Chemistry · 2025
JEE Main · 3 April 2025, Shift 1 · Q71
Given: Δ H^_sub [ C (graphite) ]=710 kJ mol^-1; Δ_C - H H^ =414 kJ mol^-1; Δ_H - H H^ =436 kJ mol^-1; Δ_C = C H^ =611 kJ mol^-1 The Δ H_f^ for CH_2=…
Given :
$$\begin{aligned}
& \Delta \mathrm{H}^{\ominus}{ }_{\text {sub }}[\mathrm{C} \text { (graphite) }]=710 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\ominus}=414 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta_{\mathrm{H}-\mathrm{H}} \mathrm{H}^{\ominus}=436 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta_{\mathrm{C}=\mathrm{C}} \mathrm{H}^{\ominus}=611 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
The $\displaystyle \Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\displaystyle \mathrm{CH}_2=\mathrm{CH}_2$ is $\displaystyle \_\_\_\_$ $\displaystyle \mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer value)
Official answer
From NTA’s final answer key for this paper.
25
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JEE Main 2025 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.