Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Question

For the reaction \( \mathrm{N_2} + 3\mathrm{H_2} \rightleftharpoons 2\mathrm{NH_3} \), the equilibrium constant \(K_c\) is 0.5 at a certain temperature. If initially, 1 mole of \(\mathrm{N_2}\) and 3 moles of \(\mathrm{H_2}\) are placed in a 1 L container, what is the equilibrium concentration of \(\mathrm{NH_3}\)?

Answer 1

Explanation:

While writing the equilibrium constant expression, the concentrations of pure solids and pure liquids are omitted because their concentrations remain constant throughout the reaction.

Reason:

• The concentration (or density) of a pure solid or pure liquid does not change with the progress of the reaction.
• Their activity is taken as unity (1).
• Since equilibrium constant expressions involve only variable concentrations, constant terms are absorbed into the value of the equilibrium constant.

Illustration:

For the reaction:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)

The equilibrium constant expression is:

K_c = [CO₂]

The solids CaCO₃ and CaO are not included in the expression.

Conclusion:

Thus, pure solids and pure liquids are ignored in equilibrium constant expressions because their concentrations remain constant and do not affect the position of equilibrium.

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Solution and Explanation

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