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Educerie · IB Diploma · Chemistry
Reactivity 2 How much, how fast and how far? · R2.3 How far? The extent of chemical change
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Describe the characteristics of physical and chemical systems at dynamic equilibrium | SL, HL | "Outline two characteristics of a system at dynamic equilibrium" (2 marks) |
| Deduce the equilibrium constant expression for a homogeneous reaction | SL, HL | "Deduce the expression for K" (1 mark) |
| Interpret the size of K, from K << 1 to K >> 1, and state that K depends on temperature | SL, HL | Paper 1A: "Which K shows the greatest extent of reaction?" |
| Relate the K values of a reaction and its reverse | SL, HL | "Calculate K for the reverse reaction at the same temperature" (1 mark) |
| Apply Le Châtelier's principle to changes in concentration, pressure and temperature, including the effect on K | SL, HL | "Predict and explain the effect of increasing the pressure on the yield" (2–3 marks) |
| Explain why a catalyst changes neither K nor the equilibrium composition | SL, HL | "State and explain the effect of a catalyst on the position of equilibrium" (2 marks) |
| Calculate Q and use it to predict the direction of reaction | HL only | "Deduce the direction in which the reaction proceeds" (2–3 marks) |
| Solve problems with K and initial and equilibrium concentrations | HL only | "Calculate the equilibrium concentration of…" (3–4 marks) |
| Use ΔG⦵ = −RT lnK | HL only | "Calculate K at 298 K from ΔG⦵" (2 marks) |
Before you start
You need rate and collision theory from Reactivity 2.2, especially that a catalyst lowers the activation energy for both directions, and the exothermic and endothermic signs of ΔH from Reactivity 1.1. For the HL sections you need moles and concentration from Reactivity 2.1, and the Gibbs energy change ΔG from Reactivity 1.4. Square brackets, [X], mean the concentration of X in mol dm⁻³.
1The idea in one paragraph
Many reactions are reversible: products can react to re-form the reactants. In a closed container the forward reaction slows as reactants are used up and the backward reaction speeds up as products build up, until the two rates are equal. From then on the concentrations stop changing, although both reactions carry on. That state is dynamic equilibrium, and where it settles is measured by the equilibrium constant, K: large K means mostly products, small K mostly reactants. K is fixed for a reaction at a given temperature. Change the concentrations, the pressure or the temperature, and the system shifts to oppose the change; only a change of temperature alters K itself.
2Dynamic equilibrium
A reversible reaction is written with the sign ⇌. Equilibrium can only be reached in a closed system, one that matter cannot enter or leave. In an open beaker a gaseous product escapes and the backward reaction never gets going.
A physical equilibrium. Put a little liquid bromine in a stoppered flask, as in Figure 1. Molecules escape from the liquid into the vapour, and vapour molecules return to the liquid. At first evaporation is faster, so the brown colour of the vapour deepens. As the vapour thickens, condensation speeds up, until the two rates are equal.
A chemical equilibrium. Hydrogen and iodine react at a high temperature to make hydrogen iodide, H₂(g) + I₂(g) ⇌ 2HI(g). Figure 2 follows the concentrations from a start of 1.00 mol dm⁻³ each of H₂ and I₂ in a sealed vessel, and Figure 3 follows the two rates.
The characteristics of any system at equilibrium, physical or chemical:
- The forward and backward rates are equal. Neither is zero: the reaction has not stopped.
- The concentrations of reactants and products are constant, but not equal to each other. In Figure 2, [HI] is seven times [H₂].
- Macroscopic properties are constant: colour, pressure, density, the level of a liquid.
- It needs a closed system.
- It can be reached from either direction. Starting from pure HI at the same temperature and the same total amount of atoms gives the same equilibrium mixture.
"Dynamic" is the word that earns the mark. Nothing appears to happen, because every molecule of HI made is balanced by one that breaks down.
