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# The equilibrium constant for the reaction ${\mathrm{CaCO}}_{3}\left(\mathrm{s}\right)⇌\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_{2}\left(\mathrm{g}\right)$ is______

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## Chemical equilibriumIn 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, $\underset{\mathrm{Calcium}\mathrm{carbonate}}{{\mathrm{CaCO}}_{3}\left(\mathrm{s}\right)}⇌\underset{\mathrm{Calcium}\mathrm{oxide}}{\mathrm{CaO}\left(\mathrm{s}\right)}+\underset{\mathrm{Carbon}\mathrm{dioxide}}{{\mathrm{CO}}_{2}\left(\mathrm{g}\right)}$, 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 ${\mathrm{CaCO}}_{3}\left(\mathrm{s}\right)⇌\mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_{2}\left(\mathrm{g}\right)$ is $\left[{\mathrm{CO}}_{2}\right]$

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