Equilibrium

Chemistry 1110 · CCEA A Level · 7 min read · Updated: Aug 31, 2026

Welcome to Equilibrium!

Hello and welcome to one of the most exciting and fundamental chapters in A2 Chemistry! In AS Chemistry, you learned about Le Chatelier's Principle and how reversible reactions can shift back and forth. Now, in A2, we take this a step further by putting actual numbers to these systems using equilibrium constants: \(K_c\) (concentrations) and \(K_p\) (partial pressures).

Don't worry if this seems a bit mathematical or tricky at first. We will break every single concept down into small, bite-sized steps with clear examples and easy-to-remember rules!

1. Dynamic Equilibrium: A Quick Refresher

Before we dive into calculations, let's remind ourselves what dynamic equilibrium really means.

Imagine you are walking up a "down" escalator. If you walk upwards at the exact same speed that the escalator is moving downwards, what happens? To someone watching from the side, you stay in the exact same spot! You are still moving, the escalator is still moving, but your overall position doesn't change.

A chemical reaction at dynamic equilibrium works the exact same way:

• The rate of the forward reaction equals the rate of the reverse reaction.
• The concentrations of reactants and products remain constant (they don't have to be equal to each other, just unchanging).
• It can only occur in a closed system (where no matter can enter or leave).

Homogeneous vs. Heterogeneous Equilibria

• Homogeneous equilibrium: All reactants and products are in the same physical state or phase. For example, all gases: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\).
• Heterogeneous equilibrium: The species are in different physical states or phases. For example, a solid and gases: \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\).

Quick Review: Dynamic equilibrium is not "static" — reactions are still happening continuously in both directions at equal rates!

2. The Equilibrium Constant in Terms of Concentration: \(K_c\)

For any general reversible reaction in a homogeneous system:
\(aA + bB \rightleftharpoons cC + dD\)

The equilibrium constant expression, \(K_c\), is written as:

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

Here, square brackets \([ \ ]\) mean equilibrium concentration in \(\text{mol dm}^{-3}\). The powers (\(a, b, c, d\)) are the balancing numbers (stoichiometry) from the chemical equation.

Golden Rule: Always put the products on top and the reactants on the bottom ("Products over Reactants").

How to Determine the Units of \(K_c\)

Units of \(K_c\) change depending on the reaction! Let's work them out step-by-step.

Example: Find the units of \(K_c\) for \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\).

Step 1: Write the \(K_c\) expression:
\(K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}\)

Step 2: Substitute the unit \(\text{mol dm}^{-3}\) into the expression:
\(\text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^3} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^4}\)

Step 3: Cancel out common terms:
\(\text{Units} = \frac{1}{(\text{mol dm}^{-3})^2} = (\text{mol dm}^{-3})^{-2} = \text{mol}^{-2}\text{dm}^{6}\)

Step-by-Step \(K_c\) Calculation (The ICE Table Method)

When solving equilibrium problems, always use the ICE layout to stay organised:

• I = Initial moles
• C = Change in moles
• E = Equilibrium moles

Worked Example:
Suppose \(2.00\text{ mol}\) of \(SO_2\) and \(2.00\text{ mol}\) of \(O_2\) are placed in a \(2.00\text{ dm}^3\) container and allowed to reach equilibrium at a given temperature:
\(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\)
At equilibrium, \(1.60\text{ mol}\) of \(SO_3\) is found. Let's calculate \(K_c\).

Step 1: Set up ICE for moles
• Initial moles: \(SO_2 = 2.00\), \(O_2 = 2.00\), \(SO_3 = 0.00\)
• Change in moles: Since \(1.60\text{ mol}\) of \(SO_3\) formed (\(+1.60\)), then according to the \(2:1:2\) molar ratio, \(SO_2\) used is \(1.60\text{ mol}\) (\(-1.60\)) and \(O_2\) used is \(0.80\text{ mol}\) (\(-0.80\)).
• Equilibrium moles:
\(n(SO_2) = 2.00 - 1.60 = 0.40\text{ mol}\)
\(n(O_2) = 2.00 - 0.80 = 1.20\text{ mol}\)
\(n(SO_3) = 1.60\text{ mol}\)

