Question Detail
Question
If the standard enthalpies of formation of HBr(g) and Br₂(g) are -36 and +31 kJ mol⁻¹ (at 298 K) respectively, what is ΔH° (298 K) in kJ mol⁻¹ for the following reaction? H₂(g) + Br₂(g) → 2HBr(g)
Correct Answer
Option B is the correct answer.
Detailed Explanation
ΔH°rxn = Σ ΔHf°(products) - Σ ΔHf°(reactants) = [2 × (-36)] - [0 + 31] = -72 - 31 = -103 kJ/mol.
Hint
ΔHf° of elements in standard state is zero. H₂(g) is standard state, so ΔHf° = 0.