Le Chatelier's Principle: Predicting Which Way Equilibrium Shifts

Quick answer: Le Chatelier's principle says that when a system at equilibrium is disturbed, the position of equilibrium shifts to partially oppose the change. Adding a substance shifts equilibrium away from it; raising the pressure shifts it toward the side with fewer gas moles; raising the temperature shifts it in the endothermic direction. Only a temperature change actually alters the value of Kc or Kp — concentration, pressure and catalysts do not.

When I first met Le Chatelier's principle I treated "shifts to oppose the change" like a magic spell I could chant for marks. It only clicked when I stopped memorising and started asking one question of every disturbance: what did I just do to this system, and which direction would undo a little bit of it? Once you have that, you can predict any shift in seconds.

This topic runs right through A-Level, AP, and Cambridge chemistry, and it's the whole reason the Haber and Contact processes work the way they do. Here's how I actually reason through it, plus the traps that cost me marks in mocks.

What the principle really says

The full statement: if a system at dynamic equilibrium is subjected to a change in concentration, pressure or temperature, the position of equilibrium moves to counteract that change. The word examiners want is counteract or oppose — not cancel. The system never fully reverses what you did; it just eases it a bit and settles at a new equilibrium.

There are only three things you can change, plus catalysts:

• Concentration of a dissolved or gaseous species
• Total pressure (only matters for gases)
• Temperature
• A catalyst — which, spoiler, does nothing to the position

One rule that saves marks: pure solids and pure liquids don't appear in the equilibrium expression, so they don't count when you're weighing up concentration or gas moles. If you're not sure why a step works, paste the equation into Explain and ask it to walk you through the shift line by line.

Concentration changes

Add more of a species, and equilibrium shifts away from it to use it up. Remove a species, and it shifts toward that side to replace it. That's the whole rule.

Take the esterification equilibrium:

CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O

Add more ethanol (a reactant) and equilibrium shifts right, making more ester. Remove water as it forms and equilibrium also shifts right to replace it. Notice Kc hasn't changed at all here; the ratio of concentrations simply resettles to the same value.

Pressure changes (gases only)

Increasing the total pressure (by squeezing the volume smaller) shifts equilibrium toward the side with fewer moles of gas, because that side takes up less space and relieves the pressure.

Count the gas moles on each side first — that single habit answers most pressure questions. If both sides have equal moles of gas, pressure has no effect at all. For example:

H2(g) + I2(g) ⇌ 2HI(g)

Two moles of gas on the left, two on the right — so changing the pressure does nothing to the position here. A favourite trap.

The subtle one: adding an inert gas (like argon) at constant volume changes nothing, because the partial pressures of the reacting gases are unchanged.

Temperature — the only change that alters K

This is the one that separates the A and B grades. To predict a temperature shift, look at the sign of ΔH for the forward reaction:

• Forward reaction exothermic (ΔH negative): raising the temperature shifts equilibrium backward (the endothermic direction).
• Forward reaction endothermic (ΔH positive): raising the temperature shifts equilibrium forward.

The logic: adding heat is like adding a "reactant" of energy, so the system shifts in whichever direction absorbs that heat — the endothermic direction. Crucially, a temperature change is the only disturbance that changes the value of Kc or Kp. If a question asks "what happens to K?", the answer is "nothing" for every change except temperature.

Fully worked example: the Contact process

The industrial equilibrium for making sulfur trioxide (on the way to sulfuric acid) is:

2SO2(g) + O2(g) ⇌ 2SO3(g), ΔH = -197 kJ/mol

Let me predict the effect of four separate changes, step by step.

1. Count and classify. Left side: 2 + 1 = 3 moles of gas. Right side: 2 moles of gas. The forward reaction is exothermic (ΔH is negative).

1. Increase the pressure. Equilibrium shifts toward fewer gas moles — from 3 on the left to 2 on the right. So it shifts right, making more SO3. Higher pressure means higher yield.

1. Increase the temperature. Equilibrium shifts in the endothermic direction, which is the reverse here. So it shifts left, yield of SO3 falls, and Kp decreases. This is why the Contact process runs at a moderate ~450°C, not a very high temperature.

1. Remove SO3 as it forms. Equilibrium shifts right to replace it — a neat way to keep pushing the reaction forward without touching temperature.

1. Add a V2O5 catalyst. No shift at all. Position and yield are unchanged; the catalyst only makes the system reach equilibrium faster.

Catalysts and the industrial compromise

Say it with me: a catalyst does not shift the position of equilibrium. It speeds up the forward and reverse reactions equally, so equilibrium arrives sooner but at exactly the same composition. Marks are routinely lost by writing that a catalyst "increases yield." It doesn't.

This is where the Haber process becomes such a clean exam story:

N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -92 kJ/mol

Four moles of gas become two, and the forward reaction is exothermic. Le Chatelier says the best yield of ammonia comes from high pressure (favours the two-mole side) and low temperature (favours the exothermic direction). But low temperature makes the reaction painfully slow. So industry picks a compromise: about 200 atm, around 450°C, and an iron catalyst to claw back the speed lost by not going colder. That compromise reasoning — yield versus rate — is a guaranteed mark earner. Want it drilled into memory? Turn these conditions into flashcards.

The mistakes that cost real marks

• Writing that a catalyst shifts equilibrium or raises yield. It never does.
• Saying Kc or Kp changes when you alter concentration or pressure. Only temperature changes K.
• Forgetting to check the moles of gas before answering a pressure question (equal moles = no effect).
• Getting the temperature direction backward. Always anchor to the sign of ΔH first.
• Counting solids or pure liquids in the mole comparison. They don't appear in the expression.

Test yourself

1. For N2(g) + 3H2(g) ⇌ 2NH3(g) (ΔH = -92 kJ/mol), state and explain the effect of decreasing the temperature on the yield of ammonia and on Kp.

1. For H2(g) + I2(g) ⇌ 2HI(g), predict what happens to the position of equilibrium when the total pressure is increased. Explain.

1. In the Contact process 2SO2(g) + O2(g) ⇌ 2SO3(g), you add a catalyst. What happens to the equilibrium yield of SO3?

Quick answers: (1) Lower temperature shifts equilibrium in the exothermic (forward) direction, so the yield of ammonia increases and Kp increases. (2) No change — both sides have two moles of gas, so pressure has no effect on the position. (3) No change to the yield; the catalyst only lets equilibrium be reached faster.

Stuck on the "explain" part of any of these? Drop the reaction into Explain and ask it to justify the shift the way a mark scheme would, then test yourself again with a fresh set of quiz questions.

FAQ

Does Le Chatelier's principle change the value of the equilibrium constant?

Only a temperature change does. Concentration, pressure and catalysts shift the position of equilibrium (or the rate) but leave Kc and Kp untouched.

Why does pressure not always affect equilibrium?

Because pressure only shifts equilibrium when the two sides have different numbers of gas moles. If the moles of gas are equal (like H2 + I2 ⇌ 2HI), squeezing the system relieves pressure equally on both sides, so nothing moves.

Does adding an inert gas shift the equilibrium?

At constant volume, no — the partial pressures of the reacting gases don't change, so the position stays put. It only matters if adding gas forces a change in volume (constant pressure), which then shifts equilibrium toward the side with more gas moles.

Why doesn't a catalyst increase yield?

It lowers the activation energy of the forward and reverse reactions by the same amount, so both speed up equally. Equilibrium is reached faster, but the final position — and therefore the yield — is identical.

In short: Ask "what did I just change, and which direction eases it?" — shift away from what you add, toward fewer gas moles under pressure, and in the endothermic direction when you heat it, remembering only temperature ever moves K.