AS Chemistry: Chemical Equilibria — Revision Notes

Condensed for the final weeks. For the full explanation, use the Chemical Equilibria study guide.

Dynamic equilibrium

Two conditions, both required:

1. The rates of the forward and reverse reactions are equal.
2. The concentrations of all species remain constant.

And it must be a closed system. “Dynamic” means the reactions have not stopped — they continue at equal rates, which is why saying “the reaction has stopped” or “the amounts are equal” both lose the mark. Constant is not the same as equal.

Equilibrium constants

For aA + bB ⇌ cC + dD:

Kc = [C]^c [D]^d / ([A]^a [B]^b)
Kp = (pC^c x pD^d) / (pA^a x pB^b)

Partial pressure: p = mole fraction × total pressure.

Only Kc and Kp tell you the position of equilibrium. Everything else — concentration, pressure, catalyst — shifts the position without changing K.

Change	Effect on K
Concentration	None
Pressure	None
Catalyst	None
Temperature	Changes K

That table is the topic’s core. A catalyst speeds up both directions equally, so equilibrium is reached sooner but the position is identical.

Solids and pure liquids are omitted from K expressions — their concentrations are effectively constant.

Worked example: Kc

At equilibrium in a 1 dm³ sealed container, H₂(g) + I₂(g) ⇌ 2HI(g) contains 0.50 mol H₂, 0.50 mol I₂ and 3.0 mol HI. Since the volume is 1 dm³, concentrations equal the mole values directly.

Kc = [HI]^2 / ([H2][I2]) = 3.0^2 / (0.50 x 0.50) = 9.0 / 0.25 = 36

Equal total moles of gas on each side (1 + 1 → 2), so the concentration units cancel exactly and Kc has no units.

Mole fraction, partial pressure and Kp

The mole fraction of a gas component A is xₐ = nₐ ÷ n(total); its partial pressure is Pₐ = xₐ × P(total). Kp is written the same way as Kc, but using partial pressures.

Worked example. A sealed vessel at total pressure 200 kPa contains an equilibrium mixture of N₂O₄(g) ⇌ 2NO₂(g) with mole fractions 0.60 for N₂O₄ and 0.40 for NO₂.

P(N2O4) = 0.60 x 200 = 120 kPa
P(NO2)  = 0.40 x 200 = 80 kPa
Kp = (P_NO2)^2 / P_N2O4 = 80^2 / 120 = 53.3 kPa

The mole count changes here (1 → 2), so the pressure units don’t cancel — Kp carries units (kPa here), found from the net order of (2 − 1) = 1.

Le Chatelier’s principle

When a system at equilibrium is subjected to a change, the position of equilibrium shifts to oppose that change.

Change	Shift
Increase concentration of a reactant	Towards products
Increase pressure	Towards the side with fewer moles of gas
Increase temperature	In the endothermic direction
Add a catalyst	No shift

If both sides have equal moles of gas, pressure has no effect — a favourite trick.

Temperature is the only change that alters K, because it changes the relative rate constants of the two directions. Raising temperature favours the endothermic direction, so if the forward reaction is exothermic, K decreases.

The industrial compromise

Both the Haber and Contact processes are exothermic with fewer moles of gas on the right, so equilibrium yield is favoured by low temperature and high pressure. Yet both run hot.

Haber:    N2 + 3H2 <=> 2NH3     450 C, 200 atm, iron catalyst
Contact:  2SO2 + O2 <=> 2SO3    450 C, 1-2 atm, V2O5 catalyst

The answer is always rate versus yield, plus cost. Low temperature would give a better yield but unacceptably slowly, so a moderate temperature is a compromise. High pressure improves yield but costs more in plant and energy — which is why the Contact process, with industrial SO₂-to-SO₃ conversion typically quoted at around 96-99.5% (depending on plant and source), runs near atmospheric pressure while the Haber process pays for 200 atm.

The catalyst does not improve yield. It only shortens the time to reach equilibrium.

Calculations

Use an ICE table — Initial, Change, Equilibrium.

• Change values are in the ratio of the stoichiometric coefficients.
• Convert moles to concentration by dividing by volume before substituting into Kc.
• Kc units vary with the equation; work them out from the expression each time rather than memorising.
• If Δn(gas) = 0, Kc has no units.

Exam traps

• Saying the forward and reverse reactions stop.
• Saying concentrations become equal rather than constant.
• Claiming a catalyst increases yield.
• Claiming pressure changes K.
• Using moles instead of concentrations in Kc.
• Including solids or pure liquids in the expression.
• Forgetting that pressure has no effect when moles of gas are equal on both sides.
• Using mole values directly in Kp instead of converting to partial pressures via mole fraction.
• Assuming K always carries units, or never does, instead of working it out from the net power difference each time.

Self-test

1. State the two conditions for dynamic equilibrium.
2. Which single factor changes the value of Kc?
3. What does a catalyst do to the position of equilibrium?
4. N₂ + 3H₂ ⇌ 2NH₃ is exothermic. Why is 450 °C used?
5. When does a pressure change have no effect on the position of equilibrium?
6. For H₂(g) + I₂(g) ⇌ 2HI(g) in a 1 dm³ container at equilibrium with 0.50 mol H₂, 0.50 mol I₂ and 3.0 mol HI, calculate Kc and state its units.
7. How is the partial pressure of a gas found from its mole fraction, and why does Kp sometimes carry units?

Answers: 1. The rates of the forward and reverse reactions are equal, and the concentrations of all species remain constant, in a closed system. 2. Temperature. 3. Nothing — it increases the rate of both directions equally, so equilibrium is reached sooner at the same position. 4. A compromise: a lower temperature would give a higher yield but the rate would be too slow to be economic. 5. When there are equal numbers of moles of gas on both sides of the equation. 6. Kc = 3.0² ÷ (0.50 × 0.50) = 9.0 ÷ 0.25 = 36, with no units, since the total moles of gas are equal on both sides (2 on each). 7. Partial pressure = mole fraction × total pressure; Kp carries units whenever the total moles of gas differ between products and reactants, since the pressure units in the expression then fail to cancel exactly.