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Equilibrium — Notes

6 sections 6 min readBeta

1. Physical and chemical equilibrium

Physical equilibria (closed system, fixed T): solid ⇌ liquid at the melting point; liquid ⇌ vapour (vapour pressure); solid ⇌ vapour; solute ⇌ solution in a saturated solution (solubility); gas ⇌ dissolved gas (Henry’s law: solubility ∝ pressure).

Chemical equilibrium is dynamic: the forward and reverse rates are equal and the macroscopic properties (concentration, colour, pressure) stay constant. It is reached from either side, needs a closed system, and is not altered in position by a catalyst, which only speeds its attainment.

Illustrative model with A ⇌ B, forward rate constant four times the reverse.

Key terms

Dynamic equilibrium

State in which forward and reverse processes continue at equal rates, so observable properties stay constant.

Closed system

A system that exchanges energy but not matter with its surroundings.

2. Law of mass action, Kc, Kp and Q

For aA+bB⇌cC+dDa\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}: Kc=[C]c[D]d[A]a[B]bK_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}. Pure solids and pure liquids are omitted (e.g. for CaCOX3(s)⇌CaO(s)+COX2(g)\ce{CaCO3(s) <=> CaO(s) + CO2(g)}, Kc=[CO2]K_c = [\text{CO}_2]).

For gases KpK_p uses partial pressures, and Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n} with Δn=ngaseous products−ngaseous reactants\Delta n = n_\text{gaseous products} - n_\text{gaseous reactants} (R=0.08314 L bar K−1mol−1R = 0.08314\ \text{L bar K}^{-1}\text{mol}^{-1} for pp in bar).

Manipulation: reverse → 1/K1/K; multiply coefficients by nn → KnK^n; add equations → K1K2K_1K_2. A large K (≫ 1) means products are favoured; small K means reactants are favoured.

Reaction quotient QQ: same form as KK with the present concentrations. Q<KQ < K: net forward; Q=KQ = K: equilibrium; Q>KQ > K: net reverse.

ICE table: Initial, Change (−ax, +cx), Equilibrium; substitute into KK and solve for xx.

Formulas

  • Relation between Kp and Kc
    Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}
  • Equilibrium constant
    Kc=[C]c[D]d[A]a[B]bK_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}

Key terms

Equilibrium constant

The value of the concentration ratio at equilibrium; depends only on temperature.

Reaction quotient (Q)

The concentration ratio of the same form as K, evaluated at any moment.

3. Le Chatelier’s principle

A system at equilibrium responds to a disturbance by partly opposing it.

  • Concentration: adding a reactant (or removing a product) shifts the equilibrium forward; the opposite shifts it back. KK is unchanged.
  • Pressure: increasing pressure (decreasing volume) shifts the equilibrium towards the side with fewer moles of gas; no effect if Δn=0\Delta n = 0. KK is unchanged.
  • Temperature: raising T favours the endothermic direction; lowering T favours the exothermic direction. This is the only change that alters KK.
  • Catalyst: speeds both directions equally; no shift, no change in KK.
  • Inert gas: at constant volume, no shift; at constant pressure the volume grows, favouring the side with more gas moles.

Haber process NX2+3 HX2⇌2 NHX3\ce{N2 + 3H2 <=> 2NH3}, ΔH∘=−92 kJ\Delta H^\circ = -92\ \text{kJ}: high pressure and low temperature favour ammonia; about 700 K, high pressure and an iron catalyst are a compromise for rate.

Effects of disturbances on equilibrium
DisturbanceShiftK changes?
Add reactantForwardNo
Increase pressureFewer gas molesNo
Raise T (exothermic reaction)BackwardYes: K falls
CatalystNoneNo

Key terms

Le Chatelier’s principle

If a system at equilibrium is disturbed, it shifts in the direction that partly cancels the disturbance.

4. Acids, bases and pH

  • Arrhenius: acid gives HX+\ce{H+}, base gives OHX−\ce{OH-} in water.
  • Brønsted–Lowry: acid = proton donor, base = proton acceptor; each has a conjugate partner (HCl/ClX−\ce{HCl}/\ce{Cl-}, NHX4X+/NHX3\ce{NH4+}/\ce{NH3}); water and HCOX3X−\ce{HCO3-} are amphoteric. A strong acid has a very weak conjugate base.
  • Lewis: acid = electron-pair acceptor (BFX3\ce{BF3}, HX+\ce{H+}), base = electron-pair donor (NHX3\ce{NH3}, OHX−\ce{OH-}).

Kw=[H+][OH−]=1.0×10−14K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} at 298 K (increases with temperature). pH=−log⁡[H+]\text{pH} = -\log[\text{H}^+], pH+pOH=14\text{pH} + \text{pOH} = 14 at 298 K. A tenfold change in [H+][\text{H}^+] is one pH unit. For a conjugate pair KaKb=KwK_aK_b = K_w, i.e. pKa+pKb=pKwpK_a + pK_b = pK_w.

Typical, approximate values at 298 K.

Formulas

  • Ionic product of water at 298 K
    Kw=[HX+][OHX−]=1.0×10−14K_w = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14}
  • Conjugate pair
    pKa+pKb=pKwpK_a + pK_b = pK_w

Key terms

Conjugate acid–base pair

Two species that differ by one proton, e.g. CH₃COOH and CH₃COO⁻.