3The equilibrium law
At equilibrium, a particular ratio of concentrations always has the same value at a given temperature, however you started. For a general homogeneous reaction:
a A + b B ⇌ c C + d D
K = [C]ᶜ[D]ᵈ ÷ [A]ᵃ[B]ᵇ
Products on top, reactants underneath, each concentration raised to the power of its coefficient in the balanced equation. This is the equilibrium law, and K is the equilibrium constant. A homogeneous equilibrium has everything in the same phase: all gases, or all in solution. That is the only kind whose K you will be asked to write.
Three examples:
| Equation | K expression |
|---|---|
| H₂(g) + I₂(g) ⇌ 2HI(g) | K = [HI]² ÷ ([H₂][I₂]) |
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | K = [NH₃]² ÷ ([N₂][H₂]³) |
| CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) | K = [CH₃COOC₂H₅][H₂O] ÷ ([CH₃COOH][C₂H₅OH]) |
Notice the difference from Reactivity 2.2. The powers in a K expression are the coefficients of the equation, because K describes the equilibrium mixture, not the route to it. The powers in a rate equation are not.
Worked example 1. Calculate K from the equilibrium concentrations in Figure 2.
In this course K is quoted without units.
4What the size of K tells you
K compares products with reactants at equilibrium, so its size is a measure of the extent of reaction. Figure 4 lays out the range.
| K | At equilibrium |
|---|---|
| K >> 1 (say above 10¹⁰) | products dominate; the reaction is effectively complete |
| K > 1 | products favoured; the position of equilibrium lies to the right |
| K = 1 | similar amounts of reactants and products (exactly equal only if the powers match) |
| K < 1 | reactants favoured; the position lies to the left |
| K << 1 (say below 10⁻¹⁰) | almost no reaction; reactants dominate |
K says nothing about how fast equilibrium is reached. A reaction with a huge K can be too slow to notice, which is the separate question of Reactivity 2.2.
K depends on temperature, and only on temperature. Every K value must be quoted with its temperature. Section 5 explains which way it moves.
The reverse reaction. Writing the equation backwards turns the expression upside down, so at the same temperature:
K(reverse) = 1 ÷ K(forward)
If K = 50.3 for H₂ + I₂ ⇌ 2HI, then for 2HI ⇌ H₂ + I₂ at the same temperature, K = 1 ÷ 50.3 = 0.0199. A reaction that favours products in one direction favours reactants when written the other way; the mixture itself is the same.
5Le Châtelier's principle
Le Châtelier's principle: if a system at equilibrium is disturbed, the position of equilibrium shifts in the direction that tends to oppose the change. "Position of equilibrium" means the composition of the mixture. "Shifts to the right" means more products form.
It is a prediction rule, not an explanation, and it only says which way. For each kind of change, ask two questions: which way does the position shift, and does K change?
Concentration. Add a reactant, and the system uses some of it up by shifting to the right. Remove a product, and it shifts right to replace some of it. Figure 5 shows extra hydrogen added to the equilibrium of Figure 2.
The new H₂ does not all react: its concentration settles higher than before, 0.44 instead of 0.22, but lower than the 0.52 just after it was added. That is what "tends to oppose" means. [I₂] falls and [HI] rises. And K is unchanged: (1.728)² ÷ (0.436 × 0.136) = 50.3. The system shifted precisely so that the ratio came back to K.
Pressure. This affects only reactions with gases, and only when the two sides have different numbers of moles of gas. Increasing the pressure (by squeezing the mixture into a smaller volume) shifts the position towards the side with fewer moles of gas, which reduces the pressure. In N₂(g) + 3H₂(g) ⇌ 2NH₃(g), four moles of gas become two, so high pressure favours ammonia. In H₂ + I₂ ⇌ 2HI, two moles become two, so pressure has no effect on the position. K is unchanged.
Temperature. This is the one change that alters K. Raising the temperature shifts the position in the endothermic direction, which absorbs heat and so opposes the rise; lowering it favours the exothermic direction. Figure 6 sketches what this does to K.
