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If enthalpies of formation of $\mathrm{C}_2 \mathrm{H}_4(\mathrm{~g}), \mathrm{CO}_2(\mathrm{~g})$ and $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ at $25^{\circ} \mathrm{C}$ and $1 \mathrm{~atm}$ pressure are 52,394 and $-286 \mathrm{~kJ} / \mathrm{mol}$ respectively, the change in enthalpy for combustion of $\mathrm{C}_2 \mathrm{H}_4$ is equal to
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Verified Answer
The correct answer is:
$-1412 \mathrm{~kJ} / \mathrm{mol}$
Enthalpy of formation of $\mathrm{C}_2 \mathrm{H}_4, \mathrm{CO}_2$ and $\mathrm{H}_2 \mathrm{O}$ are $52,-394$ and $-286 \mathrm{~kJ} / \mathrm{mol}$ respectively.
(Given)
The reaction is
$\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$
change in enthalpy,
$(\Delta H)=\Delta H_{\text {products }}-\Delta H_{\text {reactants }}$
$=2 \times(-394)+2 \times(-286)-(52+0)$
$=-1412 \mathrm{~kJ} / \mathrm{mol}$.
(Given)
The reaction is
$\mathrm{C}_2 \mathrm{H}_4+3 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$
change in enthalpy,
$(\Delta H)=\Delta H_{\text {products }}-\Delta H_{\text {reactants }}$
$=2 \times(-394)+2 \times(-286)-(52+0)$
$=-1412 \mathrm{~kJ} / \mathrm{mol}$.
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