Question

For the reaction C₂H₆ → C₂H₄ + H₂ the reaction enthalpy ΔH = _________ kJ mol$^{-1}$. (Round off to the Nearest Integer). [Given : Bond enthalpies in kJ mol$^{-1}$ : C-C : 347, C=C : 611; C-H : 414, H-H : 436]

Hint:

Instead of calculating all bonds, cancel common bonds: $\Delta H = [1(C-C) + 2(C-H)] - [1(C=C) + 1(H-H)]$.

Answer 1

Correct Answer: 128

Step 1: $\Delta H_{reaction} = \sum (\text{B.E. of reactants}) - \sum (\text{B.E. of products})$.
Step 2: Reactants ($C_2H_6$): 1 C-C bond + 6 C-H bonds. Products ($C_2H_4 + H_2$): 1 C=C bond + 4 C-H bonds + 1 H-H bond.
Step 3: $\Delta H = [347 + 6(414)] - [611 + 4(414) + 436]$. $\Delta H = [347 + 2484] - [611 + 1656 + 436] = 2831 - 2703 = 128$ kJ/mol.