Mini-Lesson

Chemical Equilibria (Kc and Kp)

This mini-lesson covers AQA 3.1.6 Chemical equilibria and 3.1.10 Kp: dynamic equilibrium, Le Chatelier’s principle, calculating K_c from equilibrium concentrations, mole fractions and partial pressures, and calculating K_p — plus what does (and does not) change the value of K.

Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Everything here is A-level standard — the maths is done properly, not skipped. Press Start when you're ready.

Equilibria · dynamic equilibrium

Dynamic equilibrium and Le Chatelier

In a closed system, a reversible reaction reaches dynamic equilibrium: the forward and reverse reactions are still happening, but at equal rates, so the concentrations stay constant (they are not equal to each other — a common error).

Le Chatelier’s principle: if a change is imposed on a system at equilibrium, the position of equilibrium shifts so as to oppose that change.

Concentration — add a reactant, or remove a product, and the equilibrium shifts right.
Pressure (gases only) — increasing pressure shifts the equilibrium to the side with fewer moles of gas. If both sides have equal moles of gas, pressure has no effect on the position.
Temperature — increasing temperature shifts the equilibrium in the endothermic direction (to absorb the heat).
Catalyst — no effect on the position at all. It speeds up forward and reverse reactions equally, so equilibrium is simply reached sooner.

The industrial compromise: in the Haber process a low temperature would give a better yield (the forward reaction is exothermic) but far too slow a rate, so around 400–450 °C and an iron catalyst are used — a deliberate compromise between yield and rate.

Sort it

N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ mol⁻¹)

Tap a change, then tap what it does to the position of equilibrium.

🟩 Shifts right (more NH₃)

🟪 Shifts left (less NH₃)

🟦 No change in position

Equilibria · K_c

The equilibrium constant K_c

For the general reaction aA + bB ⇌ cC + dD:

K_c = [C]^c [D]^d ÷ ( [A]^a [B]^b )products on top; each concentration raised to its balancing number; square brackets mean equilibrium concentration in mol dm⁻³

Rules that catch people out:

Only equilibrium concentrations go in — never the starting ones. Set up an initial / change / equilibrium table.
Solids and pure liquids are omitted (their concentration is effectively constant).
The units of K_c depend on the equation — work them out by substituting mol dm⁻³ and cancelling. If the moles cancel exactly, K_c has no units.
Only a change in temperature changes the value of K_c. Adding a catalyst or changing pressure does not.

Large K_c (>> 1) means the equilibrium lies well to the right (mostly products). Small K_c means mostly reactants.

Calculate

Your turn

1For H₂ + I₂ ⇌ 2HI, the equilibrium concentrations are [H₂] = 0.20, [I₂] = 0.20 and [HI] = 1.60 mol dm⁻³. Calculate K_c.

Hint: K_c = [HI]² ÷ ([H₂][I₂]) = 1.60² ÷ (0.20 × 0.20) = 2.56 ÷ 0.040.

Quick check

?A catalyst is added to a system at equilibrium at constant temperature. What happens to K_c?

Equilibria · partial pressures

Mole fractions and partial pressures

For gases it is easier to work with pressures. First find the mole fraction of each gas:

x(A) = moles of A ÷ total moles of gasall the mole fractions must add up to exactly 1

Then the partial pressure of that gas — the pressure it would exert alone in the same container:

p(A) = x(A) × P_totaland all the partial pressures add up to the total pressure

Worked example

At equilibrium a vessel contains 2.0 mol N₂, 1.0 mol H₂ and 2.0 mol NH₃ at a total pressure of 50 kPa.

Total moles = 2.0 + 1.0 + 2.0 = 5.0 mol

x(N₂) = 2.0/5.0 = 0.40 · x(H₂) = 1.0/5.0 = 0.20 · x(NH₃) = 2.0/5.0 = 0.40 (sum = 1 ✓)

p(N₂) = 0.40 × 50 = 20 kPa · p(H₂) = 0.20 × 50 = 10 kPa · p(NH₃) = 0.40 × 50 = 20 kPa (sum = 50 ✓)

Calculate

Your turn

2That vessel holds 2.0 mol N₂, 1.0 mol H₂ and 2.0 mol NH₃. Calculate the mole fraction of H₂.

