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Fundamentals of thermodynamics

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Question

For the reaction $C_{2} H_{6} \to C_{2} H_{4} + H_{2}$ the reaction enthalpy $Δ_{l} H = kJ {mol}^{- 1}$ (Round off to the Nearest Integer) [Given : Bond enthalpies in $K J {mol}^{- 1} : C - C = 347, C = C = 611 C - H = 414, H - H = 436]$

Moderate

Solution

$\begin{array}{l} Δ, H = [ϵ_{C - C} + 2 ϵ_{C - H}] - [ϵ_{C - C} + ϵ_{H - H}] \\ = [347 + 2 \times 414] - [611 + 436] \end{array}$
$= 128$

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