Kinetics	

	Name:	
________________________	

Class:	
________________________	

Date:	
________________________	

 

 	
Time:	209 minutes	
Marks:	205 marks	
Comments:		

Q1.Rate = k [A]2 [B]
Correct units for the rate constant in the rate equation above are
A       mol dm−3 s−1
B       mol−1 dm−3 s−1
C       mol2 dm−6 s−1
D       mol−2 dm6 s−1
(Total 1 mark)

Q2.The rate equation for the hydrogenation of ethene
C2H4(g) + H2(g)  C2H6(g)
is Rate = k[C2H4][H2]
At a fixed temperature, the reaction mixture is compressed to triple the original pressure.
What is the factor by which the rate of reaction changes?
A        6            
B        9            
C        12          
D        27          
(Total 1 mark)

Q3.This question is about the reaction between propanone and an excess of ethane-1,2-diol, the equation for which is given below.
 
In a typical procedure, a mixture of 1.00 g of propanone, 5.00 g of ethane-1,2-diol and 0.100 g of benzenesulphonic acid, C6H5SO3H, is heated under reflux in an inert solvent. Benzenesulphonic acid is a strong acid.
When the concentration of benzenesulphonic acid is doubled, the rate of the reaction doubles. It can be deduced that
A       the reaction is first order overall.
B       the reaction is third order overall.
C       the reaction is acid-catalysed.
D       units for the rate constant, k, are mol−2 dm6 s−1.
(Total 1 mark)

Q4.          Iodine and propanone react in acid solution according to the equation
I2   +   CH3COCH3   →   CH3COCH2I   +   HI
The rate equation for the reaction is found to be
rate = k [CH3COCH3][H+]
(a)     Deduce the order of reaction with respect to iodine and the overall order of reaction.
Order with respect to iodine .........................................................................
Overall order ................................................................................................
(2)

(b)     At the start of the experiment, the rate of reaction was found to be 2.00 × 10–5 mol dm–3 s–1 when the concentrations of the reactants were as shown below.
 
Reactant	Concentration / mol dm–3	
CH3COCH3	1.50	
I2	2.00 × 10–2	
H+	3.00 × 10–2	

Use these data to calculate a value for the rate constant and deduce its units.
Rate constant ...............................................................................................
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Units .............................................................................................................
(3)

(c)     How can you tell that H+ acts as a catalyst in this reaction?
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(2)

(d)     Calculate the initial rate of reaction if the experiment were to be repeated at the same temperature and with the same concentrations of iodine and propanone as in part (b) but at a pH of 1.25
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(3)
(Total 10 marks)

 

Q5.          (a)     The following data were obtained in a series of experiments on the rate of the reaction between compounds A and B at a constant temperature.
 
Experiment	Initial concentration of A/mol dm–3	Initial concentration of B/mol dm–3	Initial
rate/mol dm–3 s–1	
1	0.15	0.24	0.45 × 10–5	
2	0.30	0.24	0.90 × 10–5	
3	0.60	0.48	7.20 × 10–5	

(i)      Show how the data in the table can be used to deduce that the reaction is first-order with respect to A.
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(ii)     Deduce the order with respect to B.
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(2)

(b)     The following data were obtained in a second series of experiments on the rate of the reaction between compounds C and D at a constant temperature.
 
Experiment	Initial concentration of A/mol dm–3	Initial concentration of B/mol dm–3	Initial
rate/mol dm–3 s–1	
4	0.75	1.50	9.30 × 10–5	
5	0.20	0.10	To be calculated	

The rate equation for this reaction is
rate = k[C]2[D]
(i)      Use the data from Experiment 4 to calculate a value for the rate constant, k, at this temperature. State the units of k.
Value for k ...........................................................................................
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Units of k .............................................................................................
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(ii)     Calculate the value of the initial rate in Experiment 5.
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(4)
(Total 6 marks)

 

Q6.          The rate of the reaction between substance A and substance B was studied in a series of experiments carried out at the same temperature. In each experiment the initial rate was measured using different concentrations of A and B. These results were used to deduce the order of reaction with respect to A and the order of reaction with respect to B.
(a)     What is meant by the term order of reaction with respect to A?
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(1)

(b)     When the concentrations of A and B were both doubled, the initial rate increased by a factor of 4. Deduce the overall order of the reaction.
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(1)

(c)     In another experiment, the concentration of A was increased by a factor of three and the concentration of B was halved. This caused the initial rate to increase by a factor of nine.
(i)      Deduce the order of reaction with respect to A and the order with respect to B.
Order with respect to A .......................................................................
Order with respect to B ........................................................................

(ii)     Using your answers from part (c)(i), write a rate equation for the reaction and suggest suitable units for the rate constant.
Rate equation .....................................................................................
Units for the rate constant ...................................................................
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(4)
(Total 6 marks)

 

Q7.          (a)     The initial rate of the reaction between substances P and Q was measured in a series of experiments and the following rate equation was deduced.
rate = k[P]2[Q]
(i)      Complete the table of data below for the reaction between P and Q.
 
Experiment	Initial [P] / mol dm–3	Initial [Q] / mol dm–3	Initial rate / mol dm–3 s–1	
1	0.20	0.30	4.8 × 10–3	
2	0.10	0.10	 	
3	0.40	 	9.6 × 10–3	
4	 	0.60	19.2 × 10–3	

(ii)     Using the data from experiment 1, calculate a value for the rate constant, k, and deduce its units.
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(6)

(b)     What change in the reaction conditions would cause the value of the rate constant to change?
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(1)
(Total 7 marks)

 

Q8.          Kinetic studies enable chemists to suggest mechanisms for reactions.
(a)     The following data were obtained in a series of experiments on the rate of the reaction between compounds A and B at a constant temperature.
 
