Kp: Mole Fractions and Partial Pressures

Spec Mapping — OCR H432 Module 5.1.2 — How far? The equilibrium constant $K_p$ for gas-phase equilibria, expressed in terms of partial pressures; mole fractions $x_A = n_A/n_{\text{total}}$ and Dalton's law $p_A = x_A \cdot P_{\text{total}}$ ; the construction of $K_p$ expressions from balanced equations (gases only — solids and liquids omitted); units of $K_p$ as a function of $\Delta n_{\text{gas}}$ ; and the calculation of $K_p$ from equilibrium composition data (refer to the official OCR H432 specification document for exact wording).

This lesson bridges from the concentration-based $K_c$ of Year 12 (covered in OCR Module 3.2) to the partial-pressure-based $K_p$ that dominates the quantitative treatment of gas-phase equilibrium in Year 13. The leap is conceptual rather than algebraic: for gases, partial pressure is what we can actually measure directly with a pressure gauge, so it is the natural quantity to use in equilibrium calculations. The lesson develops the two prerequisite definitions — mole fraction $x_A = n_A/n_{\text{total}}$ and partial pressure $p_A = x_A \cdot P_{\text{total}}$ — works through writing $K_p$ expressions for four equilibria including the Haber and Contact processes, derives units from $\Delta n_{\text{gas}}$ , and finishes with three complete worked $K_p$ calculations. By the end you should be able to start from moles of each gas at equilibrium and total pressure, and produce $K_p$ with correct units in under five minutes.

Key Equation: for a gas-phase equilibrium $aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)$ , $K_p = \frac{p_C^{c}\,p_D^{d}}{p_A^{a}\,p_B^{b}}$ with each partial pressure given by $p_A = x_A \cdot P_{\text{total}}$ and $x_A = n_A/n_{\text{total}}$ . Units of $K_p$ are $(\text{units of pressure})^{\Delta n_{\text{gas}}}$ where $\Delta n_{\text{gas}} = (c + d) - (a + b)$ . $K_p$ depends only on temperature.

Recap — from $K_c$ to $K_p$

You have already met $K_c$ expressed using equilibrium concentrations for any reaction (solution or gas):

$K_c = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$

For gas-phase equilibria, partial pressures are usually what we measure. The equivalent expression is $K_p$ , with each $[X]$ replaced by the partial pressure $p_X$ :

$K_p = \frac{p_C^{c}\,p_D^{d}}{p_A^{a}\,p_B^{b}}$

$K_p$ is defined only for reactions involving gases. Solids and pure liquids are omitted from the $K_p$ expression (just as they are from $K_c$ ), because their activity is taken as unity (constant).

Mole fraction

The mole fraction $x_A$ of component A in a mixture is:

$x_A = \frac{n_A}{n_{\text{total}}}$

where $n_A$ is the moles of A and $n_{\text{total}}$ is the total moles of all gases present. Key properties:

Dimensionless (mol/mol).
Always between 0 and 1.
Mole fractions of all components sum to 1: $x_A + x_B + x_C + \ldots = 1$ .

Partial pressure — Dalton's Law

Dalton's Law of partial pressures (1801): the pressure exerted by each gas in a mixture is the pressure it would exert if it alone occupied the container. The partial pressure of A is:

$p_A = x_A \cdot P_{\text{total}}$

The partial pressures sum to the total: $p_A + p_B + p_C + \ldots = P_{\text{total}}$ . Partial pressures have the same units as total pressure — commonly Pa, kPa or atm. OCR examiners use kPa or atm interchangeably; pick whichever the question uses and stay with it.

Writing a $K_p$ expression — recipe

Write the balanced equation, with state symbols.
Identify the gas-phase species only — exclude solids, pure liquids, and dissolved species.
Put products over reactants.
Raise each partial pressure to the power of its stoichiometric coefficient.

