IB Chemistry HLTopic 5 — How Far? The Position of EquilibriumPaper 1 & 2Core skill~9 min read
The Equilibrium Law
The equilibrium law is the recipe for turning a balanced equation into an equilibrium constant expression. There is no thinking to do once you know the pattern — but there are three places students trip: which side goes on top, where the powers come from, and which species you leave out altogether.
📘 What you need to know
For the reaction aA + bB ⇌ cC + dD, the expression is K = [C]c[D]d ÷ [A]a[B]b.
Products on top, reactants underneath. Always.
The big numbers in front become the powers. Balance the equation first or every power is wrong.
Square brackets mean concentration in mol dm–3. Round brackets lose the mark.
Concentrations must be the ones at equilibrium, not the ones you started with.
Solids are left out. So are pure liquids and the water in a dilute solution.
K belongs to one specific equation. Reverse the equation and K becomes 1/K.
The pattern, once and for all
Write out the balanced equation. Put the products on top, the reactants on the bottom, and raise each concentration to the power of its balancing number. That is the whole law.
The equilibrium law
For aA + bB ⇌ cC + dD:
K = [C]c[D]d[A]a[B]b
The dashed lines are the only mapping you need to remember: the number in front of a species becomes the power on its square bracket.
🧩 Writing any K expression
Balance the equation. Non-negotiable — the powers come from here.
Cross out any solid, and any pure liquid. They never appear in the expression.
Write the fraction line. Products above, reactants below.
Copy each surviving species in square brackets, in the order it appears.
Add the powers from the balancing numbers. A coefficient of 1 needs no power.
Which species get left out
This is the part the pattern alone will not tell you. A concentration only means something if the substance can spread out and get more or less crowded. A block of solid cannot: its “concentration” is fixed by its density, so it is folded into K and never written.
GOES IN THE EXPRESSION
Gases (g)
Dissolved species (aq)
Liquids that are part of a mixture, such as ethanol and ethanoic acid in an esterification
These can genuinely be more or less concentrated, so their concentration is a real variable.
LEFT OUT COMPLETELY
Solids (s)
Pure liquids on their own, such as water in H2O(l) ⇌ H2O(g)
Water as the solvent in a dilute aqueous reaction
Their concentration cannot change in any useful way, so it is absorbed into the value of K.
Quick sanity check: grinding a solid into powder or adding a bigger lump changes nothing about the equilibrium position. That is exactly why solids are missing from the expression.
Look at the state symbols before you write a single bracket. An equilibrium with an (s) in it is the classic trap — students dutifully include it and lose the mark on an otherwise perfect expression.
K belongs to one particular equation
There is no such thing as “the K of ammonia”. K is tied to the exact equation you wrote down, including which way round it is and what the coefficients are. Change the equation and K changes in a predictable way.
What you do to the equation
What happens to K
Why, in one line
Reverse it
K becomes 1 ÷ K
Top and bottom of the fraction swap over
Double every coefficient
K becomes K2
Every power doubles, so the whole fraction is squared
Halve every coefficient
K becomes √K
Every power halves, so you take the square root
Multiply every coefficient by n
K becomes Kn
Same logic as doubling, generalised
Add two equations together
K becomes K1 × K2
The two fractions multiply, and the shared terms cancel
🧠 Remembering the reverse rule
Reversing the equation flips the fraction upside down, and flipping a fraction is the same as taking 1 over it. So a reaction with K = 100 going forwards has K = 0.01 going backwards. Big K one way always means tiny K the other way — a useful check on your answer.
(a) Everything is a gas, so nothing is left outK = [SO3]2[SO2]2[O2]O₂ has a coefficient of 1, so it gets no power(b) Watch the 3 in front of hydrogenK = [CO][H2]3[CH4][H2O]H₂O is a gas here, so it does count(c) Two solids to cross out firstK = [Cu2+][Ag+]2Cu(s) and Ag(s) never appearthe 2 in front of Ag(s) is ignored too – it vanishes with the solid
WORKED EXAMPLE
A heterogeneous equilibrium
Ammonium chloride sublimes in a sealed tube: NH4Cl(s) ⇌ NH3(g) + HCl(g). Write the expression for K, and state what happens to the equilibrium if more solid NH4Cl is added at constant temperature.
Step 1: Cross out the solidNH₄Cl(s) is not included, so the bottom of the fraction is empty.Step 2: Write the products onlyK = [NH3][HCl]Step 3: Adding more solidThe solid is not in the expression, so nothing in K can change.No shift — the gas concentrations stay exactly the sameadding more solid just gives you more unreacted solid
WORKED EXAMPLE
Rewriting the equation, rewriting K
At 373 K, for N2O4(g) ⇌ 2NO2(g), K = 0.212. Calculate K at the same temperature for
(a) 2NO2(g) ⇌ N2O4(g) and (b) ½N2O4(g) ⇌ NO2(g).
(a) This is the reverse reaction, so take 1 over KK = 10.212 = 4.7169…K = 4.72 (3 s.f.)(b) Every coefficient has been halved, so square root itK = √0.212 = 0.46043…K = 0.460 (3 s.f.)check: 0.460 squared gives 0.212 back, so the direction of the change is right
💡 Exam tip
Square brackets only. Round brackets are marked wrong because they do not mean concentration.
Write the state symbols in your own working. They are what tell you which species to delete.
If a coefficient is 1, leave the power off. Writing [O2]1 is not wrong, but it wastes time.
K has no units at IB level. Do not invent any.
When a question says “deduce the expression”, it wants the expression only — no numbers and no explanation.
If a question hands you K for one equation and asks about a rearranged one, it is testing the reverse or power rule. Spot it before you start calculating.
⚠ Common mix-up
Reactants on top. Products go on top, every time. If your K comes out as the reciprocal of the answer, this is why.
Including solids. An (s) never enters the expression, no matter how big the coefficient in front of it.
Using starting concentrations. The equilibrium law only works with equilibrium concentrations.
Multiplying instead of raising to a power. A coefficient of 2 means squared, not doubled.
Forgetting to balance first. Wrong coefficients means wrong powers means no marks anywhere downstream.
Treating water the same in every reaction. As a gas or as a reactant in a liquid mixture it counts; as the solvent in a dilute solution it does not.
Up next: The Equilibrium Constant, Kc — what the actual number tells you about how far a reaction goes, and how to calculate it from data.
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