For an exothermic forward reaction (ΔH < 0), heating shifts the position to the left and K decreases. For an endothermic forward reaction (ΔH > 0), heating shifts it to the right and K increases. The sign of ΔH is quoted for the forward reaction, so read it before you decide.
A catalyst. No shift, and K is unchanged. A catalyst lowers the activation energy of the forward and backward reactions equally (Reactivity 2.2), so it speeds both up by the same factor. Equilibrium is reached sooner, as Figure 7 shows, but its position is exactly where it would have been.
| Change | Position of equilibrium | K |
|---|---|---|
| Add a reactant or remove a product | shifts right | unchanged |
| Increase the pressure | shifts to the side with fewer moles of gas | unchanged |
| Increase the temperature | shifts in the endothermic direction | changes: falls if forward is exothermic, rises if endothermic |
| Add a catalyst | no change; equilibrium reached faster | unchanged |
A heterogeneous example. Le Châtelier's principle also works across phases. A sealed bottle of fizzy drink holds the equilibrium CO₂(g) ⇌ CO₂(aq) under a high pressure of carbon dioxide. Open it, the pressure of CO₂ above the liquid falls, and the position shifts to the left: gas comes out of solution as bubbles. Dissolving a gas is exothermic, so a warm drink holds less CO₂ than a cold one, which is why it goes flat faster.
An industrial compromise. Ammonia is made by N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH⦵ = −92 kJ mol⁻¹. Le Châtelier says: high pressure (fewer moles of gas on the right) and low temperature (exothermic forward) for the best yield. But at low temperature the rate is far too slow. So plants use a moderate temperature of roughly 400 to 450 °C, trading some yield for a useful rate, a high pressure, limited by the cost and danger of the equipment, and an iron catalyst to speed up the approach to equilibrium. Unreacted gases are recycled. A "discuss the conditions" question wants exactly this tension between yield, rate and cost.
6Reading an equilibrium graph
A concentration–time graph like Figure 5 is the Paper 1B form of this topic. Read it in this order.
- Where are the lines flat? Flat means equilibrium.
- What happens at the disturbance? A vertical jump in one line means that substance was added or removed. All lines changing smoothly with no jump means a change of temperature. All concentrations jumping together, up or down, means a change of volume, and so of pressure.
- Which way do the lines then move? Products rising means a shift to the right.
- Is K the same afterwards? Calculate it from the new flat values. Only a temperature change gives a different K.
7HLThe reaction quotient, Q
SL students can skip to section 10.
The reaction quotient, Q, has exactly the same expression as K but uses the concentrations at any moment, not necessarily at equilibrium. Comparing Q with K tells you which way a mixture must react to reach equilibrium, as Figure 8 shows.
Q < K: reaction goes forward. Q > K: reaction goes backward. Q = K: at equilibrium.
If Q is smaller than K, there is too little product for equilibrium, so the forward reaction runs faster until Q rises to K.
Worked example 2. For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), K = 280 at a certain temperature (an illustrative value). A mixture contains [SO₂] = 0.50, [O₂] = 0.20 and [SO₃] = 1.2 mol dm⁻³. Deduce the direction of reaction.
Q is also the neatest way to explain section 5. Add hydrogen to the HI equilibrium and the denominator grows, so Q drops below K and the forward reaction runs until Q = K again. K has not moved; the mixture has.
8HLCalculating equilibrium compositions
The method is always a table of initial, change and equilibrium values, often called an ICE table. Work in concentrations (convert moles by dividing by the volume), and use the coefficients for the changes. Only homogeneous equilibria are assessed, and you are not expected to solve a quadratic.
Worked example 3: from one equilibrium value to K. 0.300 mol of N₂O₄ is sealed in a 2.00 dm³ flask and warmed: N₂O₄(g) ⇌ 2NO₂(g). At equilibrium, 0.120 mol of NO₂ is present. Calculate K.
| N₂O₄ | NO₂ | |
|---|---|---|
| Initial / mol dm⁻³ | 0.150 | 0 |
| Change / mol dm⁻³ | −0.030 | +0.060 |
| Equilibrium / mol dm⁻³ | 0.120 | 0.060 |
The change row is where the coefficients do their work: NO₂ rises twice as fast as N₂O₄ falls.
