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Bond Order and Enthalpy

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Question

Calculate the enthalpy changes $(δ H)$ of the following reaction.
$2 C_{2} H_{2 (g)} + 5 O_{2 (g)} \to 4 C O_{2 (g)} + 2 H_{2} O_{(g)}$ given average bond enthalpies of various bonds, i.e $C - H, C \equiv C, O = O, C = O, O - H$ as $414, 814, 499, 724$ and $640 k J m o l^{- 1}$ respectively.

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Solution

Given Bond enthalpies $(k J m o l^{- 1}) C - H = 414; C \equiv= 814; O - H = 640; O = O = 499; C = O = 724$
$△ H_{r e a c t i o n} = \sum △ H_{p r o d u c t} - \sum △ H_{r e a c t a n t}$
$2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (g)$
$4 (C - H) + 2 (C \equiv C)$ $+ 5 (O = O)$ $8 (C = O) + 4 (O - H)$
$△ H_{r x n} = 8 (C = O) + 4 (O - H) - 5 (O = O) - 4 (C - H) - 2 (C \equiv C)$
$△ H_{r x n} = 8 \times 724 + 4 \times 6405 \times 4994 \times 4142 \times 814$
$= 5792 + 2560 - 2495 - 1656 - 1628$
$△ H_{r x n} = 2573 k J m o l^{- 1}$

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(△ H)

2 C_{2} H_{2} (g) + 5 O_{2} (g) \to 4 C O_{2} (g) + 2 H_{2} O (g)

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(△ H)

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Δ H^{o}

{C H}_{2} = {C H}_{2} + 3 O = O ⟶ 2 O = C = O + 2 H - O - H

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B o n d e n t h a l p y

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