Question

The equilibrium constant for the reaction ${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$ is______

Answer 1

Chemical equilibrium

• In the case of chemical equilibrium, the rates of the forward and reverse reactions become equal.
• Chemical equilibrium is dynamic in nature wherein concentrations of both reactant and product change constantly with equal speed.

Equilibrium constant:

• For the given reaction, $\underset{\mathrm{Calcium}\mathrm{carbonate}}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)}\rightleftharpoons \underset{\mathrm{Calcium}\mathrm{oxide}}{\mathrm{CaO}\left(\mathrm{s}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}$, the equilibrium constant can be calculated as:
• $K_{\mathrm{c}}=\frac{\left[\mathrm{CaO}\right]\left[{\mathrm{CO}}_2\right]}{\left[{\mathrm{CaCO}}_3\right]}$
• For pure solids and liquids, the value of the equilibrium constant is unity.
• So, $\left[{\mathrm{CaCo}}_3\right]=\left[{\mathrm{CO}}_2\right]=\mathrm{unity}$
• Thus, the equilibrium constant will be $K_{\mathit{c}}\mathit{=}\left[{\mathrm{CO}}_2\right]$.

Hence, the equilibrium constant for the reaction ${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$ is $\left[{\mathrm{CO}}_2\right]$