If the bond energies of H−H, Br−Br, and H−Br are 433, 192, and 364 kJ mol⁻¹ respectively, the ΔHr for the reaction:
H₂(g) + Br₂(g) → 2HBr(g) is:

Question

Accepted Answer

−103 kJ. The enthalpy change (ΔHr) is calculated using bond energies: ΔHr = Σ(Bond energies of reactants) - Σ(Bond energies of products). For H₂ + Br₂ → 2HBr, ΔHr = [433 + 192] - [2 × 364] = 625 - 728 = -103 kJ.

If the bond energies of H−H, Br−Br, and H−Br are 433, 192, and 364 kJ mol⁻¹ respectively, the ΔHr for the reaction: H₂(g) + Br₂(g) → 2HBr(g) is:

Solution

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