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Answer :
Final answer:
The ΔH° for the reaction SO₂(g) + 1/2O₂(g) → SO₃(g) is calculated using the enthalpies of formation for the reactants and the product. The result is -99.1 kJ/mol, indicating the reaction is exothermic and matching answer choice B.
Explanation:
To calculate
H° for a reaction, we use the standard enthalpies of formation (
ΔH°f) for the reactants and products involved in the reaction. The reaction in question is SO₂(g) + 1/2O₂(g)
→ SO₃(g). Using the provided enthalpies of formation, we apply Hess's law. The calculation of the enthalpy change (
ΔH°) for this reaction is as follows:
ΔH°reaction = ΔH°f[SO₃(g)] - (ΔH°f[SO₂(g)] + (1/2)ΔH°f[O₂(g)])
ΔH°reaction = (-395.2 kJ/mol) - [(-296.1 kJ/mol) + (1/2)(0 kJ/mol)]
ΔH°reaction = -395.2 kJ/mol - (-296.1 kJ/mol)
ΔH°reaction = -395.2 kJ/mol + 296.1 kJ/mol
ΔH°reaction = -99.1 kJ/mol
The calculated value is -99.1 kJ/mol, which indicates that the correct answer is B. -99.1kJ/molrxn, meaning the reaction is exothermic.
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