Experiment	Initial concentration of A/mol dm–3	Initial concentration of B/mol dm–3	Initial rate/
mol dm–3 s–1	
1	0.12	0.15	0.32 × 10–3	
2	0.36	0.15	2.88 × 10–3	
3	0.72	0.30	11.52 × 10–3	

(i)      Deduce the order of reaction with respect to A.
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(ii)     Deduce the order of reaction with respect to B.
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(2)

(b)     The following data were obtained in a series of experiments on the rate of the reaction between NO and O2 at a constant temperature.
 
Experiment	Initial concentration of NO/mol dm–3	Initial concentration of O2/mol dm–3	Initial rate/
mol dm–3 s–1	
4	5.0 × 10–2	2.0 × 10–2	6.5 × 10–4	
5	6.5 × 10–2	3.4 × 10–2	To be calculated	

The rate equation for this reaction is
rate = k[NO]2[O2]
(i)      Use the data from Experiment 4 to calculate a value for the rate constant, k, at this temperature, and state its units.
Value of k ............................................................................................
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Units of k .............................................................................................
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(ii)     Calculate a value for the initial rate in Experiment 5.
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(iii)     Using the rate equation, a scientist suggested a mechanism for the reaction which consisted of the two steps shown below.
          Step 1    NO + NO → N2O2
          Step 2    N2O2 + O2 → 2NO2
Which did the scientist suggest was the rate–determining step?
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(5)
(Total 7 marks)

 

Q9.          (a)     The following data were obtained in a series of experiments on the rate of the reaction between compounds A and B at a constant temperature.
 
Experiment	Initial concentration
of A/mol dm–3	Initial concentration
of B/mol dm–3	Initial rate/mol dm–3 s–1	
1	0.12	0.15	0.32 × 10–3	
2	0.36	0.15	2.88 × 10–3	
3	0.72	0.30	11.52 × 10–3	

(i)      Deduce the order of reaction with respect to A.
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(ii)     Deduce the order of reaction with respect to B.
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(2)

(b)     The following data were obtained in a series of experiments on the rate of the reaction between NO and O2 at a constant temperature.
 
Experiment	Initial concentration
of NO/mol dm–3	Initial concentration
of O2/mol dm–3	Initial rate/mol dm–3 s–1	
4	5.0 × 10–2	2.0 × 10–2	6.5 × 10–4	
5	6.5 × 10–2	3.4 × 10–2	To be calculated	

The rate equation for this reaction is
rate = k[NO]2[O2]
(i)      Use the data from experiment 4 to calculate a value for the rate constant, k, at this temperature, and state its units.
Value of k ............................................................................................
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Units of k .............................................................................................
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(ii)     Calculate a value for the initial rate in experiment 5.
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(4)
(Total 6 marks)

 

Q10.          (a)     The initial rate of the reaction between compounds A and B was measured in a series of experiments at a fixed temperature. The following rate equation was deduced.
rate = k[A][B]2
(i)      Complete the table of data below for the reaction between A and B.
Expt	Initial [A]
/mol dm–3	Initial [B]
/mol dm–3	Initial rate
/mol dm–3 s–1	
1	4.80 × 10–2	6.60 × 10–2	10.4 × 10–3	
2	4.80 × 10–2	3.30 × 10–2	 	
3	 	13.2 × 10–2	5.20 × 10–3	
4	1.60 × 10–2	 	10.4 × 10–3	

(ii)     Using the data for experiment 1, calculate a value for the rate constant, k, and state its units.
Calculation ..........................................................................................
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Units ....................................................................................................
(6)

(b)     State how the value of the rate constant, k, would change, if at all, if the concentration of A were increased in a series of experiments.
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(1)
(Total 7 marks)

 

Q11.          The initial rate of the reaction between the gases NO and H2 was measured in a series of experiments at a constant temperature and the following rate equation was determined.
rate = k[NO]2[H2]
(a)     Complete the table of data below for the reaction between NO and H2
 
Experiment 	Initial [NO] / mol dm–3	Initial [H2] / mol dm–3	Initial rate / mol dm–3 s–1 	
1 	3.0 × 10–3 	1.0 × 10–3	1.8 × 10–5 	
2 	3.0 × 10–3	 	7.2 × 10–5	
3 	1.5 × 10–3	1.0 × 10–3 	 	
4 	 	0.50 × 10–3 	8.1 × 10–5	

(3)

(b)     Using the data from experiment 1, calculate a value for the rate constant, k, and state its units.
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(3)
(Total 6 marks)

 

Q12.          (a)     Compound A, HCOOCH2CH2CH3, is an ester. Name this ester and write an equation for its reaction with aqueous sodium hydroxide.
Name ……….................................................................................................
Equation .......................................................................................................
(2)

(b)     The initial rate of reaction between ester A and aqueous sodium hydroxide was measured in a series of experiments at a constant temperature. The data obtained are shown below.
 
Experiment	Initial concentration of NaOH / mol dm–3	Initial concentration of A / mol dm–3	Initial rate
/ mol dm–3 s–1	
1	0.040	0.030	4.0 × 10–4	
2	0.040	0.045	6.0 × 10–4	
3	0.060	0.045	9.0 × 10–4	
4	0.120	0.060	to be calculated	

Use the data in the table to deduce the order of reaction with respect to A and the order of reaction with respect to NaOH. Hence calculate the initial rate of reaction in Experiment 4.
Order with respect to A ................................................................................
Order with respect to NaOH .........................................................................
Initial rate in Experiment 4 ............................................................................
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(3)

(c)     In a further experiment at a different temperature, the initial rate of reaction was found to be 9.0 × 10–3 mol dm–3 s–1 when the initial concentration of A was 0.020 mol dm–3 and the initial concentration of NaOH was 2.00 mol dm–3.
Under these new conditions with the much higher concentration of sodium hydroxide, the reaction is first order with respect to A and appears to be zero order with respect to sodium hydroxide.
(i)      Write a rate equation for the reaction under these new conditions.
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(ii)     Calculate a value for the rate constant under these new conditions and state its units.
Calculation ..........................................................................................
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Units ....................................................................................................

