IB DP Chemistry Reactivity 2.2: How fast? Rate equations, mechanisms and the Arrhenius equation – 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.

This set tests Reactivity 2.2.6 to 2.2.13 of the IB Diploma Programme Chemistry guide (issued February 2023, first examined in 2025). These are additional higher level understandings, so the whole set is HL only. Use R = 8.31 J K⁻¹ mol⁻¹ throughout.

For the rest of the course, open the IB DP chemistry hub or the printable checklist. SL and HL basics are in the rates and collision theory practice set.

Questions

1. Distinguish between a reaction intermediate and a transition state. [2]

2. State the molecularity of each elementary step.

(a) O₃ → O₂ + O [1] (b) NO + O₃ → NO₂ + O₂ [1] (c) O + O₂ + N₂ → O₃ + N₂ [1]

3. Mercury(II) chloride reacts with ethanedioate ions: 2HgCl₂(aq) + C₂O₄²⁻(aq) → 2Cl⁻(aq) + 2CO₂(g) + Hg₂Cl₂(s)

Starting rates were measured at one temperature.

Experiment	[HgCl₂] / mol dm⁻³	[C₂O₄²⁻] / mol dm⁻³	Starting rate / mol dm⁻³ s⁻¹
1	0.10	0.20	3.2 × 10⁻⁵
2	0.20	0.20	6.4 × 10⁻⁵
3	0.10	0.30	7.2 × 10⁻⁵
4	0.050	0.40	?

(a) Using the data, find the order in HgCl₂ and the order in C₂O₄²⁻. Show which experiments you compare. [2] (b) Give the rate equation for this reaction and its overall order. [1] (c) Calculate k, giving its units. [2] (d) Calculate the starting rate in experiment 4. [1]

4. Nitryl chloride, NO₂Cl, breaks down in the gas phase: 2NO₂Cl(g) → 2NO₂(g) + Cl₂(g). The experimental rate equation is rate = k[NO₂Cl]. A proposed mechanism is:

Step 1 (slow): NO₂Cl → NO₂ + Cl Step 2 (fast): NO₂Cl + Cl → NO₂ + Cl₂

(a) Show that the mechanism is consistent with the overall equation, and identify the intermediate. [2] (b) State the molecularity of each step. [1] (c) Explain why the mechanism is consistent with the rate equation. [1] (d) Explain why a single-step mechanism in which two NO₂Cl molecules collide can be rejected. [1]

5. For H₂(g) + I₂(g) → 2HI(g), one proposed mechanism is:

Step 1 (fast, reversible): I₂ ⇌ 2I Step 2 (slow): H₂ + 2I → 2HI

(a) Deduce the rate equation predicted by this mechanism. Explain your answer. [2] (b) State the molecularity of step 2 and comment on how likely such a step is. [2] (c) The overall reaction is slightly exothermic. Sketch an energy profile consistent with this mechanism, labelling the intermediate, the transition states, the activation energy of the rate-determining step and ΔH. [3]

6. (a) Describe the rate–concentration line for a reactant that is (i) zero, (ii) first, (iii) second order. [3] (b) The concentration of a reactant falls from 0.60 to 0.30 mol dm⁻³ in the first 40 s and from 0.30 to 0.15 mol dm⁻³ in the next 40 s. Deduce the order with respect to this reactant. [1]

7. Methyl isocyanide, CH₃NC, isomerises to CH₃CN in the gas phase. Tangents to a concentration–time graph give:

[CH₃NC] / mol dm⁻³	0.0400	0.0250	0.0100
Rate / mol dm⁻³ s⁻¹	2.4 × 10⁻⁵	1.5 × 10⁻⁵	6.0 × 10⁻⁶

(a) Deduce the order of reaction, showing your reasoning. [2] (b) Calculate k, with units. [2] (c) Find the rate at [CH₃NC] = 0.0180 mol dm⁻³. [1]

8. This question links to Reactivity 3.4.

2-iodo-2-methylpropane reacts with aqueous hydroxide ions. rate = k[(CH₃)₃CI]. When [(CH₃)₃CI] = 0.012 mol dm⁻³ and [OH⁻] = 0.050 mol dm⁻³, the rate is 3.6 × 10⁻⁵ mol dm⁻³ s⁻¹.

