Chemical Equilibrium
Chemical equilibrium describes the dynamic balance reached in reversible reactions when forward and reverse rates become equal. The PMDC MDCAT 2026 syllabus expects you to compute Kc/Kp, apply Le Chatelier's principle, calculate buffer pH, recognise the common-ion effect, and explain industrial processes such as Haber synthesis. Expect 3–4 MCQs.
Chemical Equilibrium
A reaction is at chemical equilibrium when the forward and reverse reactions occur at the same rate, so that macroscopic concentrations of reactants and products remain constant. Equilibrium is dynamic — molecules continue to interconvert, but no net change is observed.
Equilibrium constants
For the general reaction aA + bB ⇌ cC + dD:
Kc = [C]c[D]d / ([A]a[B]b). Concentrations in mol/dm3. Units depend on Δn; dimensionless when Δn = 0.
Kp = PCcPDd / (PAaPBb). Units atmΔn. Relation: Kp = Kc(RT)Δn where Δn = (c+d) − (a+b).
Self-ionisation: H2O ⇌ H+ + OH−. Kw = [H+][OH−] = 1.0 × 10−14 at 25°C. Hence pH + pOH = 14.
Reading the value of K
- K >> 1 → equilibrium lies to the right (products favoured).
- K << 1 → equilibrium lies to the left (reactants favoured).
- K depends only on temperature — not on initial concentrations or pressures.
Le Chatelier's Principle
If a system at equilibrium is subjected to a stress (change in concentration, pressure or temperature), the equilibrium shifts in the direction that opposes the stress.
Effect of changes
| Stress applied | Direction of shift | Effect on K | Example: N2 + 3H2 ⇌ 2NH3 (ΔH < 0) |
|---|---|---|---|
| Add reactant | Toward products (forward) | Unchanged | Add N2 → more NH3 |
| Add product | Toward reactants (backward) | Unchanged | Add NH3 → backward |
| Remove product | Toward products (forward) | Unchanged | Liquefy NH3 → pulls forward |
| Increase pressure | Toward side with fewer moles of gas | Unchanged | Forward (4 mol gas → 2 mol) |
| Decrease pressure | Toward side with more moles of gas | Unchanged | Backward |
| Pressure if Δngas = 0 | No shift | Unchanged | (N/A) |
| Increase T — exothermic fwd | Backward | K decreases | Less NH3 at higher T |
| Increase T — endothermic fwd | Forward | K increases | (opposite scenario) |
| Add catalyst | No shift | Unchanged | Same final composition, reached faster |
| Add inert gas, constant V | No shift | Unchanged | Partial pressures of reactive gases unchanged |
| Add inert gas, constant P | Toward side with more moles of gas | Unchanged | Backward shift (volume must rise) |
Big rules: only T changes K; catalyst never shifts equilibrium; inert gas at constant V never shifts equilibrium.
Haber's Process
Industrial synthesis of ammonia from atmospheric N2 and hydrogen produced by steam reforming. The reaction is exothermic and goes with a decrease in moles of gas, so favourable thermodynamically at high pressure and low temperature — but very slow at low T.
N2(g) + 3 H2(g) ⇌ 2 NH3(g); ΔH = −92 kJ/mol.
· Pressure: 200 atm (high P shifts equilibrium toward 2 mol of NH3).
· Temperature: 400–450°C (a compromise — lower T favours equilibrium but is too slow).
· Catalyst: finely divided iron (Fe) with Mo or Al2O3/K2O promoters.
· The unreacted N2/H2 is recycled; ammonia is condensed out.
Other important industrial equilibria
- Contact process: 2 SO2 + O2 ⇌ 2 SO3; V2O5 catalyst, ~ 450°C, 1–2 atm. (For sulphuric acid manufacture.)
- Ostwald process: 4 NH3 + 5 O2 ⇌ 4 NO + 6 H2O; Pt-Rh catalyst. (For HNO3.)
Buffer Solution
A buffer is a solution that resists change in pH on the addition of small amounts of acid or base. Buffers contain a weak acid + its conjugate base (acidic buffer) or a weak base + its conjugate acid (basic buffer).
For an acidic buffer (HA / A−):
pH = pKa + log([salt]/[acid]).
For a basic buffer (B / BH+):
pOH = pKb + log([salt]/[base]); pH = 14 − pOH.
The buffer works best when [salt] ≈ [acid] (or [base]), giving pH ≈ pKa.
How a buffer resists pH change
If you add H+: the conjugate base A− mops it up — A− + H+ → HA. If you add OH−: the weak acid HA neutralises it — HA + OH− → A− + H2O. Either way the [salt]/[acid] ratio changes only slightly, so pH stays nearly constant.
