Factors Affecting Kc Revision Notes | A Level AQA Chemistry

1. The Central Question: What Actually Changes Kc?

Here is a question that trips up a huge number of A-level students: if you add more reactant to an equilibrium mixture, does Kc change? The answer is no — and understanding why is the whole point of this lesson.

Kc is not a description of where the equilibrium sits. It is a fixed ratio that depends on one thing only: temperature. Everything else — concentration, catalysts, and pressure (for reactions where the number of moles of gas differs between sides) — can shift the equilibrium position without touching the value of Kc at all. Pressure changes do not shift the equilibrium position when the total moles of gas are equal on both sides, and in no case do they alter Kc.

This distinction between equilibrium position and equilibrium constant is one of the most heavily examined ideas in A-level physical chemistry. Nail it here and you will pick up marks that many students drop.

2. Writing and Interpreting the Kc Expression

Before examining what changes Kc, you need a firm grip on what Kc actually is. For a general equilibrium:

Notice that the concentrations in this expression are equilibrium concentrations — the values the system settles at, not the starting values. When a disturbance shifts the equilibrium position, the individual concentrations change, but their ratio (raised to the appropriate powers) returns to the same value of Kc, provided temperature is constant.

3. Concentration Changes: Position Shifts, Kc Stays Fixed

Imagine you have the Haber process equilibrium established in a sealed container: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$. You suddenly inject more $N_2$ into the system. By Le Chatelier's principle, the equilibrium position shifts to the right — the system makes more $NH_3$ to partially counteract the increase in $N_2$.

Here is the key insight: as the system shifts right, $[N_2]$ and $[H_2]$ fall and $[NH_3]$ rises. The concentrations all change — but they change in exactly the right proportions so that the ratio $\frac{[NH_3]^2}{[N_2][H_2]^3}$ returns to the same numerical value. Kc is unchanged.

This is not a coincidence. Kc is defined as the ratio at equilibrium. The system must reach a new equilibrium position where that ratio equals Kc again, because Kc is fixed by temperature alone. The equilibrium position shifts; the equilibrium constant does not.

4. Catalysts: Faster Equilibrium, Same Kc

A catalyst provides an alternative reaction pathway with a lower activation energy. More precisely, it lowers the energy of the transition state, so both the forward and reverse activation energies decrease and equilibrium is reached faster in both directions. The position of equilibrium and the value of Kc do not change.

Because both forward and reverse rates increase by the same factor, the system reaches equilibrium more quickly — but the position at which it settles, and therefore the ratio of equilibrium concentrations, is identical to the uncatalysed case. Kc is unchanged.

Think of it this way: Kc reflects the thermodynamic stability of products relative to reactants at a given temperature. A catalyst does not alter the energy of reactants or products — it only lowers the energy barrier between them. The destination is the same; the route is faster.

5. Temperature Changes: The One Factor That Alters Kc

Temperature is the only factor that changes the value of Kc. To predict the direction of that change, you need to know whether the forward reaction is exothermic or endothermic — and then apply Le Chatelier's principle to the temperature change.

Think of heat as a reactant or product in the equilibrium. For an exothermic forward reaction, you can write: reactants $\rightleftharpoons$ products + heat. Increasing temperature adds heat to the system, so the equilibrium shifts left (away from the side with heat) to oppose the change. This means fewer products and more reactants at the new equilibrium — so Kc decreases.

For an endothermic forward reaction, heat is on the reactant side: reactants + heat $\rightleftharpoons$ products. Increasing temperature shifts the equilibrium to the right, producing more products. The ratio of product to reactant concentrations increases, so Kc increases.

A worked example makes this concrete. The synthesis of ammonia is exothermic: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $\Delta H = -92 \text{ kJ mol}^{-1}$. Raising the temperature shifts the equilibrium to the left, so $[NH_3]$ falls and $[N_2]$, $[H_2]$ rise. The numerator of the Kc expression falls and the denominator rises, so Kc decreases.

Worked example: Predicting the Effect of Temperature on Kc

The decomposition of dinitrogen tetroxide is endothermic: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$, $\Delta H = +57 \text{ kJ mol}^{-1}$. Predict and explain the effect of increasing temperature on the value of Kc.

1. $\Delta H = +57 \text{ kJ mol}^{-1}$ (positive) — A positive $\Delta H$ means the forward reaction is endothermic — it absorbs heat from the surroundings.
2. Increasing temperature adds thermal energy to the system. — The system opposes this by absorbing the extra energy — it shifts in the endothermic direction, which is the forward direction (left to right).
3. $[NO_2]$ increases; $[N_2O_4]$ decreases at the new equilibrium. — More $NO_2$ is produced and $N_2O_4$ is consumed as the position shifts right.
4. $K_c = \frac{[NO_2]^2}{[N_2O_4]}$. Numerator increases, denominator decreases. — The ratio of equilibrium concentrations increases, so Kc increases.
5. Kc increases when temperature is raised for this endothermic equilibrium. — This is the exam-ready conclusion: link the sign of $\Delta H$ to the direction of shift, then link the shift to the change in the concentration ratio, and therefore Kc.

Answer: Increasing temperature shifts the equilibrium to the right (forward direction is endothermic). $[NO_2]$ increases and $[N_2O_4]$ decreases, so the value of Kc increases.

6. Comparing All Three Factors: A Unified View

Now that each factor has been treated individually, it is worth placing them side by side. The pattern is clear: only temperature changes Kc. Concentration and catalyst changes shift the equilibrium position but leave Kc intact.

This table is worth committing to memory. In an exam, if you see a question about any change to an equilibrium system, your first move should always be to ask: has the temperature changed? If yes, Kc changes. If no, Kc stays the same — whatever else has happened.

7. Exam Technique and Mark-Winning Responses

AQA examiners are very precise about the language they expect. When a question asks you to 'explain' the effect of a change on Kc, a one-word answer ('unchanged') will not earn full marks. You need a chain of reasoning that connects the change to the equilibrium shift (or lack of it) and then to the value of Kc.

For temperature questions, the chain runs: identify $\Delta H$ sign $\rightarrow$ state direction of shift $\rightarrow$ state how concentrations change $\rightarrow$ state whether Kc increases or decreases. For concentration and catalyst questions, the chain runs: state that the equilibrium position shifts (or does not shift) $\rightarrow$ state that temperature is unchanged $\rightarrow$ conclude that Kc is unchanged.

Watch out for questions that describe an industrial process and ask about the choice of temperature. The Haber process uses around 450 °C — not because this maximises Kc (a lower temperature would give a higher Kc and better yield), but because a lower temperature makes the rate too slow to be economically viable. Kc and rate are separate considerations, and examiners frequently test whether you can distinguish them.

To consolidate the full picture: Kc is a thermodynamic quantity fixed by temperature. Concentration changes and catalysts are kinetic or positional interventions — they change how quickly equilibrium is reached or where it sits, but they cannot alter the fundamental thermodynamic ratio that Kc represents. Temperature changes the energy landscape of the reaction itself, which is why it is the only factor that changes Kc.