Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction: 4 S(s) + 6 O2(g) → 4 SO3(g). Given: SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8 kJ, 2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -198.2 kJ. What is the ΔH°rxn for the target reaction?

• A. -792.8 kJ
• B. -396.4 kJ
• C. +396.4 kJ
• D. +792.8 kJ

Answer: (B)

Verified step by step guidance

1. Identify the target reaction: 4 S(s) + 6 O2(g) → 4 SO3(g).
2. Recognize that you need to manipulate the given reactions to match the target reaction. The first given reaction is SO2(g) → S(s) + O2(g) with ΔH°rxn = +296.8 kJ, and the second is 2 SO2(g) + O2(g) → 2 SO3(g) with ΔH°rxn = -198.2 kJ.
3. Reverse the first reaction to match the formation of SO2 from S and O2: S(s) + O2(g) → SO2(g). This changes the sign of ΔH°rxn to -296.8 kJ.
4. Multiply the reversed first reaction by 4 to match the target reaction's stoichiometry: 4 S(s) + 4 O2(g) → 4 SO2(g). This multiplies the ΔH°rxn by 4, resulting in -1187.2 kJ.
5. Add the modified first reaction to the second reaction (multiplied by 2 to match the stoichiometry): 4 SO2(g) + 2 O2(g) → 4 SO3(g). The ΔH°rxn for this is 2 * (-198.2 kJ) = -396.4 kJ. Combine these to find the overall ΔH°rxn for the target reaction.