Equilibrium constants

Chemistry· Unit 7: Chemical equilibria· 25 min read

1. Writing Kc Expressions★★☆☆☆⏱ 7 min

📘 Definition

Equilibrium Constant Kc

K_c

For a reversible reaction at constant temperature, (K_c) is the ratio of equilibrium concentrations of products to reactants, each raised to the power of their stoichiometric coefficient. Solids and pure liquids are always omitted, as their concentration is constant.

Example:

For (CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)), (K_c = [CO_2])

The general form for a reaction (aA + bB \rightleftharpoons cC + dD) is given below:

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

📐 Worked Example

Write the (K_c) expression for the equilibrium: (2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g))

1
All species are gaseous, so all are included in the expression.
2
Write products over reactants, raising each to the power of its stoichiometric coefficient:
3
$K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$

2. Writing Kp Expressions & Partial Pressures★★★☆☆⏱ 8 min

📘 Definition

Equilibrium Constant Kp

K_p

An equilibrium constant for gaseous systems expressed in terms of partial pressures of gases, following the same rules as (K_c): solids/liquids are omitted, and each partial pressure is raised to its stoichiometric power.

Example:

For (CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)), (K_p = p_{CO_2})

To calculate partial pressure of a gas, you first find its mole fraction, then multiply by total equilibrium pressure:

Mole fraction of A: ( \chi_A = \frac{\text{moles of A at equilibrium}}{\text{total moles of gas at equilibrium}} )
Partial pressure of A: ( p_A = \chi_A \times P_{\text{total}} )

📐 Worked Example

Write the (K_p) expression and calculate mole fractions for (N_2O_4(g) \rightleftharpoons 2NO_2(g)), where 1.0 mol (N_2O_4) dissociates to give 0.2 mol (N_2O_4) and 1.6 mol (NO_2) at equilibrium.

1
All species are gaseous, so the (K_p) expression is:
2
$K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}}$
3
Calculate total moles of gas at equilibrium: 0.2 + 1.6 = 1.8 mol
4
Calculate mole fractions:
5
Mole fraction of (N_2O_4): ( \frac{0.2}{1.8} = 0.111 )
6
Mole fraction of (NO_2): ( \frac{1.6}{1.8} = 0.889 )

3. Calculating Kc and Kp★★★☆☆⏱ 6 min

The most reliable method to find equilibrium moles is an ICE table, which tracks Initial moles, Change in moles, and Equilibrium moles for all species. Convert equilibrium moles to concentrations (for (K_c)) or partial pressures (for (K_p)) before substituting into the equilibrium expression.

📐 Worked Example

0.1 mol ethanoic acid reacts with 0.1 mol ethanol in a total volume of 250 cm³. At equilibrium, 0.067 mol ethyl acetate is formed. Calculate (K_c) for: (CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O)

1
Complete the ICE table for moles:
2
Species Initial Change Equilibrium
$C H_{3} C O O H$ 0.1 -0.067 0.033
$C_{2} H_{5} O H$ 0.1 -0.067 0.033
$C H_{3} C O O C_{2} H_{5}$ 0 +0.067 0.067
$H_{2} O$ 0 +0.067 0.067
3
Write the (K_c) expression (all species are in solution, so all included):
4
$K_c = \frac{[CH_3COOC_2H_5][H_2O]}{[CH_3COOH][C_2H_5OH]}$
5
Concentration = moles / volume V. All volume terms cancel here (Δn = 0):
6
$K_c = \frac{(\frac{0.067}{V})(\frac{0.067}{V})}{(\frac{0.033}{V})(\frac{0.033}{V})} = \frac{0.067^2}{0.033^2} \approx 4.1$
7
Units cancel out, so (K_c = 4.1) (unitless)

Species	Initial	Change	Equilibrium
$C H_{3} C O O H$	0.1	-0.067	0.033
$C_{2} H_{5} O H$	0.1	-0.067	0.033
$C H_{3} C O O C_{2} H_{5}$	0	+0.067	0.067
$H_{2} O$	0	+0.067	0.067

4. Properties of Equilibrium Constants★★☆☆☆⏱ 4 min

📘 Definition

Magnitude of K

The value of K at a given temperature indicates the position of equilibrium: if (K \gg 1), equilibrium lies far to the right (mostly products); if (K \ll 1), equilibrium lies far to the left (mostly reactants); if (K \approx 1), significant amounts of both are present.

Only temperature changes the value of (K_c) or (K_p). Changes in concentration, pressure, or adding a catalyst do not change K. The direction of change depends on the enthalpy of the reaction:

Exothermic (ΔH < 0): Increasing temperature → K decreases
Endothermic (ΔH > 0): Increasing temperature → K increases

📐 Worked Example

The Haber process (N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)) ΔH = -92 kJ mol⁻¹. What happens to (K_p) when temperature increases?

1
The forward reaction is exothermic, so heat is released when ammonia forms.
2
Increasing temperature shifts equilibrium in the endothermic (reverse) direction to oppose the change.
3
This increases the concentration of reactants and decreases products, so (K_p) decreases.

5. Common Pitfalls

Wrong move:

Including solids or pure liquids in Kc/Kp expressions

Why:

The concentration of pure solids/liquids is constant at a given temperature, so it is already incorporated into the value of K

Correct move:

Always omit all solids and pure liquids from equilibrium constant expressions, only include gaseous and aqueous species

Wrong move:

Saying K changes when concentration or pressure changes

Why:

K is only affected by temperature. Changes in concentration/pressure shift equilibrium position to keep K constant

Correct move:

State that K remains unchanged for any change other than a change in temperature

Wrong move:

Forgetting to raise concentration/partial pressure to the power of the stoichiometric coefficient

Why:

The power must match the coefficient in the balanced equation, leaving powers out gives an incorrect value for K

Correct move:

Always double check that each species is raised to the power of its stoichiometric coefficient

Wrong move:

Confusing the effect of temperature on K for exothermic and endothermic reactions

Why:

It is easy to reverse the direction of change when answering exam questions

Correct move:

Remember: Exothermic → T up = K down; Endothermic → T up = K up

6. Quick Reference Cheatsheet

Concept	Rule	Notes
Kc Expression	Products / Reactants, powers = stoichiometry; omit solids/liquids	Units = $(m o l d m^{- 3})^{Δ n}$
Kp Expression	Products / Reactants, powers = stoichiometry; omit solids/liquids	Units = $(P_{t o t a l})^{Δ n}$
Mole fraction	$χ_{A} = \frac{n _{A}}{n _{total}}$	Unitless
Partial pressure	$p_{A} = χ_{A} \times P_{total}$	Units: atm or Pa
Effect of concentration/pressure/catalyst on K	No change to K	Only position of equilibrium shifts
Effect of T increase on K (exothermic, ΔH < 0)	K decreases	Equilibrium shifts left
Effect of T increase on K (endothermic, ΔH > 0)	K increases	Equilibrium shifts right

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
Kc expression and calculation
2023 · 22
Kp from equilibrium moles
2024 · 11
Effect of temperature on Kc

What's Next

Mastering equilibrium constants is a core skill for all further equilibrium topics in A-Level Chemistry. The rules you learned here for writing expressions, using ICE tables, and interpreting the effect of temperature apply directly to acid-base equilibria, buffer solutions, solubility products, and even redox equilibria. Many exam questions combine these topics, so a solid foundation in this sub-topic will make all subsequent equilibrium topics much easier to tackle. You can now move on to applying Le Chatelier's principle to quantitative equilibrium problems, then progress to acid-base equilibria next.