IB Chemistry SLTopic 5 — The Extent of Chemical ChangePaper 1 & 2Core skill~10 min read
The Equilibrium Law
Every equilibrium settles at a particular balance of reactants and products. The equilibrium law is the rule that turns that balance into a single number — and the whole of the rest of this topic depends on you writing it out correctly.
📚 What you need to know
For aA + bB ⇌ cC + dD, the expression is K = [C]c[D]d ÷ [A]a[B]b.
Products on top, reactants underneath. Always that way round.
Each concentration is raised to the power of its balancing number.
Square brackets mean equilibrium concentration in mol dm–3. Round brackets lose the mark.
Solids are left out of the expression entirely.
An expression belongs to one specific equation. Rewrite the equation and the expression changes.
The expression itself
The equilibrium law says that, at a given temperature, one particular combination of the equilibrium concentrations always comes out the same, no matter what amounts you started with. That combination is the equilibrium constant.
The equilibrium law
for aA + bB ⇌ cC + dD K = [C]c[D]d / [A]a[B]b
Four things to get right: which side goes on top, the powers, the square brackets, and the fact that every value is measured at equilibrium.
The balancing numbers become powers, not multipliers. Three moles of hydrogen gives you [H2]3, never 3[H2]. If you have already met rate equations, keep the two apart in your head: the powers in a rate equation come from experiment, but the powers here come straight off the balanced equation.
What gets left out
Solids never appear in the expression. The reason is worth understanding rather than memorising: the “concentration” of a pure solid is fixed by its density, and you cannot change it. A big lump and a small lump of calcium carbonate have exactly the same amount of substance packed into each cubic centimetre, so the solid contributes nothing that can vary.
An equilibrium involving more than one phase like this is called heterogeneous. Cross the solids out first, then write what is left.
The same logic covers a pure liquid and a solvent present in large excess, such as the water in a dilute aqueous equilibrium — their concentrations are effectively fixed, so they are left out too. Be careful with reactions where liquids are genuinely mixed in comparable amounts, such as esterification: there every species is included.
🧩 Writing any expression
Check the equation is balanced. Wrong coefficients means wrong powers.
Cross out any solids and any pure liquid or solvent.
Put the surviving products on the top, multiplied together.
Put the surviving reactants underneath, multiplied together.
Give each one a power equal to its balancing number, and write it in square brackets.
WORKED EXAMPLE
Deduce the equilibrium constant expression for each reaction. (a) N2(g) + 3H2(g) ⇌ 2NH3(g) (b) 2SO2(g) + O2(g) ⇌ 2SO3(g) (c) Ag+(aq) + Fe2+(aq) ⇌ Ag(s) + Fe3+(aq)
(a) ammoniaK = [NH₃]² ÷ ([N₂] × [H₂]³)The 3 in front of the hydrogen becomes a cubed, not a × 3.(b) sulfur trioxideK = [SO₃]² ÷ ([SO₂]² × [O₂])Two powers of two here, one on each side.(c) silver and iron ionsK = [Fe³⁺] ÷ ([Ag⁺] × [Fe²⁺])Ag(s) is a solid, so it is left outEvery other species is aqueous, and every coefficient is 1, so no powers appear.
WORKED EXAMPLE
An equilibrium constant expression is K = [NO]2[Cl2] ÷ [NOCl]2. Deduce the balanced equation it belongs to.
Step 1 — read the bottom lineThe denominator holds the reactants. [NOCl]² means 2NOCl.Step 2 — read the top lineThe numerator holds the products: 2NO and 1Cl₂.2NOCl(g) ⇌ 2NO(g) + Cl₂(g)Check it balances: 2 N, 2 O and 2 Cl on each side. Powers and coefficients always match, so this works in either direction.
The expression belongs to the equation
There is nothing sacred about the way an equation is written. You could write the ammonia synthesis with all the coefficients doubled, or write it backwards, and it would still be a true statement about the chemistry. But each version has its own expression, and its own value of K.
Equation as written
Expression
Relationship to the first K
N2 + 3H2 ⇌ 2NH3
[NH3]2 ÷ ([N2][H2]3)
call this K
2NH3 ⇌ N2 + 3H2
[N2][H2]3 ÷ [NH3]2
1/K — the expression is flipped
½N2 + 1½H2 ⇌ NH3
[NH3] ÷ ([N2]½[H2]3/2)
√K — halving the equation square-roots it
This is why a value of K is meaningless on its own. Quote it with the equation it came from and the temperature it was measured at, or it tells the reader nothing.
At SL, treat K as having no units. You will see units quoted in some textbooks, but the IB does not require them and inventing them can cost you marks.
💡 Exam tip
Balance the equation before you write anything. A wrong coefficient guarantees a wrong power.
Use square brackets, every time. Round brackets are treated as a different quantity.
Say out loud “products over reactants” as you write it. Upside-down expressions are the most common error in the whole topic.
Scan the state symbols and delete the solids before you start.
If a coefficient is 1, no power is written — but check it really is 1.
⚠️ Common mix-up
Reactants on top. Half a mark’s worth of carelessness that ruins every calculation after it.
Multiplying by the coefficient instead of raising to that power: 2[NH3] instead of [NH3]2.
Including solids, especially the solid metal in a redox equilibrium.
Adding concentrations rather than multiplying them.
Using starting concentrations. Every value in the expression is measured at equilibrium.
Up next: The Equilibrium Constant, Kc — the expression is only the recipe. Now for the number it produces, and what a very large or very small one is really telling you.
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