For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, $\Delta$CH$\Theta$ = $-$601.70 kJ mol$-$1, the magnitude of change in internal energy for the reaction is __________ kJ. (Nearest integer)
(Given : R = 8.3 J K$-$1 mol$-$1)
Answer (integer)
600
Solution
$\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{MgO}(\mathrm{s})$
<br/><br/>
$$
\begin{aligned}
&\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT} \\\\
&\Delta \mathrm{ng}=-\frac{1}{2} \\\\
&-601.70=\Delta \mathrm{U}-\frac{1}{2}(8.3)(300) \times 10^{-3} \\\\
&\Delta \mathrm{U}=-601.70+1.245 \\\\
&\Delta \mathrm{U} \simeq-600 \mathrm{~kJ}
\end{aligned}
$$
<br/><br/>
The magnitude of change in internal energy is $600 \mathrm{~kJ}$.
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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