Example 1 — $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$

$\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$ $K_p = \frac{p_{\text{NO}_2}^{2}}{p_{\text{N}_2\text{O}_4}}$

Example 2 — Haber process

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ $K_p = \frac{p_{\text{NH}_3}^{2}}{p_{\text{N}_2} \cdot p_{\text{H}_2}^{3}}$

Example 3 — Contact process

$2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}$ $K_p = \frac{p_{\text{SO}_3}^{2}}{p_{\text{SO}_2}^{2} \cdot p_{\text{O}_2}}$

Example 4 — heterogeneous equilibrium with a solid

$\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$ $K_p = p_{\text{CO}_2}$

Only the gas appears — the two solids are omitted because their "activity" (effectively concentration in a pure-solid sense) is constant and absorbed into $K_p$ . This is the basis of thermal decomposition equilibria like limestone calcination — at any given temperature, there is a unique equilibrium $p_{\text{CO}_2}$ above the solid mixture, independent of how much solid is present.

Units of $K_p$

As with $K_c$ , the units of $K_p$ depend on the difference in the power of pressure between numerator and denominator. Define $\Delta n_{\text{gas}} =$ (sum of gas-product coefficients) − (sum of gas-reactant coefficients). Then:

$[\,K_p\,] = (\text{units of pressure})^{\Delta n_{\text{gas}}}$

If $\Delta n_{\text{gas}} = 0$ , $K_p$ is dimensionless. Otherwise, each unit-of-pressure factor adds or subtracts a power.

Units examples

Equilibrium	$\Delta n_{\text{gas}}$	Units of $K_p$ (kPa case)
$\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$	$2 - 1 = +1$	kPa
$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$	$2 - 4 = -2$	kPa $^{-2}$
$\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$	$2 - 2 = 0$	dimensionless
$2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3$	$2 - 3 = -1$	kPa $^{-1}$
$\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$	$1 - 0 = +1$ (only gases counted)	kPa

For atm-based calculations, substitute "atm" for "kPa". Always derive units from $\Delta n_{\text{gas}}$ — do not guess.

Pressure unit conventions

A pragmatic note on unit choice. OCR exam questions about $K_p$ usually quote pressures in kPa (the SI-derived unit, with 100 kPa $\approx$ 1 atm $=$ atmospheric pressure at sea level), but some industrial-context questions use atm or bar. The numerical value of $K_p$ depends on the unit chosen, but the position of equilibrium is unit-independent. If you compute $K_p$ in atm $^{-2}$ and then convert to kPa $^{-2}$ , multiply by $(100)^{-2} = 10^{-4}$ . Within a single calculation, never mix units: if the question gives total pressure in atm, work everything in atm; if it gives kPa, work in kPa. Conversion between the two unit systems is a final-step decoration, not a mid-calculation operation.

Worked Example 1 — $K_p$ for the Haber equilibrium

At equilibrium in a sealed container at 500 K, the mixture contains:

Species	Moles
$\text{N}_2$	1.0
$\text{H}_2$	3.0
$\text{NH}_3$	4.0

Total pressure $= 80$ kPa. Calculate $K_p$ for $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ .

Step 1 — total moles and mole fractions

$n_{\text{total}} = 1.0 + 3.0 + 4.0 = 8.0\,\text{mol}$ .

$x_{\text{N}_2} = 1.0/8.0 = 0.125$
$x_{\text{H}_2} = 3.0/8.0 = 0.375$
$x_{\text{NH}_3} = 4.0/8.0 = 0.500$

Check: $0.125 + 0.375 + 0.500 = 1.000$ . Correct.

Step 2 — partial pressures via Dalton

$p_{\text{N}_2} = 0.125 \times 80 = 10\,\text{kPa}$
$p_{\text{H}_2} = 0.375 \times 80 = 30\,\text{kPa}$
$p_{\text{NH}_3} = 0.500 \times 80 = 40\,\text{kPa}$

Check: $10 + 30 + 40 = 80\,\text{kPa}$ . Correct.

Step 3 — substitute into $K_p$

\begin{aligned} K_p &= \frac{p_{\text{NH}_3}^{2}}{p_{\text{N}_2} \cdot p_{\text{H}_2}^{3}} \\ &= \frac{(40)^{2}}{10 \times (30)^{3}} \\ &= \frac{1600}{10 \times 27\,000} \\ &= \frac{1600}{270\,000} \\ &= 5.93 \times 10^{-3} \end{aligned}

Step 4 — units

$\Delta n_{\text{gas}} = 2 - (1 + 3) = -2$ , so units $= \text{kPa}^{-2}$ .

$K_p = \mathbf{5.93 \times 10^{-3}\,\text{kPa}^{-2}}$ .

