IB DP Chemistry Reactivity 2.2: How fast? Rates and collision theory – 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 2.2.1 to 2.2.5 (rates, collision theory, factors affecting rate, activation energy and catalysts) from the February 2023 IB Diploma Programme Chemistry guide, whose first exams ran in 2025. All of this content is studied at both SL and HL; the HL-only understandings that follow in the guide are not tested here.

To see where this unit sits in the course, use the IB DP chemistry hub or the printable checklist.

Questions

1. (a) Define rate of reaction. [1] (b) State a suitable unit for rate when concentration is in mol dm⁻³ and time is in seconds. [1]

2. A cylinder of neon is heated from 7 °C to 147 °C. Calculate how many times greater the average kinetic energy of the neon atoms becomes, and state the relationship you used. [3]

3. Explain, using collision theory, why only a small fraction of the collisions between reactant particles in a gas lead to reaction. [3]

4. Calcium turnings react with excess water. A gas syringe measures the hydrogen given off.

Time / s	0	20	40	60	80	100	120
H₂ collected / cm³	0	18.0	31.0	40.0	46.0	49.5	51.0

(a) Calculate the average rate of reaction over the first 40 s. [1] (b) A tangent to the curve at 40 s passes through (0 s, 9.0 cm³) and (80 s, 53.0 cm³). Calculate the rate at 40 s in mol of H₂ per second. Take the molar volume of a gas under these conditions as 24.5 dm³ mol⁻¹. [3] (c) Suggest why the gradient of the curve falls with time. [2]

5. Potassium carbonate solution is poured into excess dilute hydrochloric acid in an open beaker standing on a balance. The reading is recorded every 15 s.

(a) Explain why the reading falls, and why this method would be poor for following the reaction of magnesium with an acid. [2] (b) State whether time is the independent or dependent variable in this experiment. Describe a rate experiment in which it is the other kind. [2]

6. For the reaction 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g), ammonia is used up at 2.4 × 10⁻³ mol dm⁻³ s⁻¹. Calculate the rate at which O₂ is used up and the rate at which H₂O forms. [2]

7. Predict and explain the effect on the rate of each change.

(a) Iron wool is used instead of an iron nail of the same mass, burning in pure oxygen. [2] (b) A sealed steel vessel of NO and O₂ is topped up with more of the same mixture, at constant temperature, until the pressure doubles. [2] (c) Hydrochloric acid reacting with zinc is diluted with an equal volume of water. [2]

8. (a) Sketch Maxwell–Boltzmann energy distribution curves for a gas at temperature T₁ and at a higher temperature T₂. Label the axes and mark the activation energy, Ea. [3] (b) With reference to your sketch, explain why heating speeds up the reaction. [2] (c) For one reaction, a rise from 293 K to 303 K doubles the rate. Show that the average kinetic energy goes up by less than 4%, and explain why the rate rises so much more. [2]

9. An endothermic reaction has Ea = 185 kJ mol⁻¹ and ΔH = +52 kJ mol⁻¹. With a catalyst, Ea = 110 kJ mol⁻¹.

(a) Sketch one energy profile showing the reaction with and without the catalyst. Label ΔH and both activation energies. [3] (b) Calculate the activation energy of the reverse reaction with and without the catalyst. [2] (c) Explain, using a Maxwell–Boltzmann curve, how the catalyst increases the rate. [2] (d) State and explain the effect of the catalyst on ΔH and on the position of equilibrium. [2]

10. Acidified potassium manganate(VII) is mixed with ethanedioic acid, and the time for the purple colour to disappear is recorded at different temperatures.

Temperature / °C	25	35	45	55	65
Time / s	400	200	160	56	30

(a) State two variables that must be controlled. [2] (b) Explain why 1/time is proportional to the rate in this experiment. [1] (c) Identify the anomalous result and suggest one cause. [2] (d) The thermometer used reads 2 °C too high throughout. State the type of error and how it would appear on a graph of 1/time against temperature. [2]

11. An uncapped spirit burner of ethanol can stand in air at room temperature without burning, even though ethanol combustion is strongly exothermic.

(a) Explain this observation. [2] (b) Suggest why a high activation energy is a useful property of a fuel. [1] (c) Alkanes are described as kinetically stable but thermodynamically unstable. Explain both terms. [2] (d) State the term for a biological catalyst. [1]

Answers

1. (a) How fast the concentration of a named reactant or product changes, per unit time [1] (b) mol dm⁻³ s⁻¹ [1] Examiner insight: Name it as a change in concentration per unit time; “how fast a reaction goes” restates the question and earns nothing.

2. Convert to kelvin: 280.15 K and 420.15 K [1]; ratio = 420.15/280.15 = 1.50 [1]; average kinetic energy is directly proportional to the absolute (kelvin) temperature [1] Examiner insight: Dividing the Celsius values gives 21, which is meaningless; convert to kelvin before any comparison of kinetic energies.

3. A collision must have energy at least equal to the activation energy [1]; most particles have energies below Ea, so most collisions lack enough energy [1]; the particles must also collide with the proper orientation (geometry) [1] Examiner insight: Two separate conditions are expected here; an answer about energy alone is incomplete.

