Reversible reactions, dynamic equilibrium, the equilibrium constant (Kc and Kp) and Le Chatelier's principle, applied to the Haber and Contact processes — exam-focused for the BIEK / Sindh Board paper. Read it, or open the interactive lecture and shift the equilibrium yourself.
1 — Reversible & irreversible reactions
- Irreversible reaction — proceeds in one direction only and goes to completion (e.g. burning, precipitation). Shown with →.
- Reversible reaction — proceeds in both directions; products can re-form reactants. Shown with the double arrow ⇌.
ExampleN₂(g) + 3H₂(g) ⇌ 2NH₃(g)
2 — Dynamic equilibrium
- Chemical equilibrium — the state of a reversible reaction (in a closed system) at which the rate of the forward reaction equals the rate of the reverse reaction, so the concentrations of all species stay constant.
It is called dynamic because both reactions are still happening — they just cancel out, so nothing appears to change.
3 — Characteristics of the equilibrium state
- Reached only in a closed system.
- It is dynamic — forward and reverse reactions continue at equal rates.
- The concentrations of reactants and products remain constant (but not necessarily equal).
- It can be reached from either direction.
- A catalyst speeds up the approach to equilibrium but does not change the position.
- The observable (macroscopic) properties become constant.
4 — Law of mass action & Kc
The law of mass action states that the rate of a reaction is proportional to the product of the molar concentrations of the reactants. Applying it to both directions gives the equilibrium constant.
For aA + bB ⇌ cC + dDKc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
Kc depends only on temperature. Concentrations are the equilibrium values (mol dm⁻³).
5 — Kp and its relation to Kc
For gaseous equilibria we can use partial pressures instead of concentrations:
Kp and the link to KcKp = (P_C)ᶜ(P_D)ᵈ / (P_A)ᵃ(P_B)ᵇ
Kp = Kc (RT)^Δn (Δn = moles of gaseous products − reactants)
If Δn = 0 (equal moles of gas on both sides), then Kp = Kc.
6 — The significance of K
| Value of K | Means |
|---|
| K ≫ 1 (large) | products favoured — reaction nearly complete |
| K ≈ 1 | appreciable amounts of both at equilibrium |
| K ≪ 1 (small) | reactants favoured — little product forms |
Units of Kc depend on Δn; pure solids and liquids are not included in the expression (their concentration is constant).
7 — Le Chatelier's principle
- Le Chatelier's principle — if a stress (a change in concentration, pressure or temperature) is applied to a system at equilibrium, the equilibrium shifts in the direction that opposes (relieves) the stress.
8 — Effect of changing conditions
| Change | Equilibrium shifts… |
|---|
| Increase [reactant] | forward (right), to use it up |
| Increase [product] | backward (left) |
| Increase pressure | towards the side with fewer moles of gas |
| Increase temperature | in the endothermic direction (absorbs heat) |
| Add a catalyst | no shift (reaches equilibrium faster) |
Only temperature changes K. Concentration and pressure changes shift the position but leave Kc unchanged.
9 — The Haber process (ammonia)
Haber processN₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ (exothermic)
To maximise the yield of NH₃, Le Chatelier says use high pressure (4 moles of gas → 2) and low temperature (exothermic). But a low temperature is too slow, so a compromise is used.
| Condition | Industrial value | Reason |
|---|
| Pressure | ~200 atm | high P favours NH₃ (fewer moles) |
| Temperature | ~450 °C | compromise — fast enough, decent yield |
| Catalyst | finely divided iron | speeds up equilibrium |
10 — The Contact process (sulphuric acid)
Key step2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −ve (exothermic)
High pressure and a moderate temperature (~450 °C) with a V₂O₅ catalyst give a high yield of SO₃, which is then used to make H₂SO₄. Same Le Chatelier reasoning as the Haber process.
11 — Worked numericals
Kc expression
Write Kc for 2SO₂ + O₂ ⇌ 2SO₃.
Kc = [SO₃]² / ([SO₂]²[O₂])
calculate Kc
At equilibrium [H₂]=0.5, [I₂]=0.5, [HI]=2.0 (mol/dm³) for H₂ + I₂ ⇌ 2HI. Find Kc.
Kc = [HI]²/([H₂][I₂]) = (2.0)²/(0.5×0.5) = 4/0.25 = 16
Le Chatelier
In N₂+3H₂⇌2NH₃ (exothermic), what does raising the temperature do?
Shifts towards the endothermic (reverse) direction → less NH₃, Kc decreases
12 — Exam recap
- Reversible vs irreversible; the ⇌ sign.
- Dynamic equilibrium (equal forward & reverse rates); its characteristics.
- Law of mass action; the Kc expression; Kp = Kc(RT)^Δn.
- Significance of the size of K.
- Le Chatelier's principle; effect of concentration, pressure, temperature, catalyst.
- Haber & Contact processes (conditions & reasons).