18.3 Reversible Reactions and Equilibrium

Reversible Reactions

  • Some reactions are reversible, meaning the conversion of reactants to products and vice versa occurs simultaneously.

  • Example of a reversible reaction:
    2SO2(g) + O2(g) \rightleftharpoons 2SO_3(g)

    • Forward reaction: 2SO2(g) + O2(g) \rightarrow 2SO_3(g)
    • Reverse reaction: 2SO2(g) + O2(g) \leftarrow 2SO_3(g)
  • Molecules of SO2 and O2 react to produce SO3, while SO3 decomposes into SO2 and O2.

Establishing Equilibrium

  • Chemical equilibrium is reached when the rates of the forward and reverse reactions are equal.

  • At equilibrium, concentrations of reactants and products remain constant over time.

  • Chemical equilibrium is a dynamic state where both forward and reverse reactions continue, but there is no net change in concentrations.

Concentrations at Equilibrium

  • At equilibrium, the rates of forward and reverse reactions are equal, but the concentrations of reactants and products are not necessarily equal.

  • The equilibrium position indicates the relative concentrations of reactants and products at equilibrium.

  • For a reaction A \rightleftharpoons B, if the equilibrium mixture contains 1% A and 99% B, the formation of B is favored.

  • In principle, most reactions are reversible to some extent under the right conditions.

  • If no reactants are detected, the reaction is considered to have gone to completion or to be irreversible. If no products are detected, it is considered that no reaction has taken place.

Factors Affecting Equilibrium: Le Châtelier’s Principle

  • Le Châtelier’s principle: If a stress is applied to a system in dynamic equilibrium, the system changes to relieve the stress.

  • Stresses include changes in concentration, temperature, and pressure.

Concentration

  • Changing the concentration of reactants or products disturbs the equilibrium, and the system adjusts to minimize this effect.

  • For the decomposition of carbonic acid:
    H2CO3(aq) \rightleftharpoons CO2(aq) + H2O(l)
    At equilibrium, carbonic acid is less than 1%, while carbon dioxide and water are greater than 99%.

  • Adding carbon dioxide increases the rate of the reverse reaction, shifting the equilibrium towards the reactants.

  • Removing carbon dioxide decreases the rate of the reverse reaction, shifting the equilibrium towards the products.

  • In blood, an equilibrium exists between carbonic acid, carbon dioxide, and water. During exercise, the concentration of CO_2 increases, shifting the equilibrium towards carbonic acid.

Temperature

  • Increasing temperature shifts the equilibrium in the direction that absorbs heat (endothermic direction).

  • For the reaction:
    N2(g) + 3H2(g) \rightleftharpoons 2NH_3(g) + heat
    Heating the reaction mixture shifts the equilibrium to the left (favors reactants), while cooling shifts it to the right (favors products).

Pressure

  • Changes in pressure affect equilibrium systems involving gases only if there's an unequal number of moles of gas on each side of the equation.

  • Increasing pressure favors the side with fewer gas molecules, and decreasing pressure favors the side with more gas molecules.

  • For the reaction: N2(g) + 3H2(g) \rightleftharpoons 2NH_3(g), increasing pressure shifts the equilibrium to the right, favoring ammonia production.

Catalysts and Equilibrium

  • Catalysts decrease the time it takes to reach equilibrium but do not affect the amounts of reactants and products at equilibrium.

Sample Problem 18.2

Consider the reaction:
PCl5(g) + heat \rightleftharpoons PCl3(g) + Cl_2(g)

  • Adding Cl_2 shifts the equilibrium to the left.
  • Increasing pressure shifts the equilibrium to the left (fewer gas molecules on the left side).
  • Removing heat shifts the equilibrium to the left.
  • Removing PCl_3 as it forms shifts the equilibrium to the right.

Increase in Pressure

  • For the reaction: 4HCl(g) + O2(g) \rightleftharpoons 2Cl2(g) + 2H_2O(g), an increase in pressure shifts the equilibrium to the right because there are fewer moles of gas on the right side (4 vs. 5).