Worked example 4: from K to the composition, when a square root does the work. For H₂(g) + I₂(g) ⇌ 2HI(g), K = 64.0 at a lower temperature (illustrative). 0.100 mol dm⁻³ each of H₂ and I₂ are mixed. Calculate the equilibrium concentrations.
Let x mol dm⁻³ of H₂ react. Then [H₂] = [I₂] = 0.100 − x and [HI] = 2x.
Worked example 5: when K is very small. For N₂(g) + O₂(g) ⇌ 2NO(g), take K = 1.0 × 10⁻⁵ at a high temperature (illustrative). Air at that temperature starts with [N₂] = 0.80 and [O₂] = 0.20 mol dm⁻³. Calculate the equilibrium [NO].
Because K is tiny, hardly any N₂ or O₂ reacts, so their equilibrium concentrations are almost exactly their initial ones. This is the approximation the guide expects you to understand: [reactant]initial ≈ [reactant]eqm when K is very small.
Always say that you are making the approximation, and check it afterwards.
9HLK and Gibbs energy
Reactivity 1.4 showed that ΔG becomes zero at equilibrium, and that the standard Gibbs energy change is linked to K:
ΔG⦵ = −RT lnK, with ΔG⦵ in J mol⁻¹, R = 8.31 J K⁻¹ mol⁻¹, T in K
Both numbers measure the position of equilibrium. The sign of ΔG⦵ tells you which side K favours:
| ΔG⦵ | lnK | K | Position of equilibrium |
|---|---|---|---|
| negative | positive | K > 1 | products favoured |
| zero | 0 | K = 1 | neither favoured |
| positive | negative | K < 1 | reactants favoured |
Because K sits inside a logarithm, a modest ΔG⦵ gives an enormous K or a tiny one.
Worked example 6. (a) Calculate ΔG⦵ for H₂ + I₂ ⇌ 2HI at 700 K, where K = 50.3. (b) A reaction has ΔG⦵ = +33.0 kJ mol⁻¹ at 298 K. Calculate K.
Convert kJ to J before you divide. That one slip changes K by a factor of about e¹³.
10Where marks are lost
"At equilibrium the reaction has stopped." Both reactions continue at equal rates. Say "dynamic" and "rates equal".
"At equilibrium the concentrations are equal." They are constant, not equal.
Putting reactants on top, or leaving out the powers, in a K expression. Products over reactants, each raised to its coefficient.
Saying a catalyst shifts the equilibrium or increases the yield. It speeds up both directions equally: same position, same K, reached sooner.
Saying a change in concentration or pressure changes K. Only temperature changes K. The position moves; K stays.
Applying the pressure rule without counting moles of gas. Count gas moles on each side; liquids, solids and dissolved species do not count. Equal numbers means no effect.
Getting the temperature effect backwards. Heating favours the endothermic direction. For an exothermic forward reaction that means less product and a smaller K.
HL: using moles instead of concentrations in K, or forgetting to say you made the small-K approximation. Divide by the volume first; state and check the approximation.
11Draw it right
- Concentration–time graphs: time on the x-axis, concentration on the y-axis; reactant lines falling, product lines rising, all becoming horizontal at the same moment when equilibrium is reached. The lines level off at different heights.
- Changes in step with the equation: if 1 mol of H₂ reacts to give 2 mol of HI, the HI line rises twice as far as the H₂ line falls.
- Rate–time graphs: forward rate falling, backward rate rising, both meeting at one value that is not zero, then running together.
- A disturbance: a vertical jump only in the substance added or removed; then smooth curves to new flat levels. The added substance settles between its old value and its value just after the addition.
