IB DP Chemistry Reactivity 2.3: How far? The extent of chemical change

This guide teaches Reactivity 2.3, How far? The extent of chemical change, based on the IB Diploma Programme Chemistry guide (February 2023, first assessment 2025). All seven understandings, Reactivity 2.3.1 to 2.3.7, are taught here. Understandings 2.3.1 to 2.3.4 are studied at both SL and HL; 2.3.5 to 2.3.7 belong to the additional higher level material and are HL only.

For the whole course, see the IB DP chemistry hub or print the checklist. Our free diagnostics show which topics need more work.

What this unit covers

Understanding	What you must be able to do	SL/HL
Reactivity 2.3.1	Describe a physical or chemical system at dynamic equilibrium in a closed system	SL and HL
Reactivity 2.3.2	Deduce the equilibrium constant expression, K, for a homogeneous reaction	SL and HL
Reactivity 2.3.3	Link the size of K to the extent of reaction; relate K for a reaction and its reverse; know K depends on temperature	SL and HL
Reactivity 2.3.4	Apply Le Châtelier’s principle to changes in concentration, temperature and pressure, including heterogeneous cases such as X(g) ⇌ X(aq), and say what happens to K	SL and HL
Reactivity 2.3.5	Calculate the reaction quotient, Q, and predict the direction of change	HL only
Reactivity 2.3.6	Solve problems linking K with initial and equilibrium concentrations, including the small-K approximation	HL only
Reactivity 2.3.7	Use ΔG⦵ = −RT lnK to connect Gibbs energy with the position of equilibrium	HL only

Dynamic equilibrium (2.3.1)

Many reactions are reversible: products can react to re-form reactants. In a closed system (no matter enters or leaves), a reversible change eventually reaches dynamic equilibrium.

At the start, only reactants are present, so the forward rate is high and the backward rate is zero. As the reactant concentrations drop, the forward reaction slows down; as product concentrations grow, the backward reaction speeds up. Equilibrium is reached when the two rates are equal.

Characteristics of a system at equilibrium:

• The forward and backward reactions both continue, at equal rates. That is why it is called dynamic.
• The concentrations of all reactants and products stay constant. They are not usually equal to each other.
• Macroscopic properties (colour, pressure, density) no longer change.
• It needs a closed system and is reached from either direction.

Physical equilibrium. Put a little liquid bromine in a stoppered flask at constant temperature: Br₂(l) ⇌ Br₂(g). At first, evaporation is faster than condensation, so the brown colour above the liquid deepens. As more Br₂ molecules collect in the vapour, the condensation rate rises until it equals the evaporation rate. The colour then stays constant, although molecules still cross the surface in both directions.

Chemical equilibrium. Consider the gas-phase synthesis of methanol, CO(g) + 2H₂(g) ⇌ CH₃OH(g). In a sealed vessel, the amount of methanol rises and then levels off, while CH₃OH molecules continue to form and break down.

The equilibrium law (2.3.2)

For a homogeneous reaction (all species in the same phase), written as aA + bB ⇌ cC + dD, the equilibrium law gives:

K = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

Each square-bracket term is a concentration at equilibrium, in mol dm⁻³. The numerator holds the products and the denominator the reactants; the coefficient of each species from the balanced equation becomes its exponent.

Worked example 1: writing K

(a) CO(g) + 2H₂(g) ⇌ CH₃OH(g)

K = [CH₃OH] / ([CO][H₂]²)

(b) CO₂(g) + 3H₂(g) ⇌ CH₃OH(g) + H₂O(g)

K = [CH₃OH][H₂O] / ([CO₂][H₂]³)

(c) HCOOH(l) + C₂H₅OH(l) ⇌ HCOOC₂H₅(l) + H₂O(l)

K = [HCOOC₂H₅][H₂O] / ([HCOOH][C₂H₅OH])

Here every species is in the same liquid mixture, so water appears in the expression.

The size of K (2.3.3)

K tells you how far a reaction has gone when equilibrium is reached at a given temperature.

