Le Chatelier's Principle: Shifting Equilibria Revision Notes | A Level AQA Chemistry

1. Why Equilibrium Shifts Matter

Picture a reversible reaction sitting at equilibrium — reactants converting to products at exactly the same rate as products converting back. Now disturb it. What happens next is the heart of this topic.

Le Chatelier's principle gives you a powerful shortcut: whenever you stress an equilibrium system, it fights back. The system shifts its position to partially counteract whatever you did to it. This single idea lets you predict the outcome of changing temperature, pressure, or concentration without needing to recalculate everything from scratch.

This matters enormously in industrial chemistry — the Haber process for ammonia synthesis, for example, is optimised entirely around Le Chatelier's principle. Understanding it at the mechanistic level (not just as a rule to memorise) is what separates a grade 7 answer from a grade 9 one.

2. Le Chatelier's Principle: The Core Idea

The key word in the definition is opposes. The system does not eliminate the disturbance — it reduces its effect. Think of it like a shock absorber: it cushions the impact rather than preventing it entirely.

At the molecular level, a shift occurs because the rates of the forward and reverse reactions become temporarily unequal after a disturbance. The system re-establishes equilibrium by adjusting concentrations until the two rates are equal again. Le Chatelier's principle tells you which direction that adjustment goes.

3. Effect of Concentration Changes

Consider the equilibrium for the formation of ammonia in the Haber process:

If you increase the concentration of $N_2$, the system opposes this by shifting to the right — consuming some of the added $N_2$ and producing more $NH_3$. The forward reaction rate temporarily increases, more product forms, and a new equilibrium is established with a higher proportion of $NH_3$.

Conversely, if you remove $NH_3$ as it forms (as is done industrially), the equilibrium shifts right again to replace it. This is a deliberate strategy: continuous removal of product drives the reaction towards higher yield.

Notice that concentration changes do not alter the value of $K_c$. The ratio of product to reactant concentrations at the new equilibrium is the same as before — the system adjusts quantities to maintain that ratio. Temperature is the only variable that changes $K_c$.

4. Effect of Pressure Changes on Gaseous Equilibria

Pressure changes only affect equilibria involving gases. The key is to count the total moles of gas on each side of the equation.

For the Haber process: the left side has $1 + 3 = 4$ moles of gas; the right side has $2$ moles of gas. Increasing the pressure means the system opposes this by shifting towards the side with fewer moles of gas — in this case, to the right. More $NH_3$ is produced at higher pressure.

Worked example: Predicting the Effect of Pressure on an Equilibrium

For the reaction $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, predict the effect of increasing the total pressure on the position of equilibrium.

1. $2SO_2 + O_2$: moles of gas $= 2 + 1 = 3$ — Only gaseous species count when considering pressure effects.
2. $2SO_3$: moles of gas $= 2$ — The right-hand side has fewer moles of gas.
3. Increasing pressure $\rightarrow$ system opposes by shifting towards fewer moles of gas $\rightarrow$ shifts right — Shifting right reduces the total number of gas molecules, which partially reduces the pressure — opposing the change.
4. Position of equilibrium shifts to the right; yield of $SO_3$ increases. — This is the principle exploited in the Contact process for manufacturing sulphuric acid.

Answer: Increasing pressure shifts the equilibrium to the right, increasing the yield of $SO_3(g)$, because the right-hand side has fewer moles of gas (2 vs 3).

When the number of moles of gas is equal on both sides, a pressure change has no effect on the position of equilibrium. The system has no 'preferred' side to shift towards.

5. Effect of Temperature Changes

Temperature is the most conceptually important variable because it is the only one that changes the value of $K_c$. Every other change — concentration, pressure, catalyst — shifts the position of equilibrium without altering $K_c$. Temperature changes the equilibrium constant itself.

The reasoning follows directly from Le Chatelier's principle. Think of heat as a reactant or product: if the forward reaction is exothermic, heat is released on the right-hand side. Increasing temperature adds heat to the system, so the equilibrium shifts left to absorb some of that heat — favouring the endothermic (reverse) direction.

For the Haber process ($\Delta H = -92$ kJ mol$^{-1}$, exothermic), increasing temperature shifts the equilibrium to the left. The yield of $NH_3$ falls, and $K_c$ decreases. Industrially, a compromise temperature of around 450 °C is used — high enough for a reasonable rate, but not so high that the yield becomes unacceptably low.

This is a crucial distinction to internalise: a catalyst speeds up the rate at which equilibrium is reached, but temperature actually changes where that equilibrium sits. They are fundamentally different in their effects.

6. The Role of a Catalyst

A catalyst increases the rate of both the forward and reverse reactions by the same factor. Because both rates increase equally, the position of equilibrium is unchanged — the same proportions of reactants and products are present at equilibrium.

The reaction profile above shows that a catalyst lowers the activation energy for both the forward and reverse reactions by the same amount. The energy difference between reactants and products — which determines the position of equilibrium — is completely unaffected. This is why $K_c$ does not change when a catalyst is added.

7. Bringing It Together: Exam Technique and Common Pitfalls

AQA questions on Le Chatelier's principle typically ask you to 'explain' or 'predict' — and those two command words require different responses. 'Predict' needs a direction and a brief reason. 'Explain' needs the full mechanistic chain: what changes, which rate is affected, which direction the system shifts, and what the new equilibrium looks like.

A mark-winning 'explain' answer for a concentration change follows this structure: state the change in rate (forward or reverse), state the direction of shift, and state the consequence for the concentrations of reactants and products. For temperature questions, always link the direction of shift to whether the forward reaction is exothermic or endothermic.

One final examiner point worth noting: AQA spec point 3.1.6.1.8 focuses on homogeneous equilibria when writing Kc expressions — make sure you recognise this term and understand that all species share the same state in a homogeneous system. However, Le Chatelier's principle itself applies broadly to all reversible equilibria, not only homogeneous ones. When applying pressure arguments to any equilibrium, count only the moles of gaseous species on each side, regardless of whether other species are in different phases.

Every spec requirement in this lesson connects to the same underlying logic: the system always responds to oppose a change. Master that principle at the mechanistic level — in terms of rates and concentrations — and you can handle any equilibrium question AQA sets, whether it involves familiar reactions like the Haber process or an unfamiliar industrial context you have never seen before.