JEE MAIN - Chemistry (2025 - 3rd April Morning Shift - No. 25)
Given :
$$ \begin{aligned} & \left.\Delta \mathrm{H}^{\ominus}{ }_{\text {sub }}[\mathrm{C} \text { (graphite })\right]=710 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\ominus}=436 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta_{\mathrm{C}}=\mathrm{C} \mathrm{H}^{\ominus}=611 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
The $\Delta \mathrm{H}_{\mathrm{f}} \ominus$ for $\mathrm{CH}_2=\mathrm{CH}_2$ is_________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)
Explanation
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$$\begin{aligned} & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{fl})}}=2 \times\left[\Delta \mathrm{H}_{\mathrm{sub}}^{\mathrm{o}}\right]_{\mathrm{C}_{(\mathrm{s})}}+2 \times \Delta \mathrm{H}_{\mathrm{H}-\mathrm{H}}^{\mathrm{o}}-1 \times \Delta \mathrm{H}_{\mathrm{C}=\mathrm{C}}^{\mathrm{o}}-4 \times \Delta \mathrm{H}_{\mathrm{C}-\mathrm{H}}^{\circ} } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=(2 \times 710)+(2 \times 436)-611-4 \times 414 } \\ \Rightarrow \quad & {\left[\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\right]_{\mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}}=25 \mathrm{~kJ} / \mathrm{mol} } \end{aligned}$$
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