IB Chemistry HLTopic 5 — How Far? The Position of EquilibriumPaper 1 & 2HL only | Core skill~12 min read
Equilibrium Law Problem Solving
Real exam questions rarely hand you the equilibrium concentrations. They give you what you started with, one number measured at the end, and expect you to work out the rest. The tool for that is the ICE table — and once you can fill one in without thinking, these become some of the most reliable marks on the paper.
📘 What you need to know
Square brackets need concentration in mol dm–3: divide moles by the volume in dm3.
An ICE table has three rows: Initial, Change, Equilibrium.
The Change row follows the mole ratio of the equation. Reactants get a minus, products get a plus.
Equilibrium = Initial + Change. That single line fills most of the table.
If the number of concentration terms on the top and bottom of K is the same, the volume cancels and you can use moles directly.
When K < 10–3, the change is tiny, so you may assume the reactant concentration barely alters. You must say that you have done this.
Answers go to 3 significant figures unless told otherwise.
The ICE table, laid out
Three rows, one column per species, and one rule that fills nearly all of it. The only part that needs thought is the Change row, and that comes straight off the coefficients.
The coefficient in front of B is doing the work in the Change row. Get that number wrong and every value below it is wrong, which is why it pays to write the balanced equation across the top.
🧩 The full method
Write the balanced equation across the top of your table.
Fill the Initial row with what you were given. Anything not present yet is 0.
Work out one change from the data, then scale the rest by the coefficients.
Add the two rows to get the Equilibrium row.
Convert to concentrations by dividing by the volume — unless the volume cancels.
Substitute into K and solve. Only the Equilibrium row goes in.
When the volume cancels
Here is a shortcut worth having. Every concentration is (moles ÷ volume). If the top of the K expression has the same number of concentration terms as the bottom, every volume cancels out — so you can put the moles straight in and never touch the volume at all.
Volume cancels when the term counts match
K = (nC/V)(nD/V)(nA/V)(nB/V) = nC × nDnA × nB
VOLUME CANCELS
CH3COOH + C2H5OH ⇌ ester + H2O (2 over 2)
H2 + I2 ⇌ 2HI (2 over 2)
Use moles directly. If the question does not give you a volume, this is almost always why.
VOLUME MATTERS
N2O4 ⇌ 2NO2 (2 over 1)
N2 + 3H2 ⇌ 2NH3 (2 over 4)
You must divide by the volume. The volumes do not cancel, so using moles gives the wrong K.
Quick test before you start dividing: count the concentration terms on the top and the bottom, powers included. Equal counts means the volume vanishes. Unequal means you need it, and if the question has not given it to you, re-read the question.
The shortcut when K is tiny
If K is smaller than about 10–3, hardly any reactant is converted. That means the change to the reactant concentration is so small it makes no difference to your answer, so you are allowed to ignore it — which turns a nasty quadratic into a square root.
The approximation, valid when K < 10–3
[reactant]equilibrium ≈ [reactant]initial
Say it out loud in your answer. The mark scheme wants the assumption stated and justified: “since K < 10–3, the change in reactant concentration is negligible, so 0.400 – x ≈ 0.400″. Doing the maths silently loses the mark.
Worked examples
WORKED EXAMPLE
Finding K with no volume given
Ethanoic acid and ethanol form an ester:
CH3COOH(l) + C2H5OH(l) ⇌ CH3COOC2H5(l) + H2O(l)
0.800 mol of ethanoic acid is mixed with 0.800 mol of ethanol and a trace of acid catalyst. At equilibrium, 0.250 mol of ethanoic acid remains. Calculate K.
