Chemical Equilibria

Q: Do I include pure solids in Kc or Kp expressions?

No. Pure solids and pure liquids have constant activity values, so they are omitted entirely from equilibrium constant expressions for heterogeneous systems.

Q: When does the value of the equilibrium constant K change?

K only changes when the temperature of the system changes. Changes to concentration, total pressure, or addition of a catalyst do not alter the value of K, only the position of equilibrium.

Q: What units should I use for partial pressures in Kp calculations?

Edexcel IAL Chemistry requires partial pressures for Kp to be measured in atmospheres (atm). Convert any given kPa or Pa values to atm before substituting into Kp expressions.

Edexcel International A-Level Chemistry· 13.1–13.9· 45 min read

1. Kc and Kp Expressions for Equilibrium Systems★★☆☆☆⏱ 10 min

📘 Definition

Equilibrium constant (K)

Ratio of equilibrium product concentrations/partial pressures raised to their stoichiometric coefficients, divided by equilibrium reactant concentrations/partial pressures raised to their stoichiometric coefficients.

For homogeneous systems, all species are included in K expressions. For heterogeneous systems, pure solids and pure liquids are omitted, as their concentrations/activities remain constant at a given temperature. Aqueous and gaseous species are always included.

📐 Worked Example

Write the Kc expression for the heterogeneous equilibrium: $CaCO_{3} (s) ⇌ CaO (s) + CO_{2} (g)$

1
Identify pure solid species: CaCO₃(s) and CaO(s) are omitted from the Kc expression.
2
Only the gaseous CO₂ is included, so the Kc expression is:
3
$K_c = [\text{CO}_2]$

Exam tip:

Always cross out pure solid and liquid species first when writing K expressions to avoid losing easy marks.

2. Calculating Kc and Kp with Correct Units★★★☆☆⏱ 15 min

Use ICE (Initial, Change, Equilibrium) tables to calculate equilibrium concentrations from given initial values and reaction data. For Kp, first calculate mole fractions of gaseous species, then partial pressures using $p_{A} = x_{A} \times P_{t o t a l}$ where $P_{t o t a l}$ is total system pressure in atm.

📐 Worked Example

For the equilibrium $N_{2} (g) + 3 H_{2} (g) ⇌ 2 NH_{3} (g)$ at 400°C, total pressure is 200 atm. Equilibrium mole fractions: N₂ = 0.12, H₂ = 0.36, NH₃ = 0.52. Calculate Kp and its units.

1
Calculate partial pressures of each gas:
2
$p(\text{N}_2) = 0.12 \times 200 = 24 \text{ atm}, p(\text{H}_2) = 0.36 \times 200 = 72 \text{ atm}, p(\text{NH}_3) = 0.52 \times 200 = 104 \text{ atm}$
3
Write the Kp expression:
4
$K_p = \frac{p(\text{NH}_3)^2}{p(\text{N}_2) \times p(\text{H}_2)^3}$
5
Substitute values and calculate:
6
$K_p = \frac{(104)^2}{24 \times (72)^3} \approx 1.21 \times 10^{-3}$
7
Derive units:
8
$\frac{\text{atm}^2}{\text{atm} \times \text{atm}^3} = \text{atm}^{-2}$

Exam tip:

Always derive units for K each time you calculate it, do not memorize units as they change with reaction stoichiometry.

3. Effect of Reaction Conditions on Equilibrium and K★★★☆☆⏱ 12 min

For exothermic forward reactions (ΔH negative), increasing temperature shifts equilibrium to the endothermic reverse direction, reducing the ratio of products to reactants and decreasing K. For endothermic forward reactions (ΔH positive), increasing temperature increases K, shifting equilibrium to the right.

📐 Worked Example

The forward reaction $2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g)$ has ΔH = -196 kJ mol⁻¹. State and explain the effect of increasing temperature on the value of Kp.

1
The forward reaction is exothermic, so the reverse reaction is endothermic.
2
Increasing temperature shifts equilibrium to the endothermic (left) direction to absorb excess heat.
3
This reduces the partial pressure of products relative to reactants, so the value of Kp decreases.

Exam tip:

When explaining equilibrium shifts, always link temperature changes to a change in K first, before describing the direction of shift.

4. Thermodynamic Link Between ΔStotal and K★★★★☆⏱ 10 min

📘 Definition

Entropy-equilibrium constant relationship

\Delta S_{total} = R \ln K

The total entropy change of a system and its surroundings is directly proportional to the natural logarithm of the equilibrium constant K.

A positive total entropy change means the reaction is thermodynamically feasible, so K > 1 and products are favored at equilibrium. A negative total entropy change means K < 1 and reactants are favored at equilibrium. R is given in the data booklet as 8.31 J K⁻¹ mol⁻¹.

📐 Worked Example

At 298 K, the total entropy change for a reaction is +125 J K⁻¹ mol⁻¹. Calculate the value of K at this temperature, using R = 8.31 J K⁻¹ mol⁻¹.

