Question

Consider the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) at \(100^{\circ} \mathrm{C}\). $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad K=11 $$ (a) What is \(\Delta G^{\circ}\) at that temperature? (b) What is \(K\) at \(25^{\circ} \mathrm{C}\) ? \(\left(\Delta G_{\mathrm{f}}^{\circ}\right.\) for \(\left.\mathrm{N}_{2} \mathrm{O}_{4}=97.9 \mathrm{~kJ} / \mathrm{mol}\right)\)

Short answer

In summary: (a) The standard Gibbs free energy change (ΔG°) for the dissociation of N2O4 into NO2 at 100°C is approximately -24.663 kJ/mol. (b) The equilibrium constant K for the same reaction at 25°C is approximately 1.30.

Step-by-step solution

(a) Finding the standard Gibbs free energy change (ΔG°) at 100°C

We know that ΔG° and K are related by the equation: ΔG° = -RT ln(K) Where R is the gas constant (8.314 J/mol*K), T is the temperature in Kelvin, and K is the equilibrium constant. First, convert the temperature to Kelvin: T = 100°C + 273.15 = 373.15 K Now, plug the values into the equation: ΔG° = - (8.314 J/mol*K) * 373.15 K * ln(11) ΔG° = -24662.73 J/mol Convert the value to kJ/mol: ΔG° = -24.663 kJ/mol

(b) Finding the equilibrium constant K at 25°C

To find K at the new temperature, we can use the Van't Hoff equation: $$ \frac{-\Delta H}{R}=\frac{\ln \frac{K_{2}}{K_{1}}}{\frac{1}{T_{2}}-\frac{1}{T_{1}}} $$ But we don't have the enthalpy change (ΔH) directly. To find ΔH, we can use the relationship: ΔG° = ΔH - TΔS We know the standard Gibbs free energy at 298K (25°C) for N2O4, so let's determine the value of ΔG° at 298K: ΔG°(298 K) = 97.9 kJ/mol We also know the ΔG° at 373.15K (100°C), which we found in Step 1. Now, we can set up two equations to solve for ΔH and ΔS: 1) ΔG°(298 K) = ΔH - 298.15*ΔS 2) ΔG°(373.15 K) = ΔH - 373.15*ΔS Solve these equations simultaneously to find ΔH and ΔS: ΔH = 78.750 kJ/mol ΔS = 0.2677 kJ/mol*K Now, we can use the Van't Hoff equation to find K at 25°C: $$ \frac{-\Delta H}{R}=\frac{\ln \frac{K_{2}}{K_{1}}}{\frac{1}{T_{2}}-\frac{1}{T_{1}}} $$ Plug in the values: $$ \frac{-78750 J/mol}{8.314 J/mol*K}=\frac{\ln \frac{K_{2}}{11}}{\frac{1}{298.15 K}-\frac{1}{373.15 K}} $$ Solve for K2: K2 = 1.30 So the equilibrium constant K at 25°C is approximately 1.30.