Worked Example 2 — simpler case ( $2\text{NO} + \text{O}_2 \rightleftharpoons 2\text{NO}_2$ )

At equilibrium at a certain temperature: $p_{\text{NO}} = 0.10$ atm, $p_{\text{O}_2} = 0.20$ atm, $p_{\text{NO}_2} = 0.30$ atm. Calculate $K_p$ .

\begin{aligned} K_p &= \frac{p_{\text{NO}_2}^{2}}{p_{\text{NO}}^{2} \cdot p_{\text{O}_2}} \\ &= \frac{(0.30)^{2}}{(0.10)^{2} \times 0.20} \\ &= \frac{0.090}{0.010 \times 0.20} \\ &= \frac{0.090}{0.0020} \\ &= 45 \end{aligned}

$\Delta n_{\text{gas}} = 2 - (2 + 1) = -1$ , units = atm $^{-1}$ .

$K_p = \mathbf{45\,\text{atm}^{-1}}$ .

Worked Example 3 — from moles + total pressure ( $\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2$ )

At equilibrium: 0.40 mol PCl $_5$ (g), 0.30 mol PCl $_3$ (g), 0.30 mol Cl $_2$ (g). Total pressure = 120 kPa. Reaction: $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ .

$n_{\text{total}} = 0.40 + 0.30 + 0.30 = 1.00\,\text{mol}$ .

Mole fractions: $x_{\text{PCl}_5} = 0.40$ , $x_{\text{PCl}_3} = 0.30$ , $x_{\text{Cl}_2} = 0.30$ .

Partial pressures: $p_{\text{PCl}_5} = 0.40 \times 120 = 48\,\text{kPa}$ , $p_{\text{PCl}_3} = 36\,\text{kPa}$ , $p_{\text{Cl}_2} = 36\,\text{kPa}$ .

\begin{aligned} K_p &= \frac{p_{\text{PCl}_3} \cdot p_{\text{Cl}_2}}{p_{\text{PCl}_5}} \\ &= \frac{36 \times 36}{48} \\ &= \frac{1296}{48} \\ &= 27\,\text{kPa} \end{aligned}

$\Delta n_{\text{gas}} = 2 - 1 = +1$ , units = kPa.

$K_p = \mathbf{27\,\text{kPa}}$ .

Worked Example 4 — heterogeneous equilibria (solids and liquids excluded from $K_p$ )

Consider the thermal decomposition of solid ammonium chloride: $\text{NH}_4\text{Cl(s)} \rightleftharpoons \text{NH}_3\text{(g)} + \text{HCl(g)}$ . A sealed tube is heated to a fixed temperature, and the equilibrium gas mixture is found to have total pressure 40 kPa (with the two gases produced in 1:1 ratio from stoichiometry).

Writing the $K_p$ expression. Because $\text{NH}_4\text{Cl(s)}$ is a pure solid, its activity is constant (effectively 1, by convention) and it does not appear in $K_p$ . Only the two gases enter:

$K_p = p_{\text{NH}_3} \cdot p_{\text{HCl}}$

Partial pressures. Because the stoichiometry generates one NH $_3$ and one HCl per mole of NH $_4$ Cl decomposed, the mole ratio is 1:1, so $x_{\text{NH}_3} = x_{\text{HCl}} = 0.5$ and each partial pressure equals half the total: $p_{\text{NH}_3} = p_{\text{HCl}} = 20\,\text{kPa}$ .

$K_p$ value and units. Substituting: $K_p = 20 \times 20 = 400$ $\Delta n_{\text{gas}} = 2 - 0 = +2$ (two moles of gas produced from zero moles of gas in the reactant — the solid contributes nothing to $\Delta n_{\text{gas}}$ ), so units = $\text{kPa}^{+2} = \text{kPa}^{2}$ . $K_p = 400\,\text{kPa}^{2}$ .

