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Calculate $\Delta \mathrm{H}^{\circ}$ for the reaction,
$\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+\mathrm{SO}_{3}(\mathrm{~g}) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{~g})$
given the following:
(A) $\mathrm{Na}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(\mathrm{s})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})$
$\Delta \mathrm{H}^{\circ}=-146 \mathrm{~kJ}$
(B) $\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(l)$
$\begin{array}{c}
\longrightarrow 2 \mathrm{NaOH}(\mathrm{s})+\mathrm{SO}_{3}(\mathrm{~g}) \\
\Delta \mathrm{H}^{\circ}=+418 \mathrm{~kJ}
\end{array}$
(C) $2 \mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(l)$
$\Delta \mathrm{H}^{\circ}=+259 \mathrm{~kJ}$
Options:
$\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+\mathrm{SO}_{3}(\mathrm{~g}) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{~g})$
given the following:
(A) $\mathrm{Na}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{NaOH}(\mathrm{s})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})$
$\Delta \mathrm{H}^{\circ}=-146 \mathrm{~kJ}$
(B) $\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(l)$
$\begin{array}{c}
\longrightarrow 2 \mathrm{NaOH}(\mathrm{s})+\mathrm{SO}_{3}(\mathrm{~g}) \\
\Delta \mathrm{H}^{\circ}=+418 \mathrm{~kJ}
\end{array}$
(C) $2 \mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{Na}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(l)$
$\Delta \mathrm{H}^{\circ}=+259 \mathrm{~kJ}$
Solution:
2892 Upvotes
Verified Answer
The correct answer is:
$-581 \mathrm{~kJ}$
By ' $2 A+\frac{C}{2}-B^{\prime}$, we get
$$
\begin{array}{l}
\mathrm{Na}_{2} \mathrm{O}+\mathrm{SO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \\
\Delta H=-2 \times 146+\frac{259}{2}-418 \\
\Rightarrow \Delta H=-580.5 \approx 581 \mathrm{~kJ}
\end{array}
$$
$$
\begin{array}{l}
\mathrm{Na}_{2} \mathrm{O}+\mathrm{SO}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} \\
\Delta H=-2 \times 146+\frac{259}{2}-418 \\
\Rightarrow \Delta H=-580.5 \approx 581 \mathrm{~kJ}
\end{array}
$$
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