Equilibrium Constants: Kc

Le Chatelier's principle gives qualitative predictions about equilibrium shifts, but for quantitative work we need equilibrium constants. The equilibrium constant Kc expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficient.

Writing the Kc Expression

For the general equilibrium: aA + bB ⇌ cC + dD

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Where [X] represents the equilibrium concentration of species X in mol dm⁻³.

Rules for Writing Kc Expressions

Products on top, reactants on bottom.
Each concentration is raised to the power of its coefficient in the balanced equation.
Pure solids and pure liquids are omitted from the expression (their concentrations are effectively constant).
Only species in solution (aq) or in the gas phase (g) are included.

Examples

For: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Kc = [NH₃]² / ([N₂][H₂]³)

For: CaCO₃(s) ⇌ CaO(s) + CO₂(g)

Kc = [CO₂] (the solids are omitted)

For: CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

Kc = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]) (all species are liquids in the same phase -- they are included because they form a homogeneous mixture, not pure separate liquids)

Units of Kc

The units of Kc depend on the equilibrium expression. Work them out from the expression by substituting (mol dm⁻³) for each concentration term.

Equilibrium	Kc expression	Powers top	Powers bottom	Kc units
N₂O₄ ⇌ 2NO₂	[NO₂]²/[N₂O₄]	2	1	mol dm⁻³
H₂ + I₂ ⇌ 2HI	[HI]²/([H₂][I₂])	2	2	No units
N₂ + 3H₂ ⇌ 2NH₃	[NH₃]²/([N₂][H₂]³)	2	4	mol⁻² dm⁶
2SO₂ + O₂ ⇌ 2SO₃	[SO₃]²/([SO₂]²[O₂])	2	3	mol⁻¹ dm³

If the total powers on top and bottom are equal, Kc has no units (it is dimensionless).

Calculating Kc from Equilibrium Concentrations

Worked Example 1: Direct Substitution

For: H₂(g) + I₂(g) ⇌ 2HI(g) at 450 °C

At equilibrium: [H₂] = 0.030 mol dm⁻³, [I₂] = 0.020 mol dm⁻³, [HI] = 0.140 mol dm⁻³

Kc = [HI]² / ([H₂][I₂]) = (0.140)² / (0.030 × 0.020) = 0.0196 / 0.0006 = 32.7 (no units -- 2 powers on top, 2 on bottom)

ICE Tables

When you know initial concentrations but not equilibrium concentrations, you use an ICE table (Initial, Change, Equilibrium). This is a systematic method that works for every equilibrium calculation.

Worked Example 2: Simple ICE Table

1.00 mol of N₂O₄ is placed in a 1.00 dm³ container at a certain temperature. At equilibrium, 0.40 mol of N₂O₄ has dissociated. Find Kc for:

N₂O₄(g) ⇌ 2NO₂(g)

	N₂O₄	2NO₂
Initial	1.00	0
Change	−0.40	+0.80
Equilibrium	0.60	0.80

Note: if 0.40 mol of N₂O₄ dissociates, 2 × 0.40 = 0.80 mol of NO₂ is formed (from the 1:2 ratio).

Since volume = 1.00 dm³: [N₂O₄] = 0.60 mol dm⁻³, [NO₂] = 0.80 mol dm⁻³

Kc = [NO₂]² / [N₂O₄] = (0.80)² / 0.60 = 0.64 / 0.60 = 1.07 mol dm⁻³

Units: (mol dm⁻³)² / (mol dm⁻³) = mol dm⁻³ ✓

Worked Example 3: ICE Table with Volume Adjustment

0.50 mol of H₂ and 0.50 mol of I₂ are mixed in a 2.0 dm³ flask at 450 °C. Kc = 50. Find the equilibrium concentrations.