4. (a) 31.0/40 = 0.775 cm³ s⁻¹ [1] (b) Gradient = (53.0 − 9.0)/(80 − 0) = 0.55 cm³ s⁻¹ [1]; 0.55 cm³ = 5.5 × 10⁻⁴ dm³ [1]; 5.5 × 10⁻⁴/24.5 = 2.2 × 10⁻⁵ mol s⁻¹ [1] (c) Calcium is used up, so less of its surface is exposed to water molecules [1]; collisions at the surface become less frequent, so the gradient falls [1] Examiner insight: Read gradient points off the tangent, not the table, and convert cm³ to dm³ before using the molar volume.

5. (a) Carbon dioxide is produced and escapes into the air [1]; hydrogen has a very small molar mass, so the mass lost would be too small to measure precisely [1] (b) Time is the independent variable here (readings at chosen times) [1]; it is the dependent variable when you measure the time for a fixed change, for example the time for a colour to disappear [1] Examiner insight: Name the gas in part (a); the second point is the tiny mass of hydrogen lost, not that hydrogen is flammable.

6. O₂: 2.4 × 10⁻³ × 5/4 = 3.0 × 10⁻³ mol dm⁻³ s⁻¹ [1]; H₂O: 2.4 × 10⁻³ × 6/4 = 3.6 × 10⁻³ mol dm⁻³ s⁻¹ [1] Examiner insight: Scale by the ratio of coefficients (O₂ to NH₃ is 5 : 4); inverting the ratio is the usual slip.

7. (a) Rate increases [1]; iron wool has a much larger surface area, so more iron atoms are exposed and collisions with O₂ are more frequent [1] (b) Rate increases [1]; twice as many gas particles occupy the same volume, so collisions are more frequent [1] (c) Rate decreases [1]; the acid concentration halves, so there are fewer H⁺ ions per unit volume and less frequent collisions with zinc [1] Examiner insight: Each explanation should say “more (or less) frequent collisions”; “more collisions” with no reference to time does not describe a rate.

8. (a) Axes labelled kinetic energy and number of particles, both curves starting at the origin [1]; T₂ curve with a lower peak shifted to higher energy, tails approaching but not meeting the x-axis [1]; Ea marked to the right of both peaks [1] (b) The area under the T₂ curve beyond Ea is larger, so a greater fraction of particles have E ≥ Ea [1]; there are more frequent successful collisions [1] (c) 303/293 = 1.034, a rise of about 3.4% [1]; the fraction of collisions with E ≥ Ea increases much more than the average energy, which is the main reason the rate doubles [1] Examiner insight: The two curves must enclose the same area; a hotter curve drawn taller or bigger everywhere suggests more particles and loses credit.

9. (a) Products drawn above reactants, with ΔH labelled as the upward gap [1]; uncatalysed hump with Ea measured from the reactant level [1]; lower catalysed hump between the same start and finish levels [1] (b) Uncatalysed: 185 − 52 = 133 kJ mol⁻¹ [1]; catalysed: 110 − 52 = 58 kJ mol⁻¹ [1] (c) On one curve, Ea(cat) lies to the left of Ea [1]; a larger area lies beyond Ea(cat), so a greater fraction of collisions are successful [1] (d) ΔH is unchanged, because the reactant and product energies are the same [1]; the equilibrium position is unchanged, because both barriers fall by the same amount (75 kJ mol⁻¹) and forward and backward rates rise equally [1] Examiner insight: For an endothermic reaction the reverse barrier is Ea(forward) − ΔH; adding ΔH, as you would for an exothermic reaction, is a common error.

10. (a) Any two from: volume and concentration of manganate(VII) solution; volume and concentration of ethanedioic acid; amount of acid used to acidify; total volume [1] [1] (b) Each run reaches the same fixed point (the same amount of manganate(VII) used up), so the rate is inversely proportional to the time taken [1] (c) 45 °C (160 s): the time is far longer than the trend suggests (about 100 s expected) [1]; for example, the solutions were not fully warmed to 45 °C before mixing, or cooled during the run [1] (d) Systematic error [1]; every point is displaced by the same amount along the temperature axis, so the whole line shifts but its scatter does not change [1] Examiner insight: Controlled variables must be specific; “same amounts” or “same conditions” is too vague.

11. (a) Ethanol combustion has a high activation energy [1]; at room temperature almost no collisions between ethanol and oxygen molecules have E ≥ Ea, so the rate is negligible until a flame supplies energy [1] (b) The fuel can be stored and transported safely without igniting until it is lit [1] (c) Kinetically stable: reaction is very slow because Ea is high (strong C–C and C–H bonds) [1]; thermodynamically unstable: the products of combustion are lower in energy, so ΔH is negative [1] (d) Enzyme [1] Examiner insight: Keep “how fast” (Ea) separate from “how much energy is released” (ΔH); answers that mix the two rarely earn full credit.

Where marks are usually lost

• Reading a rate as a single value off a graph instead of the gradient of a tangent.
• Taking gradient points from the curve or data table rather than from the drawn tangent.
• Forgetting to convert cm³ to dm³ before using a molar volume.
• Ignoring the stoichiometric ratio when asked for the rate of a different species.
• Explaining a temperature effect only by “more collisions”, with no mention of Ea.
• Drawing a second Maxwell–Boltzmann curve for a catalyst, or a hotter curve with a larger area.
• Changing ΔH when adding the catalysed pathway to an energy profile.

Next steps

• Go back over the revision notes
• Re-read the study guide
• 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 2.2, How fast? The rate of chemical change, understandings 2.2.1 to 2.2.5, which are common to SL and HL.