Hint: x(H₂) = 1.0 ÷ (2.0 + 1.0 + 2.0).

Calculate

Your turn

3The total pressure is 50 kPa. Using your mole fraction, calculate the partial pressure of H₂.

kPa

Hint: p(H₂) = x(H₂) × P_total = 0.20 × 50.

Equilibria · K_p

The equilibrium constant K_p

K_p has exactly the same form as K_c, but uses partial pressures. It applies to homogeneous gaseous systems (everything a gas).

K_p = p(NH₃)² ÷ ( p(N₂) × p(H₂)³ )for N₂ + 3H₂ ⇌ 2NH₃ — products on top, each raised to its balancing number

Worked example — using the partial pressures we just found

p(N₂) = 20 kPa, p(H₂) = 10 kPa, p(NH₃) = 20 kPa

K_p = 20² ÷ (20 × 10³) = 400 ÷ 20 000 = 0.020 kPa⁻²

Units: kPa² ÷ (kPa × kPa³) = kPa² ÷ kPa⁴ = kPa⁻²

Same rule as K_c: the value of K_p changes only with temperature. Raising the pressure changes the individual partial pressures and shifts the position, but the value of K_p stays the same — the partial pressures rearrange themselves to keep it constant.

Calculate

Your turn

4Using p(N₂) = 20 kPa, p(H₂) = 10 kPa and p(NH₃) = 20 kPa, calculate K_p for N₂ + 3H₂ ⇌ 2NH₃.

kPa⁻²

Hint: K_p = p(NH₃)² ÷ (p(N₂) × p(H₂)³) = 400 ÷ (20 × 1000).

Match it

Match the term to its meaning

Tap a term on the left, then its meaning on the right.

Term

Meaning

Quick check

?For N₂ + 3H₂ ⇌ 2NH₃ (ΔH = −92 kJ mol⁻¹), the temperature is increased. What happens?

Quick check

?Which change would increase the value of K_c for an endothermic forward reaction?

Quick check

?Argon (an inert gas) is added to an equilibrium mixture at constant volume. What happens to the position of equilibrium?

Equilibria · exam traps

ICE tables and the changes that do nothing

Most K_c questions give you starting moles and one equilibrium quantity. Build an ICE table — Initial, Change, Equilibrium:

Worked example — A + B ⇌ C + D in a 1.00 dm³ flask

Initial: 1.00 mol A, 1.00 mol B, 0 C, 0 D

At equilibrium, 0.25 mol of A remains, so the change is −0.75 for A — and therefore −0.75 for B, and +0.75 for C and D.

Equilibrium: [A] = 0.25, [B] = 0.25, [C] = 0.75, [D] = 0.75 mol dm⁻³

K_c = (0.75 × 0.75) ÷ (0.25 × 0.25) = 0.5625 ÷ 0.0625 = 9.0 (no units — they cancel)

Divide moles by the volume to get concentrations before substituting — unless the volume is 1 dm³, or unless the moles cancel top and bottom.
Only a temperature change changes the value of K. Catalysts, pressure changes and adding inert gas do not.

Quick check

?For H₂ + I₂ ⇌ 2HI, what are the units of K_c?

Calculate

Your turn

5In a 1.00 dm³ flask, 1.00 mol A and 1.00 mol B react: A + B ⇌ C + D. At equilibrium 0.25 mol of A remains. Calculate K_c.

Hint: Change = −0.75 for A and B, +0.75 for C and D. K_c = (0.75 × 0.75) ÷ (0.25 × 0.25).

Recap

The big ideas to know

Dynamic equilibrium: closed system · forward rate = reverse rate · concentrations constant, not equal

Le Chatelier: the position shifts to oppose the change (concentration, pressure, temperature)

K_c: [products]^coefficients ÷ [reactants]^coefficients — solids and pure liquids are left out

Mole fraction: x = moles of that gas ÷ total moles of gas (all the mole fractions add to 1)

Partial pressure: p = mole fraction × total pressure (all partial pressures add to the total)

What changes K: ONLY temperature. Catalysts and pressure changes never change K — they only change the rate or the position

That is the whole of AQA 3.1.6 and 3.1.10. Press Finish.

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