Step 2: Convert moles to concentrations (\(\text{concentration} = \frac{\text{moles}}{\text{volume in dm}^3}\))
\([SO_2] = \frac{0.40}{2.00} = 0.20\text{ mol dm}^{-3}\)
\([O_2] = \frac{1.20}{2.00} = 0.60\text{ mol dm}^{-3}\)
\([SO_3] = \frac{1.60}{2.00} = 0.80\text{ mol dm}^{-3}\)

Step 3: Calculate \(K_c\)
\(K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]} = \frac{(0.80)^2}{(0.20)^2 \times 0.60} = \frac{0.64}{0.040 \times 0.60} = \frac{0.64}{0.024} = 26.67\text{ mol}^{-1}\text{dm}^3\)

Common Mistake to Avoid: Forgetting to divide equilibrium moles by volume before putting them into \(K_c\)! (Note: If the number of moles on both sides of the balanced equation is identical, the volume terms cancel out, but it is always best practice to calculate concentrations first).

3. The Equilibrium Constant in Terms of Partial Pressure: \(K_p\)

When all reactants and products are gases, it is much more convenient to measure pressure instead of concentration. This gives us \(K_p\).

Understanding Mole Fraction and Partial Pressure

Let's imagine a pizza shared among friends:

• Mole Fraction (\(x_A\)): The fraction of the total moles of gas that is gas \(A\).
\(x_A = \frac{\text{moles of gas } A}{\text{total moles of gas}}\)
Note: The sum of all mole fractions always equals \(1.0\).

• Partial Pressure (\(p_A\)): The pressure that a single gas in a mixture would exert if it occupied the entire container by itself.
\(p_A = x_A \times P_{\text{total}}\)
Where \(P_{\text{total}}\) is the total pressure of the system.

Writing the \(K_p\) Expression

For the gas-phase reaction: \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\)

\(K_p = \frac{(pC)^c(pD)^d}{(pA)^a(pB)^b}\)

Crucial Tip: Use round brackets \(( \ )\) or simply \(pA\), NOT square brackets \([ \ ]\). Square brackets mean concentration (\(\text{mol dm}^{-3}\)), whereas \(p\) represents partial pressure (e.g. \(\text{kPa}\) or \(\text{atm}\)).

Worked Example: Calculating \(K_p\)

Consider the equilibrium: \(PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)\)
At a total pressure of \(150\text{ kPa}\), the equilibrium mixture contains \(0.20\text{ mol}\) of \(PCl_5\), \(0.80\text{ mol}\) of \(PCl_3\), and \(0.80\text{ mol}\) of \(Cl_2\).

Step 1: Calculate Total Moles
\(\text{Total moles} = 0.20 + 0.80 + 0.80 = 1.80\text{ mol}\)

Step 2: Calculate Mole Fractions
\(x(PCl_5) = \frac{0.20}{1.80} = 0.111\)
\(x(PCl_3) = \frac{0.80}{1.80} = 0.444\)
\(x(Cl_2) = \frac{0.80}{1.80} = 0.444\)

Step 3: Calculate Partial Pressures (\(p = x \times P_{\text{total}}\))
\(p(PCl_5) = 0.111 \times 150\text{ kPa} = 16.67\text{ kPa}\)
\(p(PCl_3) = 0.444 \times 150\text{ kPa} = 66.67\text{ kPa}\)
\(p(Cl_2) = 0.444 \times 150\text{ kPa} = 66.67\text{ kPa}\)

Step 4: Substitute into \(K_p\) Expression and find units
\(K_p = \frac{p(PCl_3) \times p(Cl_2)}{p(PCl_5)} = \frac{66.67 \times 66.67}{16.67} = 266.7\text{ kPa}\)

Units: \(\frac{\text{kPa} \times \text{kPa}}{\text{kPa}} = \text{kPa}\)

Key Takeaway: Partial pressure is simply a gas's "share" of the total pressure based on its mole fraction!

