Question Detail
Question
If the standard enthalpies of formation of HBr(g) and Br(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 ΔH°_f(products) - Σ ΔH°_f(reactants). For H₂(g) and Br₂(g), ΔH°_f = 0 (elements in standard state). ΔH°_rxn = [2 × (-36)] - [0 + 31] = -72 - 31 = -103 kJ mol⁻¹.
Hint
Remember that the standard enthalpy of formation for elements in their standard state is zero.