JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 14)
For the reaction
$$\mathrm{H}_{2} \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})$$
$$\Delta U=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$$ at $$27^{\circ} \mathrm{C}$$.
The enthalpy change for the above reaction is ($$-$$) __________ $$\mathrm{kJ} \,\mathrm{mol}^{-1}$$ [nearest integer]
Given: $$\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$.
Answer
57
Explanation
$\mathrm{H}_{2} \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})$
$\Delta \mathrm{U}=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$
$$ \begin{aligned} \Delta \mathrm{H} &=\Delta \mathrm{U}+\Delta \mathrm{n}_{g} \mathrm{RT} \\ &=-59.6+\frac{1 \times 8.314 \times 300}{1000} \\ &=-57.10 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
$\Delta \mathrm{U}=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$
$$ \begin{aligned} \Delta \mathrm{H} &=\Delta \mathrm{U}+\Delta \mathrm{n}_{g} \mathrm{RT} \\ &=-59.6+\frac{1 \times 8.314 \times 300}{1000} \\ &=-57.10 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
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