Let's solve the problem **step by step**. ## **Given Reaction** \[ 2 \text{FeS}_2(s) + \frac{11}{2} \text{O}_2(g) \rightarrow \text{}_2\text{O}_3(s) +4 \textSO}_2) \] We are to **ermine the entpy of reaction 550°C** (which is 823.15 K). --- ## **Step 1: Standard Enthalpy of Reaction at 298.15 K** \[ \Delta H^\circ_\text{rxn, 298} = \left[ \sum n_p \Delta H_f^\circ(\text{products}) \right] - \left[ \sum n_r \Delta H_f^\circ(\text{reactants}) \right] \] Values from the table (in kJ/mol): - FeS₂(s): -167.36 - Fe₂O₃(s): -824.20 - SO₂(g): -296.81 - O₂(g): Plug in the coefficients: \[ \Delta H^\circ_\text{rxn, 298} = \left[1 \times (-824.20) + 4 \times (-296.81)\right] - \left[2 \times (-167.36) + 5.5 \times \right] \] \[ = \left(-824.20 + (-1187.24)\right) - (-334.72) \] \[ = (-2011.44) - (-334.72) \] \[ = -2011.44 + 334.72 \] \[ = -1676.72 \text{ kJ} \] --- ## **Step 2: Find the Enthalpy at 550°C (823.15 K)** ### **a. Apply Kirchhoff's Law:** \[ \Delta H^\circ_\text{rxn, T_2} = \Delta H^\circ_\text{rxn, T_1} + \int_{T_1}^{T_2} \Delta C_p \, dT \] Assume \(C_p\) is constant over this range: \[ \Delta H^\circ_\text{rxn, 823.15} = \Delta H^\circ_\text{rxn, 298.15} + \Delta C_p \cdot (823.15 - 298.15) \] --- ### **b. Calculate the Change in Heat Capacity (\(\Delta C_p\)):** From the table (\(C_{p,m}\) in J/mol·K): - FeS₂(s): 62.19 - Fe₂O₃(s): 103.9 - SO₂(g): 39.90 - O₂(g): 29.4 \[ \Delta C_p = \left[1 \times 103.9 + 4 \times 39.90\right] - \left[2 \times 62.19 + 5.5 \times 29.4\right] \] Calculate each part: **Products:** \[ 1 \times 103.9 = 103.9 \] \[ 4 \times 39.90 = 159.6 \] \[ 103.9 + 159.6 = 263.5 \] **Reactants:** \[ 2 \times 62.19 = 124.38 \] \[ 5.5 \times 29.4 = 161.7 \] \[ 124.38 + 161.7 = 286.08 \] \[ \Delta C_p = 263.5 - 286.08 = -22.58 \text{ J/K} \] --- ### **c. Calculate the Correction Term** \[ \Delta T = 823.15 - 298.15 = 525 \text{ K} \] \[ \Delta C_p \cdot \Delta T = -22.58 \times 525 = -11,359.5 \text{ J} \] Convert to kJ: \[ -11,359.5 \text{ J} = -11.36 \text{ kJ} \] --- ### **d. Calculate Final Enthalpy** \[ \Delta H^\circ_\text{rxn, 823.15} = -1676.72 \text{ kJ} + (-11.36 \text{ kJ}) \] \[ = -1688.08 \text{ kJ} \] --- ## **Final Answer** ### **The enthalpy of reaction at 550°C (823.15 K) is:** \[ \boxed{-1688 \text{ kJ}} \] (rounded to 4 significant figures; you can report as \(-1688 \text{ kJ}\) or \(-1688.1 \text{ kJ}\) if more precision is desired).