Chemical Equilibrium & Le Chatelier's Principle

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A reaction that can proceed in both the forward and reverse directions. Shown by the double arrow (⇌). Example: N₂ + 3H₂ ⇌ 2NH₃. Products can react to reform reactants.

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The state in a reversible reaction where the forward reaction rate equals the reverse reaction rate. Concentrations of reactants and products remain constant — but both reactions are still occurring simultaneously.

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The system must be closed (no substances can enter or leave). The reaction must be reversible. Equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction.

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For the reaction aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ where [ ] = molar concentrations at equilibrium. Kc is constant at a given temperature. Pure solids and liquids are excluded.

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Kc >> 1: Equilibrium lies to the right — products favoured. Kc << 1: Equilibrium lies to the left — reactants favoured. Kc ≈ 1: Significant amounts of both reactants and products present.

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Kp is the equilibrium constant expressed in terms of partial pressures of gases (in atm or Pa). Used for gas-phase reactions. Related to Kc by: Kp = Kc(RT)^Δn where Δn = moles of gaseous products − moles of gaseous reactants.

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Q has the same expression as Kc but uses current concentrations (not equilibrium). Q < Kc: Reaction proceeds forward (more products form). Q > Kc: Reaction proceeds backward. Q = Kc: System is at equilibrium.

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"If a system at equilibrium is subjected to a change (stress), the system will shift its equilibrium position in the direction that partially counteracts the applied change." — Henri Louis Le Chatelier (1884).

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The system shifts to the right (forward reaction) to use up the added reactant and partially reduce the disturbance. More products are formed. Kc itself does not change (temperature is constant).

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The system shifts toward the side with fewer moles of gas to reduce the pressure. Example: N₂ + 3H₂ ⇌ 2NH₃ (4 mol gas → 2 mol gas) — increased pressure shifts equilibrium right. No effect if Δn = 0.

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The system shifts in the endothermic direction (absorbs heat) to counteract the increase. This is the only change that alters the value of Kc. Exothermic reactions: Kc decreases. Endothermic reactions: Kc increases.

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A catalyst increases the rates of both forward and reverse reactions equally — it does NOT shift the equilibrium position and does NOT change Kc. It only helps the system reach equilibrium faster.

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Industrial synthesis of ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol. Conditions: Temperature ~450°C (compromise), Pressure ~200 atm, Iron catalyst. High pressure favours NH₃ (fewer gas moles); low temperature favours NH₃ (exothermic) but slows reaction.

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Low temperature gives higher yield (exothermic reaction — Le Chatelier) but the rate is too slow to be economical. High temperature gives faster rate but lower equilibrium yield. ~450°C gives acceptable rate and reasonable yield.

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Industrial production of sulfuric acid (H₂SO₄). Key equilibrium step: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJ/mol. Conditions: 450°C, 1–2 atm, V₂O₅ catalyst. SO₃ then reacts with H₂SO₄ to form oleum, then diluted with water.

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All reactants and products are in the same phase (e.g., all gases or all in aqueous solution). Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) — all gases.

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Reactants and products are in different phases. Pure solids and liquids are excluded from the Kc expression (their concentrations are constant). Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g) — Kc = [CO₂] only.

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It does NOT affect the equilibrium. The partial pressures and concentrations of reacting gases are unchanged. The total pressure increases but the equilibrium position stays the same.

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The equilibrium constant for the dissolution of a sparingly soluble ionic compound. For AB(s) ⇌ A⁺(aq) + B⁻(aq): Ksp = [A⁺][B⁻]. Small Ksp = very insoluble. Used to predict whether a precipitate will form.

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The reduction in solubility of an ionic compound when a solution already contains one of its ions. Adding a common ion shifts equilibrium to the left (Le Chatelier) — less compound dissolves. Used in buffer solutions and qualitative analysis.

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Strong electrolyte: Completely dissociates in water (strong acids, strong bases, most salts). Kc is very large. Weak electrolyte: Partially dissociates — establishes equilibrium (weak acids, weak bases). Kc is small.

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The system shifts toward the side with more moles of gas to increase pressure. The opposite of increasing pressure. For the Haber Process, decreasing pressure shifts equilibrium left — less NH₃ produced.

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For a weak acid HA ⇌ H⁺ + A⁻ with Ka: the degree of dissociation α ≈ √(Ka/C) at low concentrations. As concentration C decreases, the degree of dissociation increases — dilution favours ionization (Le Chatelier).

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Heat can be treated as a product: A + B ⇌ C + D + heat. Increasing temperature adds heat (product) → equilibrium shifts LEFT → fewer products → Kc decreases. Cooling increases Kc for exothermic reactions.

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ΔG° = −RT ln Kc. If Kc > 1: ΔG° < 0 (products favoured). If Kc < 1: ΔG° > 0 (reactants favoured). This links thermodynamics and equilibrium — the equilibrium constant is a thermodynamic quantity.