Amphoteric species

A species that can act as both an acid and a base.

5. Weak electrolytes, common ion effect, hydrolysis and buffers

Weak acid HA: Ka=cα2/(1−α)≈cα2K_a = c\alpha^2/(1-\alpha) \approx c\alpha^2; α=Ka/c\alpha = \sqrt{K_a/c} (rises on dilution); [H+]=Kac[\text{H}^+] = \sqrt{K_ac}. Weak base: [OH−]=Kbc[\text{OH}^-] = \sqrt{K_bc}. Use the approximation only when α≪1\alpha \ll 1 (below about 5%).

Common ion effect: adding a common ion suppresses the ionisation of a weak electrolyte (and the solubility of a sparingly soluble salt).

Hydrolysis: salt of strong acid + strong base: neutral; weak acid + strong base: basic (Kh=Kw/KaK_h = K_w/K_a, [OH−]=Khc[\text{OH}^-] = \sqrt{K_hc}); strong acid + weak base: acidic; weak acid + weak base: depends on KaK_a and KbK_b.

Buffers resist a change in pH on adding small amounts of acid or base. Acidic buffer: weak acid + its salt; basic buffer: weak base + its salt. Henderson: pH=pKa+log⁡([salt]/[acid])\text{pH} = pK_a + \log([\text{salt}]/[\text{acid}]); at equal concentrations pH = pKapK_a; useful range about pKa±1pK_a \pm 1.

Formulas

  • Henderson equation
    pH=pKa+log⁡[salt][acid]\text{pH} = pK_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}
  • Hydrolysis constant of the anion of a weak acid
    Kh=KwKaK_h = \dfrac{K_w}{K_a}

Key terms

Degree of ionisation

Fraction of the dissolved electrolyte that is ionised at equilibrium.

Buffer solution

A solution of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists changes in pH.

Hydrolysis

Reaction of the ions of a salt with water that changes the pH.

6. Solubility product

For AXxBXy(s)⇌x AXy++y BXx−\ce{A_xB_y(s) <=> xA^{y+} + yB^{x-}} in a saturated solution, Ksp=[Ay+]x[Bx−]yK_{sp} = [\text{A}^{y+}]^x[\text{B}^{x-}]^y. With molar solubility ss: AB, Ksp=s2K_{sp} = s^2; AB2\text{AB}_2 or A2B\text{A}_2\text{B}, Ksp=4s3K_{sp} = 4s^3; A3B\text{A}_3\text{B}, Ksp=27s4K_{sp} = 27s^4.

A common ion lowers solubility: for AgCl in cc mol L⁻¹ NaCl, s=Ksp/cs = K_{sp}/c.

Precipitation: compare the ionic product QspQ_{sp} (using concentrations after mixing) with KspK_{sp}: Qsp>KspQ_{sp} > K_{sp}, precipitate forms; Qsp=KspQ_{sp} = K_{sp}, saturated; Qsp<KspQ_{sp} < K_{sp}, no precipitate. Values used: Ksp(AgCl)=1.8×10−10K_{sp}(\text{AgCl}) = 1.8 \times 10^{-10}, Ksp(BaSO4)=1.1×10−10K_{sp}(\text{BaSO}_4) = 1.1 \times 10^{-10}, Ksp(Ag2CrO4)=1.1×10−12K_{sp}(\text{Ag}_2\text{CrO}_4) = 1.1 \times 10^{-12} at 298 K.

Solubility products of three salts
SaltKsp (298 K)
AgCl1.8×10−101.8 \times 10^{-10}
BaSO₄1.1×10−101.1 \times 10^{-10}
Ag₂CrO₄1.1×10−121.1 \times 10^{-12}

Formulas

  • Ksp and molar solubility
    Ksp=s2  (AB),Ksp=4s3  (AB2)K_{sp} = s^2 \;(\text{AB}), \quad K_{sp} = 4s^3 \;(\text{AB}_2)

Key terms

Solubility product (Ksp)

Product of the equilibrium concentrations of the ions of a sparingly soluble salt in its saturated solution, each raised to its coefficient.

Ionic product (Qsp)

The same product evaluated for any given solution; compared with Ksp to predict precipitation.

Exam tips

Common mistakes

  • Including pure solids or liquids in K, or using concentrations of solids in Q.
  • Saying that pressure, a catalyst or concentration changes the value of K (only temperature does).
  • Using Kp = Kc for reactions with Δn ≠ 0, or the wrong R for the pressure unit.
  • Forgetting the coefficient in the ICE change row (+2x for 2HI), or the two OH⁻ in Ba(OH)₂.
  • Using √(Ka c) when α is not small, or pH = −log[OH⁻].
  • Inverting salt and acid in the Henderson equation.
  • Forgetting dilution when two solutions are mixed in a precipitation question.

How marks are usually awarded

  • 1-mark items: K expression, direction of shift, conjugate base or pH.
  • Numericals (2–3 marks): expression (½–1), substitution (1), answer with unit and sign (½–1).
  • Reasoning (2–3 marks): the principle used (Le Chatelier, common ion, hydrolysis) and the argument.