What is Heterogeneous Equilibria? Definition, Factors, Application & Class 11 Notes

A state in chemical equilibrium that consists of more than one phase is called heterogeneous equilibrium. The example of a heterogeneous equilibrium is, chemical equilibrium between water vapour and liquid water in a closed container. 

Also Read: NCERT Solutions | Class 11 Chemistry NCERT Solutions

Key Features of Heterogeneous Equilibria 

1. Equilibrium is available in two different phases
2. The concentration of pure solid and liquid is not included in the expression due to constant concentration. Only gas or aqueous phases are included.

Example:  

${\mathrm{CaCO}}_3\left(s\right)\rightleftharpoons \mathrm{CaO}\left(s\right)+{\mathrm{CO}}_2\left(g\right)$

 In the equilibrium expression, solid calcium carbonate and calcium oxide are excluded.

The equilibrium constant is based on CO₂ (gas). $K{c}_{}=\left[{\mathrm{CO}}_2\right]$

Heterogeneous Equilibrium Key Points: 

1. Heterogeneous equilibrium is used in geological reactions, industrial processes, and environmental systems.
2. To find the equilibrium constant, it is important to know which substance is included or excluded.

To calculate the equilibrium constant in a heterogeneous system, the concentration or partial pressure of a gaseous or aqueous substance is required. Pure solids and liquids are excluded from the expression due to constant concentration.

General Form of the Reaction: $aA\left(s\right)+bB\left(g\right)\rightleftharpoons cC\left(l\right)+dD\left(g\right)$

Equilibrium Expression:

$K_c=\frac{{\left[D\right]}^d}{{\left[B\right]}^b}$

• [A] and [C] are not included because A is solid and C is a pure liquid.

Summary:

In chemical equilibrium, a state where two opposite physical processes occur at the same rate, with no net change in the system, is called equilibrium in physical change. It is familiar in phase changes such as melting, freezing, boiling, and evaporation. 

Important Links:

Example of Physical Equilibria

1. Solid ⇌ Liquid (Melting / Freezing)
   Example: Ice to water: At 0°C and ice melts into water and water freezes into ice at the same rate.
2. Liquid ⇌ Gas (Evaporation / Condensation)
Example: Water ⇌ Water Vapour: Water evaporates and condenses at an equal rate in a closed container.

3. Solid ⇌ Solution ( Dissolution / Crystallization)
Example: ${\text{NaCl}}_{\text{(s)}}\rightleftharpoons {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$ . Salt dissolved and crystallizes at the same rate in the saturated solution.

Factors affecting Heterogeneous Equilibrium are:

• Temperature: The equilibrium constant is affected by the change in temperature. In an endothermic reaction, a rise in temperature favours the product. For an exothermic reaction, a rise in temperature favours the reactants.
• Pressure (for gaseous components): When the pressure is increased, it favours the side with fewer gas molecules. In case of low pressure, it will favour the side with more gas molecules.
• Concentration (for gaseous and aqueous solution): According to Le Chatelier's Principle, when we add or remove gaseous or aqueous components, the equilibrium shifts to resist the change.

Problem 1: For ${\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{C}\mathrm{a}\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{C}\mathrm{O}}_2\left(\mathrm{g}\right),K_p=0.1$ bar at 1000 K . Find $p_{{\mathrm{C}\mathrm{O}}_2}$ at equilibrium.

$K_p=p_{{\mathrm{C}\mathrm{O}}_2}=0.1\mathrm{b}\mathrm{a}\mathrm{r}$

Problem 2: For $3\mathrm{F}\mathrm{e}\left(\mathrm{s}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{F}\mathrm{e}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)+4{\mathrm{H}}_2\left(\mathrm{g}\right),K_p=16$ at 800 K . Initial: $p_{{\mathrm{H}}_2\mathrm{O}}=1$ bar, no   ${\mathrm{H}}_2$ .Find $p_{{\mathrm{H}}_2}$ at equilibrium.

Let $p_{{\mathrm{H}}_2}=4x,p_{{\mathrm{H}}_2\mathrm{O}}=1-4x$ .

$K_p=\frac{(4x{)}^4}{(1-4x{)}^4}=16,\mathrm{}\frac{4x}{1-4x}=2,\mathrm{}x=0.125,\mathrm{}{p}_{{\mathrm{H}}_2}=0.5\text{bar}$

Problem 3: For ${\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{C}\mathrm{a}\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{C}\mathrm{O}}_2\left(\mathrm{g}\right),K_p=0.1$ a bar at 1000 K . Convert to $K_c$ . $\left(R=0.0831\right.$ bar $\left.\mathrm{L}\mathrm{m}\mathrm{o}\mathrm{l}{}^{-1}{\mathrm{K}}^{-1}\right)$  

$\mathrm{\Delta }n=1,\mathrm{}{K}_p=K_c\left(RT\right),\mathrm{}{K}_c=\frac{K_p}{RT}=\frac{0.1}{0.0831\times 1000}\approx 1.2\times {10}^{-3}\mathrm{m}\mathrm{o}\mathrm{l}{\mathrm{L}}^{-1}$

Problem 4: For ${\mathrm{N}\mathrm{H}}_4\mathrm{H}\mathrm{S}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{N}\mathrm{H}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right),K_p=0.01$ bar ${}^2$ at 298 K. Find the total pressure at equilibrium.

$p_{{\mathrm{N}\mathrm{H}}_3}=p_{{\mathrm{H}}_2\mathrm{S}}=x,\mathrm{}{K}_p=x^2=0.01,\mathrm{}x=0.1\mathrm{b}\mathrm{a}\mathrm{r}$

Total pressure $=p_{{\mathrm{N}\mathrm{H}}_3}+p_{{\mathrm{H}}_2\mathrm{S}}=0.2$  bar.

• Pitfall: Including solids/liquids in $K$ Tip: Omit pure phases.
• Pitfall: Miscalculating $\mathrm{\Delta }n$ . Tip: Count only gaseous moles.
• Pitfall: Ignoring external gas effects. Tip: Use Le Chatelier’s principle.
• Tip: Practice ICE tables for gaseous equilibria.
• Tip: Memorize $R=0.0831$ bar $\mathrm{L}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$  for $K_p$ conversions.