IB Chemistry SLTopic 5 — The Extent of Chemical ChangePaper 1 & 2Core idea~14 min read
Le Chatelier’s Principle
You know where a reaction settles. Now the useful part: how to move it somewhere better. One sentence covers every case, and the whole skill is learning to apply it without waving your hands.
📚 What you need to know
If a change is made to a system at equilibrium, the position of equilibrium shifts so as to counteract that change.
Concentration: add a substance and the equilibrium shifts away from it; remove one and it shifts towards it.
Pressure: increasing it shifts the equilibrium to the side with fewer gas molecules.
Temperature: increasing it shifts the equilibrium in the endothermic direction.
Catalysts change nothing except the time taken to arrive.
Only a change in temperature changes the value of K.
Industrial processes use compromise conditions, balancing yield against rate and cost.
Position of equilibrium
The phrase “position of equilibrium” just means the relative amounts of reactants and products in the mixture. A shift to the right means more product and less reactant; a shift to the left means the reverse.
Le Chatelier’s principle
if a change is made to a system at equilibrium, the position of equilibrium moves to minimise that change
The word to hold on to is oppose. The system is not trying to help you, and it does not restore things completely — it pushes back against whatever you did and settles somewhere new. Every prediction in this topic is the same two-step thought: what did I change, and which direction partly undoes it?
Changing a concentration
Add more of a reactant and, for a moment, the ratio of products to reactants is too small to be the equilibrium value. Collisions between reactant particles become more frequent, the forward reaction speeds up, and product builds until the original ratio is restored. That restored ratio is K, unchanged.
Continuously removing a product is the industrial chemist’s favourite trick: the equilibrium keeps shifting right and never gets the chance to settle.
Diluting an aqueous equilibrium with water looks like a concentration change, but if there are the same number of aqueous species on each side it dilutes everything equally, the ratio is untouched, and nothing shifts. Count the species before you answer.
Changing the pressure
Pressure only matters where gases are involved, because only gases are compressible enough for it to make any difference. Squeeze a gas mixture into a smaller volume and every concentration rises; the system responds by reducing the total number of gas molecules, which lowers the pressure again.
Count only the gas molecules. Solids and solutions are left out of the count entirely, so a solid on one side does not tip the balance.
Do not go looking for a “high pressure side” and a “low pressure side”. Compare the number of moles of gas written on each side of the equation, and that is the whole calculation. If they are the same, pressure changes nothing at all — a favourite one-mark question.
Changing the temperature
Heating a system supplies energy. The equilibrium opposes that by shifting in whichever direction absorbs energy, which is the endothermic one. Cooling does the opposite: the exothermic direction is favoured because it releases energy to replace what you removed.
Check the sign of ΔH for the forward reaction first, then decide which way is endothermic. Everything else follows from that one reading.
Temperature is also the only change that alters K itself. Look at what happens to an endothermic reaction on heating: the products increase and the reactants decrease, so the numerator of the expression grows while the denominator shrinks, and the value of K must rise. For an exothermic reaction the same reasoning gives a fall.
Catalysts
A catalyst speeds up the forward and reverse reactions by the same factor. The two rates therefore become equal at exactly the same composition as before — the system simply gets there sooner.
A catalyst has no effect on the position of equilibrium and no effect on K. It cannot improve a yield. Writing that a catalyst increases the amount of product is one of the quickest ways to lose a mark in this topic.
Everything on one page
Change
Which way does the equilibrium shift?
Value of K
Increase concentration of a reactant
right, to use it up
unchanged
Decrease concentration of a reactant
left, to replace it
unchanged
Increase concentration of a product
left, to use it up
unchanged
Remove a product as it forms
right, to replace it
unchanged
Increase pressure (decrease volume)
towards fewer gas molecules
unchanged
Decrease pressure (increase volume)
towards more gas molecules
unchanged
Increase temperature
in the endothermic direction
changes
Decrease temperature
in the exothermic direction
changes
Add a catalyst
no shift — equilibrium is reached faster
unchanged
🧩 Answering any Le Chatelier question
Name the change. Concentration, pressure, temperature or catalyst?
Find the relevant feature. Which species changed, how many gas molecules on each side, or what is the sign of ΔH.
