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JEE Main 2026 (Online) 21st January Morning Shift
Numerical
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$$ \begin{aligned} &\text { Use the following data : }\\ &\begin{array}{|c|c|c|} \hline \text { Substance } & \frac{\Delta_f \mathrm{H}^{\ominus}(500 \mathrm{~K})}{\mathrm{kJ} \mathrm{~mol}^{-1}} & \frac{\mathrm{~S}^{\ominus}(500 \mathrm{~K})}{\mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\ \hline \mathrm{AB}(\mathrm{~g}) & 32 & 222 \\ \hline \mathrm{~A}_2(\mathrm{~g}) & 6 & 146 \\ \hline \mathrm{~B}_2(\mathrm{~g}) & x & 280 \\ \hline \end{array} \end{aligned} $$

One mole each of $\mathrm{A}_2(\mathrm{~g})$ and $\mathrm{B}_2(\mathrm{~g})$ are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K .

$$ \mathrm{A}_2(\mathrm{~g})+\mathrm{B}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{AB}(\mathrm{~g}) $$

The value of $x\left(\mathrm{in} \mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\_\_\_\_$ . (Nearest integer)

(Given : $\log \mathrm{K}=2.2 \quad \mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ )

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2
JEE Main 2025 (Online) 8th April Evening Shift
Numerical
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Resonance in $\mathrm{X}_2 \mathrm{Y}$ can be represented as

JEE Main 2025 (Online) 8th April Evening Shift Chemistry - Thermodynamics Question 35 English

The enthalpy of formation of $X_2Y$ $ \left(X = X(g) + \frac{1}{2} Y = Y(g) \rightarrow X_2Y(g) \right) $ is 80 kJ mol$^{-1}$. The magnitude of resonance energy of $X_2Y$ is __ kJ mol$^{-1}$ (nearest integer value).

Given: Bond energies of $X \equiv X$, $X = X$, $Y = Y$ and $X = Y$ are 940, 410, 500, and 602 kJ mol$^{-1}$ respectively.
valence $X$: 3, $Y$: 2

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JEE Main 2025 (Online) 3rd April Evening Shift
Numerical
+4
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A perfect gas ( 0.1 mol ) having $\overline{\mathrm{C}}_v=1.50 \mathrm{R}$ (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is $(-)$___________$J$ (nearest integer)

[Given: $\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ ]

JEE Main 2025 (Online) 3rd April Evening Shift Chemistry - Thermodynamics Question 25 English
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4
JEE Main 2025 (Online) 3rd April Evening Shift
Numerical
+4
-1
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A sample of n -octane $(1.14 \mathrm{~g})$ was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is $5 \mathrm{~kJ} \mathrm{~K}^{-1}$. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K . The magnitude of the heat of combustion of octane at constant volume is__________ $\mathrm{kJ} \mathrm{mol}^{-1}$ (nearest integer).
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