The enthalpies of formation of $\mathrm{Al}_{2} \mathrm{O}_{3}$ and $\mathrm{Cr}_{2} \mathrm{O}_{3}$ are…
- $-2730 \mathrm{~kJ}$
- $-462 \mathrm{~kJ}$
- $-1365 \mathrm{~kJ}$
- $+2730 \mathrm{~kJ}$
Solution
$\cdots \cdots$
$2 \mathrm{Cr}+\frac{3}{2} \mathrm{O}_{2} ightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}, \Delta \mathrm{H}=-1134 \mathrm{~kJ}$
$\mathrm{By}(\mathrm{i})-(\mathrm{ii})$
$2 \mathrm{Al}+\mathrm{Cr}_{2} \mathrm{O}_{3} ightarrow 2 \mathrm{Cr}+\mathrm{Al}_{2} \mathrm{O}_{3}, \Delta \mathrm{H}=-462 \mathrm{~kJ}$
Asked in: JEE-TOPICTESTS-CHEMISTRY