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 (integer)
57
Solution
$\mathrm{H}_{2} \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})$
<br/><br/>
$\Delta \mathrm{U}=-59.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$
<br/><br/>
$$
\begin{aligned}
\Delta \mathrm{H} &=\Delta \mathrm{U}+\Delta \mathrm{n}_{g} \mathrm{RT} \\
&=-59.6+\frac{1 \times 8.314 \times 300}{1000} \\
&=-57.10 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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