IX. Chemical Equilibrium

Key focus of this chapter: equilibrium constant

This chapter focuses on equilibrium constant and gives concise summaries of the important things about Le Chatelier’s principle, acid and base equilibrium, pH, and solubility product constant in more detail.

A. Reversible and irreversible reactions

Reversible reaction
Irreversible reaction

B. Equilibrium state

: a state of balance in a reversible reaction, which is reached when the forward and reverse reactions have the same reaction rate (velocity) and the concentrations of reactant and product are no longer changing.

Fig. 1 Equilibrium state of reaction rate and concentration over time

Forward and reverse reactions continue at equal rates at equilibrium.
Reactant and product concentrations remain constant; they need not be equal.

C. The equilibrium constant, K

1. K_eq

For a specified reaction, K depends on temperature and is independent of initial concentrations and a catalyst.
Pure solids and pure liquids have unit activity and are omitted from the usual equilibrium expression. Dissolved species and gases are included.
Q < K means net forward reaction; Q > K means net reverse reaction; Q = K means equilibrium, for any value of K. K > 1 or K < 1 describes the relative equilibrium preference, not whether the system is at equilibrium.
Rate-law exponents generally must be measured experimentally; the derivation K = k_f/k_r using stoichiometric exponents applies to an elementary reversible step.

2. K_cand K_p

Fig. 2 Factors affecting equilibrium and reaction rate. Temperature can change K and the equilibrium composition and also affects rate constants. A catalyst speeds approach to equilibrium but does not change K, ΔG°, or the equilibrium composition at a fixed temperature.

D. Le Chatelier’s principle

: when an equilibrium system is disturbed, its composition changes in a direction that counteracts the disturbance. The gas-pressure rules below refer to changing volume at a fixed temperature; adding an inert gas at constant volume does not shift ideal-gas equilibrium.

Changes of conditions	Direction of reactions
Concentration
Pressure (volume) & Mole
Temperature & ΔH

Qs/ Which of the following shifts the equilibrium in the forward direction in the reaction shown?

A. Adding WO₃

B. Adding H₂

C. Removing W

D. Removing H₂O

E. Both B and D

ANS: E Adding WO₃ or removing W does not affect the direction of equilibrium, because they are solid. When adding H₂ or removing H₂O, the equilibrium moves in a forward direction to relieve the stress of the different concentrations between the reactants and products.

Qs/ Which of the following is the reverse direction of the equilibrium from the reaction shown?

A. Increasing pressure

B. Decreasing pressure

C. Increasing volume

D. Removing Y

E. A and C

ANS: A When the pressure is increased, the equilibrium moves from big moles of the product (2Y + 2Z) to small moles of the reactant (3X).

Qs/ Which of the following shifts the equilibrium in the forward direction in the reaction shown?

A. Adding X

B. Removing Y

C. Decreasing pressure

D. Increasing temperature

E. All are correct

ANS: E When the reaction is endothermic (ΔH >0), the direction of the equilibrium is shifted forward by an increase in temperature.

E. Acid and base equilibrium

1. Acid and base theory

	Acid	Base

2. Acid and base strength

a. Acid strength

b. Acids and bases in a solution

A proton-transfer equilibrium favors the weaker acid and weaker base.
The salt from stoichiometric neutralization of a strong acid and weak base generally gives an acidic solution; a weak acid and strong base generally gives a basic solution. Excess acid or base changes this conclusion.
Mg(OH)₂ is sparingly soluble; distinguish the extent of dissolution from dissociation of the dissolved portion.

Strong acids	• 100% dissociation of H
Weak acids	• Partial dissociation of H

Strong acids and strong bases
Strong acids and weak bases Weak acids and strong bases Weak acids and weak bases

	with halogens
	with soluble compound anions

	with 1A group
	with 2A group

3. Equilibrium constant, K_a, K_b, and K_w

a. K_a (acid-dissociation constant) and K_b (base-dissociation constant)

: equilibrium constants for acid and base ionization.

They depend on temperature; larger K_a or K_b corresponds to a stronger acid or base in the same solvent.
For a conjugate acid–base pair at the same temperature, K_aK_b = K_w.

b. K_w (ion-product constant for water)

: K_w = [H₃O⁺][OH⁻] in the dilute-solution approximation. At 25 °C, K_w ≈ 1.0 × 10⁻¹⁴. Its value varies with temperature.

Equilibrium constants	Equilibrium Equations

Qs/ Calculate the value of K_a when 0.002M of HOBr is 0.1% dissociated in a solution.

Qs/ At 25 °C, K_a of HNO₂ is 4.5 × 10⁻⁴. Calculate K_b of its conjugate base, NO₂⁻.

F. pH

For dilute aqueous solutions, pH ≈ −log[H₃O⁺] and pH + pOH = pK_w. At 25 °C, pK_w ≈ 14 and neutral pH ≈ 7; neutral pH changes with temperature. Acidic or basic pH is determined by the ion concentrations, not by a universal comparison of K_a with K_b.

Qs/ Calculate the pH of a solution containing [H⁺] = 0.000035 M.

pH = −log(3.5 × 10⁻⁵) = 4.46 (4.5 to one decimal place).

Qs/ For 0.005 M CH₃COOH with K_a = 1.8 × 10⁻⁵, calculate pH.

The quadratic equation x²/(0.005 − x) = 1.8 × 10⁻⁵ gives [H₃O⁺] ≈ 2.91 × 10⁻⁴ M and pH ≈ 3.54. The approximation x ≈ 3.0 × 10⁻⁴ M gives 6% ionization, slightly exceeding the usual 5% guideline.

G. Solubility product constant, K_sp

: the equilibrium constant for dissolution of a sparingly soluble ionic solid into its ions in a saturated solution. Compare the ion product Q_sp with K_sp to predict precipitation.

Larger K_sp implies larger molar solubility only when comparing salts with the same dissolution stoichiometry under comparable conditions; conductivity and freezing point also depend on ion concentrations and properties.

Conditional comparison for the diagram above: saturated, dilute solutions; same solvent, temperature and dissolution stoichiometry; no common ions or side reactions; comparable ionic charges and mobilities. Higher Ksp then gives higher molar solubility and ion concentration. Conductivity rises only under those ion conditions; freezing-point lowering depends on total dissolved-particle molality.

Qs/ What is the expression for K_sp of Ca₃(PO₄)₂ in a solution?

Qs/ Calculate K_sp for a saturation of BaF₂when the concentration of Ba²⁺ is 2.0 × 10^-5 M.