Value of K	Extent of reaction at equilibrium
K ≪ 1	Almost no reaction; mixture is nearly all reactants
K < 1	Reactants predominate
K = 1	Neither side is strongly favoured
K > 1	Products predominate
K ≫ 1	Reaction goes almost to completion; mixture is nearly all products

K says nothing about how fast equilibrium is reached. A reaction with a very large K can still be extremely slow.

K depends on temperature. Each value of K applies only at one temperature. Concentration, pressure and catalysts do not change K; temperature does.

Reverse reactions. Writing the equation the other way round turns the expression upside down, so at the same temperature:

K(reverse) = 1 / K(forward)

Worked example 2: reversing an equation

At a certain temperature, K for CO(g) + 2H₂(g) ⇌ CH₃OH(g) is 40.0. Find K for CH₃OH(g) ⇌ CO(g) + 2H₂(g) at the same temperature.

K(reverse) = 1 / 40.0 = 0.0250

The forward value is above 1, so methanol is favoured; the reverse value is below 1, which says the same thing from the other direction.

Le Châtelier’s principle (2.3.4)

Le Châtelier’s principle: if you change the conditions of a system at equilibrium, its position moves in the direction that partly undoes your change.

Change	Shift in position	Effect on K
Add a reactant, or remove a product	Towards products	None
Add a product, or remove a reactant	Towards reactants	None
Increase pressure (decrease volume)	Towards the side with fewer gas molecules	None
Raise temperature	In the endothermic direction	Rises if forward is endothermic; falls if forward is exothermic
Add a catalyst	No shift	None

A catalyst lowers the barrier for both directions, so both rates rise by the same factor. The mixture arrives at equilibrium sooner, but its final composition is identical.

Why concentration changes leave K alone. Suppose extra reactant is injected into an equilibrium mixture. The value of the expression drops below K, so the forward reaction outpaces the backward one until the ratio climbs back to K. The new mixture holds more product, yet K is the same number.

Why temperature changes K. For an exothermic forward reaction, raising the temperature favours the backward (endothermic) reaction. Product concentrations fall, reactant concentrations rise, and the ratio, K, gets smaller.

Worked example 3: applying the principle

Chlorine can be made by oxidising hydrogen chloride with oxygen over a catalyst:

4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g) forward reaction exothermic

(a) Pressure increased by compressing the mixture. Left side: 5 gas molecules; right side: 4. The position shifts to the right, making more Cl₂, because that reduces the number of gas molecules. K is unchanged.

(b) Temperature raised. The forward reaction is exothermic, so the backward (endothermic) reaction is favoured. Less Cl₂ is present at the new equilibrium and K decreases.

(c) Extra O₂ is pumped in. The position shifts to the right to use some of it up, so more Cl₂ forms. K is unchanged.

(d) Catalyst added. No change in the equilibrium amounts; equilibrium is reached faster.

Heterogeneous equilibria: X(g) ⇌ X(aq)

The guide notes that the principle also applies when species are in different phases, such as a gas dissolving in water. Take a closed bottle of water with oxygen above it:

O₂(g) ⇌ O₂(aq) (dissolving is exothermic)

• More O₂ pressure above the water: the position shifts right, so more oxygen dissolves.
• Warmer water: the endothermic direction (O₂ leaving solution) is favoured, so less oxygen stays dissolved. This is one reason warm water holds less dissolved oxygen.

The reaction quotient, Q (2.3.5, HL only)

Q has the same form as the K expression, but you put in the concentrations at any moment, not just at equilibrium. Comparing Q with K tells you which way the mixture will move.

Comparison	What happens
Q < K	Too few products: net forward reaction until Q = K
Q = K	Already at equilibrium
Q > K	Too many products: net backward reaction until Q = K

Worked example 4: predicting direction

2NO(g) + Br₂(g) ⇌ 2NOBr(g). At a given temperature, K = 20.0. A mixture contains [NO] = 0.200, [Br₂] = 0.100 and [NOBr] = 0.300 mol dm⁻³.

Q = [NOBr]² / ([NO]²[Br₂]) = 0.300² / (0.200² × 0.100) = 0.0900 / 0.00400 = 22.5

Q > K, so the reaction proceeds in the backward direction: some NOBr breaks down until Q falls to 20.0.

