What is Equilibrium Class 11: Definition and Key Concepts in Chemistry

Definition of Chemical Equilibrium in Class 11 Chemistry

Chemical equilibrium is the condition in which the rate of the forward reaction equals the rate of the backward reaction in a reversible chemical process. At this point, the concentrations of reactants and products remain constant over time, although both reactions continue to occur. This dynamic balance is a key topic in Class 11 NCERT Chemistry.

Key points:

• Equilibrium is achieved in a closed system.
• It does not mean the concentrations of reactants and products are equal, but their rates of change are zero.
• The reaction is reversible.

For example, in the reaction:

$$ ext{N}_2(g) + 3 ext{H}_2(g) \rightleftharpoons 2 ext{NH}_3(g)$$

At equilibrium, the rate of formation of ammonia equals the rate of its decomposition back into nitrogen and hydrogen.

Understanding the Equilibrium Constant ($K_c$) and Its Significance

The equilibrium constant, $K_c$, expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients.

For a general reaction:

$$aA + bB \rightleftharpoons cC + dD$$

The equilibrium constant is given by:

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

Where:

• Square brackets denote molar concentrations.
• $K_c$ is temperature-dependent.

Interpretation of $K_c$ values:

Example:

For the reaction:

$$ ext{H}_2(g) + ext{I}_2(g) \rightleftharpoons 2 ext{HI}(g)$$

If at equilibrium $[H_2] = 0.2$ M, $[I_2] = 0.2$ M, and $[HI] = 0.6$ M, then:

$$K_c = \frac{[HI]^2}{[H_2][I_2]} = \frac{(0.6)^2}{0.2 \times 0.2} = \frac{0.36}{0.04} = 9$$

This indicates the reaction favors product formation.

Dynamic Nature of Equilibrium: Forward and Backward Reactions

Equilibrium is dynamic, meaning both forward and backward reactions continue to occur at equal rates. This balance results in no net change in the concentrations of reactants and products.

Characteristics of dynamic equilibrium:

• Reaction rates of forward and reverse reactions are equal.
• Concentrations remain constant but not necessarily equal.
• Occurs only in a closed system.

For example, in the synthesis of ammonia:

$$ ext{N}_2(g) + 3 ext{H}_2(g) \rightleftharpoons 2 ext{NH}_3(g)$$

At equilibrium:

$$\text{Rate}_{forward} = \text{Rate}_{backward}$$

This dynamic balance is crucial to understanding how chemical systems behave under different conditions.

Le Chatelier’s Principle: Predicting Changes in Equilibrium

Le Chatelier’s Principle states that if a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, the system adjusts itself to counteract the disturbance and restore a new equilibrium.

Common disturbances and responses:

• Change in concentration: Adding reactants shifts equilibrium to products.
• Change in pressure: Increasing pressure shifts equilibrium toward fewer gas molecules.
• Change in temperature: For exothermic reactions, increasing temperature shifts equilibrium to reactants; for endothermic, to products.

Example:

For the reaction:

$$ ext{N}_2(g) + 3 ext{H}_2(g) \rightleftharpoons 2 ext{NH}_3(g) + ext{heat}$$

• Increasing pressure favors ammonia formation (fewer gas molecules).
• Increasing temperature shifts equilibrium toward nitrogen and hydrogen (endothermic direction).

Le Chatelier’s Principle helps predict how conditions affect yield.

Types of Equilibrium: Homogeneous vs Heterogeneous

Equilibrium can be classified based on the phases of reactants and products:

• Homogeneous Equilibrium: All reactants and products are in the same phase.

Example:

$$ ext{H}_2(g) + ext{I}_2(g) \rightleftharpoons 2 ext{HI}(g)$$

• Heterogeneous Equilibrium: Reactants and products are in different phases.

Example:

$$ ext{CaCO}_3(s) \rightleftharpoons ext{CaO}(s) + ext{CO}_2(g)$$

In heterogeneous equilibrium, solids and liquids do not appear in the equilibrium constant expression because their concentrations remain constant.

Worked Example: Calculating Equilibrium Concentrations

Consider the reaction:

$$ ext{N}_2(g) + 3 ext{H}_2(g) \rightleftharpoons 2 ext{NH}_3(g)$$

Initial concentrations:

• $[N_2] = 1.0$ M
• $[H_2] = 3.0$ M
• $[NH_3] = 0$ M

At equilibrium, $[NH_3] = 0.5$ M. Calculate the equilibrium concentrations of $N_2$ and $H_2$ and the value of $K_c$.

Solution:

• Since 2 moles of $NH_3$ are formed, $N_2$ decreases by 0.25 M (1:2 ratio).
• $H_2$ decreases by 0.75 M (3:2 ratio).

So,

• $[N_2]_{eq} = 1.0 - 0.25 = 0.75$ M
• $[H_2]_{eq} = 3.0 - 0.75 = 2.25$ M

Calculate $K_c$:

$$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.5)^2}{(0.75)(2.25)^3} = \frac{0.25}{0.75 \times 11.39} = \frac{0.25}{8.54} = 0.0293$$

This $K_c$ value indicates the equilibrium position favors reactants slightly.