For H_2 + I_2 2HI at equilibrium, [H_2] = 0.10, [I_2] = 0.10, [HI] = 0.80\ mol dm^-3. Calculate K_c. [2 marks]

Pearson Edexcel Edexcel 2020-style practice: For 2SO_2(g) + O_2(g) 2SO_3(g) at equilibrium, a vessel at a total pressure of 300\ kPa contains 0.60\ mol SO_3, 0.20\ mol SO_2 and 0.20\ mol O_2. Calculate K_p and give its units.

Find mole fractions, then partial pressures, then substitute into K_p. Total moles = 0.60 + 0.20 + 0.20 = 1.00\ mol (1). Partial pressures (mole fraction × total): p(SO_3) = 0.60 × 300 = 180, p(SO_2) = 0.20 × 300 = 60, p(O_2) = 0.20 × 300 = 60\ kPa (1). K_p = p(SO_3)^2p(SO_2)^2\,p(O_2) = 180^260^2 × 60 = 32400216000 = 0.15 (2). Units: power = 2 - 2 - 1 = -1, so units are kPa^-1 (1).

Pearson Edexcel Edexcel 2018-style practice: 1.00\ mol of N_2O_4 was placed in a 10.0\ dm^3 flask and allowed to reach equilibrium: N_2O_4(g) 2NO_2(g). At equilibrium 0.40\ mol of N_2O_4 had dissociated. Calculate K_c and state its units.

Build an ICE table in moles, convert to concentrations, and substitute. Equilibrium moles: N_2O_4 = 1.00 - 0.40 = 0.60; NO_2 = 2 × 0.40 = 0.80\ mol (1). Concentrations (divide by 10.0\ dm^3): [N_2O_4] = 0.060, [NO_2] = 0.080\ mol dm^-3 (1). K_c = [NO_2]^2[N_2O_4] = (0.080)^20.060 = 0.00640.060 = 0.107\ mol dm^-3 (1). Units: power = 2 - 1 = 1, so units are mol dm^-3 (1).

EnglandChemistrySyllabus dot point

How do we calculate equilibrium constants and use them quantitatively?

The equilibrium constants Kc and Kp, calculating them from equilibrium amounts and partial pressures, mole fractions, and the effect of changing conditions on their values.

An Edexcel 9CH0 Topic 16 answer covering the equilibrium constants Kc and Kp, mole fractions and partial pressures, calculating equilibrium constants, and the effect of conditions on their values.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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What this topic is asking

Edexcel Topic 16 wants you to write and calculate the equilibrium constants $K_c$ and $K_p$ , use mole fractions and partial pressures, derive the correct units in each case, and explain how changing conditions affects their values.

The answer

Kc and Kp expressions

Mole fractions and partial pressures

The mole fraction ( $x$ ) of a gas is the moles of that gas divided by the total moles of gas. The partial pressure of that gas is its mole fraction multiplied by the total pressure:

Calculating the constants

The reliable method is an ICE (Initial, Change, Equilibrium) table:

Write the initial amounts in moles.
Use the stoichiometry and the given change to find the equilibrium moles of every species.
For $K_c$ , divide each equilibrium amount by the volume to get a concentration. For $K_p$ , find the total moles, then each mole fraction, then each partial pressure.
Substitute into the expression and evaluate.
Derive the units from the powers: (sum of product powers) minus (sum of reactant powers), attaching that many factors of $\text{mol dm}^{-3}$ (for $K_c$ ) or pressure unit (for $K_p$ ).

Working out the units

Counting the powers is essential. For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ the power is $2 - 4 = -2$ , so $K_c$ has units $\text{mol}^{-2}\,\text{dm}^6$ and $K_p$ has units $\text{kPa}^{-2}$ . For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}$ the power is $2 - 2 = 0$ , so both constants are dimensionless.