(iii)     Suggest why the order of reaction with respect to sodium hydroxide appears to be zero under these new conditions.
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(6)

(d)     A naturally-occurring triester, shown below, was heated under reflux with an excess of aqueous sodium hydroxide and the mixture produced was then distilled. One of the products distilled off and the other was left in the distillation flask.

(i)      Draw the structure of the product distilled off and give its name.
Structure

 

 
Name ..................................................................................................

(ii)     Give the formula of the product left in the distillation flask and give a use for it.
Formula ...............................................................................................
Use ......................................................................................................
(4)
(Total 15 marks)

 

Q13.          The hydrolysis of methyl propanoate was studied in acidic conditions at 25°C and the rate equation was found to be
rate = k[CH3CH2COOCH3][H+]
(a)     Use the data below to calculate the value of the rate constant, k, at this temperature.
Deduce its units.
Initial rate of reaction / 
mol dm–3 s–1	Initial concentration of methyl propanoate / mol dm–3	Initial concentration of hydrochloric acid / mol dm–3	
1.15 × 10–4	0.150	0.555	

Rate constant ...............................................................................................
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Units .............................................................................................................
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(3)

(b)     The reaction in part (a) was repeated at the same temperature, but water was added so that the volume of the reaction mixture was doubled. Calculate the initial rate of reaction under these conditions.
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(1)

(c)     A third experiment was carried out at a different temperature. Some data from this experiment are shown in the table below.
Initial rate of reaction / mol dm–3 s–1	Value of rate constant at this different temperature	Initial methyl propanoate / mol dm–3	
4.56 × 10–5	8.94 × 10–4	0.123	

Calculate the initial pH of the reaction mixture. Give your answer to two decimal places.
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(3)
(Total 7 marks)

 

Q14.          Propanone and iodine react in acidic conditions according to the following equation.
CH3COCH3 + I2 → ICH2COCH3 + HI
A student studied the kinetics of this reaction using hydrochloric acid and a solution containing propanone and iodine. From the results the following rate equation was deduced.
rate = k[CH3COCH3][H+]
(a)     Give the overall order for this reaction.
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(1)

(b)     When the initial concentrations of the reactants were as shown in the table below, the initial rate of reaction was found to be 1.24 × 10–4 mol dm–3 s–1.
 
 	initial concentration / mol dm–3	
CH3COCH3	4.40	
I2	5.00 × 10–3	
H+	0.820	

Use these data to calculate a value for the rate constant, k, for the reaction and give its units.
Calculation ...................................................................................................
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Units .............................................................................................................
(3)

(c)     Deduce how the initial rate of reaction changes when the concentration of iodine is doubled but the concentrations of propanone and of hydrochloric acid are unchanged.
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(1)

(d)     The following mechanism for the overall reaction has been proposed.

Use the rate equation to suggest which of the four steps could be the rate-determining step. Explain your answer.
Rate-determining step .................................................................................
Explanation ..................................................................................................
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(2)

(e)     Use your understanding of reaction mechanisms to predict a mechanism for Step 2 by adding one or more curly arrows as necessary to the structure of the carbocation below.

(1)
(Total 8 marks)

 

Q15.Hydrogen peroxide is a powerful oxidising agent. Acidified hydrogen peroxide reacts with iodide ions to form iodine according to the following equation.
H2O2(aq) + 2H+(aq) + 2I−(aq) → I2(aq) + 2H2O(l)
The initial rate of this reaction is investigated by measuring the time taken to produce sufficient iodine to give a blue colour with starch solution.
A series of experiments was carried out, in which the concentration of iodide ions was varied, while keeping the concentrations of all of the other reagents the same. In each experiment the time taken (t) for the reaction mixture to turn blue was recorded.
The initial rate of the reaction can be represented as (), and the initial concentration of iodide ions can be represented by the volume of potassium iodide solution used.
A graph of log10 () on the y-axis against log10 (volume of KI(aq)) is a straight line. The gradient of this straight line is equal to the order of the reaction with respect to iodide ions.
The results obtained are given in the table below. The time taken for each mixture to turn blue was recorded on a stopclock graduated in seconds.
 
 	Expt.	Volume of
KI(aq) / cm3	log10 (volume of
KI(aq))	Time / s	log10 ()	
 	1	5	0.70	71	−1.85	
 	2	8	0.90	46	−1.66	
 	3	10	1.00	37	−1.57	
 	4	15	1.18	25	−1.40	
 	5	20	1.30	19	−1.28	
 	6	25	1.40	14	−1.15	

(a)     Use the results given in the table to plot a graph of log10 () on the y-axis against log10 (volume of KI(aq)).
Draw a straight line of best fit on the graph, ignoring any anomalous points.
 
