5.5 Kinetics II

1a9 marks

Gentian violet is a purple crystalline solid used as an antifungal treatment.

It can be synthesised from dimethylphenylamine, C₆H₅N(CH₃)₂.

The dimethylphenylamine used in the synthesis can be made by the stepwise reaction of phenylamine with chloromethane.

Step 1

2 C_{6} H_{5} {NH}_{2} + {CH}_{3} Cl \to C_{6} H_{5} NH ({CH}_{3}) + C_{6} H_{5} {NH}_{3}^{+} {Cl}^{-}

Step 2 $2 C_{6} H_{5} NH ({CH}_{3}) + {CH}_{3} Cl \to C_{6} H_{5} N {({CH}_{3})}_{2} + C_{6} H_{5} {NH}_{2}^{+} ({CH}_{3}) {Cl}^{-}$

The reaction mechanism for Step 1 between phenylamine and chloromethane is the same as that in the reaction between ammonia and chloromethane.

What is the reaction type and mechanism in Step 1?

(1)

☐	A	electrophilic addition
☐	B	electrophilic substitution
☐	C	nucleophilic addition
☐	D	nucleophilic substitution

ii)

Draw the mechanism for the reaction in Step 1.
Include curly arrows, and relevant lone pairs and dipoles.

(4)

iii)

Describe, in outline, how a sample of a solid, such as gentian violet, is purified by recrystallisation.
Specific details of the solvent used are not required.

(4)

How did you do?

7 marks

The rate constant for the reaction between a solution of gentian violet and aqueous sodium hydroxide was determined at different temperatures.

Temperature (T) /K	1/ Temparature (1/T) /K^-1	Rate constant, k / dm³ mol^-1 s^-1	In k
283.5	3.53 x 10^-3	2.71 x 10^-3	—5.91
287.5	3.48 x 10^-3	3.55 x 10^-3
291.5		4.75 x 10^-3	—5.35
295.0	3.39 x 10^-3	6.10 x 10^-3	—5.10
298.5	3.35 x 10^-3	7.60 x 10^-3	—4.88

Complete the data in the table.

(1)

ii)

Plot a graph and use it to determine the activation energy for the reaction in kJ mol^-1.

You should include the value and units of the gradient of the line.

The Arrhenius equation can be shown as

In k = - \frac{E_{a}}{R} \times \frac{1}{T} + constant

(6)

q8bii-9cho-al-2-june-2019-qp-edexcel-a-level-chem

Gradient

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Activation energy

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3a1 mark

This question is about the reaction kinetics of an ‘iodine clock’ reaction.

One example of an ‘iodine clock’ reaction that involves the iodate(V) ions and iodide ions in acidic solution is

${IO}_{3}^{-} (aq) + 5 I^{-} (aq) + 6 H^{+} (aq) \to 3 I_{2} (aq) + 3 H_{2} O (l)$

State why the order of reaction with respect to iodide ions cannot be five, even though 5 mol of iodide ions are shown in the equation.

How did you do?

7 marks

A series of experiments was carried out by a student to determine the order of reaction with respect to iodate(V) ions. The concentrations of the iodide ions and the acid were in large excess and the volume of the iodate(V) solution was varied.
The total volume of the reaction mixture was kept constant by the addition of suitable volumes of deionised water.
The following results were obtained:

Experiment Number	1	2	3	4	5	6
Volume of iodate(V) solution / cm³	10.0	7.0	5.0	3.0	2.0	1.0
Time (t) / s	180	260	357	606	900	800
(1000/t) / s^-1	5.56				1.11	1.25

In experiment 6, the student forgot to add deionised water to keep the total volume the same for each experiment.
State why the total volume should be kept the same.

(1)

ii)

Complete the table and use the results from experiments 1, 2, 3, 4 and 5 to plot a graph of 1000/t against volume of iodate(V) ions.

(4)

q9bii-9cho-al-2-june-2018-qp-edexcel-a-level-chem

iii)

Deduce the order of reaction with respect to the iodate(V) ions. Justify your answer.

(2)

How did you do?

4 marks

A different version of the ‘iodine clock’ reaction involves mixing hydrogen peroxide with aqueous solutions of potassium iodide, sodium thiosulfate and starch.

The main reaction is

H₂O₂(aq) + 2I^-( aq) + 2H⁺(aq) → I₂(aq) + 2H₂O (l)

The reaction is first order with respect to hydrogen peroxide and iodide ions but zero order with respect to hydrogen ions.

In one experiment, the following data were obtained:

Reactants	Initial concentration/mol dm^-3
H₂O₂(aq)	1.50 × 10^-3
I^-(aq)	2.10 × 10^-3
H⁺(aq)	2.10 × 10^-3

Initial rate = 1.24 × 10^-3 mol dm^-3 s^-1

Write the rate equation and hence deduce the value of the rate constant, k, from these data. Include units and give your answer to an appropriate number of significant figures.

