Kp and Partial Pressures Revision Notes | A Level AQA Chemistry

1. Why Partial Pressures Matter

When gases mix, each one contributes to the total pressure of the system. The Haber process, the Contact process, and countless industrial reactions all involve gas-phase equilibria — and to describe those equilibria quantitatively, you need Kp.

Think of a gas mixture like a crowd in a room. Each person (each gas) pushes against the walls independently. The total pressure is the sum of all those individual pushes — and each individual push is called a partial pressure.

Kp is the equilibrium constant calculated from those partial pressures for a system at constant temperature. It plays the same role for gas-phase reactions that Kc plays for reactions in solution — but the two are not interchangeable, and the AQA specification requires you to work confidently with Kp specifically.

2. Mole Fractions and Partial Pressures

Once you have the mole fraction, deriving the partial pressure is one multiplication away. The partial pressure of component A is:

Notice that the units of partial pressure are the same as the units of total pressure — typically kPa or atm at A-level. The mole fraction carries no units, so it does not change the unit of the result.

Worked example: Calculating Mole Fractions and Partial Pressures

A gas mixture contains 0.60 mol nitrogen ($\text{N}_2$), 0.20 mol oxygen ($\text{O}_2$), and 0.20 mol argon ($\text{Ar}$). The total pressure is 100 kPa. Calculate the partial pressure of each gas.

1. $n_{\text{total}} = 0.60 + 0.20 + 0.20 = 1.00 \text{ mol}$ — Add all moles together first. Here the total is conveniently 1.00 mol, which makes the mole fractions numerically equal to the individual mole values.
2. $\chi_{\text{N}_2} = \dfrac{0.60}{1.00} = 0.60$ $\chi_{\text{O}_2} = \dfrac{0.20}{1.00} = 0.20$ $\chi_{\text{Ar}} = \dfrac{0.20}{1.00} = 0.20$ — Divide each component's moles by the total. Check: 0.60 + 0.20 + 0.20 = 1.00. Mole fractions must sum to exactly 1.
3. $p_{\text{N}_2} = 0.60 \times 100 = 60 \text{ kPa}$ $p_{\text{O}_2} = 0.20 \times 100 = 20 \text{ kPa}$ $p_{\text{Ar}} = 0.20 \times 100 = 20 \text{ kPa}$ — Multiply each mole fraction by the total pressure. Units of partial pressure match units of total pressure.
4. $60 + 20 + 20 = 100 \text{ kPa} = P_{\text{total}}$ ✓ — The partial pressures must sum to the total pressure. If they do not, recheck your mole fractions.

Answer: $p_{\text{N}_2} = 60$ kPa, $p_{\text{O}_2} = 20$ kPa, $p_{\text{Ar}} = 20$ kPa

3. Constructing the Kp Expression

Now that you can calculate partial pressures, you can build the Kp expression. The logic mirrors Kc exactly: products over reactants, each raised to the power of its stoichiometric coefficient from the balanced equation.

Three rules govern every Kp expression you write. First, only gaseous species appear in the Kp expression; pure solids and liquids are omitted because their activities are taken as 1. Second, the stoichiometric coefficients from the balanced equation become powers in the expression. Third, the units of Kp depend on the overall change in moles of gas.

Here is where it gets interesting: the units of Kp depend on the stoichiometry of the reaction. You must derive units each time rather than assuming Kp is always dimensionless.

4. Worked Kp Calculation

Worked example: Why only gases appear in Kp

Explain why solids and liquids are omitted from a Kp expression.

1. Kp is written in terms of partial pressures. — Partial pressure is defined only for gases.
2. Pure solids and liquids have constant activity, taken as 1. — They therefore do not affect the numerical value of the expression.
3. Include gases only; omit pure solids and liquids. — This rule works for both homogeneous and heterogeneous equilibria so long as Kp is expressed with gas-phase species only.

Writing the expression is one skill; calculating a numerical value for Kp is another. The process always follows the same sequence: find moles at equilibrium, calculate mole fractions, convert to partial pressures, then substitute into the Kp expression.

