The standard enthalpies of formation of $\mathrm{Al}_{2} \mathrm{O}_{3}$ and $\mathrm{CaO}$ are $-1675 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $-635 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
For the reaction
$3 \mathrm{CaO}+2 \mathrm{Al} \rightarrow 3 \mathrm{Ca}+\mathrm{Al}_{2} \mathrm{O}_{3}$ the standard reaction enthalpy $\Delta_{\mathrm{r}} \mathrm{H}^{0}=$ $\mathrm{kJ}$. (Round off to the Nearest Integer).
Given reaction:
$3 \mathrm{CaO}+\mathrm{Al} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+3 \mathrm{Ca}$
Now, $\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=\Sigma \Delta_{\mathrm{f}} \mathrm{H}^{\circ}{ }_{\text {Products }}-\Sigma \Delta_{\mathrm{f}} \mathrm{H}^{\circ}{ }_{\text {Reactants }}$
$=[1 \times(-1675)+3 \times 0]-[3 \times(-635)+2 \times 0]$
$=+230 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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