6.6 Applications of Equilibrium Constants – Class 11 Chemistry

Before discussing the applications of equilibrium constants, it is important to summarize their key features:

1. The expression for the equilibrium constant is valid only when concentrations of reactants and products have reached equilibrium.
2. The value of the equilibrium constant is independent of the initial concentrations of the reactants and products.
3. The equilibrium constant is temperature-dependent, having a unique value for a particular reaction at a given temperature.
4. The equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction.
5. If a chemical equation is multiplied or divided by an integer, the new equilibrium constant is the original constant raised to the power of that integer.

The equilibrium constant (K) can be used to:

• Predict the extent of a reaction.
• Determine the direction of a reaction.
• Calculate equilibrium concentrations of reactants and products.

Predicting the Extent of a Reaction

The magnitude of the equilibrium constant gives an idea about how far a reaction proceeds before reaching equilibrium. A large K₍c₎ value indicates a high concentration of products at equilibrium, whereas a small K₍c₎ value suggests that reactants are largely unconverted.

General Interpretation

• If K₍c₎ > 10³, products predominate — the reaction nearly goes to completion.
• If K₍c₎ < 10⁻³, reactants predominate — the reaction hardly proceeds.
• If 10⁻³ < K₍c₎ < 10³, appreciable concentrations of both reactants and products are present.

Examples

1. H₂(g) + Cl₂(g) ⇌ 2HCl(g) at 300 K
   K₍c₎ = 4.0 × 10³¹ → reaction proceeds almost completely to form HCl.

2. H₂(g) + Br₂(g) ⇌ 2HBr(g) at 300 K
   K₍c₎ = 5.4 × 10⁸ → reaction proceeds nearly to completion.

3. H₂O(g) ⇌ H₂(g) + ½O₂(g) at 500 K
   K₍c₎ = 4.1 × 10⁻⁴⁸ → reaction hardly proceeds.

4. N₂(g) + O₂(g) ⇌ 2NO(g) at 298 K
   K₍c₎ = 4.8 × 10⁻³¹ → negligible product formation.

5. HI(g) ⇌ ½H₂(g) + ½I₂(g) at 700 K
   K₍c₎ = 57.0 → both reactants and products are present in appreciable amounts.

Reaction Quotient (Q)

The reaction quotient (Q) helps predict the direction in which a reaction will proceed before equilibrium is achieved.
For a general reaction:
aA + bB ⇌ cC + dD

Q = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

• If Q = K, the reaction is at equilibrium.
• If Q < K, the reaction proceeds forward (toward products).
• If Q > K, the reaction proceeds backward (toward reactants).

Example

For the reaction:
A + B ⇌ C + D
Given: K₍c₎ = 1 × 10⁻³, [A] = [B] = 3 × 10⁻⁴ M, [C] = [D] = 3 × 10⁻³ M

Q = [C][D] / [A][B] = (3 × 10⁻³ × 3 × 10⁻³) / (3 × 10⁻⁴ × 3 × 10⁻⁴) = 100
Since Q > K, the reaction will proceed in the reverse direction.

Predicting the Direction of a Reaction

The equilibrium constant helps determine whether a reaction will proceed forward or backward under given conditions.

For the gaseous reaction:
H₂(g) + I₂(g) ⇌ 2HI(g)
At 700 K, K₍c₎ = 57.0

If [H₂] = 0.10 M, [I₂] = 0.20 M, and [HI] = 0.40 M:
Q = [HI]² / [H₂][I₂] = (0.40)² / (0.10 × 0.20) = 8.0
Since Q < K, the reaction will proceed forward (toward HI formation).

This is where the direction of reaction depends on Q vs K comparison.

Calculating Equilibrium Concentrations

When initial concentrations are known, equilibrium concentrations can be determined using the equilibrium constant expression.

Stepwise Method

1. Write the balanced chemical equation.
2. Tabulate initial and equilibrium concentrations.
3. Substitute equilibrium values into the K expression.
4. Solve for unknowns (usually by quadratic equation).
5. Verify results by substituting back into K expression.

Example 1: Nitrogen Dioxide Formation

Given: 13.8 g of N₂O₄ in a 1 L vessel at 400 K
Molar mass (N₂O₄) = 92 g mol⁻¹

Moles of gas = 13.8 / 92 = 0.15 mol
Initial pressure = 0.15 × 0.083 × 400 = 4.98 bar

Reaction: N₂O₄(g) ⇌ 2NO₂(g)
Let x bar be the dissociation pressure.
At equilibrium:
pN₂O₄ = 4.98 − x, pNO₂ = 2x

K₍p₎ = (pNO₂)² / pN₂O₄ = (2x)² / (4.98 − x)
Given K₍p₎ = 0.15 bar, solving gives x = 0.33, hence
pNO₂ = 0.66 bar, pN₂O₄ = 4.65 bar.

Example 2: Decomposition of PCl₅

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
Given K₍c₎ = 1.8 at 400 K, initial [PCl₅] = 3.0 M.

Let x mol/L of PCl₅ dissociate:
At equilibrium: [PCl₅] = (3 − x), [PCl₃] = x, [Cl₂] = x

K₍c₎ = [PCl₃][Cl₂] / [PCl₅] = x² / (3 − x) = 1.8
Solving gives x = 1.59 M
Thus, [PCl₃] = [Cl₂] = 1.59 M, [PCl₅] = 1.41 M.

NEET & JEE Key Points

• Large K₍c₎ indicates a reaction goes nearly to completion; small K₍c₎ means incomplete reaction.
• The reaction quotient (Q) determines whether the reaction shifts forward or backward.
• Use ICE (Initial, Change, Equilibrium) tables to find equilibrium concentrations.
• Numerical problems based on K₍p₎–K₍c₎ relations are commonly asked in competitive exams.

FAQs

Q1. What does a large value of K signify?

It means the reaction proceeds almost completely toward product formation.

Q2. What does Q represent in equilibrium?

Q is the reaction quotient that predicts the direction of reaction before equilibrium is achieved.

Q3. What does it mean if Q = K?

It indicates that the system has already reached equilibrium.

Q4. How do we calculate equilibrium concentrations?

By substituting known values into the K expression and solving algebraically, often using quadratic equations.

Q5. Why is temperature crucial for equilibrium constants?

Because equilibrium constants vary with temperature but remain unchanged by catalyst or pressure variations.

Conclusion

Understanding the applications of equilibrium constants enables us to predict reaction behavior, determine the direction of shift, and compute equilibrium compositions. These principles form the backbone of chemical thermodynamics and are crucial for mastering NEET and JEE Chemistry.