Given : ${\mathrm{ΔH}}_{\text{sub }}^{⊖}[\mathrm{C}\text{ (graphite) }]=710\mathrm{kJ}{\mathrm{mol}}^{-1}\ {\Delta }_{\mathrm{C}-\mathrm{H}}{\mathrm{H}}^{⊖}=414\mathrm{kJ}{\mathrm{mol}}^{-1}\ {\Delta }_{\mathrm{H}-\mathrm{H}}{\mathrm{H}}^{⊖}=436\mathrm{kJ}{\mathrm{mol}}^{-1}\ {\Delta }_{\mathrm{C}=\mathrm{C}}{\mathrm{H}}^{⊖}=611\mathrm{kJ}{\mathrm{mol}}^{-1}$ The $\Delta {\mathrm{H}}_{\mathrm{f}}^{⊖}$ for ${\mathrm{CH}}_{2}={\mathrm{CH}}_{2}$ is ________ ${\mathrm{kJmol}}^{-1}$ (nearest integer value) [JEE Main 2025, 3 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: a) 25. ${\Delta \mathrm{H}}_{\mathrm{r}}^{⊖}=2(710)+2\times 436−611−4(414)\ =1420+872−611−1656\ =25\mathrm{kJ}{\mathrm{mole}}^{−1}$

Given : \({\mathrm{ΔH}}_{\text{sub }}^{⊖}[\mathrm{C}\text{ (graphite) }]=710\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\Delta }_{…

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