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Thermodynamics

# Calculate the enthalpy change ($\Delta H$) of the following reaction:  $2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(g)$ given average bond energies of various bonds, i.e, $C-H,\: CC,\: O=O,\: C=O,\: O-H$ as 414, 814, 499, 724 and 640 kJ/mol respectively.

$\begin {array} {1 1} (A)\;-1861\: kJ & \quad (B)\;2321\: kJ \\ (C)\;1362\: kJ & \quad (D)\;– 2321\: kJ \end {array}$

$\Delta H$ = [Total energy required] – [total energy released]
$= [4 \Delta H_{C-H} + 2 \Delta H_{CC} + 5 \Delta H_{O=O}] – [8 \Delta H_{C=O} + 4 \Delta H_{O-H}]$
$= [4 \times 414 + 2 \times 810 + 5 \times 499] – [8 \times 724 + 4 \times 460]$
$= 5771 – 7632 = -1861\: kJ$
Ans : (A)