Calculate $\Delta H^{\circ}$ for the reaction, $\mathrm{Na}_2…
Calculate $\Delta H^{\circ}$ 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 :
+823 kJ
–581 kJ
–435 kJ
+531 kJ
Solution
By ' $2 A+\frac{C}{2}-B$ ', we get
$\begin{aligned}
& \mathrm{Na}_2 \mathrm{O}+\mathrm{SO}_3 \longrightarrow \mathrm{Na}_2 \mathrm{SO}_4 \\
& \Delta H=-2 \times 146+\frac{259}{2}-418 \\
& \text { or } \quad \Delta H=-580.5 \approx-581 \mathrm{~kJ}
\end{aligned}$
