Given the following reactions:2S (s) + 3O2 (g) → 2SO3 (g) ΔH = -790 kJS (s) + O2 (g) → SO2 (g) ΔH = -297 kJWhat is the enthalpy change (ΔH) for the reaction in which sulfur dioxide is oxidized to sulfur trioxide: 2SO2 (g) + O2 (g) → 2SO3 (g)?

• A. -196 kJ
• B. -486 kJ
• C. -790 kJ
• D. -593 kJ

Answer: (B)

Verified step by step guidance

1. Identify the target reaction: 2SO2 (g) + O2 (g) → 2SO3 (g). We need to find the enthalpy change (ΔH) for this reaction.
2. Use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction. We will use the given reactions to construct the target reaction.
3. Write the given reactions: (1) 2S (s) + 3O2 (g) → 2SO3 (g) with ΔH = -790 kJ and (2) S (s) + O2 (g) → SO2 (g) with ΔH = -297 kJ.
4. Reverse reaction (2) to get SO2 (g) → S (s) + O2 (g), which changes the sign of ΔH to +297 kJ. Multiply this reversed reaction by 2 to match the stoichiometry of the target reaction: 2SO2 (g) → 2S (s) + 2O2 (g) with ΔH = 2 * 297 kJ.
5. Add the modified reaction (2) to reaction (1) to obtain the target reaction: 2SO2 (g) + O2 (g) → 2SO3 (g). Calculate the total ΔH by adding the ΔH values of the modified reactions.