4. Factors Affecting Equilibrium and Equilibrium Constants

Students often get confused between what shifts the position of equilibrium and what changes the actual value of \(K\). Here is the golden rule to remember for your exams:

ONLY TEMPERATURE changes the numerical value of \(K_c\) and \(K_p\).

1. Effect of Temperature

To predict what happens, look at the sign of \(\Delta H\) for the forward reaction:

• For an Exothermic Reaction (\(\Delta H < 0\)):
- Increasing temperature shifts equilibrium to the left (endothermic direction to absorb added heat).
- Reactant amounts increase, product amounts decrease \(\implies\) value of \(K\) decreases.
- Decreasing temperature shifts equilibrium to the right \(\implies\) value of \(K\) increases.

• For an Endothermic Reaction (\(\Delta H > 0\)):
- Increasing temperature shifts equilibrium to the right (to absorb added heat).
- Product amounts increase, reactant amounts decrease \(\implies\) value of \(K\) increases.
- Decreasing temperature shifts equilibrium to the left \(\implies\) value of \(K\) decreases.

2. Effect of Concentration and Pressure

• Changing concentration or pressure shifts the position of equilibrium (according to Le Chatelier's Principle) to keep the ratio of products to reactants constant.
• The value of \(K_c\) and \(K_p\) DOES NOT CHANGE.

3. Effect of a Catalyst

• A catalyst speeds up both the forward and reverse reactions equally by providing an alternative reaction pathway with a lower activation energy (\(E_a\)).
• Therefore, a catalyst allows equilibrium to be reached faster.
• A catalyst has NO effect on the position of equilibrium and NO effect on the value of \(K_c\) or \(K_p\).

Memory Trick: "Temperature alters the Constant; Catalysts and Pressure keep it Constant!"

5. Summary Checklist for Revision

Before sitting your exam, make sure you can confidently:

• Write the expression for \(K_c\) and \(K_p\) for any balanced equation.
• Deduce the correct units for \(K_c\) and \(K_p\) by algebraic cancellation.
• Use an ICE table to calculate equilibrium quantities from initial amounts.
• Calculate mole fractions and partial pressures given total pressure and mole numbers.
• State clearly that only a change in temperature alters the value of \(K_c\) or \(K_p\).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are the key characteristics of a chemical reaction at dynamic equilibrium?

The rate of the forward reaction equals the rate of the reverse reaction, the concentrations of reactants and products remain constant, and the system is closed.

What is the difference between a homogeneous equilibrium and a heterogeneous equilibrium?

In a homogeneous equilibrium, all species are in the same physical state; in a heterogeneous equilibrium, species are in different physical states.

How is the equilibrium constant in terms of concentration, \(K_c\), expressed for a general reversible reaction?

\(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\) for the reaction \(aA + bB \rightleftharpoons cC + dD\).

How are the units of \(K_c\) determined for a given chemical equilibrium?

By substituting \(\text{mol dm}^{-3}\) into the \(K_c\) expression for each concentration term and simplifying algebraically.

What are the three steps in the ICE method used for equilibrium calculations?

Listing Initial moles, calculating Change in moles using stoichiometry, and finding Equilibrium moles before converting to concentrations.

How are mole fraction and partial pressure calculated in a gaseous mixture?

Mole fraction is \(x_A = \frac{\text{moles of } A}{\text{total moles}}\), and partial pressure is \(p_A = x_A \times P_{\text{total}}\).

How is the equilibrium constant in terms of partial pressures, \(K_p\), written?

\(K_p = \frac{(pC)^c(pD)^d}{(pA)^a(pB)^b}\) for the reaction \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\).

What is the only factor that alters the numerical value of an equilibrium constant (\(K_c\) or \(K_p\))?

Temperature.

How does changing temperature affect the numerical value of \(K\) for exothermic and endothermic reactions?

For an exothermic reaction (\(\Delta H < 0\)), increasing temperature decreases \(K\); for an endothermic reaction (\(\Delta H > 0\)), increasing temperature increases \(K\).

How does the addition of a catalyst affect dynamic equilibrium and the value of \(K_c\) / \(K_p\)?

A catalyst increases the rates of the forward and reverse reactions equally, reaching equilibrium faster without changing the equilibrium position or the value of \(K\).

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