Homogeneous Equilibria: Definition, Key Features, Problems & Class 12 Notes

A chemical equilibrium where all the reactants and products are in the same phase is called homogeneous equilibrium. For example, consider:

$N2\left(g\right)+3H2\left(g\right)<=>2NH3\left(g\right)$

Here, all components are gases, making it homogeneous. The equilibrium constant $K_c$ is the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient. For a reaction $aA+bB<=>cC+dD$  : $K_c=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B{]}^b}$

For gaseous reactions, $K_p$ use partial pressures:

$K_p=\frac{{\left(p_C\right)}^c{\left(p_D\right)}^d}{{\left(p_A\right)}^a{\left(p_B\right)}^b}$

These are related by: $K_p=K_c(RT{)}^{\mathrm{\Delta }n}$

where,

$\mathrm{\Delta }n$  is the change in moles of gas (products minus reactants), $R=0.0831$  bar $\mathrm{L}\mathrm{m}\mathrm{o}\mathrm{l}{}^{-1}{\mathrm{K}}^{-1}$ , and $T$ is in Kelvin. This framework, outlined in is essential for tackling JEE numericals.

The equilibrium constant in a gaseous system is used to calculate the ratio of the concentrations or partial pressures of products to reactants at equilibrium. It is used to say the direction and extent of a chemical reaction.

Read More: NCERT Solutions | Class 12 Chemistry NCERT Solutions

There are two main forms of equilibrium constant:
Kc - molar concentration
Kp - partial pressure

For Gaseous Reactions

aA(g)+bB(g)⇌cC(g)+dD(g)

Equilibrium constant in terms of concentration (Kc):

$\begin{array}{l}\Rightarrow K_{{}_c}=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}\end{array}$

Equilibrium constant in terms of Pressure (Kp):

$K_p=\frac{{\left(P_C\right)}^c{\left(P_D\right)}^d}{{\left(P_A\right)}^a{\left(P_B\right)}^b}$

Where, pA, pB, pC and pD are the partial pressures of the substances A, B, C, and D, respectively. If we assume that the gases are ideal, then the equation of an  ideal gas is:

$pV=nRT\text{or}p=\frac{nRT}{V}$

• p is pressure
• V is volume in cubic meters 
• n is the number of moles of gas
• T temperature in Kelvin
• n/V molar concentration

Homogeneous equilibria have various characteristics:

1. All reactants and products are in the same phase.
2. The concentration remains constant over time because the forward and reverse reaction occurs at the same time.
3. The ratio of concentrations of products to reactants is constant at a given temperature.
4. Homogeneous equilibrium takes place in a reversible reaction, which is indicated by a double arrow ⇌.
5. In exothermic reactions equilibrium constant (K) decreases with heat, while in endothermic reactions, K increases with heat.
6. According to Le Chatelier's principle, applies change in temperature, concentration, or pressure can shift the equilibrium position to restore balance. 
7. In homogeneous equilibrium, gaseous or aqueous solutions are included. Solids and liquids are ignored.

Important Links:

Homogeneous equilibria are an important topic in the chapter Equilibrium.NCERT explains the homogeneous equilibria using reactions like ammonia synthesis. Here, we have added the JEE-style twist by analyzing the extent of reaction using the hydrogen-iodine system.

$H2\left(g\right)+I2\left(g\right)<=>2HI\left(g\right)$

According to NCERT, $K_c\approx 46.4$ at 731 K, which indicates significant formation of HI. The JEE twist reverses the starting condition: starting with only HI, the reaction becomes:

$2HI\left(g\right)<=>H2\left(g\right)+I2\left(g\right)$

Here, $K_c^{\text{'}}=1/K_c\approx 0.0216$ , suggesting HI remains dominant at equilibrium.

NCERT aslo covers the relationship between $K_p-K_c$ . For ammonia synthesis,

$\mathrm{\Delta }n=2-(1+3)=-2$ ; so

$K_p=K_c(RT{)}^{-2}$

A JEE problem: $K_c=3.55\times {10}^2$ at 500 K, compute $K_p$ .

Solution:

$RT=$ $0.0831\times 500=41.55$ , so $(RT{)}^{-2}\approx 5.79\times {10}^{-4}$ .

