Lesson 7.2: Equilibrium Law and the Equilibrium Constant (12 Chem)

Chemical Equilibrium

Equilibrium Law and the Equilibrium Constant

The equilibrium law, also known as the law of mass action, is a mathematical description of a chemical system at equilibrium. It allows chemists to predict the concentrations of reactants and products in a closed system once equilibrium is reached. The equilibrium constant, K, is a numerical value that defines the equilibrium law for a given system and is specific to a particular reaction at a given temperature.

• Equilibrium Law: For a general reaction: $aA + bB \rightleftharpoons cC + dD$, the equilibrium law is given by: $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• Equilibrium Constant (K): The value of K is constant for a given reaction at a specific temperature, regardless of the initial concentrations of reactants and products.

• Units: The units of K vary depending on the reaction, so K is usually reported without units.

Analyzing the Equilibrium Law

Experimental data can be used to calculate the equilibrium constant for a reaction. For example, the formation of hydrogen iodide gas from hydrogen and iodine can be represented as:

• Balanced equation: $H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)$

• Equilibrium law: $K = \frac{[HI]^2}{[H_2][I_2]}$

• Experimental trials show that K remains constant (e.g., 49.8) for all trials at the same temperature.

Temperature Dependence of K

The equilibrium constant is dependent on temperature. For example, the synthesis of ammonia from nitrogen and hydrogen:

• Balanced equation: $N_2(g) + 3 H_2(g) \rightleftharpoons 2 NH_3(g)$

• K decreases as temperature increases, indicating less product formation at higher temperatures.

Writing Equilibrium Law Equations

To write an equilibrium law equation:

1. Identify reactants and products and their coefficients from the balanced equation.

2. Place products in the numerator and reactants in the denominator, using coefficients as exponents.

Example: $4 NH_3(g) + 7 O_2(g) \rightleftharpoons 4 NO_2(g) + 6 H_2O(g)$ $K = \frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}$

Calculating K for Forward and Reverse Reactions

The equilibrium constant for the reverse reaction is the reciprocal of the forward reaction's K.

• Forward: $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$

• Reverse: $K_r = \frac{[N_2][H_2]^3}{[NH_3]^2}$

• $K_r = 1/K$

Equilibrium Constant and Reaction Rate

At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. For an elementary reaction:

• Forward rate: $k_f [A][B]$

• Reverse rate: $k_r [C][D]$

• At equilibrium: $k_f [A][B] = k_r [C][D]$

• $\frac{k_f}{k_r} = \frac{[C][D]}{[A][B]} = K$

Types of Equilibria

Homogeneous vs. Heterogeneous Equilibrium

Homogeneous equilibrium occurs when all reactants and products are in the same state (e.g., all gases). Heterogeneous equilibrium involves reactants and products in different states (e.g., solids, liquids, gases).

• For heterogeneous equilibria, concentrations of pure solids and liquids are omitted from the equilibrium law equation because they are constant.

Example: $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ $K = [CO_2]$

Example: $2 H_2O(l) \rightleftharpoons 2 H_2(g) + O_2(g)$ $K = [H_2]^2[O_2]$ (water omitted because it is a pure liquid)

Sample Problems for Heterogeneous Equilibria

• Decomposition of solid ammonium chloride: $NH_4Cl(s) \rightleftharpoons NH_3(g) + HCl(g)$ $K = [NH_3][HCl]$

• Production of sodium carbonate: $2 NaHCO_3(s) \rightleftharpoons Na_2CO_3(s) + CO_2(g) + H_2O(g)$ $K = [CO_2][H_2O]$

The Magnitude of the Equilibrium Constant

Interpreting K Values

The magnitude of K indicates the equilibrium position:

K <<<1	K ≈ 1	K >>> 1
Equilibrium lies far to the left (reactants favored)	Products and reactants have similar concentrations at equilibrium.	Equilibrium lies far to the right (products favored).

Example: $2 CO(g) + O_2(g) \rightleftharpoons 2 CO_2(g)$ K = $3.3 \times 10^{91}$ (products strongly favored)

Example: $NO_2(g) + NO(g) \rightleftharpoons N_2O(g) + O_2(g)$ K = 0.915 (products and reactants similar)

Example: $2 H_2O(g) \rightleftharpoons 2 H_2(g) + O_2(g)$ K = $7.3 \times 10^{-10}$ (reactants strongly favored)

Magnitude of K	Equilibrium Position
K >>> 1	Far to the right (favours products)
K ≈ 1	Products ≈ Reactants
K <<< 1	Far to the left (favours reactants)

Summary of Key Concepts

• An equilibrium law equation mathematically represents a reversible chemical reaction.

• The equilibrium constant, K, is the ratio of product concentrations to reactant concentrations at equilibrium, independent of initial concentrations but dependent on temperature.

  • Why? If the initial concentrations of the reactants are said amount, then the products are initially zero to start. K represents the E (equilibrium) / ratio between reactants and products on an ICE table.

• Homogeneous equilibrium: all substances in the same state; heterogeneous equilibrium: substances in different states.

• Concentrations of pure solids and liquids are not included in equilibrium expressions!!

• The magnitude of K reflects the equilibrium position.

Practice Problems

• Write equilibrium law equations for various reactions.

• Calculate equilibrium constants from given concentrations.

• Interpret the significance of K values for predicting product yield.