1
Rearrange the equation to isolate ln K:
2
$\ln K = \frac{\Delta S_{total}}{R} = \frac{125}{8.31} \approx 15.04$
3
Exponentiate both sides to solve for K:
4
$K = e^{15.04} \approx 3.4 \times 10^6$
5
This very large K value indicates the reaction proceeds almost to completion at 298 K.

5. Predicting Extent of Reaction Using K★★☆☆☆⏱ 8 min

The magnitude of K directly tells you how far a reaction proceeds at equilibrium, before the forward and reverse reaction rates are equal. This avoids needing to calculate equilibrium concentrations for quick estimations of reaction feasibility.

📐 Worked Example

The Kc value for the dissociation of ethanoic acid in water at 298 K is 1.7 × 10⁻⁵. State what this tells you about the extent of dissociation.

1
The Kc value is much smaller than 1, so reactants (undissociated ethanoic acid) are heavily favored at equilibrium.
2
Only a very small fraction of ethanoic acid molecules dissociate in water at 298 K.

6. Common Pitfalls

Wrong move:

Including pure solids/liquids in Kc/Kp expressions

Why:

Pure solids and liquids have constant activity, so they do not affect the equilibrium ratio

Correct move:

Omit all pure solid and liquid species when writing K expressions

Wrong move:

Using initial concentrations instead of equilibrium concentrations in K calculations

Why:

K is defined only for values measured at the equilibrium state

Correct move:

Use ICE tables to calculate equilibrium concentrations/partial pressures before substituting into K expressions

Wrong move:

Stating that pressure or concentration changes alter the value of K

Why:

Only temperature affects the value of K; other conditions shift equilibrium position to maintain the same K value

Correct move:

Always specify that K only changes when the temperature of the system changes

Wrong move:

Using kPa instead of atm for partial pressures in Kp calculations

Why:

Edexcel IAL specification requires Kp partial pressures to be measured in atm

Correct move:

Convert any given pressure units to atm before calculating Kp values

Wrong move:

Forgetting to raise concentration/partial pressure values to their stoichiometric coefficients

Why:

Exponents in K expressions match the stoichiometric ratios of the balanced reaction equation

Correct move:

Double check that each species in the K expression is raised to the power of its coefficient from the balanced equation

7. Quick Reference Cheatsheet

Concept	Rule/Formula	Key Exam Note
Kc Expression	$K_{c} = \frac{[ products ] ^{coeff}}{[ reactants ] ^{coeff}}$	Omit pure solids/liquids, use equilibrium concentrations in mol dm⁻³
Kp Expression	$K_{p} = \frac{p ( products ) ^{coeff}}{p ( reactants ) ^{coeff}}$	$p_{A} = x_{A} \times P_{t o t a l}$ , partial pressure units in atm
Effect on K	Only temperature changes K value	Concentration, pressure, catalysts do not alter K, only equilibrium position
Temperature & K	Exothermic forward: T↑ → K↓; Endothermic forward: T↑ → K↑	Explain shifts via change in K first to gain full marks
ΔStotal & K	$Δ S_{t o t a l} = R ln K$ , R = 8.31 J K⁻¹ mol⁻¹	$Δ S_{t o t a l} > 0 \to K > 1$ , products favored at equilibrium
Extent of Reaction	$K >> 1$ : ~complete; $K << 1$ : ~no reaction; $K \approx 1$ : both reactants and products present	Units of K are derived per reaction, no universal units exist

8. Frequently Asked

Do I include pure solids in Kc or Kp expressions?

No. Pure solids and pure liquids have constant activity values, so they are omitted entirely from equilibrium constant expressions for heterogeneous systems.

When does the value of the equilibrium constant K change?

K only changes when the temperature of the system changes. Changes to concentration, total pressure, or addition of a catalyst do not alter the value of K, only the position of equilibrium.

What units should I use for partial pressures in Kp calculations?

Edexcel IAL Chemistry requires partial pressures for Kp to be measured in atmospheres (atm). Convert any given kPa or Pa values to atm before substituting into Kp expressions.

Going deeper

What's Next

Now that you have mastered quantitative chemical equilibrium for Edexcel IAL Chemistry Unit 4, you are ready to move to the next core Unit 4 topic: ionic equilibria, including Kw, Ka, pH calculations and buffer systems. This topic builds directly on your understanding of equilibrium constants, applying the same K calculation rules to weak acid and base dissociation reactions. You will also encounter these equilibrium concepts again in Unit 6, where you will use K values to interpret experimental results from equilibrium practical investigations. Solidifying your understanding of Kc/Kp calculations and the temperature dependence of K will also help you tackle higher-mark extended response questions in your Unit 4 exam, which frequently ask you to combine equilibrium concepts with thermodynamic data to explain reaction behavior.