(5)

(b)     Determine the gradient of the line you have drawn. Give your answer to two decimal places. Show your working.
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(3)

(c)     Deduce the order of reaction with respect to iodide ions.
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(1)

(d)     A student carried out the experiment using a flask on the laboratory bench. The student recorded the time taken for the reaction mixture to turn blue. State one way this method could be improved, other than by repeating the experiment or by improving the precision of time or volume measurements. Explain why the accuracy of the experiment would be improved.
Improvement .................................................................................................
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Explanation ....................................................................................................
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(2)
(Total 11 marks)

Q16.The initial rate of the reaction between gases D and E was measured in a series of experiments at a constant temperature. The results are shown in the table.
 
 	Expt	Initial [D] / mol dm–3	Initial [E] / mol dm–3	Initial rate / mol dm–3 s–1	
 	1	1.25 × 10–2	5.81 × 10–1	1.16 × 10–2	
 	2	1.88 × 10–2	8.73 × 10–1	3.92 × 10–2	
 	3	1.88 × 10–2	1.75	1.57 × 10–1	

(a)     Deduce the order of reaction with respect to D and the order with respect to E.
Order with respect to D .......................................................................................
Order with respect to E .......................................................................................
Space for working ...............................................................................................
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(2)

(b)     Suggest why initial rates of reaction are used to determine these orders rather than rates of reaction at other times during the experiments.
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(1)

(c)     State how the initial rate is obtained from a graph of the concentration of the product against time.
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(2)
(Total 5 marks)

Q17.          A reaction mechanism is a series of steps by which an overall reaction may proceed.
The reactions occurring in these steps may be deduced from a study of reaction rates.
Experimental evidence about initial rates leads to a rate equation. A mechanism is then proposed which agrees with this rate equation.
Ethanal dimerises in dilute alkaline solution to form compound X as shown in the following equation.
2CH3CHO → CH3CH(OH)CH2CHO
X
A chemist studied the kinetics of the reaction at 298 K and then proposed the following rate equation.
Rate = k [CH3CHO][OH–]
(a)     Give the IUPAC name of compound X.
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(1)

(b)     The initial rate of the reaction at 298K was found to be 2.2 × 10–3 mol dm–3 s–1 when the initial concentration of ethanal was 0.10 mol dm–3 and the initial concentration of sodium hydroxide was 0.020 mol dm–3.
Calculate a value for the rate constant at this temperature and give its units.
Calculation ...................................................................................................
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Units .............................................................................................................
(3)

(c)     The sample of X produced consists of a racemic mixture (racemate). Explain how this racemic mixture is formed.
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(2)

(d)     A three-step mechanism has been proposed for this reaction according to the following equations.
 
Step 1		
Step 2		
Step 3		

(i)      Using the rate equation, predict which of the three steps is the rate-determining step. Explain your answer.
Rate-determining step ........................................................................
Explanation .........................................................................................
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(2)

(ii)     Deduce the role of ethanal in Step 1.
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(1)

(iii)     Use your knowledge of reaction mechanisms to deduce the type of reaction occurring in Step 2.
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(1)

(iv)    In the space below draw out the mechanism of Step 2 showing the relevant curly arrows.

 
 
(2)

(e)     In a similar three-step mechanism, one molecule of X reacts further with one molecule of ethanal. The product is a trimer containing six carbon atoms.
Deduce the structure of this trimer.

 

 
 
(1)
(Total 13 marks)

 

Q18.The initial rate of the reaction between two gases P and Q was measured in a series of experiments at a constant temperature. The following rate equation was determined.
rate = k[P]2[Q]
(a)     Complete the table of data below for the reaction between P and Q.
 
Experiment	Initial [P] /mol dm–3	Initial [Q] /mol dm–3	Initial rate /mol dm–3 s–1	
1	0.20	0.30	1.8 = 10–3	
2	0.40	0.60	 	
3	0.60	 	5.4 = 10–3	
4	 	0.90	12.2 = 10–3	

(Space for working) .......................................................................................
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(3)

(b)     Use the data from Experiment 1 to calculate a value for the rate constant k and deduce its units.
Calculation .....................................................................................................
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Units ...............................................................................................................
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(3)

(c)     Consider the graphs E, F, G and H below.

Write in the box below the letter of the graph that shows how the rate constant k varies with temperature.

(1)
(Total 7 marks)

 

Q19.          (a)     In the presence of the catalyst rhodium, the reaction between NO and H2 occurs according to the following equation.
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
The kinetics of the reaction were investigated and the rate equation was found to be
rate = k[NO]2[H2]
The initial rate of reaction was 6.2 × 10–6 mol dm–3 s–1 when the initial concentration of NO was 2.9 × 10–2 mol dm–3 and the initial concentration of H2 was 2.3× 10–2 mol dm–3.
(i)      Calculate the value of the rate constant under these conditions and give its units.
Calculation ..........................................................................................
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Units ....................................................................................................
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(3)

(ii)     Calculate the initial rate of reaction if the experiment is repeated under the same conditions but with the concentrations of NO and of H2 both doubled from their original values.
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(1)

(b)     Using the rate equation and the overall equation, the following three-step mechanism for the reaction was suggested. X and Y are intermediate species.
Step 1           NO  +  NO       X
Step 2              X  +  H2         Y
Step 3              Y  +  H2         N2 + 2H2O
Suggest which one of the three steps is the rate-determining step.
Explain your answer.
Rate-determining step..................................................................................
Explanation ..................................................................................................
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(Extra space) ...............................................................................................
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(2)
(Total 6 marks)

Q20.          The rate of hydrolysis of an ester X (HCOOCH2CH2CH3) was studied in alkaline conditions at a given temperature. The rate was found to be first order with respect to the ester and first order with respect to hydroxide ions.
(a)     (i)      Name ester X.
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(1)

(ii)     Using X to represent the ester, write a rate equation for this hydrolysis reaction.
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(1)

