Kp Calculations and Le Chatelier's Principle Revision Notes | A Level AQA Chemistry

1. Why Kp Matters — and Where It Fits

You already know that reversible reactions reach a dynamic equilibrium where the rates of the forward and reverse reactions are equal. At A-level, you need to go further: you need to quantify that equilibrium using a constant. For gas-phase equilibria, that constant is Kp.

Think of Kp as a snapshot of the equilibrium mixture — it tells you the ratio of products to reactants at equilibrium, expressed through the pressures each gas exerts. A large Kp means the equilibrium lies to the right (products favoured); a small Kp means it lies to the left (reactants favoured).

The key insight that runs through this entire lesson: Kp only changes when temperature changes. Pressure and concentration changes shift the position of equilibrium; a catalyst affects only the rate of attainment of equilibrium. None of these alter the value of Kp. Keep that distinction sharp — examiners test it repeatedly.

2. Building the Kp Expression

Before you can calculate Kp, you need to know how to write its expression. For a general homogeneous gas-phase equilibrium:

The partial pressure of each gas is found in two stages. First, calculate the mole fraction of that gas. Then multiply by the total pressure of the mixture.

A common source of errors is forgetting to derive units. Write the Kp expression, substitute the pressure units in place of each partial pressure, cancel where possible, and read off the remaining unit. This approach works for any equilibrium and removes the need to memorise units for specific reactions.

3. Kp Calculations — A Step-by-Step Method

Every Kp calculation follows the same four-stage method: find equilibrium moles, calculate mole fractions, convert to partial pressures, then substitute into the Kp expression. Work through this method carefully with the example below.

Worked example: Calculating Kp for the Haber Process Equilibrium

At a certain temperature and a total pressure of 200 kPa, an equilibrium mixture of the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ contains 0.50 mol $N_2$, 1.50 mol $H_2$, and 1.00 mol $NH_3$. Calculate Kp and state its units.

1. $n_{\text{total}} = 0.50 + 1.50 + 1.00 = 3.00 \text{ mol}$ — Add up all moles of gas present at equilibrium — not the initial moles.
2. $\chi_{N_2} = \frac{0.50}{3.00} = 0.1\overline{6}$ $\chi_{H_2} = \frac{1.50}{3.00} = 0.50$ $\chi_{NH_3} = \frac{1.00}{3.00} = 0.3\overline{3}$ — Each mole fraction is that component's moles divided by the total. Check: mole fractions must sum to 1.
3. $p_{N_2} = 0.1\overline{6} \times 200 = 33.3 \text{ kPa}$ $p_{H_2} = 0.50 \times 200 = 100 \text{ kPa}$ $p_{NH_3} = 0.3\overline{3} \times 200 = 66.7 \text{ kPa}$ — Multiply each mole fraction by the total pressure. Carry enough decimal places to avoid rounding errors in the final step.
4. $K_p = \frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$ — Products over reactants, each raised to its stoichiometric coefficient.
5. $K_p = \frac{(66.7)^2}{(33.3) \times (100)^3}$ $= \frac{66.7 \times 66.7}{33.3 \times 100 \times 100 \times 100}$ $= \frac{4448.9}{33300000}$ $= 1.34 \times 10^{-4}$ — Expand each power longhand: 66.7 × 66.7 = 4448.9; 100³ = 100 × 100 × 100 = 1 000 000; 33.3 × 1 000 000 = 33 300 000.
6. $\text{Units} = \frac{\text{kPa}^2}{\text{kPa} \times \text{kPa}^3} = \frac{\text{kPa}^2}{\text{kPa}^4} = \text{kPa}^{-2}$ — Δn = 2 − 4 = −2, confirming units of kPa⁻². Always show the unit derivation — it is a separate mark on the mark scheme.

Answer: $K_p = 1.34 \times 10^{-4} \text{ kPa}^{-2}$

Notice how the small value of Kp here tells you the equilibrium lies to the left — the Haber process does not give a high yield of ammonia at equilibrium under these conditions. That is exactly why industrial chemists manipulate conditions, which brings us to Le Chatelier's Principle.

4. Le Chatelier's Principle — Pressure and Concentration Changes

Le Chatelier's Principle gives you a systematic way to predict how an equilibrium responds to a disturbance. The system always shifts to oppose the change imposed on it. For gas-phase equilibria, the two most important disturbances are changes in pressure and changes in concentration (or the addition of more of one reactant or product).

When you increase the total pressure on a gas-phase equilibrium, the system shifts towards the side with fewer moles of gas — this reduces the total number of gas molecules and so partially counteracts the pressure increase. If both sides have the same number of moles of gas, a pressure change has no effect on the position of equilibrium.

The table makes one pattern unmistakable: only a temperature change alters Kp. Every other disturbance shifts the equilibrium position to restore the same ratio of partial pressures — and so Kp remains constant.

Here is a concrete example. For the Contact Process equilibrium $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$, there are 3 moles of gas on the left and 2 on the right. Increasing the pressure shifts the equilibrium to the right, increasing the yield of $SO_3$. The mole fractions change, the partial pressures change, but when you substitute the new values into the Kp expression, the ratio comes out to exactly the same Kp value as before.

5. Temperature Changes — Shifting Position and Changing Kp

Temperature is the one variable that actually changes the value of Kp. To predict the direction of change, you need to know whether the forward reaction is exothermic or endothermic. The equilibrium shifts in the direction that absorbs the added heat energy — this is Le Chatelier's Principle applied to temperature.

For an exothermic forward reaction (negative ΔH): increasing temperature shifts the equilibrium to the left (towards reactants), because the reverse endothermic reaction absorbs the extra heat. The proportion of products decreases, so Kp decreases.

For an endothermic forward reaction (positive ΔH): increasing temperature shifts the equilibrium to the right (towards products), because the forward endothermic reaction absorbs the extra heat. The proportion of products increases, so Kp increases.

Apply this to the Haber process: $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, ΔH = −92 kJ mol⁻¹. The forward reaction is exothermic. Increasing temperature shifts the equilibrium left, reducing the yield of ammonia and decreasing Kp. This is why industrial ammonia production uses a compromise temperature of around 450 °C — high enough for an acceptable rate, but not so high that the yield becomes negligible.

6. Catalysts and the Equilibrium Constant

A catalyst provides an alternative reaction pathway with a lower activation energy. This speeds up both the forward and reverse reactions by the same factor — so the system reaches equilibrium more quickly, but the equilibrium position itself is unchanged.

Because the equilibrium position does not change, the ratio of partial pressures at equilibrium is identical with or without a catalyst. Kp therefore remains unchanged. A catalyst affects the rate of attainment of equilibrium, not the value of the equilibrium constant.

The reaction profile above shows that a catalyst lowers the activation energy for both the forward and reverse reactions equally. The energy difference between reactants and products — and therefore ΔH — is unchanged. This is why the equilibrium position and Kp are unaffected.

7. Exam Technique and Common Mistakes

Kp questions follow predictable patterns on AQA papers. The most reliable way to secure full marks is to work methodically through the four-stage calculation method every time, show all intermediate values, and always derive units explicitly. Examiners award separate marks for the expression, the partial pressures, the numerical value, and the units — so even a slip in arithmetic does not cost you all four marks if your method is clear.

Bridge: pressure changes can alter the equilibrium position when the total moles of gas differ, but the value of Kp itself changes only if the temperature changes.