H₂(g) + I₂(g) ⇌ 2HI(g)

Initial concentrations: [H₂]₀ = 0.50/2.0 = 0.25 mol dm⁻³, [I₂]₀ = 0.25 mol dm⁻³

Let x = change in [H₂] at equilibrium:

	H₂	I₂	2HI
Initial	0.25	0.25	0
Change	−x	−x	+2x
Equilibrium	0.25 − x	0.25 − x	2x

Kc = (2x)² / ((0.25 − x)(0.25 − x)) = 4x² / (0.25 − x)² = 50

Taking the square root (since both sides are perfect squares):

2x / (0.25 − x) = √50 = 7.07

2x = 7.07(0.25 − x) = 1.768 − 7.07x

9.07x = 1.768

x = 0.195 mol dm⁻³

[H₂] = [I₂] = 0.25 − 0.195 = 0.055 mol dm⁻³

[HI] = 2 × 0.195 = 0.39 mol dm⁻³

Check: Kc = (0.39)² / (0.055 × 0.055) = 0.152 / 0.00303 = 50.2 ≈ 50 ✓

Worked Example 4: ICE Table with Percentage Dissociation

2.00 mol of PCl₅ is placed in a 4.00 dm³ vessel. At equilibrium, 30% has dissociated.

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

30% of 2.00 mol = 0.60 mol dissociated.

	PCl₅	PCl₃	Cl₂
Initial (mol)	2.00	0	0
Change (mol)	−0.60	+0.60	+0.60
Equilibrium (mol)	1.40	0.60	0.60

Convert to concentrations (divide by 4.00 dm³):

[PCl₅] = 1.40/4.00 = 0.350 mol dm⁻³

[PCl₃] = 0.60/4.00 = 0.150 mol dm⁻³

[Cl₂] = 0.60/4.00 = 0.150 mol dm⁻³

Kc = [PCl₃][Cl₂] / [PCl₅] = (0.150)(0.150) / 0.350 = 0.0225 / 0.350 = 0.0643 mol dm⁻³

What the Value of Kc Tells You

Large Kc (≫ 1): equilibrium lies to the right -- products predominate.
Small Kc (≪ 1): equilibrium lies to the left -- reactants predominate.
Kc ≈ 1: significant amounts of both reactants and products present.

Kc value	Equilibrium position	Practical meaning
> 10¹	Far to the right	Reaction essentially goes to completion
10² -- 10⁹	To the right	Products predominate
0.01 -- 100	Neither extreme	Significant amounts of both
10⁻⁹ -- 10⁻²	To the left	Reactants predominate
< 10⁻⁹	Far to the left	Essentially no reaction

The Reaction Quotient Qc

Before a system reaches equilibrium, you can calculate the reaction quotient Qc using the same expression as Kc but with current (non-equilibrium) concentrations.

If Qc < Kc: too many reactants, too few products. The reaction shifts right to reach equilibrium.
If Qc > Kc: too many products, too few reactants. The reaction shifts left.
If Qc = Kc: the system is at equilibrium.

What Changes Kc?

Only temperature changes Kc. Changes in concentration, pressure, or adding a catalyst do not change the value of Kc.

For an exothermic forward reaction: increasing T decreases Kc.
For an endothermic forward reaction: increasing T increases Kc.

When concentration or pressure changes disturb an equilibrium, the concentrations adjust until they once again satisfy the (unchanged) Kc expression.

Common Mistakes to Avoid

Including solids or pure liquids in the Kc expression. Only include aqueous and gaseous species. CaCO₃(s) is omitted.
Forgetting to convert moles to concentrations. ICE tables often start in moles. You MUST divide by volume (in dm³) before substituting into the Kc expression.
Getting stoichiometry wrong in the ICE table. If 2 mol of product forms per mol of reactant consumed, the change row must reflect this ratio.
Quoting Kc without units (when it has units). Always derive and include units unless they cancel to give a dimensionless quantity.
Thinking Kc changes when you add more reactant. It doesn't. Q changes (becomes < Kc), and the system shifts to restore Kc.

Summary

Kc = products/reactants, each raised to the power of their stoichiometric coefficient. Pure solids and liquids are excluded. ICE tables provide a systematic way to calculate equilibrium concentrations. Only temperature changes Kc. A large Kc indicates products favoured; a small Kc indicates reactants favoured. Units must be derived from the specific expression.

