Equilibrium and Le Chatelier's principle

Dynamic means nothing has stopped

A reaction at equilibrium looks finished and is not. Both reactions are still running - forward and reverse - and they are running at the same rate, so the amount of everything stays the same while the molecules themselves keep changing places. That is the whole of what dynamic means, and almost every mark in this topic depends on you having understood it rather than memorised it.

Two things follow immediately. The concentrations are constant, but they are not equal: a system can sit at equilibrium with almost no product in it. And nothing about equilibrium says the reaction went far. A reaction that barely happens still reaches equilibrium, and reaches it quickly.

If you can only keep one sentence from this page, keep this one: at equilibrium the rates are equal, not the amounts.

What K tells you, and what it does not

The equilibrium constant is products over reactants, each raised to its coefficient. Pure solids and pure liquids are left out, because their concentration does not change as the reaction proceeds - there is no meaningful way for a lump of solid to become more concentrated.

A large K means the position of equilibrium lies well to the right: when the system settles, there is much more product than reactant. A small K means the opposite. That is all it tells you. It says nothing about speed - a reaction with an enormous K can take a century - and nothing about how much you will actually get out of a particular flask, which depends on what you put in.

The one thing that changes K is temperature. Not concentration, not pressure, not a catalyst. When a question asks you to explain why the yield changed and the temperature was changed, you are being asked about K itself; when the temperature was held constant, you are being asked about position only, and K has not moved at all.

• Solids and pure liquids do not appear in the expression.
• K has no units in the way you are asked to use it at this level; do not invent them.
• Reversing the reaction inverts K. Doubling the coefficients squares it.
• Only temperature changes K.

Le Chatelier as reasoning, not as a slogan

The principle is usually quoted as: a system at equilibrium responds to a change by partially opposing it. That sentence is true and it earns you nothing on its own, because the question is always which change, and why that produces that response.

Reason it in three steps, every time. First, name what the change did to the rates - adding a reactant makes the forward reaction faster, because collisions between reactant particles just became more frequent. Second, say what that does to the amounts while the two rates are unequal. Third, say where it settles: the forward rate falls as reactant is used up, the reverse rate rises as product builds, and they meet again at a new position.

Written that way, the direction falls out of the reasoning instead of being asserted. An answer that says the system shifts right to use up the added substance is describing the outcome as if it were the cause, and it is the single most common way to lose marks in this topic.

Pressure, volume and the trap in the middle

Reducing the volume of a gaseous system raises every concentration at once. The side with fewer moles of gas is the one that relieves that, so the position moves that way. This is the part students can usually recite.

The trap is the question where both sides have the same number of moles of gas. Changing the volume then changes nothing about the position at all - both rates rise by the same factor. Check the moles of gas on each side before you answer, every single time, and say in your answer that you checked. Examiners are asking that question precisely because it separates the students who reasoned from the students who recited.

The second trap is adding an inert gas at constant volume. The total pressure goes up and nothing else does - the partial pressures of the reacting gases are unchanged, so the rates are unchanged, so the position is unchanged. If the volume is allowed to change instead, you are back to the ordinary case.

Temperature is the only one that moves K

Treat heat as if it were a substance on one side of the equation. For an exothermic forward reaction, heat is a product: raise the temperature and the system moves back towards reactants, and K gets smaller. For an endothermic forward reaction it is the other way round.

This is why industrial conditions are always a compromise, and why that compromise is worth a mark or two whenever it comes up. A lower temperature may give a better position but a rate so slow the process is useless; a higher temperature gives you the product sooner and less of it. The catalyst does not resolve this by moving the position, because a catalyst speeds both directions equally and moves nothing. What it does is let you run at a lower temperature and still finish, which is an argument about rate and cost, not about equilibrium.

Where it usually goes wrong

Four faults account for most of the lost marks, and all four are habits rather than gaps in knowledge.

• Saying the system shifts to oppose the change, without ever mentioning rates. The direction is the conclusion; the rates are the argument.
• Treating a catalyst as if it changed the position or the yield. It changes neither.
• Forgetting to check the moles of gas before answering a volume question.
• Answering a temperature question without saying that K itself has changed - which is the one thing that makes temperature different from everything else.

What to practise next

Take any equilibrium question you have already done and rewrite your answer in the three steps above: what happened to the rates, what happened to the amounts, where it settled. Do that five times and the reasoning stops being something you assemble under pressure.

Then try the awkward versions deliberately - equal moles of gas on both sides, an inert gas added at constant volume, a catalyst introduced. Those are the ones written to find out whether you understand the topic or have learned the shape of the usual answer.

Check yourself

Three questions on what is above. Have a go before you open them - reading an answer you have not tried to give is the version of this that does nothing.