(a) Calculate k, with units. [2] (b) State the new rate if [OH⁻] is doubled and [(CH₃)₃CI] is halved at the same time. [1]

1-iodopropane reacts with aqueous hydroxide ions. rate = k[CH₃CH₂CH₂I][OH⁻]. When [CH₃CH₂CH₂I] = 0.020 and [OH⁻] = 0.030 mol dm⁻³, the rate is 2.4 × 10⁻⁶ mol dm⁻³ s⁻¹.

(c) Calculate k, with units. [2] (d) Calculate the rate when [CH₃CH₂CH₂I] = 0.040 and [OH⁻] = 0.045 mol dm⁻³. [1]

9. The rate constant of a reaction was measured at four temperatures.

T / K	290	305	320	335
k / mol⁻¹ dm³ s⁻¹	0.229	0.749	2.19	5.82

(a) Calculate 1/T and ln k for the 290 K and 335 K data. [2] (b) All four points lie on a straight line. Determine its gradient. [2] (c) Determine the activation energy in kJ mol⁻¹. [2] (d) Determine the Arrhenius factor, A, with units. [2] (e) Deduce the overall order of the reaction. [1]

10. A first-order reaction has k = 1.6 × 10⁻⁴ s⁻¹ at 298 K and k = 1.50 × 10⁻³ s⁻¹ at 318 K.

(a) Describe how the rate constant changes with temperature. [1] (b) Calculate Ea from the relationship ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). [3] (c) Calculate k at 308 K. [2] (d) State what the Arrhenius factor, A, takes into account. [1]

Answers

1. An intermediate is produced by one step and consumed by a later step; it sits in a trough on the energy profile [1]. A transition state is the highest-energy arrangement within one step, bonds part-broken and part-formed; it sits at a peak and cannot be isolated [1]. [2] Examiner insight: Give a defining point for each species; describing only one earns half the credit at most.

2. (a) Unimolecular [1] (b) Bimolecular [1] (c) Termolecular: three particles meet, with N₂ taking away energy [1] Examiner insight: Count only the reacting particles in the step itself.

3. (a) Experiments 1 and 2: [HgCl₂] doubles, [C₂O₄²⁻] fixed, rate doubles, so first order in HgCl₂ [1]. Experiments 1 and 3: [C₂O₄²⁻] rises 1.5-fold and the rate 2.25-fold, which is 1.5², so second order in C₂O₄²⁻ [1]. (b) rate = k[HgCl₂][C₂O₄²⁻]²; third order overall [1]. (c) k = 3.2 × 10⁻⁵ ÷ (0.10 × 0.20²) [1] = 8.0 × 10⁻³ mol⁻² dm⁶ s⁻¹ [1]. (d) 8.0 × 10⁻³ × 0.050 × 0.40² = 6.4 × 10⁻⁵ mol dm⁻³ s⁻¹ [1]. Examiner insight: State which experiments you compared and which concentration stayed fixed; an order quoted with no comparison shown is hard to credit.

4. (a) Adding the steps: 2NO₂Cl + Cl → 2NO₂ + Cl₂ + Cl; Cl cancels to give the overall equation [1]. The intermediate is Cl [1]. (b) Step 1 is unimolecular; step 2 is bimolecular [1]. (c) The slow step involves only one NO₂Cl, so the predicted rate equation is rate = k[NO₂Cl], which matches experiment [1]. (d) Two NO₂Cl in the only step would give rate = k[NO₂Cl]², second order, which does not match the first-order experimental result [1]. Examiner insight: Justify “consistent” by linking the slow step’s species to the experimental rate equation.

5. (a) The slow step uses H₂ and two I atoms; each pair of I atoms comes from one I₂ in the fast step before it [1]. So rate = k[H₂][I₂] [1]. (b) Step 2 is termolecular [1]. Three particles meeting at once with enough energy and the right orientation is rare, so this step is slow [1]. (c) Two humps with the intermediate (2I + H₂) in the trough between them [1]; second peak higher than the first, with Ea of the rate-determining step marked from reactants to that peak [1]; products slightly below reactants, ΔH shown as a downward arrow [1]. Examiner insight: An intermediate drawn at a peak, or a tallest hump for the fast step, contradicts the mechanism and loses that credit.

6. (a) (i) Zero order: a horizontal line, as rate does not depend on concentration [1]. (ii) First order: a straight line through the origin [1]. (iii) Second order: an upward curve through the origin [1]. (b) Each 40 s halves the concentration: a constant half-life, so first order [1]. Examiner insight: Name the axes you mean: zero order is horizontal on a rate–concentration graph but a sloping straight line on a concentration–time graph.

