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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover Reactivity 2.3, How far? The extent of chemical change (understandings 2.3.1 to 2.3.7), from the February 2023 IB Diploma Programme Chemistry guide, whose first exams ran in 2025. Questions 1 to 7 test content for both SL and HL. Questions 8 to 12 test understandings 2.3.5 to 2.3.7, which are HL only, and are labelled so.

To see where this unit sits in the course, use the IB DP chemistry hub or the printable checklist. Values of K given below apply only to these questions; use them as stated.

Questions

1. State two characteristics of a chemical system at dynamic equilibrium. [2]

2. A small volume of ethanol is placed in a stoppered flask kept at 25 °C. After a while, the pressure of ethanol vapour above the liquid stops changing. Explain, in terms of rates, how this equilibrium is established. [3]

3. (a) Deduce the equilibrium constant expression for 2NO(g) + 2H₂(g) ⇌ N₂(g) + 2H₂O(g). [1] (b) Deduce the equilibrium constant expression for Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq). [1] (c) Write the expression for the reverse of the reaction in (b). [1]

4. Four reactions, W to Z, have these values of K at 298 K: W, 5.0 × 10⁻²⁰; X, 0.40; Y, 1.0; Z, 8.5 × 10¹⁵.

(a) Describe the composition of the equilibrium mixture for each reaction. [4] (b) Calculate K at 298 K for the reverse of reaction X. [1]

5. Ethyne and hydrogen reach equilibrium with ethane in a sealed vessel: C₂H₂(g) + 2H₂(g) ⇌ C₂H₆(g). The forward reaction is exothermic. For each change, state and explain the effect on the amount of C₂H₆ at equilibrium, and state the effect on the value of K.

(a) More hydrogen is added at constant temperature. [2] (b) The temperature is raised. [2] (c) The gas mixture is squeezed into half its original volume, with no change in temperature. [2] (d) A catalyst is added. [2]

6. A closed flask contains aqueous ammonia with ammonia gas above it: NH₃(g) ⇌ NH₃(aq). Dissolving ammonia in water releases heat.

(a) Predict and explain the effect on the concentration of dissolved ammonia when more ammonia gas is pumped into the flask. [2] (b) State the effect of warming the flask on the concentration of dissolved ammonia. [1]

7. For a gas-phase reaction, K = 4.6 × 10³ at 300 K and K = 2.1 at 500 K. Deduce, with a reason, whether the forward reaction is exothermic or endothermic. [3]

8. (HL only) For N₂(g) + O₂(g) ⇌ 2NO(g), K = 4.0 × 10⁻⁴ at a set high temperature. At one moment a mixture at that temperature contains [N₂] = 0.50, [O₂] = 0.20 and [NO] = 0.0050 mol dm⁻³. Calculate Q, and deduce what happens to [NO] as the mixture moves to equilibrium. [4]

9. (HL only) 0.600 mol PCl₃ and 0.500 mol Cl₂ are sealed in a 5.00 dm³ vessel and heated: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g). At equilibrium the vessel holds 0.150 mol PCl₅. Calculate K at this temperature. [5]

10. (HL only) For 2BrCl(g) ⇌ Br₂(g) + Cl₂(g), K = 0.0625 at a set temperature. Pure BrCl at 0.300 mol dm⁻³ is allowed to reach equilibrium. Calculate the equilibrium concentrations of all three species. [5]

11. (HL only) For Cl₂(g) ⇌ 2Cl(g), use K = 2.5 × 10⁻⁷ at 1200 K. Chlorine at an initial concentration of 0.400 mol dm⁻³ is heated to 1200 K.

(a) Calculate [Cl] at equilibrium, stating the approximation you use. [3] (b) Show that the approximation is reasonable. [1] (c) Calculate ΔG⦵ for the reaction at 1200 K. Use R = 8.31 J K⁻¹ mol⁻¹. [2]

12. (HL only) An isomerisation has ΔG⦵ = −12.4 kJ mol⁻¹ at 350 K.

(a) Calculate K at 350 K. Use R = 8.31 J K⁻¹ mol⁻¹. [2] (b) Describe the equilibrium mixture. [1] (c) Explain why any reaction with K < 1 has a positive ΔG⦵. [1]

Answers

1. Choose from: forward and backward rates are equal; concentrations of reactants and products stay constant; both reactions continue; the system is closed. First correct characteristic [1]; second correct characteristic [1] Examiner insight: “Concentrations are equal” is wrong and contradicts an otherwise correct answer; write “constant”, not “equal”.

2. At first only evaporation happens, at a constant rate set by the temperature [1]; as ethanol molecules build up in the vapour, the rate of condensation increases [1]; equilibrium is reached when the rate of condensation equals the rate of evaporation, so the vapour pressure no longer changes [1] Examiner insight: Name both processes and compare their rates; “the liquid stops evaporating” contradicts the dynamic nature of equilibrium.

3. (a) K = [N₂][H₂O]² / ([NO]²[H₂]²) [1] (b) K = [[Ag(NH₃)₂]⁺] / ([Ag⁺][NH₃]²) [1] (c) K = [Ag⁺][NH₃]² / [[Ag(NH₃)₂]⁺] [1] Examiner insight: Each coefficient becomes a power; writing 2[NO] or 2[NH₃] instead of [NO]² or [NH₃]² loses the mark.

