1
JEE Main 2025 (Online) 3rd April Morning Shift
Numerical
+4
-1
Change Language

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)

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2
JEE Main 2025 (Online) 28th January Evening Shift
Numerical
+4
-1
Change Language

Consider the following data :

Heat of formation of $\mathrm{CO}_2(\mathrm{g})=-393.5 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$

Heat of formation of $\mathrm{H}_2 \mathrm{O}(\mathrm{l})=-286.0 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}$

Heat of combustion of benzene $=-3267.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The heat of formation of benzene is __________ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest integer)

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3
JEE Main 2025 (Online) 28th January Morning Shift
Numerical
+4
-1
Change Language

The formation enthalpies, $\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\mathrm{H}_{(\mathrm{g})}$ and $\mathrm{O}_{(\mathrm{g})}$ are 220.0 and $250.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively, at 298.15 K , and $\Delta \mathrm{H}_{\mathrm{f}}^{\ominus}$ for $\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})}$ is $-242.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at the same temperature. The average bond enthalpy of the $\mathrm{O}-\mathrm{H}$ bond in water at 298.15 K is _______ $\mathrm{kJ} \mathrm{~mol}^{-1}$ (nearest integer).

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4
JEE Main 2025 (Online) 24th January Morning Shift
Numerical
+4
-1
Change Language

Standard entropies of $\mathrm{X}_2, \mathrm{Y}_2$ and $\mathrm{XY}_5$ are 70, 50 and $110 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature in Kelvin at which the reaction

$$\frac{1}{2} \mathrm{X}_2+\frac{5}{2} \mathrm{Y}_2 \rightleftharpoons \mathrm{XY}_5 \Delta \mathrm{H}^{\ominus}=-35 \mathrm{~kJ} \mathrm{~mol}^{-1}$$

will be at equilibrium is __________ (Nearest integer)

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