(iii)     When the initial concentration of X was 0.024 mol dm–3 and the initial concentration of hydroxide ions was 0.035 mol dm–3, the initial rate of the reaction was 
8.5 × 10–5 mol dm–3 s–1.
Calculate a value for the rate constant at this temperature and give its units.
Calculation ..........................................................................................
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Units ....................................................................................................
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(3)

(iv)    In a second experiment at the same temperature, water was added to the original reaction mixture so that the total volume was doubled.
Calculate the initial rate of reaction in this second experiment.
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(1)

(v)     In a third experiment at the same temperature, the concentration of X was half that used in the experiment in part (a) (iii) and the concentration of hydroxide ions was three times the original value.
Calculate the initial rate of reaction in this third experiment.
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.............................................................................................................
(1)
(vi)    State the effect, if any, on the value of the rate constant k when the temperature is lowered but all other conditions are kept constant. Explain your answer.
Effect …...............................................................................................
Explanation .........................................................................................
(2)

(b)     Compound A reacts with compound B as shown by the overall equation
A + 3B → AB3
The rate equation for the reaction is
rate = k[A][B]2
A suggested mechanism for the reaction is
Step 1     A    + B → AB
Step 2     AB  + B → AB2
Step 3     AB2 + B → AB3
Deduce which one of the three steps is the rate-determining step.
Explain your answer.
Rate-determining step .................................................................................
Explanation ..................................................................................................
......................................................................................................................
(2)
(Total 11 marks)

Q21.Gases P and Q react as shown in the following equation.
 
2P(g)	+	2Q(g)		R(g)	+	S(g)	

The initial rate of the reaction was measured in a series of experiments at a constant temperature. The following rate equation was determined.
rate = k[P]2[Q]

(a)     Complete the table of data for the reaction between P and Q.
 
Experiment	Initial [P] / mol dm–3	Initial [Q] / mol dm–3	Initial rate / mol dm–3 s–1	
1	2.5 × 10–2	1.8 × 10–2	5.0 × 10–5	
2	7.5 × 10–2	1.8 × 10–2	 	
3	5.0 × 10–2	 	5.0 × 10–5	
4	 	5.4 × 10–2	4.5 × 10–4	

(Space for working) .......................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)

(b)     Use the data from Experiment 1 to calculate a value for the rate constant (k) at this temperature. Deduce the units of k.
Calculation .....................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
Units ...............................................................................................................
........................................................................................................................
(3)
(Total 6 marks)

 

Q22.This question involves the use of kinetic data to calculate the order of a reaction and also a value for a rate constant.
(a)     The data in this table were obtained in a series of experiments on the rate of the reaction between compounds E and F at a constant temperature.
 
 	Experiment	Initial concentration 
of E / mol dm −3	Initial concentration 
of F / mol dm −3	Initial rate of reaction 
/ mol dm−3 s−1	
 	1	0.15	0.24	0.42 × 10−3	
 	2	0.45	0.24	3.78 × 10−3	
 	3	0.90	0.12	7.56 × 10−3	

(i)      Deduce the order of reaction with respect to E.
...............................................................................................................
(Space for working) ...............................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)

(ii)     Deduce the order of reaction with respect to F.
...............................................................................................................
(Space for working) ...............................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)

(b)     The data in the following table were obtained in two experiments on the rate of the reaction between compounds G and H at a constant temperature.
 
 	Experiment	Initial concentration 
of G / mol dm−3	Initial concentration 
of H / mol dm−3	Initial rate of reaction 
/ mol dm−3 s−1	
 	4	3.8 × 10−2	2.6 × 10−2	8.6 × 10−4	
 	5	6.3 × 10−2	7.5 × 10−2	To be calculated	

The rate equation for this reaction is
rate = k[G]2[H]
(i)      Use the data from Experiment 4 to calculate a value for the rate constant k at this temperature. Deduce the units of k.
Calculation ............................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Units ......................................................................................................
...............................................................................................................
(3)

(ii)     Calculate a value for the initial rate of reaction in Experiment 5.
...............................................................................................................
...............................................................................................................
(1)
(Total 6 marks)

Q23.(a)    The data in the following table were obtained in two experiments about the rate of the reaction between substances B and C at a constant temperature.
 
 	Experiment	Initial concentration 
of B / mol dm−3	Initial concentration 
of C / mol dm−3	Initial rate / mol dm−3 s−1	
 	1	4.2 × 10−2	2.6 × 10−2	8.4 × 10−5	
 	2	6.3 × 10−2	7.8 × 10−2	To be calculated	

The rate equation for this reaction is known to be
rate = k[B]2[C]
(i)      Use the data from Experiment 1 to calculate a value for the rate constant k at this temperature and deduce its units.
Calculation ............................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Units ......................................................................................................
...............................................................................................................
(Extra space) ........................................................................................
...............................................................................................................
(3)

(ii)     Calculate a value for the initial rate in Experiment 2.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)

(b)     The data in the following table were obtained in a series of experiments about the rate of the reaction between substances D and E at a constant temperature.
 
 	Experiment	Initial concentration 
of D / mol dm−3	Initial concentration 
of E / mol dm−3	Initial rate /mol dm−3 s−1	
 	3	0.13	0.23	0.26 × 10−3	
 	4	0.39	0.23	2.34 × 10−3	
 	5	0.78	0.46	9.36 × 10−3	

(i)      Deduce the order of reaction with respect to D.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)

(ii)     Deduce the order of reaction with respect to E.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)

(c)     The compound (CH3)3CBr reacts with aqueous sodium hydroxide as shown in the folfollowing equation.
(CH3)3CBr + OH−   (CH3)3COH + Br−
This reaction was found to be first order with respect to (CH3)3CBr but zero order with respect to hydroxide ions.
The following two-step process was suggested.
Step 1 (CH3)3CBr   (CH3)3C+ + Br−
Step 2 (CH3)3C+ + OH−   (CH3)3COH
(i)      Deduce the rate-determining step in this two-step process.
...............................................................................................................
(1)

(ii)     Outline a mechanism for this step using a curly arrow.

