Given the reaction: 2 H_2S(g) + 3 O_2(g) → 2 SO_2(g) + 2 H_2O(g), what is the standard enthalpy change (ΔH°) for this reaction? (Use the following standard enthalpies of formation: ΔH_f°(H_2S(g)) = -20.6 kJ/mol, ΔH_f°(O_2(g)) = 0 kJ/mol, ΔH_f°(SO_2(g)) = -296.8 kJ/mol, ΔH_f°(H_2O(g)) = -241.8 kJ/mol)

• A. +535.2 kJ
• B. -1107.6 kJ
• C. +1107.6 kJ
• D. -535.2 kJ

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: \$2 \mathrm{H_2S}(g) + 3 \mathrm{O_2}(g) \rightarrow 2 \mathrm{SO_2}(g) + 2 \mathrm{H_2O}(g)$.
2. Recall the formula for the standard enthalpy change of a reaction using standard enthalpies of formation: \(\Delta H^\circ_{\text{reaction}} = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants})\).
3. Calculate the total enthalpy of formation for the products by multiplying the standard enthalpy of formation of each product by its coefficient and summing: \$2 \times \Delta H^\circ_f(\mathrm{SO_2}(g)) + 2 \times \Delta H^\circ_f(\mathrm{H_2O}(g))$.
4. Calculate the total enthalpy of formation for the reactants similarly: \$2 \times \Delta H^\circ_f(\mathrm{H_2S}(g)) + 3 \times \Delta H^\circ_f(\mathrm{O_2}(g))$.
5. Subtract the sum of the reactants' enthalpies from the sum of the products' enthalpies to find \(\Delta H^\circ_{\text{reaction}}\): \(\Delta H^\circ = [2 \times (-296.8) + 2 \times (-241.8)] - [2 \times (-20.6) + 3 \times 0]\) (do not calculate the final value here).