Thus: $K_p=3.55\times {10}^2\times 5.79\times {10}^{-4}\approx 0.206{\mathrm{b}\mathrm{a}\mathrm{r}}^{-2}$

For reactions like: $CO\left(g\right)+H2O\left(g\right)<=>CO2\left(g\right)+H2\left(g\right)$

In JEE Main, problems based on equilibrium topics focus on equilibrium calculations and shifts. Important strategies for solving the homogeneous equilibrium problems are:

1. Write an accurate expression for $K_c$ or $K_p$ , matching stoichiometry.
2. Create an ICE tale for concentration problems.
3. For $K_p=K_c(RT{)}^{\mathrm{\Delta }n}$ , calculate the exponent $\mathrm{\Delta }n$ correctly correctly.
4. To simplify quadratics, use approximations for small $K_c$ .

5. Check $K_p$ uses the bar unit.

Also Read: NCERT Solution for Class 11 Chapter 6 

Problem 1: For $2NO\left(g\right)+O2\left(g\right)<=>2NO2\left(g\right)$ at 800 K , the equilibrium concentrations are $\left[\mathrm{N}\mathrm{O}\right]=3.0\times {10}^{-3}\mathrm{M},\left[\mathrm{O}2\right]=4.2\times {10}^{-3}\mathrm{M},\left[\mathrm{N}\mathrm{O}2\right]=2.8\times {10}^{-3}\mathrm{M}$ . Find $K_c$ .

$K_c=\frac{[NO2{]}^2}{\left[NO{]}^2\right[O2]}=\frac{{\left(2.8\times {10}^{-3}\right)}^2}{{\left(3.0\times {10}^{-3}\right)}^2\left(4.2\times {10}^{-3}\right)}\approx 207.4$

Problem 2: For $2\mathrm{N}\mathrm{O}\mathrm{C}\mathrm{l}\left(g\right)<=>2\mathrm{N}\mathrm{O}\left(g\right)+Cl2\left(g\right)$ at $1069\mathrm{K},K_c=3.75\times {10}^{-6}$ . Find $K_p$ . 

$\mathrm{\Delta }n=3-2=1,\mathrm{}{K}_p=K_c\left(RT\right)=3.75\times {10}^{-6}\times (0.0831\times 1069)\approx 0.033$

Problem 3: A 10 L vessel at 725 K contains 1 mol each of $H2O\left(\mathrm{g}\right)$ and $CO\left(\mathrm{g}\right)$ . At equilibrium, $40\mathrm{\%}$  of $H2O$  reacts via $H2O\left(g\right)+CO\left(g\right)<=>H2\left(g\right)+CO2\left(g\right)$ . Find $K_c$ . (NCERT Exercise 6.14, p. 209)

$\text{Initial:}\left[H2O\right]=\left[CO\right]=0.1\mathrm{M},\mathrm{}\text{Reacted:}0.4\times 0.1=0.04\mathrm{M}$

Equilibrium: $\left[H2O\right]=\left[CO\right]=0.06\mathrm{M},\mathrm{}\left[H2\right]=\left[CO2\right]=0.04\mathrm{M}$

$K_c=\frac{\left[H2\right]\left[CO2\right]}{\left[H2O\right]\left[CO\right]}=\frac{(0.04{)}^2}{(0.06{)}^2}\approx 0.444$

Problem 4: For $N2\left(g\right)+O2\left(g\right)<=>2NO\left(g\right)$ at $2000\mathrm{K},K_c=4.1\times {10}^{-4}$ . If 1 mol each of ${\mathrm{N}}_2$ ${\mathrm{O}}_2$ and are in a 2 L vessel, find [NO] at equilibrium.

Let $\left[\mathrm{N}\mathrm{O}\right]=2\mathrm{x},\left[{\mathrm{N}}_2\right]=\left[{\mathrm{O}}_2\right]=0.5-\mathrm{x}$ . Then: $\begin{array}{c}{K}_c=\frac{(2x{)}^2}{(0.5-x{)}^2}=4.1\times {10}^{-4}\\ \frac{4x^2}{(0.5-x{)}^2}=4.1\times {10}^{-4}\end{array}$

Since $K_c$  is small, $x\ll 0.5$ , so . Thus: $4x^2\approx 4.1\times {10}^{-4}\times 0.25$

$x^2\approx 2.5625\times {10}^{-5},\mathrm{}x\approx 0.00506,\mathrm{}\left[NO\right]=2x\approx 0.0101\mathrm{M}$

• Pitfall: Miscalculating $\mathrm{\Delta }n$ . Tip: Count only gaseous species for $\mathrm{\Delta }n$ .
• Pitfall: Using concentrations for $K_p$ . Tip: Convert via $p=cRT$
• Pitfall: Misinterpreting equilibrium shifts.
• Tip: Practice ICE tables with varied initial conditions for JEE scenarios.
• Tip: Memorize $R=0.0831$ bar $\mathrm{L}{\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$  for $K_p$ .