Calculating Kp for the Haber process

At equilibrium a vessel at $200\ \text{kPa}$ total pressure contains $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ with $1.0\ \text{mol}$ $\text{N}_2$ , $3.0\ \text{mol}$ $\text{H}_2$ and $2.0\ \text{mol}$ $\text{NH}_3$ . Find $K_p$ and its units.

step 1: Total moles

$n_{\text{total}} = 1.0 + 3.0 + 2.0 = 6.0\ \text{mol}$ .

step 2: Mole fractions

$x(\text{N}_2) = 1.0/6.0 = 0.167$ ; $x(\text{H}_2) = 3.0/6.0 = 0.500$ ; $x(\text{NH}_3) = 2.0/6.0 = 0.333$ .

step 3: Partial pressures (mole fraction $\times 200\ \text{kPa}$ )

$p(\text{N}_2) = 33.3$ ; $p(\text{H}_2) = 100$ ; $p(\text{NH}_3) = 66.7\ \text{kPa}$ .

step 4: Substitute

K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2)\,p(\text{H}_2)^3} = \frac{66.7^2}{33.3 \times 100^3} = \frac{4449}{3.33 \times 10^7} = 1.34 \times 10^{-4}

step 5: Units

Power $= 2 - (1 + 3) = -2$ , so units are $\text{kPa}^{-2}$ . Thus $K_p = 1.34 \times 10^{-4}\ \text{kPa}^{-2}$ .

Examples in context

Example 1. Optimising ammonia yield. Because the Haber reaction has $K_p$ units of $\text{kPa}^{-2}$ and a power of $-2$ , increasing the total pressure increases the partial pressure terms unevenly, driving the position towards ammonia (the side with fewer gas moles). Calculating $K_p$ at fixed temperature confirms that pressure changes the partial pressures and the position but not the value of $K_p$ , which is the quantitative backing for the qualitative Le Chatelier argument used in industry.

Example 2. Esterification by Kc. The reaction $\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{ester} + \text{H}_2\text{O}$ has $K_c \approx 4$ at room temperature, and because there are two moles on each side, $K_c$ is dimensionless and the volume cancels. This is why students can calculate equilibrium amounts of ester from initial moles alone, without knowing the flask volume, a frequent Edexcel calculation that follows directly from the unit analysis above.

Try this

Q1. For $H_2 + I_2 \rightleftharpoons 2HI$ at equilibrium, $[H_2] = 0.10$ , $[I_2] = 0.10$ , $[HI] = 0.80\ \text{mol dm}^{-3}$ . Calculate $K_c$ . [2 marks]

Cue. $K_c = \frac{(0.80)^2}{0.10 \times 0.10} = 64$ (no units, as they cancel).

Q2. State what affects the value of $K_p$ . [1 mark]

Cue. Only a change in temperature.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20205 marksFor

2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})

at equilibrium, a vessel at a total pressure of

300\ \text{kPa}

contains

0.60\ \text{mol}

\text{SO}_3

0.20\ \text{mol}

\text{SO}_2

and

0.20\ \text{mol}

\text{O}_2

. Calculate

K_p

and give its units.

Show worked answer →

Find mole fractions, then partial pressures, then substitute into $K_p$ .

Total moles $= 0.60 + 0.20 + 0.20 = 1.00\ \text{mol}$ (1).

Partial pressures (mole fraction $\times$ total): $p(\text{SO}_3) = 0.60 \times 300 = 180$ , $p(\text{SO}_2) = 0.20 \times 300 = 60$ , $p(\text{O}_2) = 0.20 \times 300 = 60\ \text{kPa}$ (1).

$K_p = \dfrac{p(\text{SO}_3)^2}{p(\text{SO}_2)^2\,p(\text{O}_2)} = \dfrac{180^2}{60^2 \times 60} = \dfrac{32400}{216000} = 0.15$ (2).

Units: power $= 2 - 2 - 1 = -1$ , so units are $\text{kPa}^{-1}$ (1).

Edexcel 20184 marks

1.00\ \text{mol}

\text{N}_2\text{O}_4

was placed in a

10.0\ \text{dm}^3

flask and allowed to reach equilibrium:

\text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g})

. At equilibrium

0.40\ \text{mol}

\text{N}_2\text{O}_4

had dissociated. Calculate

K_c

and state its units.

Show worked answer →

Build an ICE table in moles, convert to concentrations, and substitute.

Equilibrium moles: $\text{N}_2\text{O}_4 = 1.00 - 0.40 = 0.60$ ; $\text{NO}_2 = 2 \times 0.40 = 0.80\ \text{mol}$ (1).

Concentrations (divide by $10.0\ \text{dm}^3$ ): $[\text{N}_2\text{O}_4] = 0.060$ , $[\text{NO}_2] = 0.080\ \text{mol dm}^{-3}$ (1).

$K_c = \dfrac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} = \dfrac{(0.080)^2}{0.060} = \dfrac{0.0064}{0.060} = 0.107\ \text{mol dm}^{-3}$ (1).

Units: power $= 2 - 1 = 1$ , so units are $\text{mol dm}^{-3}$ (1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)