JEE MAIN - Chemistry (2022 - 28th June Evening Shift - No. 13)
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
600
Explanation
$\mathrm{Mg}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{MgO}(\mathrm{s})$
$$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT} \\\\ &\Delta \mathrm{ng}=-\frac{1}{2} \\\\ &-601.70=\Delta \mathrm{U}-\frac{1}{2}(8.3)(300) \times 10^{-3} \\\\ &\Delta \mathrm{U}=-601.70+1.245 \\\\ &\Delta \mathrm{U} \simeq-600 \mathrm{~kJ} \end{aligned} $$
The magnitude of change in internal energy is $600 \mathrm{~kJ}$.
$$ \begin{aligned} &\Delta \mathrm{H}=\Delta \mathrm{U}+\Delta \mathrm{ngRT} \\\\ &\Delta \mathrm{ng}=-\frac{1}{2} \\\\ &-601.70=\Delta \mathrm{U}-\frac{1}{2}(8.3)(300) \times 10^{-3} \\\\ &\Delta \mathrm{U}=-601.70+1.245 \\\\ &\Delta \mathrm{U} \simeq-600 \mathrm{~kJ} \end{aligned} $$
The magnitude of change in internal energy is $600 \mathrm{~kJ}$.
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