Examples
- Acidic buffer: CH3COOH + CH3COONa, pH ≈ 4.74.
- Basic buffer: NH4OH + NH4Cl, pH ≈ 9.25.
- Biological buffer: H2CO3/HCO3− in blood, pH 7.4.
- Phosphate buffer: H2PO4−/HPO42−, pH 7.2 (intracellular).
Common Ion Effect
The common-ion effect is the suppression of the ionisation of a weak electrolyte (or the dissolution of a sparingly soluble salt) when a strong electrolyte providing an ion in common with it is added. It is a direct application of Le Chatelier's principle.
Example 1 — weak acid
CH3COOH ⇌ CH3COO− + H+. Adding CH3COONa increases [CH3COO−], pushing the equilibrium back to the left. The acid ionises less, [H+] falls, pH rises slightly. This is exactly how an acidic buffer works.
Example 2 — sparingly soluble salt
AgCl(s) ⇌ Ag+(aq) + Cl−(aq); Ksp = [Ag+][Cl−]. Adding NaCl increases [Cl−]; the equilibrium shifts left and AgCl precipitates — its solubility decreases. Used in qualitative analysis to ensure complete precipitation.
Solubility Products
For the dissolution of a sparingly soluble salt MxAy at saturation, the solubility-product constant Ksp is defined by:
MxAy(s) ⇌ x My+(aq) + y Ax−(aq); Ksp = [My+]x[Ax−]y. The activity of the solid is unity, so it doesn't appear.
Examples
- AgCl: Ksp = [Ag+][Cl−] = 1.8 × 10−10 → molar solubility s = √Ksp = 1.34 × 10−5 M.
- BaSO4: Ksp = 1.1 × 10−10; s ≈ 1.05 × 10−5 M.
- PbI2: Ksp = [Pb2+][I−]2 = 4s3; s = (Ksp/4)1/3.
- Ag2CrO4: Ksp = [Ag+]2[CrO42−] = 4s3.
Predicting precipitation (ion product Q vs Ksp)
- If Q < Ksp → unsaturated, no precipitate.
- If Q = Ksp → saturated.
- If Q > Ksp → supersaturated, precipitate forms until Q = Ksp.
Worked MCQs
Five MCQs that capture the high-yield testing patterns for this chapter. Read the explanation even when you get the answer right — that's where the deeper concept lives.
Q1. For the reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g); ΔH < 0, the yield of NH3 is increased by:
Forward direction has fewer gas moles (4 → 2), so high pressure shifts equilibrium right by Le Chatelier. The reaction is exothermic, so high T would shift it left; inert gas at constant V doesn't affect partial pressures, so no shift.
Q2. The pH of a buffer prepared by mixing 0.1 M acetic acid with 0.1 M sodium acetate (pKa = 4.74) is:
pH = pKa + log([salt]/[acid]) = 4.74 + log(1) = 4.74. When the salt and acid concentrations are equal, the pH equals the pKa — the buffer is at its point of maximum capacity.
Q3. Which of the following does NOT shift the position of an equilibrium?
A catalyst lowers the activation energy of forward and reverse reactions equally, so it speeds both up by the same factor and the system reaches the same equilibrium faster. K is unchanged, position is unchanged.
Q4. If Ksp of AgCl at 25°C is 1.6 × 10−10, its molar solubility (mol/L) in pure water is approximately:
For a 1:1 salt, Ksp = s2, so s = √Ksp = √(1.6 × 10−10) = 1.26 × 10−5 mol/L.
Q5. Adding NaCl to a saturated solution of AgCl will:
NaCl supplies Cl−, the common ion. By Le Chatelier, the AgCl(s) ⇌ Ag+ + Cl− equilibrium shifts to the left, precipitating more AgCl and reducing its solubility. Ksp itself depends only on temperature, so it is unchanged.
Quick Recap
- Kc uses concentrations; Kp uses partial pressures; Kp = Kc(RT)Δn; Kw = 10−14.
- Le Chatelier: equilibrium shifts to oppose change. Catalyst — never shifts.
- Haber: N2 + 3 H2 ⇌ 2 NH3; 200 atm, 450°C, Fe catalyst, exothermic.
- Buffer pH (acidic): pH = pKa + log([salt]/[acid]); blood buffer = H2CO3/HCO3−.
- Common-ion effect = suppression of ionisation/solubility by a shared ion (Le Chatelier).
- Ksp = product of ion concentrations at saturation; Ksp depends only on T. Q > Ksp → precipitate.