 
(1)
(Total 8 marks)

Q24.Gases A and B react as shown in the following equation.
2A(g) + B(g)  C(g) + D(g)
The initial rate of the reaction was measured in a series of experiments at a constant temperature. The following rate equation was determined.
rate = k[A]2
An incomplete table of data for the reaction between A and B is shown in the table.
 
 	Experiment	Initial [A] / mol dm−3	Initial [B] / mol dm−3	Initial rate / mol dm−3 s−1	
 	1	4.2 × 10−3	2.8 × 10−3	3.3 × 10−5	
 	2	7.9 × 10−3	2.8 × 10−3	 	
 	3	 	5.6 × 10−3	1.8 × 10−4	

(a)     Use the data from Experiment 1 to calculate a value for the rate constant, k, at this temperature.
Deduce the units of k.
Calculation ......................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
Units ...............................................................................................................
........................................................................................................................
(3)

(b)     Use your value of k from (a) to complete the table for the reaction between A and B.
(If you have been unable to calculate an answer for (a), you may assume a value of 2.3. This is not the correct answer.)
(2)

(c)     The reaction is zero order with respect to B.
State the significance of this zero order for the mechanism of the reaction.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(Total 6 marks)

Q25.(a)     The table shows the results of three experiments to investigate the rate of reaction between compounds A and B dissolved in a given solvent.
All three experiments were carried out at the same temperature.
 
 	 	Experiment 1	Experiment 2	Experiment 3	
 	Initial concentration of A / mol dm–3	1.60 × 10–2	2.40 × 10–2	3.60 × 10–2	
 	Initial concentration of B / mol dm–3	4.20 × 10–2	6.30 × 10–2	6.30 × 10–2	
 	Initial rate /mol dm–3 s–1	8.00 × 10–5	1.80 × 10–4	4.05 × 10–4	

(i)      Deduce the order of reaction with respect to A.
Tick (✓) one box.
 
 	Order of reaction
with respect to A	Tick 
(✓)	
 	0	 	
 	1	 	
 	2	 	

(1)

(ii)     Deduce the order of reaction with respect to B.
Tick (✓) one box.
 
 	Order of reaction
with respect to B	Tick
(✓)	
 	0	 	
 	1	 	
 	2	 	

(1)

(b)     The reaction between two different compounds, C and D, is studied at a given temperature.
The rate equation for the reaction is found to be
rate = k[C][D]2
(i)      When the initial concentration of C is 4.55 × 10–2 mol dm–3 and the initial concentration of D is 1.70 × 10–2 mol dm–3, the initial rate of reaction is 6.64 × 10–5 mol dm–3 s–1.
Calculate the value of the rate constant at this temperature and deduce its units.
Calculation ............................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Units of rate constant ............................................................................
...............................................................................................................
(3)

(ii)     The experiment in part (i) is repeated at the same temperature but after the addition of extra solvent so that the total volume of the mixture is doubled.
Deduce the new initial rate of reaction.
...............................................................................................................
...............................................................................................................
(1)
(Total 6 marks)

Q26.Butadiene dimerises according to the equation
2C4H6  C8H12
The kinetics of the dimerisation are studied and the graph of the concentration of a sample of butadiene is plotted against time. The graph is shown below.

(a)     Draw a tangent to the curve when the concentration of butadiene is 0.0120 mol dm−3.
(1)

(b)     The initial rate of reaction in this experiment has the value 4.57 × 10−6 mol dm−3 s−1.
Use this value, together with a rate obtained from your tangent, to justify that the order of the reaction is 2 with respect to butadiene.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(5)
(Total 6 marks)

Q27.This question involves the use of kinetic data to deduce the order of a reaction and calculate a value for a rate constant.
The data in Table 1 were obtained in a series of experiments on the rate of the reaction between compounds A and B at a constant temperature.
Table 1
 
 	Experiment	Initial concentration of A / mol dm−3	Initial concentration of B / mol dm−3	Initial rate
/ mol dm−3 s−1	
 	1	0.12	0.26	2.10 × 10−4	
 	2	0.36	0.26	1.89 × 10−3	
 	3	0.72	0.13	3.78 × 10−3	

(a)     Show how these data can be used to deduce the rate expression for the reaction between A and B.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)

The data in Table 2 were obtained in two experiments on the rate of the reaction between compounds C and D at a constant temperature.
Table 2
 
 	Experiment	Initial concentration of C / mol dm−3	Initial concentration of D / mol dm−3	Initial rate
/ mol dm−3 s−1	
 	4	1.9 × 10−2	3.5 × 10−2	7.2 × 10−4	
 	5	3.6 × 10−2	5.4 × 10−2	To be calculated	

The rate equation for this reaction is
rate = k[C]2[D]

(b)     Use the data from experiment 4 to calculate a value for the rate constant, k, at this temperature. Deduce the units of k.

 

k = ............................... Units = ...............................
(3)

(c)     Calculate a value for the initial rate in experiment 5.

 

Initial rate = ............................... mol dm−3 s−1
(1)

(d)     The rate equation for a reaction is
rate = k[E]
Explain qualitatively why doubling the temperature has a much greater effect on the rate of the reaction than doubling the concentration of E.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)

(e)     A slow reaction has a rate constant k = 6.51 × 10−3 mol−1 dm3 at 300 K.
Use the equation ln k = ln A – Ea / RT to calculate a value, in kJ mol−1, for the activation energy of this reaction.
The constant A = 2.57 × 1010 mol−1 dm3.
The gas constant R = 8.31 J K−1 mol−1.

