moles NO_2$NO_2$) to equilibrium (E) is the entire vertical distance. The drop from B to A is 5.40$5.40$, and A to E is 0.84$0.84$. So drop from B to E is - (5.40 + 0.84) = -6.24text kJ mol^-1$- (5.40 + 0.84) = -6.24\text{ kJ mol}^{-1}$. Statement E is true. ### Step 1: Final Conclusion Only statements D and E are correct. ### Pattern Recognition The free energy of mixing creates a minimum 'dip' (Equilibrium) below both pure reactants and pure products, preventing either forward or reverse reactions from going absolutely to 100% completion. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics Class 11 Chemistry: Equilibrium