For the reaction: $C_{2} H_{6} \to_{}^{} C_{2} H_{4} + H_{2}$ . The reaction enthalpy $∆ rH =$ __________ $kJ {mol}^{- 1}$ . (Round off to the Nearest Integer).
[Given: Bond enthalpies in $kJ {mol}^{- 1}; C - C : 347; C = C : 611; C - H : 414; H - H : 436$ ]

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Solution

The above question can be solved as:
Step 1:
$C_{2} H_{6} \to C_{2} H_{4} + H_{2} ΔH = ? ?$
Step 2:
$ΔH = [(E_{C - C} + 6 E_{C - H}) - (E_{C = C} + 4 E_{C - H} + E_{H - H}) ΔH = [347 + 6 \times 414] - (611 + 4 \times 414 + 433] = 2831 - 2700 = 131 kJ / mol$
Hence, the reaction enthalpy $∆ rH$ is $131 kJ {mol}^{- 1}$

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