Consider the reaction: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ where \(P_{\mathrm{NO}_{2}}=0.29\) atm and $P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm} .$ For this reaction at these conditions, \(\Delta G=-1000 \mathrm{J}\) and \(\Delta G^{\circ}=6000 \mathrm{J}\) . Which of the following statements about this reaction is(are) true? a. The reverse reaction is spontaneous at these conditions. b. At equilibrium, \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}\) will be greater than 1.6 \(\mathrm{atm}\) . c. The value of K for this reaction is greater than 1. d. The maximum amount of work this reaction can produce at these conditions is –6000 J. e. The reaction is endothermic.

The spontaneity of a reaction can be determined using the value of ΔG. If ΔG is negative, the reaction is spontaneous, and if ΔG is positive, the reaction is non-spontaneous. For this reaction, \(\Delta G=-1000 \mathrm{J}\). Since ΔG is negative, the reaction is spontaneous.

Let's calculate the reaction quotient (Q) using the given partial pressures of the reactants and products. $$Q = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$$ Plug in the given values: $$Q = \frac{(0.29 \mathrm{atm})^2}{1.6 \mathrm{atm}}$$ $$Q = 0.0521$$

We can find the equilibrium constant K using the formula: $$\Delta G = \Delta G^{\circ} + RT\ln{Q} $$ where R is the gas constant (8.314 J/(mol∙K)) Since ΔG < 0, we can rewrite the formula: $$\Delta G^{\circ} - \Delta G = - RT\ln{Q} $$ And we can find K using ΔG° and the formula: $$\Delta G^{\circ} = - RT\ln{K} $$ Now plug in the values: $$6000\mathrm{J} = 1000\mathrm{J} + 8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T\ln{Q} $$ $$5000\mathrm{J} = 8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T\ln{Q} $$ Now, we can find the value of K using the relation: $$\ln{K} = \frac{\Delta G^{\circ}}{- RT} $$ Plug in the values: $$\ln{K} = \frac{6000\mathrm{J}}{-8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T} $$ We need to find the temperature first to get the value of K. To find the temperature, we use equation $$5000\mathrm{J} = 8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot T\ln{0.0521} $$ $$ T = \frac{5000\mathrm{J}}{{-8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot \ln{0.0521}} = 236.82 \,\mathrm{K}$$ Now plug back the Temperature and find K: $$\ln{K} = \frac{6000\mathrm{J}}{-8.314\frac{\mathrm{J}}{\mathrm{mol} \cdot\mathrm{K}}\cdot 236.82 \,\mathrm{K}} $$ $$\ln{K} = -2.86$$ $$K = e^{-2.86} = 0.057$$

a. The reverse reaction is spontaneous at these conditions. False. Since ΔG is negative, the forward reaction is spontaneous. b. At equilibrium, \(P_{N_{2}O_{4}}\) will be greater than 1.6 atm. False. Since K < 1, the value of \(P_{N_{2}O_{4}}\) will be less than 1.6 atm at equilibrium. c. The value of K for this reaction is greater than 1. False. We calculated the value of K to be 0.057, which is less than 1. d. The maximum amount of work this reaction can produce at these conditions is –6000 J. False. The maximum amount of work that can be produced is given by ΔG, which is -1000 J. e. The reaction is endothermic. False. Since ΔG° is positive and K < 1, the reaction is exothermic. Based on our analysis, all of the given statements are false.