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SO2 + 1/2 O2 → SO3 ΔH = ΔHf°(SO3) − ΔHf°(SO2) = −98.2 + 298.2 = 200 kJ/mole

1. (a) 200 kJ/mole
2. (b) 200 kJ/mole
3. (c) 200 kJ/mole
4. (d) 200 kJ/mole

Correct answer: (a) 200 kJ/mole

Solution

The enthalpy change (ΔH) is calculated using the given formula and values: ΔH = ΔHf°(SO3) − ΔHf°(SO2) = −98.2 + 298.2 = 200 kJ/mole. The correct answer is 200 kJ/mole.

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