Given the bond enthalpies I-Cl (209),H-H (435),H-I (297),and H-Cl (431)in kJ/mol,compute ΔH° in kJ/mol for: ICl₃(g)+ 2 H₂(g)→ HI(g)+ 3 HCl(g).

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Quizplus · Accepted Answer

To calculate ΔH°, use the bond enthalpy method: ΔH° = Σ(bond enthalpies of reactants) - Σ(bond enthalpies of products). For reactants: I-Cl (3 bonds) + H-H (2 bonds) = 3(209) + 2(435) = 627 + 870 = 1497 kJ/mol. For products: H-I (1 bond) + H-Cl (3 bonds) = 297 + 3(431) = 297 + 1293 = 1590 kJ/mol. ΔH° = 1497 - 1590 = -93 kJ/mol.

Given the Bond Enthalpies I-Cl (209),H-H (435),H-I (297),and H-Cl (431)in