Answer:
ΔH3 = 1/2 (629) - ΔH^0
Explanation:
Given data:
Bond energy of H2 = ΔH1 = 436 Kj/mol
Bond energy of Br2 = ΔH2 = 193 Kj/mol
To find:
Let bond energy of HBr = ΔH3 = ?
Equation:
H2 + Br2 → 2HBr
enthalpy of formation of HBr = ΔH1 + ΔH3 - 2(ΔH3)
ΔH^0 = 436 + 193 - 2(ΔH3)
(436 + 193) - ΔH^0 = 2(ΔH3)
ΔH3 = 1/2 (629) - ΔH^0