State the direction of the shift, left or right.
Justify it by saying what the shift opposes.
Say what would be observed, or what happens to the yield, if the question asks.
WORKED EXAMPLE
N2O4 is colourless and NO2 is dark brown. For N2O4(g) ⇌ 2NO2(g), ΔH = +57 kJ mol–1. State and explain the colour change observed when the sealed tube is (a) placed in iced water, (b) compressed to half its volume.
(a) cooling the tubeThe forward reaction is endothermic, so the reverse reaction is exothermic. Cooling favours the exothermic direction.shifts left, the colour fadesLess NO₂ and more colourless N₂O₄. K also falls, because this is a temperature change.(b) halving the volumeThere is 1 mol of gas on the left and 2 mol on the right, so the equilibrium shifts left to reduce the pressure.shifts left, but the colour deepensThe catch: squashing the gas concentrates everything first, so the tube darkens immediately, then pales a little as the equilibrium shifts. The final colour is still darker than at the start. K is unchanged.
WORKED EXAMPLE
For the Contact process, 2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = –196 kJ mol–1. State the conditions that would give the highest possible yield, and explain why the temperature actually used is far higher.
Temperature for maximum yieldThe forward reaction is exothermic, so a low temperature shifts the equilibrium right.as low as possiblePressure for maximum yield3 mol of gas on the left, 2 mol on the right, so a high pressure shifts the equilibrium right.high pressureWhy the real temperature is highAt a low temperature the reaction is far too slow to be useful, and the catalyst works poorly. A moderate temperature of roughly 450 °C sacrifices some yield to gain an acceptable rate.a compromise between yield and rateIn practice the yield is already very high near atmospheric pressure, so plants do not pay for high pressure equipment. Good chemistry is not always the answer with the best K.
WORKED EXAMPLE
Predict, with a reason, the effect of each change. (a) Adding water to the equilibrium Ce4+(aq) + Fe2+(aq) ⇌ Ce3+(aq) + Fe3+(aq) (b) Adding a catalyst to N2(g) + 3H2(g) ⇌ 2NH3(g) (c) Increasing the pressure on H2(g) + I2(g) ⇌ 2HI(g)
(a) diluting the ionsno shiftTwo aqueous species on each side, so the water dilutes both sides equally and the ratio is unchanged.(b) adding a catalystno shift, no change in yieldBoth rates increase by the same factor, so equilibrium is reached sooner at exactly the same composition.(c) raising the pressureno shift2 mol of gas on the left and 2 mol on the right. All three concentrations rise, and the expression is unaffected because the powers balance.
Equilibria that are not all in one phase
The principle applies just as well when the phases differ. A sealed bottle of fizzy drink holds an equilibrium between dissolved and gaseous carbon dioxide, at a pressure well above atmospheric.
In the bottle
CO2(g) ⇌ CO2(aq)
Open the cap and gaseous carbon dioxide escapes, so its pressure drops sharply. The equilibrium shifts to replace it, dissolved gas comes out of solution, and you see the bubbles. Leave the bottle open long enough and the drink goes flat — the system has become open, so it runs to completion rather than settling.
💡 Exam tip
Write “the position of equilibrium shifts to the left/right”. Vague phrases like “it goes the other way” earn nothing.
Always give the reason in terms of opposing the change, not just the direction.
For pressure, count the gas molecules on each side and say the numbers out loud in your answer.
For temperature, quote the sign of ΔH and name the endothermic direction.
Remember that a shift right does not mean K has increased — unless you changed the temperature.
For industrial processes, use the phrase compromise conditions and mention both yield and rate.
⚠️ Common mix-up
“A catalyst increases the yield.” It never does. It only saves time.
Applying pressure arguments to solids and solutions. Only gas molecules are counted.
Forgetting the equal-moles case, where a pressure change does nothing at all.
Assuming heating always improves the yield. For an exothermic reaction it makes it worse.
Saying the equilibrium shifts right so K goes up. Only temperature moves K.
Thinking the change is cancelled out. The system opposes it partly; the new position is never the old one.
Up next: Measuring Reaction Rates — how far a reaction goes is only half the question. The compromise conditions in the Contact process only make sense once you can talk properly about the other half: how fast.
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