The big idea: At a given temperature, the concentrations of reactants and products at dynamic equilibrium always settle into a fixed ratio. That ratio is the equilibrium constant, Kc.
It is a single number that tells you how far a reversible reaction goes before it reaches equilibrium — whether the mixture ends up mostly products or mostly reactants.
How to write the expression: Kc is products over reactants, and each concentration is raised to the power of its balancing coefficient:
- products (right-hand side) go on top - reactants (left-hand side) go on the bottom - the big number in front of each species becomes a power
Use square brackets [ ] to mean 'the equilibrium concentration of', in mol dm⁻³.
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Start from the balanced equation. The general rule for aA + bB ⇌ cC + dD is:
- the equilibrium constant (in terms of concentration)
- equilibrium concentrations of the products (mol dm⁻³)
- equilibrium concentrations of the reactants (mol dm⁻³)
- the balancing coefficients — each becomes a power
Coefficients become powers: The most common mistake is forgetting the powers. A coefficient of 2 in the equation becomes a squared term in Kc; a coefficient of 3 becomes a cubed term. A coefficient of 1 needs no power.
(Pure solids and pure liquids are left out — only gases and dissolved species appear. In the SL course you will almost always be given an all-gas or all-aqueous equilibrium.)
Worked example — writing Kc for ammonia synthesis
Write the equilibrium constant expression, Kc, for the synthesis of ammonia: N2(g) + 3H2(g) ⇌ 2NH3(g).
Solution
- Products on top, reactants on the bottom. The product is NH3; the reactants are N2 and H2.
- Raise each to its coefficient. NH3 has a 2 (squared), H2 has a 3 (cubed), N2 has a 1 (no power):
Final answer
Kc = [NH₃]² / ([N₂][H₂]³).
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The magnitude of Kc tells you which side the equilibrium favours. Because products are on top, a large Kc means lots of product; a small Kc means little product.
| Value of Kc | What it tells you | Position of equilibrium |
|---|---|---|
| Kc >> 1 (large) | products are favoured | lies to the right — mostly products |
| Kc ≈ 1 | comparable amounts of each | roughly in the middle |
| Kc << 1 (small) | reactants are favoured | lies to the left — mostly reactants |
Reverse reaction, reciprocal K: If you reverse an equation, products and reactants swap, so the new constant is the reciprocal of the old one:
Kreverse = 1 / Kforward.
So if a forward reaction has Kc = 100 (products favoured), the reverse reaction has Kc = 1/100 = 0.01 (reactants favoured).
Only temperature changes Kc: Kc is fixed at a given temperature. Changing concentration, pressure or adding a catalyst shifts the position but leaves Kc unchanged. Only a temperature change alters its value:
- endothermic forward reaction → raising T increases Kc - exothermic forward reaction → raising T decreases Kc
So you can deduce the sign of ΔH from how Kc changes with temperature.
How this is tested: R2.3.3 shows up in two reliable shapes.
- Write the expression: a state/deduce part on Paper 2 (or a Paper 1A MCQ picking the correct expression) — get the products on top and every coefficient as a power. - Calculate Kc: a Paper 1B / Paper 2 part giving the equilibrium amounts in a known volume — convert each amount to a concentration (mol dm⁻³) first, then substitute.
A deduce part may also link the trend of Kc with temperature to whether the reaction is exo- or endothermic.
Score it cleanly: (1) Products over reactants, each raised to its coefficient. (2) To calculate, turn every amount (mol) into a concentration with c = n/V first. (3) Quote Kc to the significant figures asked, and remember it has no fixed unit at this level — just state the number.
IB-style question — write the expression (a)
Hydrogen iodide forms by H2(g) + I2(g) ⇌ 2HI(g). (a) Write the equilibrium constant expression, Kc, for this reaction. [1]
How to score the mark
- Mark 1 — the expression. Product HI on top (coefficient 2 → squared); reactants H2 and I2 on the bottom (coefficient 1 each):
Final answer
Kc = [HI]² / ([H₂][I₂]).
IB-style question — calculate Kc (b)
(b) At equilibrium a 2.0 dm³ vessel contains 0.40 mol H2, 0.40 mol I2 and 1.6 mol HI. Calculate the value of Kc.
Solution
- Convert each amount to a concentration — formula first (c = n/V, V = 2.0 dm³):
- Substitute into the expression from part (a):
- Work it out:
Final answer
Kc = 16 (Kc > 1, so products are favoured).