Question

For a reaction,
$CaC{o}_{3\left(s\right)}\to Ca{O}_{(}s)+C{O}_{2\left(g\right)}$
$△\int H^0\left(CaO\right)$ = $-635.1kJmo{l}^{-1}$
$△\int H^0\left(C{O}_2\right)$ = $-393.5kJmo{l}^{-1}$ and
$△\int H^0\left(CaC{O}_3\right)$ = $-1206.9kJmo{l}^{-1}$
Which of the following is a correct statement?

• A. A large amount of heat is evolved during the decomposition of $CaC{O}_3$
• B. Decomposition of $CaC{O}_3$ is an endothermic process and heat is provided for decomposition.
• C. The amount of heat evolved cannot be calculated from the data provided.
• D. ${△}_r{H}^0$ = $\sum △\int H^0\left(reactants\right)-\sum △\int H^0\left(products\right)$

Answer 1

Answer: (B)

Decomposition of $CaC{O}_3$ is an endothermic process and heat is provided for decomposition.

$CaC{O}_3\left(s\right)⟶CaO\left(s\right)+C{O}_2\left(g\right)\therefore \Delta H_{decomposition}=\sum {\Delta }_f{H}_{products}^{\circ }-\sum {\Delta }_f{H}_{reactants}^{\circ }={\Delta }_f{H}^{\circ }\left(CaO\right(s\left)\right)+{\Delta }_f{H}^{\circ }\left(C{O}_2\right(g\left)\right)-{\Delta }_f{H}^{\circ }\left(CaC{O}_3\right(s\left)\right)=-635.1-393.5+1206.9\therefore \Delta H_{decomposition}ofCaC{O}_3=178.3kJ{mol}^{-1}$

Therefore, $\Delta H$ is positive for decomposition of $CaC{O}_3$, so it is anendothermic process and heat is provided for decomposition.