Calculating equilibrium composition (2.3.6, HL only)

Use an ICE table (Initial, Change, Equilibrium) and work in concentrations. The changes must follow the mole ratio in the equation. The guide says quadratic equations are not expected and only homogeneous equilibria are assessed.

Worked example 5: finding K from one equilibrium amount

0.400 mol CO and 1.00 mol H₂ are sealed in a 2.00 dm³ vessel. At equilibrium, 0.120 mol CH₃OH is present. Calculate K.

	CO	H₂	CH₃OH
Initial / mol	0.400	1.00	0
Change / mol	−0.120	−0.240	+0.120
Equilibrium / mol	0.280	0.760	0.120
Equilibrium / mol dm⁻³	0.140	0.380	0.0600

K = 0.0600 / (0.140 × 0.380²) = 0.0600 / 0.020216 = 2.97

Worked example 6: finding concentrations from K

SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g), K = 2.25 at a set temperature. SO₂ and NO₂ both start at 0.200 mol dm⁻³. Let x = [SO₃] = [NO] at equilibrium.

K = x² / (0.200 − x)² = 2.25

Both sides are perfect squares, so take the square root: x / (0.200 − x) = 1.50

x = 0.300 − 1.50x, so 2.50x = 0.300 and x = 0.120

At equilibrium: [SO₃] = [NO] = 0.120 mol dm⁻³; [SO₂] = [NO₂] = 0.0800 mol dm⁻³.

Worked example 7: the small-K approximation

COCl₂(g) ⇌ CO(g) + Cl₂(g), K = 4.0 × 10⁻⁸ at a set temperature. [COCl₂] starts at 0.250 mol dm⁻³.

K is very small, so very little COCl₂ breaks down: take [COCl₂]eqm ≈ [COCl₂]initial = 0.250.

x² / 0.250 = 4.0 × 10⁻⁸, so x² = 1.0 × 10⁻⁸ and x = 1.0 × 10⁻⁴ mol dm⁻³ = [CO] = [Cl₂]

Check: 1.0 × 10⁻⁴ is 0.04% of 0.250, so ignoring it was fine. The same idea is used for weak acids in Reactivity 3.1.

Gibbs energy and K (2.3.7, HL only)

Both ΔG⦵ and K measure the position of equilibrium. They are linked by:

ΔG⦵ = −RT lnK

The guide states this equation is given in the data booklet. Work with ΔG⦵ in J mol⁻¹, temperature in kelvin and R = 8.31 J K⁻¹ mol⁻¹.

ΔG⦵	lnK	K	Equilibrium mixture
Negative	Positive	> 1	Products favoured
Zero	0	1	Neither favoured
Positive	Negative	< 1	Reactants favoured

Worked example 8: both directions

(a) ΔG⦵ = −5.70 kJ mol⁻¹ at 298 K. Find K.

lnK = −ΔG⦵ / RT = 5700 / (8.31 × 298) = 2.302, so K = e^2.302 = 9.99 (about 10)

(b) K = 2.5 × 10⁻⁴ at 500 K. Find ΔG⦵.

ΔG⦵ = −8.31 × 500 × ln(2.5 × 10⁻⁴) = +34.5 kJ mol⁻¹ (to 3 s.f.). Positive, matching K < 1.

The background on ΔG is in the entropy and spontaneity guide.

Common errors

• Saying concentrations are equal at equilibrium. It is the rates that are equal.
• Saying the reactions stop at equilibrium.
• Using coefficients as multipliers instead of powers in K.
• Claiming that adding a reactant or raising pressure changes K. Only temperature does.
• Stating a catalyst “shifts equilibrium to the right”.
• Using moles instead of concentrations when the vessel is not 1 dm³.
• Comparing Q with K the wrong way round.
• Putting kJ, or °C, into ΔG⦵ = −RT lnK.

Where to go next

Try the practice questions and use the revision notes for quick review. Rates and catalysts are covered in the How fast? study guide.

Official syllabus

Aligned with the IB’s DP Chemistry guide (International Baccalaureate; issued February 2023; first assessment 2025). This page follows Reactivity 2.3, How far? The extent of chemical change, understandings 2.3.1 to 2.3.4 (SL and HL) and 2.3.5 to 2.3.7 (HL only).