A-Level Deep Dive: Equilibrium Constants: Kc

Spec mapping

Edexcel 9CH0 specification, Topic 10 — Equilibrium I and Topic 11 — Equilibrium II require candidates to write expressions for the equilibrium constant $K_c$ in terms of equilibrium concentrations of products over reactants raised to their stoichiometric coefficients, to derive the units of $K_c$ from the rate-equation form, and to use ICE (initial / change / equilibrium) tables to compute $K_c$ from given initial concentrations and equilibrium data (refer to the official specification document for exact wording). Candidates must also know that $K_c$ is temperature-dependent only — pressure, concentration, and catalyst changes do not affect $K_c$ (only Le Chatelier-style position changes). Examined principally on Paper 2 (Topics 11–19) and Paper 3 (synoptic with Core Practicals). The data booklet does not list $K_c$ expressions — they must be constructed from the balanced equation.

Worked example with full mark scheme

Question (8 marks):

For the equilibrium $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ , a 1.00 dm³ flask is filled with 1.00 mol H₂ and 1.00 mol I₂ at 731 K and allowed to reach equilibrium. At equilibrium, 1.560 mol of HI is present.

(a) Construct an ICE table showing initial, change and equilibrium amounts. (3)

(b) Write the expression for $K_c$ and calculate its value with units. (3)

(c) State, with reasoning, the effect on $K_c$ of (i) doubling the initial [H₂]; (ii) raising the temperature (forward reaction is exothermic). (2)

Solution with mark scheme:

(a) M1 — initial moles: H₂ = 1.00, I₂ = 1.00, HI = 0. M1 — change: HI is formed at $+x$ where the stoichiometric coefficient is 2; H₂ and I₂ each decrease by $x/2$ . From "1.560 mol of HI at equilibrium": HI change = +1.560, so $x = 1.560$ and H₂, I₂ each decrease by 0.780. A1 — equilibrium moles: H₂ = 0.220, I₂ = 0.220, HI = 1.560. (Volume = 1.00 dm³, so concentrations are numerically identical to amounts here.)

Species	Initial / mol	Change	Equilibrium / mol
H₂	1.00	−0.780	0.220
I₂	1.00	−0.780	0.220
HI	0	+1.560	1.560

(b) M1 — $K_c = [\text{HI}]^2 / ([\text{H}_2][\text{I}_2])$ . M1 — substitute: $K_c = (1.560)^2 / (0.220 × 0.220) = 2.434 / 0.0484 = 50.3$ . A1 — units: numerator is (mol dm⁻³)², denominator is (mol dm⁻³)², so units cancel — $K_c$ is dimensionless for this equilibrium.

(c) (i) M1 — $K_c$ depends only on temperature; concentration changes shift the position but do not affect the value of $K_c$ . So doubling initial [H₂] leaves $K_c$ unchanged.

(ii) M1 — forward exothermic, so raising T favours the reverse direction (Le Chatelier). The equilibrium constant decreases — fewer products at higher T, so $K_c$ smaller.

Total: 8 marks (M5 A3, split as shown).

Specimen question modelled on the Edexcel 9CH0 Paper 2 format

Question (6 marks): For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ , at 100 °C in a 1.00 dm³ flask: equilibrium [N₂O₄] = 0.040 mol dm⁻³, [NO₂] = 0.080 mol dm⁻³.

(a) Write the expression for $K_c$ and calculate its value with units. (3)

(b) The flask is then halved in volume, so all concentrations momentarily double. State the new value of the reaction quotient $Q$ , and predict the direction of shift. (3)

Mark scheme decomposition by AO:

Part	Mark	AO	Earned by
(a) M1	1	AO1	$K_c = [\text{NO}_2]^2 / [\text{N}_2\text{O}_4]$ .
(a) M1	1	AO2	Substitute: $K_c = (0.080)^2 / 0.040 = 0.00640 / 0.040 = 0.160$ .
(a) A1	1	AO2	Units: (mol dm⁻³)² / (mol dm⁻³) = mol dm⁻³, so $K_c = 0.160$ mol dm⁻³.
(b) M1	1	AO2	After halving volume: [N₂O₄] = 0.080, [NO₂] = 0.160.
(b) M1	1	AO2	$Q = (0.160)^2 / 0.080 = 0.0256 / 0.080 = 0.320$ mol dm⁻³.
(b) A1	1	AO3	$Q = 0.320 > K_c = 0.160$ , so position shifts left (toward N₂O₄) until $Q = K_c$ again. (Consistent with Le Chatelier: increased P favours fewer-moles-of-gas side.)

Total: 6 marks split AO1 = 1, AO2 = 4, AO3 = 1.

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