Let's solve the problem step by step: --- ## **a. When is the reaction spontaneous?** ### **Step 1: Write the Gibbs Free Energy Equation** \[ \Delta G = \Delta H - T\Delta S \] - **\(\Delta H^\circ_{rxn} = -72.48 \text{ kJ} = -72480 \text{ J}\)** - **\(\Delta S^\circ_{rxn} = -324.8 \text{ J/K}\)** A reaction is spontaneous when \(\Delta G < \). --- ### **Step 2: Set up the inequality for spontaneity** \[ \Delta G < \implies \Delta H - T\Delta S < \] \[ -72480 - T(-324.8) < \] \[ -72480 + 324.8T < \] \[ 324.8T < 72480 \] \[ T < \frac{72480}{324.8} \] \[ T < 223.2 \text{ K} \] #### **Explanation:** - The reaction is spontaneous **only when the temperature is less than 223.2 K**. --- ## **b. Calculate \(\Delta G\) and \(K_{eq}\) at 137°C** ### **Step 1: Calculate \(\Delta G\) at 137°C** - \(T = 137^\circ C = 137 + 273.15 = 410.15\) K \[ \Delta G = \Delta H - T\Delta S \] \[ \Delta G = -72480\ \text{J} - (410.15\ \text{K})(-324.8\ \text{J/K}) \] \[ \Delta G = -72480 + 133282.072 \] \[ \Delta G = 60799.1\ \text{J} = 60.8\ \text{kJ} \] #### **Explanation:** - The Gibbs free energy at 137°C is **+60.8 kJ** (non-spontaneous at this temperature). --- ### **Step 2: Calculate the equilibrium constant (\(K_{eq}\))** \[ \Delta G = -RT \ln K_{eq} \] \[ 60,799.1\ \text{J} = -(8.314\ \text{J/mol·K})(410.15\ \text{K}) \ln K_{eq} \] \[ 60,799.1 = -3411.5 \ln K_{eq} \] \[ \ln K_{eq} = \frac{60,799.1}{-3411.5} = -17.83 \] \[ K_{eq} = e^{-17.83} \approx 1.80 \times 10^{-8} \] #### **Explanation:** - The equilibrium constant at 137°C is **\(K_{eq} \approx 1.80 \times 10^{-8}\)**, indicating the reaction heavily favors reactants. --- ## **Final Answers** ### **a.** The reaction is spontaneous when **\(T < 223.2\) K**. ### **b.** At 137°C (**410.15 K**): - **\(\Delta G = +60.8 \text{ kJ}\)** - **\(K_{eq} = 1.80 \times 10^{-8}\)**