Given :
$\begin{aligned}<br />& \Delta \mathrm{H}_{\text {sub }}^{\ominus}[\mathrm{C}(\text { graphite })]=710 \mathrm{~kJ} \mathrm{~mol}^{-1} \\<br>& \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\<br>& \Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\Theta}=436 \mathrm{~kJ} \mathrm{~mol}^{-1} \\<br>& \Delta_{\mathrm{C}=\mathrm{C}} \mathrm{H}^{\Theta}=611 \mathrm{~kJ} \mathrm{~mol}^{-1}<br />\end{aligned}$
The $\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\mathrm{CH}_2=\mathrm{CH}_2$ is _____ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)

Solution

$\begin{aligned} & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{(4)}}=2 \times\left[\Delta \mathrm{H}_{\mathrm{sub}}^{\mathrm{o}}\right]_{\mathrm{C}_{(\mathrm{s})}}+2 \times \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}^{\mathrm{o}}-1 \times \Delta \mathrm{H}_{\mathrm{C}=\mathrm{C}}^{\mathrm{o}}-4 \times \Delta \mathrm{H}_{\mathrm{CH}}^{\mathrm{o}} } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=(2 \times 710)+(2 \times 436)-611-4 \times 414 } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=25 \mathrm{~kJ} / \mathrm{mol} }\end{aligned}$

Hence, the correct answer is 25.