- A temperature change: no jumps, just smooth movement of every line to new flat levels.
- A catalyst: the same final levels, reached sooner. Never a different final level.
- HL: in an ICE table, label the units, keep changes in the ratio of the coefficients, and show the conversion from moles to concentration.
12Try it
Marks in brackets. Answers and marker's notes are at the end.
Q1. For the reaction 2NO₂(g) ⇌ N₂O₄(g), K = 8.8 at a certain temperature (illustrative).
(a) Deduce the equilibrium constant expression. 1 mark
(b) Calculate K for N₂O₄(g) ⇌ 2NO₂(g) at the same temperature, and state what it shows about the position of this equilibrium. 2 marks
Q2. Methanol is made industrially: CO(g) + 2H₂(g) ⇌ CH₃OH(g), ΔH⦵ = −91 kJ mol⁻¹.
(a) Predict and explain the effect of increasing the pressure on the equilibrium yield of methanol. 2 marks
(b) Predict the effect of increasing the temperature on the yield of methanol and on the value of K. 2 marks
(c) Explain why a catalyst is used even though it has no effect on the yield. 1 mark
Q3. 0.200 mol dm⁻³ of NO₂ is sealed in a flask at constant temperature and the reaction 2NO₂(g) ⇌ N₂O₄(g) is followed. At 10 min a change is made.
| Time / min | 0 | 2 | 4 | 6 | 8 | 10 | 10 (just after) | 12 | 14 | 16 | 18 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| [NO₂] / mol dm⁻³ | 0.200 | 0.140 | 0.112 | 0.102 | 0.100 | 0.100 | 0.160 | 0.132 | 0.122 | 0.119 | 0.119 |
| [N₂O₄] / mol dm⁻³ | 0 | 0.030 | 0.044 | 0.049 | 0.050 | 0.050 | 0.050 | 0.064 | 0.069 | 0.071 | 0.071 |
(a) State the time at which equilibrium was first reached, and calculate K at this temperature. 2 marks
(b) Identify the change made at 10 min. 1 mark
(c) Explain, using Le Châtelier's principle, the changes in concentration after 10 min, and show that K is unchanged. 3 marks
Q4 (HL). For H₂(g) + I₂(g) ⇌ 2HI(g), K = 50.3 at 700 K. A mixture at 700 K contains [H₂] = 0.10, [I₂] = 0.20 and [HI] = 0.60 mol dm⁻³. Determine the direction in which the reaction proceeds. 3 marks
Q5 (HL). 0.400 mol of PCl₅ is heated in a sealed 2.00 dm³ flask. At equilibrium 0.120 mol of Cl₂ is present. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Calculate K. 3 marks
Q6 (HL). A reaction has K = 2.5 × 10⁻³ at 500 K. Calculate ΔG⦵ at 500 K, in kJ mol⁻¹, and state what its sign tells you about the equilibrium mixture. 3 marks
13In one breath
A reversible reaction in a closed system reaches dynamic equilibrium when the forward and backward rates are equal: both reactions continue, and the concentrations and every visible property stay constant, though not equal to each other; physical changes such as evaporation behave the same way. For a homogeneous reaction the equilibrium law gives K = products over reactants, each raised to its coefficient. K measures the extent of reaction: K >> 1 is effectively complete, K << 1 barely starts, K = 1 is balanced. It depends only on temperature, and the reverse reaction has K equal to 1 divided by the forward K. Le Châtelier's principle: a disturbed equilibrium shifts to oppose the change. More reactant or less product shifts it right; higher pressure shifts it to fewer moles of gas; higher temperature shifts it in the endothermic direction, and that alone changes K, down for an exothermic reaction and up for an endothermic one. A catalyst changes neither the position nor K; it only gets there sooner. HL: Q has the form of K with any concentrations; Q < K goes forward, Q > K goes backward. Equilibrium problems use an initial–change–equilibrium table in concentrations; with a perfect square take the square root, and when K is tiny take reactant concentrations as unchanged and say so. ΔG⦵ = −RT lnK links the two measures: negative ΔG⦵ means K > 1 and products favoured.
