A Level · Chemistry 9701 · Physical chemistry

Reaction kinetics: A Level Chemistry questions with answers

Free Reaction kinetics questions for A Level Chemistry (9701), part of Physical chemistry. Each question has the answer and the marking points an examiner would look for.

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Reaction kinetics practice questions

Question 1Structured · 10 marksA* challenge

The initial rate of the reaction A + B → products is measured at constant temperature.

experiment[A] / mol dm−3[B] / mol dm−3initial rate / mol dm−3 s−1
10.100.102.5 × 10−5
20.200.101.0 × 10−4
30.100.202.5 × 10−5
Table 1.1
(a)

Deduce the order with respect to A and to B. Explain your reasoning.

[3]
(b)

Write the rate equation and calculate the rate constant, with units.

[3]
(c)

Calculate the initial rate when [A] = 0.30 mol dm−3 and [B] = 0.30 mol dm−3.

[1]
(d)

Suggest a two-step mechanism consistent with your rate equation, identifying the rate-determining step.

[2]
(e)

For a different first-order reaction, k = 3.5 × 10−4 s−1. Calculate its half-life.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)order 2 w.r.t. A, 0 w.r.t. B
A doubled (exp 1→2): rate × 4, order 2 B1
B doubled (exp 1→3): rate × 1, order 0 B1
reasoning shown B1
3
(b)rate = k[A]2; k = 2.5 × 10−3 mol−1 dm3 s−1
correct rate equation B1
2.5 × 10−3 A1
correct units B1
3
(c)2.3 × 10−4
2.3 × 10−4 B1
1
(d)slow step involves the species in the rate equation in those numbers; fast second step completes the reaction
rate-determining step contains 2 A only B1
second (fast) step gives the overall equation B1
2
(e)1980 s
0.693 ÷ k = 1980 s B1
1
Question 2Structured · 4 marksA* challenge

The half-life of a first-order reaction is constant.

(a)

Explain how a concentration–time graph shows that a reaction is first order.

[2]
(b)

A first-order reaction has a half-life of 45 s. Calculate the rate constant.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)constant half-life
successive half-lives are equal B2
2
(b)0.0154 s⁻¹
k = ln 2 / t½ C1
0.0154 s−1 A1
2
Question 3Structured · 9 marks

Dinitrogen pentoxide decomposes at 338 K: 2N₂O₅(g) → 4NO₂(g) + O₂(g). Fig. 3.1 shows how its concentration changes with time.

01002003001020304050time / s[N₂O₅] / 10⁻³ mol dm⁻³
Fig. 3.1
(a)

Use Fig. 3.1 to determine two successive half-lives, and show that the reaction is first order with respect to N₂O₅.

[3]
(b)

Use your answer to (a) to calculate the rate constant, k, with units.

[2]
(c)

Calculate the initial rate of decrease of [N₂O₅].

[1]
(d)

Explain, in terms of the Boltzmann distribution, why a small rise in temperature greatly increases the rate.

[2]
(e)

Suggest a practical method for following the concentration of NO₂ during the reaction.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)both about 86.6 s; constant half-life
first half-life 86.6 s (± 10 s) B1
second half-life the same (± 10 s) B1
constant half-life shows first order B1
3
(b)8.0 × 10−3 s−1
k = ln 2 ÷ t½ C1
8.0 × 10−3 s−1 (ft) A1
2
(c)3.20 × 10−4 mol dm−3 s−1
rate = k[N₂O₅] = 8.0 × 10−3 × 0.040 = 3.20 × 10−4 B1
1
(d)more molecules with E ≥ Ea; more successful collisions
the distribution shifts so a much larger fraction of molecules have energy ≥ Ea B1
so the frequency of successful collisions increases greatly B1
2
(e)colorimetry (NO₂ brown)
colorimetry, since NO₂ is brown (absorbance proportional to concentration) B1
1
Question 4Structured · 10 marks

The decomposition of hydrogen peroxide catalysed by iodide ions is first order with respect to H₂O₂. With iodide ions in large excess, the observed (pseudo-first-order) rate constant is k′ = 1.50 × 10−4 s−1 at 298 K.

(a)

Write the rate equation if the reaction is also first order with respect to I⁻, and state the overall order.

[2]
(b)

Calculate the half-life of H₂O₂ under these conditions (I⁻ in large excess).

[2]
(c)

Calculate the initial rate when [H₂O₂] = 0.100 mol dm−3.

[1]
(d)

Describe how a concentration–time graph is used to show that a reaction is first order.

[2]
(e)

The mechanism is: H₂O₂ + I⁻ → H₂O + IO⁻ (slow); H₂O₂ + IO⁻ → H₂O + O₂ + I⁻ (fast). Explain how this is consistent with the rate equation and identify the catalyst and the intermediate.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)rate = k[H₂O₂][I⁻]; second order
rate = k[H₂O₂][I⁻] B1
overall order 2 B1
2
(b)4620 s
t½ = ln 2 / k′ M1
4620 s A1
2
(c)1.50 × 10−5 mol dm−3 s−1
rate = k′[H₂O₂] = 1.50 × 10−5 mol dm−3 s−1 B1
1
(d)constant successive half-lives
measure successive half-lives from the curve B1
constant half-life shows first order B1
2
(e)slow step involves H₂O₂ and I⁻ once each; I⁻ catalyst (regenerated); IO⁻ intermediate
the rate-determining step involves one H₂O₂ and one I⁻, matching the rate equation B1
I⁻ is the catalyst (used and regenerated) B1
IO⁻ is the intermediate B1
3
Question 5Structured · 4 marks

The rate equation for a reaction is rate = k[A][B]2.

(a)

State the order with respect to each reactant and the overall order.

[2]
(b)

Deduce the units of k.

[1]
(c)

State the effect on the rate of doubling [B] and halving [A].

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)1, 2, 3
first order in A, second order in B B1
overall third order B1
2
(b)dm⁶ mol⁻² s⁻¹
dm6 mol−2 s−1 B1
1
(c)×2
rate doubles B1
1
Question 6Structured · 11 marks

The reaction between peroxodisulfate ions and iodide ions is slow but is catalysed by Fe²⁺ or Fe³⁺ ions.
S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂

(a)

Explain why the uncatalysed reaction has a high activation energy.

[2]
(b)

Write two equations to show how Fe²⁺ catalyses the reaction.

[2]
(c)

Classify this catalysis as homogeneous or heterogeneous and explain why transition metal ions are good catalysts.

[2]
(d)

Describe a colorimetric method to follow the rate of formation of iodine.

[3]
(e)

In a clock reaction, a small fixed amount of sodium thiosulfate and some starch are added to the mixture. Explain why the time for the blue-black colour to appear can be used as a measure of initial rate.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)both reactants negative; repel
both reactants are negative ions B1
they repel each other so collisions with enough energy are rare B1
2
(b)S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺; 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺ B1
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ B1
2
(c)homogeneous; variable oxidation states
homogeneous (same phase) B1
transition metal ions have variable oxidation states so they can accept and donate electrons readily B1
2
(d)colorimeter with filter; absorbance measured at intervals; calibration curve to convert to concentration
use a colorimeter with a suitable (blue) filter B1
record absorbance at regular time intervals B1
use a calibration curve to convert absorbance into [I₂] B1
3
(e)only a small, fixed amount of reaction occurs before the colour appears, so concentrations stay close to initial values; rate ∝ 1/t
only a small fraction of the reactants is used before the colour appears, so concentrations are still close to their initial values (required) B1
the same fixed amount of iodine is formed each time, so rate ∝ 1/t B1
2

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