Equilibrium Constants

  • Chemists express the equilibrium position using a numerical value called the equilibrium constant (K_{eq}).

  • For a general reaction:
    aA + bB \rightleftharpoons cC + dD

  • The equilibrium constant expression is:
    K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}

  • The value of K_{eq} depends on temperature.

  • A large K{eq} indicates that products are more common at equilibrium, while a small K{eq} indicates that reactants are more common.

  • Large value of K_{eq} (e.g. 3.1 \times 10^{11}) indicates mostly product at equilibrium.

  • Small value of K_{eq} (e.g. 3.1 \times 10^{-11}) indicates mostly reactant at equilibrium.

  • Intermediate value of K_{eq} (e.g. 0.15 or 50) indicates significant amounts of both reactant and product at equilibrium.

Sample Problem 18.3

  • For the reaction: N2O4(g) \rightleftharpoons 2NO_2(g)

  • Given: [N2O4] = 0.0045 mol/L and [NO_2] = 0.030 mol/L

  • K{eq} = \frac{[NO2]^2}{[N2O4]} = \frac{(0.030 mol/L)^2}{(0.0045 mol/L)} = 0.20

Sample Problem 18.4

  • For the reaction: H2(g) + I2(g) \rightleftharpoons 2HI(g)

  • Initial concentrations: [H2] = 1.00 mol/L, [I2] = 1.00 mol/L

  • Equilibrium concentration: [HI] = 1.56 mol/L

  • Let x be the amount of H2 and I2 consumed. 2x = 1.56 mol, so x = 0.780 mol.

  • Equilibrium concentrations: [H2] = [I2] = 1.00 - 0.780 = 0.22 mol/L

  • K{eq} = \frac{[HI]^2}{[H2][I_2]} = \frac{(1.56 mol/L)^2}{(0.22 mol/L)(0.22 mol/L)} = 5.0 \times 10^1

Sample Problem 18.5

  • For the reaction: 2BrCl(g) \rightleftharpoons Br2(g) + Cl2(g)

  • Keq = 11.1 and [Cl_2] = 4.00 mol/L

  • Since equal numbers of moles of Br2 and Cl2 are formed, [Br_2] = 4.00 mol/L

  • K{eq} = \frac{[Br2][Cl2]}{[BrCl]^2}, so [BrCl]^2 = \frac{[Br2][Cl2]}{K{eq}} = \frac{(4.00 mol/L)(4.00 mol/L)}{11.1} = 1.44 mol^2/L^2

  • [BrCl] = \sqrt{1.44 mol^2/L^2} = 1.20 mol/L

  • For the reaction: H2(g) + Cl2(g) \rightleftharpoons 2HCl(g)

  • Equilibrium concentrations: [HCl] = 1.76 \times 10^{-2} mol/L, [H2] = [Cl2] = 1.60 \times 10^{-3} mol/L

  • K{eq} = \frac{[HCl]^2}{[H2][Cl_2]} = \frac{(1.76 \times 10^{-2} mol/L)^2}{(1.60 \times 10^{-3} mol/L)(1.60 \times 10^{-3} mol/L)} = 121

Key Concepts

  • Stresses that upset equilibrium: changes in concentration, temperature, and pressure.

  • At equilibrium, forward and reverse reaction rates are equal, with no net change in reactant and product concentrations.

Key Equation

  • K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Glossary Terms

  • Reversible reaction: Conversion of reactants to products and products to reactants occur simultaneously.

  • Chemical equilibrium: State of balance where forward and reverse reaction rates are equal; no net change in reactant and product amounts.

  • Equilibrium position: Relative concentrations of reactants and products at equilibrium; indicates whether reactants or products are favored.

  • Le Châtelier’s principle: When a stress is applied to a system in dynamic equilibrium, the system changes to relieve the stress.

  • Equilibrium constant: Ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power equal to the number of moles in the balanced equation.