Answers
Q1. (a) K = [N₂O₄] ÷ [NO₂]². (b) K = 1 ÷ 8.8 = 0.11 (0.114). K < 1, so as this equation is written the position of equilibrium lies to the left: N₂O₄ is favoured. It is the same mixture as in (a); reversing the equation only inverts K. 1 for the expression with the square; A1 for 0.11; 1 for linking K < 1 to the reactant side being favoured. A value of 8.8 or −8.8 for (b) scores 0.
Q2. (a) The yield increases. There are 3 mol of gas on the left and 1 mol on the right, so an increase in pressure shifts the position to the right, the side with fewer moles of gas, which reduces the pressure. (b) The forward reaction is exothermic, so raising the temperature shifts the position to the left, in the endothermic direction: the yield decreases and K decreases. (c) The catalyst increases the rate of both reactions, so equilibrium is reached faster and methanol is made more quickly at a lower, more economical temperature. for (a), 1 for the prediction, 1 for the moles-of-gas reason; for (b), 1 for yield falling, 1 for K falling; for (c), 1 for a faster rate or equilibrium reached sooner. "K increases with pressure" scores 0 wherever it appears.
Q3. (a) 8 min: the concentrations are constant from then on. K = [N₂O₄] ÷ [NO₂]² = 0.050 ÷ (0.100)² = 5.0. (b) NO₂ was added: its concentration jumps from 0.100 to 0.160 while [N₂O₄] does not change at that moment. (c) Adding NO₂ disturbs the equilibrium; the position shifts to the right to use up some of the added NO₂, so [NO₂] falls and [N₂O₄] rises until equilibrium is re-established at 16 min. [NO₂] settles at 0.119, above its old value but below 0.160, so the change is only partly opposed. New K = 0.071 ÷ (0.119)² = 5.0, unchanged because the temperature is constant. for (a), 1 for 8 min, 1 for K = 5.0; for (b), 1 for NO₂ added, with the jump as evidence; for (c), 1 for the shift right using up NO₂, 1 for the direction of both concentration changes, 1 for recalculating K = 5.0 and linking it to constant temperature.
Q4. Q = [HI]² ÷ ([H₂][I₂]) = (0.60)² ÷ (0.10 × 0.20) = 0.36 ÷ 0.020 = 18. Q < K (18 < 50.3), so there is too little HI: the reaction proceeds forward, [HI] increases and [H₂] and [I₂] decrease until Q = 50.3. M1 for the correct Q expression with substitution, A1 for 18, R1 for the forward direction with the comparison stated. A direction without a calculated Q scores 0.
Q5. Initial [PCl₅] = 0.400 ÷ 2.00 = 0.200 mol dm⁻³. [Cl₂] at equilibrium = 0.120 ÷ 2.00 = 0.060 mol dm⁻³, so [PCl₃] = 0.060 and [PCl₅] = 0.200 − 0.060 = 0.140 mol dm⁻³. K = (0.060 × 0.060) ÷ 0.140 = 0.026 (0.0257). M1 for converting to concentrations, M1 for the equilibrium concentration of PCl₅ from the 1 : 1 ratio, A1 for 0.026. Using moles throughout gives 0.051 and scores M1 only.
Q6. ΔG⦵ = −RT lnK = −8.31 × 500 × ln(2.5 × 10⁻³) = −8.31 × 500 × (−5.99) = +2.49 × 10⁴ J mol⁻¹ = +24.9 kJ mol⁻¹. Positive ΔG⦵ corresponds to K < 1: at equilibrium the reactants are favoured. M1 for substitution into ΔG⦵ = −RT lnK, A1 for +24.9 kJ mol⁻¹ with the sign, A1 for reactants favoured. An answer in J mol⁻¹ labelled as kJ mol⁻¹ loses the A1.
Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section R2.3 How far? The extent of chemical change. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.