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)

Question

Quizplus · Accepted Answer

ΔH° = Σ(ΔH°(bonds broken)) - Σ(ΔH°(bonds formed))

Bonds broken: 
1 I-Cl bond = 209 kJ/mol 
2 H-S bonds = 2(435 kJ/mol) = 870 kJ/mol

Bonds formed: 
1 H-I bond = 297 kJ/mol 
3 H-Cl bonds = 3(431 kJ/mol) = 1293 kJ/mol

ΔH° = (209 kJ/mol + 870 kJ/mol) - (297 kJ/mol + 1293 kJ/mol) 
ΔH° = -93 kJ/mol

Therefore, the answer is C.

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