7. (a) From 0.0400 to 0.0100 mol dm⁻³ the concentration is divided by 4 and the rate is divided by 4 (2.4 × 10⁻⁵ to 6.0 × 10⁻⁶) [1]; rate ∝ [CH₃NC], so first order [1]. (b) k = 2.4 × 10⁻⁵ ÷ 0.0400 [1] = 6.0 × 10⁻⁴ s⁻¹ [1]. (c) 6.0 × 10⁻⁴ × 0.0180 = 1.08 × 10⁻⁵ mol dm⁻³ s⁻¹ [1]. Examiner insight: Confirming the order with a third data point catches misreads.

8. (a) k = 3.6 × 10⁻⁵ ÷ 0.012 [1] = 3.0 × 10⁻³ s⁻¹ [1]. (b) [OH⁻] is not in the rate equation, so only the halving counts: 1.8 × 10⁻⁵ mol dm⁻³ s⁻¹ [1]. (c) k = 2.4 × 10⁻⁶ ÷ (0.020 × 0.030) [1] = 4.0 × 10⁻³ mol⁻¹ dm³ s⁻¹ [1]. (d) 4.0 × 10⁻³ × 0.040 × 0.045 = 7.2 × 10⁻⁶ mol dm⁻³ s⁻¹ [1]. Examiner insight: The two k values must carry different units because the overall orders differ.

9. (a) For 290 K, 1/T is 3.448 × 10⁻³ K⁻¹ and ln k is −1.474 [1]. For 335 K, 1/T is 2.985 × 10⁻³ K⁻¹ and ln k is 1.761 [1]. (b) gradient = (1.761 − (−1.474)) ÷ (2.985 × 10⁻³ − 3.448 × 10⁻³) [1] = 3.235 ÷ (−4.632 × 10⁻⁴) = −6.98 × 10³ K [1]. (c) Ea = −gradient × R = 6.984 × 10³ × 8.31 [1] = 5.80 × 10⁴ J mol⁻¹ = 58.0 kJ mol⁻¹ [1]. (d) ln A = ln k − gradient × (1/T) = 1.761 + 6.984 × 10³ × 2.985 × 10⁻³ = 22.61 [1]; A = e^22.61 ≈ 6.6 × 10⁹ mol⁻¹ dm³ s⁻¹ [1]. (e) k has units mol⁻¹ dm³ s⁻¹, so the reaction is second order overall [1]. Examiner insight: Graph readings vary slightly, so answers of about 57 to 59 kJ mol⁻¹ are reasonable; a negative Ea shows the gradient’s sign was mishandled.

10. (a) k increases as temperature increases, roughly exponentially [1]. (b) ln(1.50 × 10⁻³ ÷ 1.6 × 10⁻⁴) = 2.238 [1]; 1/298 − 1/318 = 2.111 × 10⁻⁴ K⁻¹ [1]; Ea = 8.31 × 2.238 ÷ 2.111 × 10⁻⁴ = 8.81 × 10⁴ J mol⁻¹ = 88.1 kJ mol⁻¹ [1]. (c) ln(k₃₀₈ ÷ 1.6 × 10⁻⁴) = (88 100 ÷ 8.31) × (1/298 − 1/308) = 1.155 [1]; k₃₀₈ = 1.6 × 10⁻⁴ × e^1.155 = 5.1 × 10⁻⁴ s⁻¹ [1]. (d) The frequency of collisions with the proper orientation [1]. Examiner insight: R = 8.31 is in joules, so turn Ea into J mol⁻¹ first; working in kJ throws the answer out by a factor of 1000.

Where marks are usually lost

• Taking orders from the coefficients in the overall equation (question 3 has 2HgCl₂ but is first order in HgCl₂).
• Missing or wrong units on k.
• Drawing the intermediate at a peak.
• Using °C instead of K, or mixing kJ and J, in the Arrhenius equation.
• Dropping the minus sign from the gradient, so Ea comes out negative.

Next steps

• Recap with the revision notes
• Revisit any weak area in the study guide
• Check yourself with all free 10-minute diagnostics
• Browse other units via the IB DP chemistry course hub
• Book a free trial class
• Mark your progress on the printable checklist

Official syllabus

Source: the Chemistry guide for the Diploma Programme (International Baccalaureate, February 2023; first assessment 2025). Every question targets Reactivity 2.2 understandings 2.2.6 to 2.2.13, which are HL only.