Calculate the enthalpy change (delta H) for the cracking reaction C2H6(g) -> C2H4(g) + H2(g) given bond energies: C-C = 348 kJ/mol, C=C = 614 kJ/mol, C-H = 413 kJ/mol, and H-H = 436 kJ/mol.
Answer and explanation
Correct answer: B. +124 kJ/mol
- Step 1: Bonds broken in C2H6: 1 C-C (348) + 2 C-H (2 413 = 826). Total broken = 348 + 826 = 1174 kJ/mol. Step 2. Bonds formed: 1 C=C (614) + 1 H-H (436). Total formed = 614 + 436 = 1050 kJ/mol. Step 3. delta H = Bonds broken - Bonds formed = 1174 - 1050 = +124 kJ/mol (endothermic). Answer: +124 kJ/mol. Why not the others:* -124 kJ/mol reverses the sign; cracking requires net thermal input.
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