Worked example: Full Kp Calculation: Haber Process

At equilibrium, a mixture contains 0.50 mol $\text{N}_2$(g), 1.50 mol $\text{H}_2$(g), and 1.00 mol $\text{NH}_3$(g). The total pressure is 200 kPa. Calculate Kp for the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$.

1. $K_p = \dfrac{(p_{\text{NH}_3})^2}{(p_{\text{N}_2})(p_{\text{H}_2})^3}$ — Products over reactants, with stoichiometric coefficients as powers. NH3 has coefficient 2, H2 has coefficient 3.
2. $n_{\text{total}} = 0.50 + 1.50 + 1.00 = 3.00 \text{ mol}$ — Sum all equilibrium moles before calculating mole fractions.
3. $\chi_{\text{N}_2} = \dfrac{0.50}{3.00} = 0.1\overline{6}$ $\chi_{\text{H}_2} = \dfrac{1.50}{3.00} = 0.50$ $\chi_{\text{NH}_3} = \dfrac{1.00}{3.00} = 0.3\overline{3}$ — Check: 0.1667 + 0.50 + 0.3333 = 1.00. Carry unrounded values forward to avoid rounding errors in the final answer.
4. $p_{\text{N}_2} = \dfrac{0.50}{3.00} \times 200 = 33.3 \text{ kPa}$ $p_{\text{H}_2} = \dfrac{1.50}{3.00} \times 200 = 100.0 \text{ kPa}$ $p_{\text{NH}_3} = \dfrac{1.00}{3.00} \times 200 = 66.7 \text{ kPa}$ — Multiply each mole fraction by 200 kPa. Sanity check: 33.3 + 100.0 + 66.7 = 200.0 kPa. ✓
5. $K_p = \dfrac{(66.7)^2}{(33.3)(100.0)^3}$ $= \dfrac{66.7 \times 66.7}{33.3 \times 100.0 \times 100.0 \times 100.0}$ $= \dfrac{4448.9}{33{,}300{,}000}$ $= 1.34 \times 10^{-4}$ — Expand each power longhand: 66.7 × 66.7 = 4448.89; 100.0 × 100.0 × 100.0 = 1,000,000; 33.3 × 1,000,000 = 33,300,000. Divide to get the numerical value.
6. $\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2$ \text{Units} = \text{kPa}^{-2}$ — Units of Kp = (pressure unit) raised to the power of delta-n-gas. Here delta-n-gas = -2, so units are kPa^-2.

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

5. Temperature, Pressure, and the Value of Kp

Here is one of the most important ideas in this topic: Kp is constant at constant temperature. Change the temperature and Kp changes. Change the pressure, add a catalyst, or alter concentrations — Kp stays the same.

This connects directly to Le Chatelier's principle, which you met earlier. When you increase pressure on a gas-phase equilibrium, the position of equilibrium shifts — but the value of Kp does not change. The shift in equilibrium position is the system's way of maintaining the same Kp under new conditions.

Temperature is different because it changes the relative energies of reactants and products. For an exothermic forward reaction, increasing temperature shifts equilibrium towards reactants, meaning the numerator of the Kp expression decreases relative to the denominator — so Kp decreases. For an endothermic forward reaction, increasing temperature increases Kp.

6. Exam Technique and Common Mistakes

Bridge: pressure changes can alter the equilibrium position by changing individual partial pressures, but the value of Kp remains fixed at constant temperature because the equilibrium ratio required by the expression is restored.

AQA Kp questions follow predictable patterns. Most multi-step calculations award method marks at each stage, so showing clear working — even if your final answer is wrong — protects marks. The most common place students lose marks is in the units, not the arithmetic.

A useful cross-check in Kp calculations is dimensional thinking. If the reaction produces more moles of gas than it consumes, Kp carries positive powers of pressure units; if it consumes more than it produces, the units carry negative powers. This provides a fast exam check that your expression and substitution are structurally sound before you commit to the final answer.