AQA A-Level Chemistry Paper 2, 2021: Question 10
12 marks · Medium difficulty · State/Explain/Numerical
Rates of Reaction and Activation Energy This question explores the kinetics of the iodine–propanone reaction in acid, including experimental techniques to determine the order of reaction with respect to iodine. Students analyse a graph of iodine concentration over time to identify zero-order behaviour, then calculate activation energy using the Arrhenius equation and a graph of ln k against 1/T.
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Question text
10 This question is about rates of reaction.
Iodine and propanone react together in an acid-catalysed reaction
CH3COCH3(aq) + I2(aq) → CH3COCH2I(aq) + HI(aq)
A student completed a series of experiments to determine the order of reaction with
respect to iodine.
Method
• Transfer 25 cm3 of 1.0 mol dm–3 propanone solution into a conical flask.
• Add 10 cm3 of 1.0 mol dm–3 HCl(aq)
• Add 25 cm3 of 5.0 × 10–3 mol dm–3 I (aq) and start a timer.
• At intervals of 1 minute, remove a 1.0 cm3 sample of the mixture and add each
sample to a separate beaker containing an excess of NaHCO3(aq)
• Titrate the contents of each beaker with a standard solution of
sodium thiosulfate and record the volume of sodium thiosulfate used.
10.1 Suggest why the 1.0 cm3 portions of the reaction mixture are added to an excess of
NaHCO3 solution.
[2 marks]
10.2 Suggest why the order of this reaction with respect to propanone can be ignored in
this experiment.
[2 marks]
The volume of sodium thiosulfate solution used in each titration is proportional to the
concentration of iodine in each beaker.
Table 5 shows the results of the experiment.
Table 5
Time / minutes Volume of sodium thiosulfate solution
/ cm3
*25* 1 41
2 35
3 24
4 22
5 16
6 10
10.3 Use the results in Table 5 to draw a graph of volume of sodium thiosulfate solution
against time.
Draw a line of best fit.
[3 marks]
10.4 Explain how the graph shows that the reaction is zero-order with respect to iodine in
the reaction between propanone and iodine.
[2 marks]
10.5 The Arrhenius equation can be written as
−Ea
In k = + In A
RT
Figure 8 shows a graph of ln k against for the reaction
T
2 HI(g) → H2(g) + I2(g)
Figure 8
Use Figure 8 to calculate a value for the activation energy (E ), in kJ mol−1, for
a
this reaction.
The gas constant R = 8.31 J K−1 mol−1
[3 marks]
E kJ mol−1
a
Mark scheme
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Question Answers Additional Comments/Guidelines Mark
The sodium hydrogencarbonate solution neutralises the acid M1
(catalyst)
10.1
So stops the reaction M2
The concentration/amount of propanone is much larger than/200 M1
times larger than the concentration/amount of iodine
10.2 M2
Concentration of propanone is (almost) constant The change in concentration in propanone is
negligible
– A-LEVEL CHEMISTRY – –
10.3
M1
M2
M3
Suitable axes (plotted points must take up at least half of the grid)
For all points correctly plotted to ± 1/ small square
For straight line of best fit which avoids the anomalous plot
– A-LEVEL CHEMISTRY – –
The graph is a straight line / has a constant gradient M1
10.4 So the rate of reaction does not change as the concentration (of Correct rate vs conc graph scores M2
iodine) changes / the iodine is being used up at a constant rate. M2
Gradient = (−14.1 − −2.8) / (0.00180 − 0.00128) Allow -21330 to -22130 M1
= −11.3 / 0.00052
= −21731
10.5
Gradient = −Ea / R
−E = their answer x 8.31 ( = 180583 J mol-1) M2
a
E = M2 ÷ 1000 (= 181 kJ mol-1) M3
a
How to answer it
Study Guide: Reaction Kinetics & the Arrhenius Equation
Q1. Why Add the Mixture to Excess NaHCO3?
Reason: NaHCO3 neutralises the acid catalyst
This stops the reaction instantly by removing the H+ ions needed for the reaction to proceed.
Q2. Ignoring Propanone Concentration
Why? The amount of propanone is much larger than the amount of iodine.
This means its concentration is effectively constant – any change is negligible over the course of the experiment.
Q3. Graph: Volume of Sodium Thiosulfate vs Time
The graph should be a straight line with a constant gradient.
This shows that the rate of iodine loss is constant, meaning the rate does not depend on iodine concentration → zero order with respect to iodine.
Q4. Activation Energy from ln k vs 1/T Graph
Gradient = –Ea/R
Given: gradient = –21731 and R = 8.31
Ea = –gradient × R = 21731 × 8.31 = 180583 J mol–1 = 181 kJ mol–1
Topics
Physical Chemistry · Required Practicals · 3.1.9 Rate Equations · Required Practical 7: Measuring the rate of reaction by an initial rate method
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.