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Consider the equation below.

$CaCO _3(s) \longleftrightarrow CaO ( s )+ CO _2(g)$



What is the equilibrium constant expression for the given reaction?


A$K _{ eq }=\left[ CO _2\right]$
B$K _{ eq }=\frac{\left[ CaO _3\right]}{\left[ CaCO _3\right]}$
C$K _{e q}=\frac{1}{\left[ CO _2\right]}$
D$K _{e q}=\frac{\left[ CO _2\right]\left[ CaO _3\right]}{\left[ CaCO _3\right]}$

Verified Answer :

  • The equilibrium constant expression for the reaction \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\) accounts for the stoichiometric coefficients of all products and reactants.
  • Activities of pure solid substances, such as \(CaCO_3(s)\) and \(CaO(s)\), are taken as 1 and excluded from the final expression.
  • Only the gaseous product \(CO_2\) remains in the expression, resulting in \(K_{eq} = [CO_2]\).
  • The final answer is \(\boxed{a}\).

Explanation

  1. Understanding the Equilibrium Constant Expression

The equilibrium constant \(K_{eq}\) for a chemical reaction \(aA + bB \rightleftharpoons cC + dD\) is defined by the concentrations of products divided by the concentrations of reactants, with each substance raised to the power of its stoichiometric coefficient:
\[K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
In the reaction \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\), we have one reactant (\(CaCO_3\)) and two products (\(CaO\) and \(CO_2\)). However, the law of mass action specifies that the activities of pure solids and pure liquids are defined as constant and are assigned a value of 1. Consequently, these terms are omitted from the equilibrium expression.

  1. Deriving the Expression and Identifying the Solution

Given the reaction \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\), both \(CaCO_3\) and \(CaO\) are solids. Applying the rule for chemical equilibrium, we substitute their activities with 1:

  • Activity of \(CaCO_3(s) = 1\)
  • Activity of \(CaO(s) = 1\)
  • Activity of \(CO_2(g) = [CO_2]\) (representing the gas concentration)

Substituting these values into the general expression:
\[K_{eq} = \frac{[CaO] \cdot [CO_2]}{[CaCO_3]} = \frac{1 \cdot [CO_2]}{1} = [CO_2]\]
Thus, the equilibrium constant for this reaction depends only on the concentration of the gaseous carbon dioxide. Comparing this result with the provided options, option 'a' is the correct expression.

Examples

In industrial chemistry, the decomposition of limestone (\(CaCO_3\)) into quicklime (\(CaO\)) and carbon dioxide (\(CO_2\)) is a fundamental process for cement production. Because the solids remain in the reaction vessel without changing their activity, engineers monitor the pressure of the \(CO_2\) gas to control the extent of the reaction and ensure optimal production efficiency. This simplified equilibrium expression allows for easier calculation of process parameters in large-scale manufacturing.

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