4. (a) W: K ≪ 1, essentially only reactants; reaction hardly occurs [1]; X: K < 1, reactants predominate [1]; Y: K = 1, neither reactants nor products favoured [1]; Z: K ≫ 1, reaction goes almost to completion, essentially all products [1] (b) 1 / 0.40 = 2.5 [1] Examiner insight: Describe the mixture, not the speed; “Z reacts quickly” earns nothing, because K gives no information about rate.

5. (a) Amount of C₂H₆ increases: the position shifts right to use up some of the added H₂ [1]; K unchanged [1] (b) Amount of C₂H₆ decreases: raising the temperature favours the endothermic, backward reaction [1]; K decreases [1] (c) Amount of C₂H₆ increases: halving the volume raises the pressure, and the position shifts right, to the side with fewer gas molecules (1 rather than 3) [1]; K unchanged [1] (d) No change in the amount of C₂H₆: a catalyst speeds up forward and backward reactions equally, so equilibrium is only reached sooner [1]; K unchanged [1] Examiner insight: For each change give direction, a reason and the K effect; a bare “shifts right” with no reason seldom gains full credit.

6. (a) Concentration of dissolved ammonia increases [1]; the higher concentration (partial pressure) of NH₃(g) shifts the position to the right, so more gas dissolves [1] (b) It decreases (the endothermic direction, ammonia leaving solution, is favoured) [1] Examiner insight: Use the information given; the heat released on dissolving tells you which direction is endothermic.

7. K is smaller at the higher temperature, so there is less product at equilibrium at 500 K [1]; raising the temperature favours the endothermic direction, which here must be the backward reaction [1]; so the forward reaction is exothermic [1] Examiner insight: State the link explicitly: a K that falls as T rises means an exothermic forward reaction; a one-word answer does not show the reasoning.

8. Q = [NO]² / ([N₂][O₂]) [1]; Q = 0.0050² / (0.50 × 0.20) = 2.5 × 10⁻⁴ [1]; Q < K [1]; net forward reaction, so [NO] increases until Q = K [1] Examiner insight: Square [NO]; without the power Q comes out as 0.050, which is greater than K and reverses the conclusion.

9. At equilibrium: PCl₃ = 0.600 − 0.150 = 0.450 mol and Cl₂ = 0.500 − 0.150 = 0.350 mol [1]; concentrations: [PCl₃] = 0.0900, [Cl₂] = 0.0700, [PCl₅] = 0.0300 mol dm⁻³ [1]; K = [PCl₅] / ([PCl₃][Cl₂]) [1]; K = 0.0300 / (0.0900 × 0.0700) [1] = 4.76 [1] Examiner insight: Divide by 5.00 dm³ before substituting; using moles here gives 0.952, five times too small.

10. Let [Br₂] = [Cl₂] = x, so [BrCl] = 0.300 − 2x [1]; K = x² / (0.300 − 2x)² = 0.0625 [1]; square root: x / (0.300 − 2x) = 0.250 [1]; x = 0.0750 − 0.500x, so x = 0.0500 [1]; [Br₂] = [Cl₂] = 0.0500 mol dm⁻³; [BrCl] = 0.200 mol dm⁻³ [1] Examiner insight: Two BrCl are used for each Br₂ formed; writing 0.300 − x in the change row is the commonest slip.

11. (a) K is very small, so assume [Cl₂]eqm ≈ [Cl₂]initial = 0.400 mol dm⁻³ [1]; [Cl]² = 2.5 × 10⁻⁷ × 0.400 = 1.0 × 10⁻⁷ [1]; [Cl] = 3.2 × 10⁻⁴ mol dm⁻³ [1] (b) Cl₂ used = half of [Cl] = 1.6 × 10⁻⁴ mol dm⁻³, which is about 0.04% of 0.400, so [Cl₂] is effectively unchanged [1] (c) ΔG⦵ = −8.31 × 1200 × ln(2.5 × 10⁻⁷) [1] = +152 kJ mol⁻¹ [1] Examiner insight: State the approximation in words before using it; also remember Cl₂ dissociates 1 : 2, so the Cl₂ used is half of [Cl].

12. (a) lnK = 12 400 / (8.31 × 350) = 4.263 [1]; K = e^4.263 = 71.0 [1] (b) K > 1, so products predominate in the equilibrium mixture [1] (c) The natural log of a number below 1 is negative, so −RT lnK is positive [1] Examiner insight: Convert −12.4 kJ to −12 400 J; leaving it in kJ gives K ≈ 1.004 and the wrong description.

Where marks are usually lost

• Describing equilibrium concentrations as “equal” when the correct word is “constant”.
• Using coefficients as multipliers, not powers, in K and Q expressions.
• Stating that K changes with concentration or pressure; only temperature changes it.
• Giving Le Châtelier shifts with no reason, or no comment on K.
• Counting moles of all species, not gas molecules only, when judging a pressure change.
• Claiming a catalyst raises the equilibrium yield.
• (HL only) Substituting moles into K for a vessel larger than 1 dm³.
• (HL only) Ignoring the mole ratio in the change row of an ICE table.
• (HL only) Using the small-K approximation without stating or checking it.
• (HL only) Putting kJ or °C into ΔG⦵ = −RT lnK.

Next steps

• Go back over the revision notes
• Re-read the study guide
• Link rates and catalysts with the How fast? practice questions
• See the IB DP chemistry hub and print the checklist
• Browse all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Every item here is based on the International Baccalaureate’s Chemistry guide for the Diploma Programme (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).