 

Activation energy = ...............................
(2)
(Total 12 marks)

Q28.The reaction between propanone and iodine in the presence of hydrochloric acid was studied at a constant temperature.
CH3COCH3   +   I2   →   CH3COCH2I   +   HI
The following rate equation was deduced.
rate = k [CH3COCH3][H+]
(a)     Suggest why the order with respect to iodine is zero.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
(1)

(b)     In an experiment the initial concentrations of propanone, iodine and hydrochloric acid were as shown in the table. The initial rate of reaction in this experiment was 8.64 × 10–7 mol dm–3 s–1.
 
 	 	Initial concentration / mol dm–3	
 	CH3COCH3	5.82 × 10–2	
 	I2	1.78 × 10–3	
 	H+	4.76 × 10–1	

Use the data in the table and the rate equation to calculate a value for the rate constant at this temperature.
Give units with your answer.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
(2)

(c)     A series of experiments was carried out using concentrations of propanone approximately 100 times the concentrations of iodine and hydrochloric acid.
Suggest the rate equation under these conditions.
Explain your answer.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
(2)
(Total 5 marks)

Q29.          (a)     The following table shows the results of three experiments carried out at the same temperature to investigate the rate of the reaction between compounds P and Q.
 
 	Experiment 1	Experiment 2 	Experiment 3	
Initial concentration of P/mol dm–3	0.50	0.25	0.25	
Initial concentration of Q/mol dm–3	0.36	0.36	0.72	
Initial rate/mol dm–3 s–1	7.6 × 10–3	1.9 × 10–3	3.8 × 10–3	

Use the data in the table to deduce the order with respect to P and the order with respect to Q.
Order with respect to P ................................................................................
Order with respect to Q ................................................................................
(2)

(b)     In a reaction between R and S, the order of reaction with respect to R is one, the order of reaction with respect to S is two and the rate constant at temperature T1 has a value of 4.2 × 10–4 mol–2 dm6 s–1.
(i)      Write a rate equation for the reaction. Calculate a value for the initial rate of reaction when the initial concentration of R is 0.16 mol dm–3 and that of S is 
0.84 mol dm–3.
Rate equation ....................…..............................................................
Calculation ..........................................................................................

(ii)     In a second experiment performed at a different temperature, T2, the initial rate of reaction is 8.1 × 10–5 mol dm–3s–1 when the initial concentration of R is 0.76 mol dm–3 and that of S is 0.98 mol dm–3. Calculate the value of the rate constant at temperature T2.
.............................................................................................................
.............................................................................................................
.............................................................................................................

(iii)     Deduce which of T1 and T2 is the higher temperature.
.............................................................................................................
(6)
(Total 8 marks)

 

Q30.The equation and rate law for the reaction of substance P with substance Q are given below.
2P + Q → R + S
        rate = k[P]2[H+]
Under which one of the following conditions, all at the same temperature, would the rate of reaction be slowest?
[P] / mol dm−3         pH
A               0.1                    0
B                1                     2
C                3                     3
D               10                    4
(Total 1 mark)

 

M1.D
[1]

M2.B
[1]

M3.C
[1]

M4.          (a)     Order with respect to iodine: 0 (1)
Overall order: 2 (1)
2

(b)     Rate constant: k =  = 4.4(4) × 10–4 (1)
Units: mol–1 dm3 s–1 (1)
3

(c)     Appears in rate equation (1)
OR implied by mention of concentration or order
          does not appear in (stoichiometric / overall) equation (1)
2

(d)     pH = –log10 [H+] (1)
      = 1.25
[H+] = 0.056(2) (1)
           rate = (4.44 × 10–4) × (1.50) × (0.0562)
 = 3.75 × 10–5 (1) (mol dm3 s–1)
(3.7 — 3.8)
Can score all 3 conseq on k from part (b)
3
[10]

 

M5.          (a)     (i)      (Experiment 1 → 2) [A] doubled, ([B] constant,)
rate doubled (1)
stated or shown numerically

(ii)     2 (1)
or shown as ... [B]2
2

(b)     (i)      k =  = 1.1(0) × 10–4
            (1)                        (1)
units of k: mol–2 dm6 s–1 (1)

(ii)     rate = (1.10 × 10–4) × (0.20)2 × (0.10)
       = 4.4(1) × 10–7 (mol dm–3 s–1)
                 (1) for the answer
Ignore units
Conseq on (i)
Upside down expression for k scores zero in (i) for 9073
but rate = 9073 × (0.2)2 × (0.1) = 36(.3)
conseq scores (1) in (ii)
4
[6]

 

M6.          (a)     Power (or index or shown as x in [ ]x) of concentration term
(in rate equation) (1)
1

(b)     2 (1)
1

(c)     (i)      Order with respect to A: 2 (1)
         Order with respect to B: 0 (1)
(ii)     Rate equation: (rate =) k [A]2 (1)
Allow conseq on c(i)
         Units for rate constant: mol–1 dm3 s–1 (1)
conseq on rate equation
4
[6]

          Organic points
(1)     Curly arrows: must show movement of a pair of electrons,
i.e. from bond to atom or from lp to atom / space
e.g.

