What is the enthalpy change associated with the reaction of 25.0 g of SO2 in excess oxygen, given the reaction: 2 SO2(g) + O2(g) → 2 SO3(g) with ΔH° = -198.2 kJ?

• A. -198.2 kJ
• B. -99.1 kJ
• C. -77.2 kJ
• D. -396.4 kJ

Answer: (C)

Verified step by step guidance

1. First, determine the molar mass of SO2. The molar mass of sulfur (S) is approximately 32.07 g/mol, and the molar mass of oxygen (O) is approximately 16.00 g/mol. Therefore, the molar mass of SO2 is 32.07 g/mol + 2(16.00 g/mol) = 64.07 g/mol.
2. Next, calculate the number of moles of SO2 in 25.0 g. Use the formula: \( \text{moles of SO2} = \frac{\text{mass of SO2}}{\text{molar mass of SO2}} \). Substitute the known values: \( \text{moles of SO2} = \frac{25.0 \text{ g}}{64.07 \text{ g/mol}} \).
3. The balanced chemical equation shows that 2 moles of SO2 react with 1 mole of O2 to form 2 moles of SO3, with an enthalpy change (ΔH°) of -198.2 kJ for the reaction as written. This means that -198.2 kJ is released when 2 moles of SO2 react.
4. Determine the enthalpy change for the moles of SO2 calculated in step 2. Use the proportion: \( \text{enthalpy change} = \left( \frac{\text{moles of SO2}}{2 \text{ moles}} \right) \times (-198.2 \text{ kJ}) \).
5. Finally, calculate the enthalpy change using the values obtained in the previous steps. This will give you the enthalpy change associated with the reaction of 25.0 g of SO2 in excess oxygen.