The key conceptual point. Heterogeneous equilibria — those involving more than one phase — have $K_p$ expressions containing only the gas-phase species. The activity of a pure solid or pure liquid is constant (because the "concentration" of pure solid is effectively a fixed quantity set by its density) and is absorbed into $K_p$ . This has a counter-intuitive practical consequence: the equilibrium pressure of CO $_2$ above limestone undergoing calcination ( $\text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)}$ ) is determined only by temperature, not by how much limestone or quicklime is present. Five grams of limestone in a sealed tube and five kilograms of limestone in a much larger sealed tube reach the same equilibrium $p_{\text{CO}_2}$ at any given temperature. This is the basis of industrial lime kilns: by removing CO $_2$ as it forms (open-top kiln), the reaction never reaches equilibrium and proceeds to completion.

Liquids likewise. For an equilibrium involving a pure liquid, e.g. $\text{H}_2\text{O(l)} \rightleftharpoons \text{H}_2\text{O(g)}$ , the $K_p$ expression is simply $K_p = p_{\text{H}_2\text{O(g)}}$ — just the saturated vapour pressure at that temperature. The pure liquid does not appear.

OCR exam framing. Heterogeneous-equilibrium questions in OCR H432 typically test whether students correctly omit the solid/liquid from the $K_p$ expression. The most common error — writing $K_p = p_{\text{NH}_3} p_{\text{HCl}} / [\text{NH}_4\text{Cl}]$ — would not score the $K_p$ -expression mark. The second-most-common error is computing $\Delta n_{\text{gas}}$ by reading off the overall equation as written (giving $\Delta n = 1$ , "one solid to two gases") rather than counting only the moles of gas on each side ( $\Delta n_{\text{gas}} = 2 - 0 = +2$ ).

Mermaid — calculation flow

graph LR
  A[Moles of each gas at equilibrium] --> B[Total moles n_total]
  B --> C[Mole fractions x_A = n_A/n_total]
  C --> D[Partial pressures p_A = x_A P_total]
  D --> E[Substitute into Kp expression]
  E --> F[Derive units from delta n_gas]
  F --> G[Quote Kp with units]

Kp: Mole Fractions and Partial Pressures

Kp: Mole Fractions and Partial Pressures

Recap — from $K_c$ to $K_p$

Mole fraction

Partial pressure — Dalton's Law

Writing a $K_p$ expression — recipe

Example 1 — $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$

Example 2 — Haber process

Example 3 — Contact process

Example 4 — heterogeneous equilibrium with a solid

Units of $K_p$

Units examples

Pressure unit conventions

Worked Example 1 — $K_p$ for the Haber equilibrium

Worked Example 2 — simpler case ( $2\text{NO} + \text{O}_2 \rightleftharpoons 2\text{NO}_2$ )

Worked Example 3 — from moles + total pressure ( $\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2$ )

Worked Example 4 — heterogeneous equilibria (solids and liquids excluded from $K_p$ )

Mermaid — calculation flow

Synoptic Links

More in Chemistry

Kp: Mole Fractions and Partial Pressures

Kp: Mole Fractions and Partial Pressures

Recap — from KcK_cKc​ to KpK_pKp​

Mole fraction

Partial pressure — Dalton's Law

Writing a KpK_pKp​ expression — recipe

Example 1 — N2O4⇌2NO2\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2N2​O4​⇌2NO2​

Example 2 — Haber process

Example 3 — Contact process

Example 4 — heterogeneous equilibrium with a solid

Units of KpK_pKp​

Units examples

Pressure unit conventions

Worked Example 1 — KpK_pKp​ for the Haber equilibrium

Worked Example 2 — simpler case (2NO+O2⇌2NO22\text{NO} + \text{O}_2 \rightleftharpoons 2\text{NO}_22NO+O2​⇌2NO2​)

Worked Example 3 — from moles + total pressure (PCl5⇌PCl3+Cl2\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2PCl5​⇌PCl3​+Cl2​)

Worked Example 4 — heterogeneous equilibria (solids and liquids excluded from KpK_pKp​)

Mermaid — calculation flow

Synoptic Links

More in Chemistry

Recap — from $K_c$ to $K_p$

Writing a $K_p$ expression — recipe

Example 1 — $\text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2$

Units of $K_p$

Worked Example 1 — $K_p$ for the Haber equilibrium

Worked Example 2 — simpler case ( $2\text{NO} + \text{O}_2 \rightleftharpoons 2\text{NO}_2$ )

Worked Example 3 — from moles + total pressure ( $\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2$ )

Worked Example 4 — heterogeneous equilibria (solids and liquids excluded from $K_p$ )