IB Chemistry SLTopic 5 — The Extent of Chemical ChangePaper 1 & 2Core skill~12 min read
The Equilibrium Constant, Kc
The expression is just a recipe. The number it produces is the useful part: one value that tells you, at a glance, whether a reaction is worth running at all — and one that most students read far too much into.
📚 What you need to know
Kc is the value of the expression when the concentrations are the equilibrium ones.
K >> 1: mostly products, the reaction goes nearly to completion.
K << 1: mostly reactants, the reaction barely proceeds.
K ≈ 1: significant amounts of both are present at equilibrium.
K is constant at a given temperature. Only a change in temperature changes its value.
For the reverse reaction, K′ = 1/K.
K tells you how far, never how fast.
Reading the size of K
Look at where K comes from and the interpretation almost writes itself. Products sit on the top of the fraction and reactants on the bottom. A big number can only mean a big numerator, which means a lot of product. A tiny number means the opposite.
Notice that even at the extremes the other side never quite disappears. That is why “goes to completion” is a description, not a fact.
Value of K
Position of equilibrium
The mixture contains
K << 1
far to the left
almost entirely reactants; the reaction hardly proceeds
K < 1
to the left
reactants are favoured
K = 1
balanced between the two
significant amounts of both reactants and products
K > 1
to the right
products are favoured
K >> 1
far to the right
almost entirely products; the reaction is effectively complete
Calculating Kc
Two rules cover almost every calculation you will be asked to do. Use equilibrium values, never starting values, and use concentrations, never amounts in moles. If a question gives you moles and a volume, dividing is your first job, not your last.
Do this first
concentration = amount (mol) ÷ volume (dm3)
WORKED EXAMPLE
A 2.00 dm3 sealed flask at equilibrium contains 0.60 mol H2, 0.40 mol I2 and 3.20 mol HI. Calculate Kc. H2(g) + I2(g) ⇌ 2HI(g)
Step 1 — write the expressionK = [HI]² ÷ ([H₂] × [I₂])Step 2 — turn moles into concentrations[H₂] = 0.60 ÷ 2.00 = 0.300[I₂] = 0.40 ÷ 2.00 = 0.200[HI] = 3.20 ÷ 2.00 = 1.600Step 3 — substituteK = (1.600)² ÷ (0.300 × 0.200) = 2.560 ÷ 0.0600K = 42.7K is much greater than 1, so at this temperature the mixture is mostly hydrogen iodide.
WORKED EXAMPLE
At equilibrium a 4.00 dm3 vessel contains 0.800 mol N2O4 and 1.20 mol NO2. Calculate Kc, then calculate Kc for the reverse reaction at the same temperature. N2O4(g) ⇌ 2NO2(g)
Step 1 — concentrations[N₂O₄] = 0.800 ÷ 4.00 = 0.200[NO₂] = 1.20 ÷ 4.00 = 0.300Step 2 — substituteK = (0.300)² ÷ 0.200 = 0.0900 ÷ 0.200K = 0.450Step 3 — the reverse reactionK′ = 1 ÷ 0.450 = 2.22K′ = 2.22 for 2NO₂ ⇌ N₂O₄Here the volume genuinely matters, because there are more gas molecules on the right than on the left. Do not skip the division.
In the hydrogen iodide example the volumes happen to cancel, because there are two gas molecules on each side. Students notice this once and then assume it always happens. It does not. Convert to concentrations every single time and you never have to think about it.
Turning the reaction round
Flip an equation and the expression turns upside down, so the new constant is the reciprocal of the old one. This is not a rule to memorise so much as something you can see by writing the two expressions next to each other.
Reversing the equation
K′ = 1 / K
Worth knowing: if you multiply an entire equation by n, the constant is raised to the power n. Double the equation and K is squared; halve it and you take the square root. The reciprocal rule is really just the case where n = –1.
What does not change K
Adding more of a reactant, squashing a gas mixture into a smaller volume, or dropping in a catalyst will all disturb an equilibrium and shift its position. None of them changes the value of K. The system responds precisely so that the same ratio is restored.
Only temperature changes K, because only temperature changes the underlying balance rather than nudging the mixture along a fixed one.
Change made
Does the position shift?
Does K change?
Add or remove a reactant or product
Yes
No
Change the pressure or volume (gases)
Yes, unless both sides have equal gas moles
No
Add a catalyst
No — equilibrium is simply reached sooner
No
Change the temperature
Yes
Yes
How far is not how fast
This is the idea that separates a good answer from a vague one. K measures the extent of a reaction — where it finishes up. It says nothing whatever about the rate — how long it takes to get there.
A mixture of hydrogen and oxygen has an enormous K for forming water, yet it can sit in a flask indefinitely. Extent and rate are decided by completely different things.
WORKED EXAMPLE
Three reactions have the following equilibrium constants at the same temperature. State what each value tells you about the equilibrium mixture, and comment on how long each reaction takes. Reaction A: K = 3.4 × 10–21 Reaction B: K = 1.8 Reaction C: K = 6.0 × 1015
Reaction Amostly reactantsK is very much smaller than 1, so the denominator dwarfs the numerator. Hardly any product forms.Reaction Bplenty of bothK is close to 1, so reactants and products are present in comparable amounts. This is not the same as saying they are equal.Reaction Cmostly productsK is very much larger than 1, so the reaction is effectively complete — although a trace of reactant always remains.How long do they take?impossible to sayK describes the destination, not the journey. Nothing in these values says anything about rate.
💡 Exam tip
Convert moles to concentrations first. Marks are lost here more often than anywhere else in the calculation.
Answer “mostly products” or “mostly reactants” rather than “the reaction works” — the examiner wants the composition of the mixture.
Justify your statement using the expression: a large K means the numerator, and so the products, must be large.
Always attach a temperature to a value of K, and remember that only temperature can change it.
If asked about the reverse reaction, take the reciprocal — do not recalculate from scratch.
At SL, quote K with no units.
⚠️ Common mix-up
Putting moles straight into the expression. If a volume is given, it is given for a reason.
Using starting concentrations instead of equilibrium ones.
Reading a large K as “a fast reaction”. It means a high yield, however long that takes.
Thinking K = 1 means exactly 50 : 50. It means comparable amounts of each.
Claiming that adding more reactant increases K. It shifts the position; K is untouched.
Forgetting the power when a coefficient is 2 or 3 — the single biggest source of wrong numerical answers.
Up next: Le Chatelier’s Principle — you can now say where an equilibrium sits. The last step is learning how to push it somewhere more useful, and knowing when pushing will not work.
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