Heterogeneous Equilibrium

Heterogeneous equilibrium occurs when the reactants and products of a chemical reaction exist in different phases while maintaining a balance between the forward and reverse reactions.

This type of equilibrium involves substances in distinct states of matter, such as solids, liquids, and gases. Examples include gas-liquid equilibrium, where gases dissolve in liquids, or gas-solid equilibrium, where gases are adsorbed onto solid surfaces.

An  example is the equilibrium between water vapor and liquid water in a closed container, demonstrating how multiple phases can coexist in a balanced state.

In heterogeneous equilibrium, the concentrations of reactants and products may not be uniform throughout the system due to multiple phases.

1.0Equilibrium Constant

The equilibrium constant (K) expresses the ratio of the concentrations of products to reactants at equilibrium and helps determine a reaction's chemical behavior.

At a constant temperature, the rate constants for the forward and reverse reactions remain unchanged. The ratio of the forward reaction rate constant to the reverse reaction rate constant defines the equilibrium constant (K).

The general expression for the equilibrium constant is:

$K=\frac{[C{]}^c\cdot [D{]}^d}{[A{]}^a\cdot [B{]}^b}$

Where:

• [A], [B], [C], and [D] represent the molar concentrations of reactants and products at equilibrium.
• a, b, c, and d are their respective stoichiometric coefficients in the balanced chemical equation.

2.0Equilibrium Constant in Heterogeneous Systems

The equilibrium constant expression for heterogeneous reactions differs from that of homogeneous reactions. For example, in the thermal dissociation of calcium carbonate:

CaCO₃(s)⇌CaO(s)+CO₂(g)

Since the molar concentrations of solids remain constant throughout the reaction, only the gaseous component (CO₂) appears in the equilibrium expression:

Kc=[CO₂]

Similarly, in terms of partial pressure:

K_p=p_CO2

This implies that, at a given temperature, the partial pressure of CO₂ remains constant at equilibrium with CaO and CaCO₃.

For another example:

CO₂(g)+C(s)⇌2CO(g)

The equilibrium constant is expressed as:

$K_c=\frac{[CO{]}^2}{[C{O}_2]}$

3.0Equilibrium in Physical Changes

Equilibrium can also exist in physical processes, particularly during phase transitions. These include:

1. Solid-Liquid Equilibrium
2. Liquid-Gas Equilibrium
3. Solid-Gas Equilibrium

1. Solid-Liquid Equilibrium

An example of solid-liquid equilibrium is the coexistence of ice and water at 0°C and 1 atm in a thermally insulated container. Here, the rate of ice melting equals the rate of water freezing:

• H₂O(s)⇌H₂O(l)

2. Liquid-Gas Equilibrium

The equilibrium between a liquid and its vapor can be observed in a closed container where water evaporates and condenses at the same rate:

• H₂O(l)⇌H₂O(g)

3. Solid-Gas Equilibrium

Certain substances, such as ammonium chloride, iodine, and camphor, undergo sublimation, transitioning directly from solid to gas without passing through the liquid state. Their equilibrium can be represented as:

• I₂(s)⇌I₂(g)
• NH₄Cl(s)⇌NH₄Cl(g)
• Camphor(s)⇌Camphor(g)

These phase transitions also establish dynamic equilibrium, where the sublimation rate matches the deposition rate.

4.0Solved Example

Q. T in homogeneous equilibrium value of K_p for the reaction, CO₂ (g) + C (s) ⇆ 2CO (g), is 3.0 at 1000 K. If initially P_CO2 = 0.48 bar and P_CO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂.

Solution.

Given:

• K_p=3.0 at 1000K1000K1000K
• P_CO2initial =0.48 bar
• P_COinitial =0 bar
• Solid graphite is present (not included in K_p).

Step 1: Define Changes in Partial Pressures

Let the change in CO₂  pressure be x, so at equilibrium:

P_CO2=(0.48−x)

Since 1 mole of CO₂ produces 2 moles of CO,

 P_CO=2x

Step 2: Express K_p in Terms of x

$K_p=\frac{(P_{CO}{)}^2}{P_{C{O}_2}}$

$3.0=\frac{(2x{)}^2}{(0.48-x)}$

$3.0(0.48-x)=4x^2$

$1.44-3x=4x^2$

$4x^2+3x-1.44=0$

Step 3: Solve for x Using Quadratic Formula

$x=\frac{-3\pm \sqrt{(3{)}^2-4(4)(-1.44)}}{2(4)}$

$x=\frac{-3\pm \sqrt{9+23.04}}{8}$

$x=\frac{-3\pm \sqrt{32.04}}{8}$

$x=\frac{-3\pm 5.66}{8}$

$x=\frac{-3+5.66}{8}=\frac{2.66}{8}=0.3325$

Step 4: Calculate Equilibrium Partial Pressures

P_CO2=0.48−0.3325=0.1475 bar

P_CO=2(0.3325)=0.665 bar

Final answer

P_CO2=0.15 bar (rounded)

P_CO =0.67 bar (rounded)