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H2(g) + Br2(g) → 2HBr(g). ΔH° = (B.E)reactant − (B.E)product.

1. ΔH° = (433 + 192) − (2 × 364)
2. ΔH° = (625 − 728) = −103 kJ
3. ΔH° = (433 + 192) − 103
4. ΔH° = (625 − 728) = −103 kJ mol⁻¹

Correct answer: ΔH° = (433 + 192) − (2 × 364)

Solution

The enthalpy change (ΔH°) is calculated using bond enthalpies: ΔH° = Σ(B.E of reactants) − Σ(B.E of products). Here, the bond enthalpies of H-H and Br-Br are 433 kJ/mol and 192 kJ/mol, respectively, and for H-Br, it is 364 kJ/mol. Substituting these values gives ΔH° = (433 + 192) − (2 × 364).

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