(2)

ii)

Explain the purpose of the starch present in the reaction mixture when starch is neither in the rate equation, nor in the reaction equation.

(2)

How did you do?

4 marks

Another ‘iodine clock’ reaction produced data that enabled the following graph of ln k against 1/T to be drawn.

q9d-9cho-al-2-june-2018-qp-edexcel-a-level-chem

The Arrhenius equation can be expressed as

In $k = - \frac{E_{a}}{R} \times [\frac{1}{T}] + constant$

From the gradient of the graph, determine the activation energy, E_a , for this reaction.

Include a sign and units in your answer.

(3)

ii)

Give a reason for the point at ln k = −7 not being included in the line drawn on the graph.

(1)

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1a1 mark

This question is about the compound potassium bromate, KBrO₃ .

These bromate ions react with bromide ions in acidic solution.

BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l)

Explain, in terms of oxidation numbers, whether or not this is a disproportionation reaction.

(2)

ii)

What is the overall order of this reaction?

(1)

☐	A	3
☐	B	6
☐	C	12
☐	D	cannot tell from this information

How did you do?

3 marks

Potassium bromate decomposes on heating.

2KBrO₃ → 2KBr + 3O₂

Calculate the maximum volume of oxygen, in dm³, measured at room temperature and pressure (r.t.p.), that could be produced from the complete decomposition of 5.20 g of potassium bromate.
[Molar volume of gas at r.t.p. = 24.0 dm³ mol⁻¹]

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2a2 marks

This question is about carbon monoxide, CO, which is a toxic and colourless gas used widely in the chemical industry.

Draw a dot-and-cross diagram of a molecule of carbon monoxide.

Use dots (•) for the carbon electrons and crosses (×) for the oxygen electrons.

How did you do?

2 marks

Carbon monoxide can be made by the thermal decomposition of sodium ethanedioate.

Na₂C₂O₄ → Na₂CO₃ + CO

Calculate the atom economy, by mass, for the production of carbon monoxide in this reaction.

How did you do?

4 marks

Carbon monoxide can also be made by the thermal decomposition of ethanal, CH₃CHO, in the gas phase.

CH₃CHO (g) → CH₄(g) + CO (g)

This reaction was carried out at two different temperatures, and all other variables were kept constant.

Temperature / K	Rate / mol dm⁻³ s⁻¹	1/Temperature (1/T) / K⁻¹	ln rate
700	0.0108	1.43 × 10⁻³
850	4.90		1.59

Complete the data in the table.

(1)

ii)

Calculate the activation energy, E_a, for the reaction without plotting a graph.
Include a sign and units in your answer.

The Arrhenius equation may be written as

In rate = − $\frac{E_{a}}{R} \times \frac{1}{T}$ + constant [R = 8.31 J mol⁻¹K⁻¹]

(3)

How did you do?

7 marks

Haemoglobin (Hb) found in red blood cells reacts almost irreversibly with carbon monoxide.
Initial rate experiments were carried out to investigate the effect of the concentrations of Hb and CO on the rate of this reaction.

Experiment	[Hb] / mol dm⁻³	[CO] / mol dm^–3	Rate / mol dm⁻³s⁻¹
1	2.09 × 10^–6	1.40 × 10^–6	8.20 × 10^–7
2	4.18 × 10^–6	1.40 × 10^–6	1.64 × 10^–6
3	3.26 × 10^–6	2.80 × 10^–6	2.56 × 10^–6

Deduce the order of reaction with respect to haemoglobin.

(1)

ii)

Determine the order with respect to carbon monoxide using your answer to (d)(i) and the data in the table.

Justify your answer.

(2)

iii)

Write the rate equation for this reaction using your answers to (d)(i) and (d)(ii).

(1)

iv)

Calculate the rate constant, k, for the reaction, using the data from Experiment 1 and the rate equation from (d)(iii).
Include units in your answer.

(3)

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Medium
Hard

Temperature / K	Rate constant / s⁻¹
328	1.50×10⁻³
338	4.87×10⁻³

[OH⁻] / mol dm⁻³	Time / s
0.00100	39
0.00200	31
0.00300	23
0.00400	16
0.00500	8

Time / s	[RBr] / mol dm^–3
0	0.100
30	0.065
60	0.042
90	0.028
120	0.019
150	0.014

Time /s	Partial pressure of ammonia / kPa
0	0.350
100	0.335
200	0.319
300	0.303
400	0.287
500	0.271

Edexcel A Level Chemistry

Topic Questions

5.5 Kinetics II

Experiment	[NO] / moldm^-3	[Cl₂] / moldm^-3	Rate /moldm^-3s¹
1	0.122	0.241	1.09 x 10^-2
2		0.482	8.72 x 10^-2
3	0.366		4.91 x 10^-2