Step 1: Find the change in the acidchange = 0.250 – 0.800 = –0.550 molStep 2: The ratio is 1:1:1:1, so every change is 0.550ethanol also falls by 0.550; ester and water each rise by 0.550Step 3: Complete the equilibrium rowacid 0.250, ethanol 0.800 – 0.550 = 0.250, ester 0.550, water 0.550Step 4: Two terms on top, two on the bottom, so the volume cancelsK = 0.550 × 0.5500.250 × 0.250 = 0.30250.0625K = 4.84 (3 s.f.)no volume in the question was the clue that it cancels
Amount / mol
CH3COOH
C2H5OH
CH3COOC2H5
H2O
Initial
0.800
0.800
0.000
0.000
Change
–0.550
–0.550
+0.550
+0.550
Equilibrium
0.250
0.250
0.550
0.550
WORKED EXAMPLE
Finding concentrations from K, using the square-root trick
0.100 mol of H2 and 0.100 mol of I2 are sealed in a 1.00 dm3 vessel at 700 K, where K = 56.3.
H2(g) + I2(g) ⇌ 2HI(g)
Calculate the equilibrium concentration of HI.
Step 1: ICE table with x as the amount of H2 reactingInitial: 0.100, 0.100, 0. Change: –x, –x, +2x. Equilibrium: (0.100 – x), (0.100 – x), 2x. Volume is 1.00 dm³, so moles and concentrations are the same numbers.Step 2: Substitute into K56.3 = (2x)2(0.100 – x)2Step 3: Both sides are perfect squares — take the square root√56.3 = 2x0.100 – x so 7.5033 = 2x0.100 – xStep 4: Rearrange0.75033 – 7.5033x = 2x0.75033 = 9.5033x so x = 0.078955Step 5: Read off the answer[HI] = 2x = 2 × 0.078955 = 0.15791[HI] = 0.158 mol dm–3 (3 s.f.)check: [H2] = [I2] = 0.0210, and 0.158² / 0.0210² gives 56.3 back
WORKED EXAMPLE
Using the small-K approximation
Phosgene decomposes: COCl2(g) ⇌ CO(g) + Cl2(g), K = 2.50 × 10–6 at 600 K. A vessel is filled with COCl2 at 0.400 mol dm–3. Calculate the equilibrium concentration of CO.
Step 1: ICE tableInitial: 0.400, 0, 0. Change: –x, +x, +x. Equilibrium: (0.400 – x), x, x.Step 2: State the approximation and why it is allowedK is smaller than 10⁻³, so x is negligible compared with 0.400. Therefore 0.400 – x ≈ 0.400.Step 3: Substitute2.50 × 10–6 = x × x0.400 = x20.400Step 4: Solve for xx2 = 2.50 × 10–6 × 0.400 = 1.00 × 10–6x = √(1.00 × 10–6) = 1.00 × 10–3[CO] = 1.00 × 10–3 mol dm–3x is 0.25% of 0.400, well under 5%, so the assumption was safe
🧠 Only one row goes into K
Students fill in a beautiful ICE table and then substitute the Initial row into K out of habit. The equilibrium law only ever accepts equilibrium values. Draw a box around the bottom row of your table before you start substituting — it is a two-second habit that saves whole questions.
💡 Exam tip
Always draw the table, even for a simple 1:1 reaction. Examiners award method marks for it, and it stops sign errors.
Let x be the amount of the species with a coefficient of 1 where you can. It keeps the other entries tidy.
Check whether the volume cancels before you start dividing. It often saves a whole step.
If the equation gives you a perfect square on both sides, square root it rather than expanding into a quadratic.
Keep full calculator precision in the working and round only at the end.
Whenever you use the small-K approximation, write the sentence that justifies it. It is a separate mark.
Finish by substituting your answers back into K. If you do not get the K you were given, something went wrong.
⚠ Common mix-up
Substituting the initial values into K. Only the equilibrium row is allowed in.
Ignoring the coefficients in the Change row. A species with a 2 in front of it changes by 2x, not x.
Sign errors. Reactants go down (minus), products go up (plus). Every time.
Using moles when the volume does not cancel. Count the terms first.
Forgetting to convert cm3 to dm3. Divide by 1000.
Using the small-K shortcut when K is not small. Above 10–3 the approximation breaks and you need the full quadratic.
Answering with x when the question asked for a product concentration. If the change was +2x, the answer is 2x.
Up next: The Equilibrium Constant and Gibbs Energy (HL) — the equation that links how far a reaction goes to whether it is spontaneous at all.
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