Le Chatelier's principle with examples

TL;DR:

• If you disturb a system at equilibrium, it shifts to partly undo the change. That is Le Chatelier's principle.

• Compare the reaction quotient Q with K: if Q < K the reaction shifts toward products, if Q > K toward reactants.

• Concentration, pressure and temperature can shift equilibrium. A catalyst speeds both directions equally and shifts nothing.

Many reactions do not go all the way to products. They reach a balance where forward and reverse reactions run at the same rate, and the amounts stop changing. That balance is called chemical equilibrium. What happens when you poke it is the subject of Le Chatelier's principle.

The principle in one sentence

When a system at equilibrium is disturbed by a change in concentration, pressure or temperature, it shifts in the direction that partly counteracts the change, and settles at a new equilibrium.

The principle gives the direction of the shift. It does not say by how much, and it does not claim the change is cancelled completely. For the amount you need the equilibrium constant.

Q versus K: the reason behind the rule

For aA + bB ⇌ cC + dD, the equilibrium constant is K = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), measured at equilibrium. The reaction quotient Q has the same form, but uses whatever concentrations are present right now.

• Q = K: at equilibrium, no net change.

• Q < K: too many reactants, so the reaction shifts toward products.

• Q > K: too many products, so the reaction shifts toward reactants.

Worked example: adding reactant

Take H₂ + I₂ ⇌ 2 HI and, for this example, say K = 50 at the chosen temperature. It is a made-up round value near the real one at several hundred degrees. At equilibrium, [H₂] = [I₂] = 0.10 M. Then [HI] = √(K × [H₂][I₂]) = √(50 × 0.010) = √0.50 = 0.707 M.

Now suddenly raise [H₂] to 0.20 M by adding more.

1. Compute Q right after the change: Q = (0.707)² / (0.20 × 0.10) = 0.50 / 0.020 = 25.

2. Compare: Q = 25 < K = 50, so there are too many reactants. The reaction shifts right, making more HI.

3. Let x mol/L of H₂ react: [H₂] = 0.20 − x, [I₂] = 0.10 − x, [HI] = 0.707 + 2x.

4. Set the new balance equal to K: (0.707 + 2x)² = 50 (0.20 − x)(0.10 − x). Expand to 46x² − 17.83x + 0.50 = 0.

5. Solve and keep the root that does not make a concentration negative: x ≈ 0.030.

New equilibrium: [H₂] ≈ 0.170 M, [I₂] ≈ 0.070 M, [HI] ≈ 0.768 M. Check K: 0.768² / (0.170 × 0.070) = 0.590 / 0.0119 ≈ 50. It works, to rounding.

Notice the principle at work. The added H₂ is not fully removed, since it fell only from 0.20 to 0.170. But the system moved the right way, using up H₂ and I₂ and making HI. The I₂ dropped, and the HI rose from 0.707 to 0.768.

Pressure and temperature

Pressure matters only when gases are involved and the number of gas molecules changes. In N₂ + 3 H₂ ⇌ 2 NH₃, 4 mol of gas become 2 mol. Compressing the mixture raises pressure, and the system shifts toward fewer gas molecules, which is toward ammonia. In H₂ + I₂ ⇌ 2 HI there are 2 mol on each side, so pressure changes nothing.

Temperature is different, because it changes the value of K itself. Treat heat as a reactant or product. The ammonia reaction is exothermic, releasing about 92 kJ per mole of reaction as written, so heat is a product. Raising the temperature shifts it toward reactants and lowers K. Cooling shifts it toward ammonia, but cold reactions are slow, so industry compromises between rate and yield.

Why catalysts do not shift equilibrium

A catalyst provides an easier path for the reaction, lowering the energy barrier for the forward and reverse directions equally. Both rates rise by the same factor, so the ratio of products to reactants at equilibrium, K, does not change. The catalyst only helps the system reach equilibrium sooner.

Paper chemistry: The examples here are calculations. Reactions involving hydrogen, ammonia or high pressure need trained supervision and proper equipment. Do not try to compress, heat or seal gas mixtures at home.

How to check your answer

• Q against K. Compute Q first. Its relation to K tells you the direction before any algebra.

• Plug back in. New concentrations should give K.

• No negatives. Reject any root that gives a negative concentration.

• Partial. The shift should reduce, but not erase, the disturbance.

Mistakes that cost marks

• Saying adding reactant raises K. Only temperature changes K.

• Applying a pressure argument when gas moles are equal on both sides.

• Claiming a catalyst increases yield at equilibrium.

• Forgetting solids and pure liquids do not appear in K.

• Treating the shift as complete.

Where you will use it next

Equilibrium shows up in buffers, solubility and industrial chemistry. It links to rates in what speeds up a chemical reaction, and to concentration scales in pH and the logarithmic scale.

Frequently asked questions

Does the system ever go back exactly to where it was?

Not usually. After adding reactant, the new equilibrium has different concentrations, though K is the same at a given temperature.

What if I remove a product as it forms?

Then Q drops below K and the reaction keeps shifting toward products. Removing product is a common way to push a reaction to completion.