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Equilibrium - Heterogeneous Equilibria

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A heterogeneous equilibrium is defined as a state of equilibrium in a system where the reactants and products are present in two or more different phases (e.g., solid, liquid, or gas).

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For any pure solid or pure liquid, the molar concentration (molarity) is constant because it is the ratio of its density to its molar mass: [Solid/Liquid]=DensityMolar Mass[ \text{Solid/Liquid} ] = \frac{\text{Density}}{\text{Molar Mass}}. Since density and molar mass are constant at a given temperature, these terms are incorporated into the equilibrium constant.

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In the equilibrium constant expression (KcK_c or KpK_p), the activities of pure solids and pure liquids are taken as 11 and are omitted from the final equation.

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The equilibrium constant for a heterogeneous reaction depends only on the concentrations or partial pressures of the gaseous or aqueous components.

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If a substance is present as a solute in a solution (e.g., aqaq), its concentration varies and must be included in the KcK_c expression.

📐Formulae

For CaCO3(s)⇌CaO(s)+CO2(g):Kc=[CO2]\text{For } CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g): K_c = [CO_2]

Kp=PCO2K_p = P_{CO_2}

For Ag2O(s)+2HNO3(aq)⇌2AgNO3(aq)+H2O(l):Kc=[AgNO3]2[HNO3]2\text{For } Ag_2O(s) + 2HNO_3(aq) \rightleftharpoons 2AgNO_3(aq) + H_2O(l): K_c = \frac{[AgNO_3]^2}{[HNO_3]^2}

General Rule: [A(s)]=1,[A(l)]=1\text{General Rule: } [A(s)] = 1, [A(l)] = 1

Kp=Kc(RT)Δng where Δng is the change in moles of gaseous species only.K_p = K_c(RT)^{\Delta n_g} \text{ where } \Delta n_g \text{ is the change in moles of gaseous species only.}

💡Examples

Problem 1:

Write the equilibrium constant expressions KcK_c and KpK_p for the following reaction: 3Fe(s)+4H2O(g)⇌Fe3O4(s)+4H2(g)3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g)

Solution:

Kc=[H2]4[H2O]4K_c = \frac{[H_2]^4}{[H_2O]^4} and Kp=(PH2)4(PH2O)4K_p = \frac{(P_{H_2})^4}{(P_{H_2O})^4}

Explanation:

The substances Fe(s)Fe(s) and Fe3O4(s)Fe_3O_4(s) are in the solid phase. In heterogeneous equilibria, the concentrations of pure solids are constant and are omitted from the equilibrium constant expression. Only the gaseous species H2OH_2O and H2H_2 are included.

Problem 2:

For the decomposition of ammonium carbamate: NH2COONH4(s)⇌2NH3(g)+CO2(g)NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g), if the total pressure at equilibrium is PP, find KpK_p.

Solution:

Kp=4×(P3)3=4P327K_p = 4 \times \left( \frac{P}{3} \right)^3 = \frac{4P^3}{27}

Explanation:

Let the partial pressure of CO2CO_2 be pp. From stoichiometry, the partial pressure of NH3NH_3 will be 2p2p. Total pressure P=2p+p=3pP = 2p + p = 3p, so p=P3p = \frac{P}{3}. The equilibrium constant is Kp=(PNH3)2(PCO2)=(2p)2(p)=4p3K_p = (P_{NH_3})^2(P_{CO_2}) = (2p)^2(p) = 4p^3. Substituting pp, we get 4(P3)3=4P3274(\frac{P}{3})^3 = \frac{4P^3}{27}. The solid reactant is excluded.

Problem 3:

Consider the equilibrium: H2O(l)⇌H2O(g)H_2O(l) \rightleftharpoons H_2O(g). Write the KcK_c expression.

Solution:

Kc=[H2O(g)]K_c = [H_2O(g)]

Explanation:

Water in the liquid phase H2O(l)H_2O(l) is a pure liquid, so its concentration is taken as 11. The equilibrium constant depends only on the concentration of the water vapor.