(2)     Structures
penalise sticks (i.e. ) once per paper

Penalise once per paper
          allow CH3– or –CH3 or  or CH3
    or   H3C–

 

M7.          (a)     (i)      Experiment 2: 0.4(0) × 10–3 (1)
Experiment 3: 0.15 (1)
Experiment 4: 0.28 (1)
(ii)     k =  = 0.4(0) mol–2 dm6 s–1
                        (1)                   (1)          (1)
6

(b)     (change in) temperature (1)
1
[7]

 

M8.          (a)     (i)      2
1
(ii)     0
1

(b)     (i)      rate/[NO2]2[O2]
1
13
1
mol dm–3
1

(ii)     1.9 × 10–3
1
(iii)     Step 2
1
[7]

 

M9.          (a)     (i)      2 (1)
(ii)     0 (1)                                                                                                        2

(b)     (i)      Value of k: k =  =  = 13
Units of k: mol–2 dm6 s–1 (1)
(ii)     rate = 13 (6.5 × 10–2)2 (3.4 × 10–2)
       = 1.9 × 10–3        (mol dm–3 s–1) (1)
If k wrong, the mark in (ii) may be gained conseq for their 
k × 1.437 × 10–4
4
[6]

 

M10.          (a)     (i)      Experiment 2     2.60 × 10–3
1
Experiment 3     0.60 × 10–2
1
Experiment 4     11.4 × 10–2
1

(ii)     k = 
1
         = 49.7
(Allow 49.8 and 50)
1
         mol–2 dm6 s–1
1

(b)     No change
1
[7]

 

M11.          (a)     exp2      4.0 × 10–3
1
exp3      0.45 × 10–5
1
exp4      9.0 × 10–3
1

(b)     
1
2000
1
mol–2 dm6 s–1
1
[6]

 

M12.          (a)     propyl methanoate;
HCOOC3H7 + OH– → HCOO– + C3H7OH
1
OR
HCOOC3H7 + NaOH → HCOONa + C3H7OH;
1

(b)     order wrt A = 1;
1
order wrt NaOH = 1;
1
Initial rate in Exp 4 = 2.4 × 10–3;
1

(c)     (i)      r(ate) = k[A]
OR
r(ate) = k[A][NaOH]0;
(penalise missing [ ] but mark on)
(penalise missing [ ] once per paper)
(if wrong order, allow only units mark conseq on their rate eqs)
(penalise ka or kw etc)
1

(ii)     ;
1
= 0.45;
1
         s–1;
l

(iii)     (large) excess of OH– or [OH–] is large/high;
1
[OH–] is (effectively) constant
OR
[A] is the limiting factor                (Q of L mark)
1

(d)     (i)
          
1
          propan(e)-1,2,3-triol
          OR
          1,2,3-propan(e)triol
          OR
          Glycerol;
1

(ii)     CH3(CH2)16COONa  or  C17H35COONa or C18H35O2Na;
(ignore 3 in front of formula but not if indicating trimer)
1
(not just anion and penalise Na shown as covalently bonded) soap - 
allow with detergent but not detergent alone;
1
[15]

 

M13.          (a)     k = rate/[CH3CH2COOCH3][H+]
1
or
 = 
 = 1.38 × 10–3 to 1.4 × 10–3
1
mol–1 dm3 s–1
1

(b)     ans = rate constant × (½ × 0.150) × (½ × 0.555)
ignore units
      = rate constant × 0.0208
2.88 × 10–5                                     (1.38 × 10–3 gives 2.87 × 10–5)
Allow 2.87 – 2.91 × 10–5 (1.4 × 10–3 gives 2.91 × 10–5)
1

(c)     [H+] = rate/ k[CH3COOCH2CH3] 
1
= 
= 0.415 (0.4146)
1
pH = 0.38 mark independently
[H+] = 0.41 gives pH = 0.39
1
[7]

 

M14.          (a)     2 or two or second
1

(b)     k = 
mark is for insertion of numbers into a correctly rearranged rate equ, k = etc
if upside down, (or use of I2 data) score only units mark
1
= 3.44 × 10–5 (min 3sfs)
1
mol–1 dm3 s–1
any order
1

(c)     no change or no effect or stays the same or 1.24 × 10–4
1

(d)     1 or 2 or 1 and 2
if wrong no further mark but mark on from no answer
1
rate equ doesn’t involve I2 or only step which includes 2
species in rate equ
1

(e)     

any second arrow loses the mark
1
[8]

 

M15.(a)      Log (1 / time) on the y-axis + log (vol) on x-axis
If axes unlabelled use data to decide that log (1 / time) is on the y-axis
1

Sensible scales
Lose this mark if the plotted points do not cover at least half of the paper
Lose this mark if the graph plot goes off the squared paper
Lose this mark if plots a non-linear / broken scale
Lose this mark if uses an ascending y-axis of negative numbers
1

Plots points correctly ± one square
1

Line through the points is smooth
Lose this mark if the candidate’s line is doubled
1

Line through the points is best fit – ignores last point
Must recognise that point at 25 cm3 is an anomaly
If wrong graph, mark consequentially on anomaly if correctly plotted.
A kinked graph loses smooth and best fit marks
1

(b)     Uses appropriate x and y readings
Allow taken from table or taken or drawn on graph
Must show triangle on graph or such as 
1

Correctly calculates gradient 0.95 ± 0.02
Ignore positive or negative sign
Correct answer only with no working scores this mark
1

Answer given to 2 decimal places
1

(c)     First order or order is 1
Allow consequential answer from candidate’s results
1

(d)     Thermostat the mixture / constant temperature / use a water bath
or Colorimeter / uv-visible spectrometer / light sensor to monitor colour change
1

Reaction / rate affected by temperature change
or Eliminates human error in timing / more accurate time of colour change
1
[11]

M16.(a)     Order wrt D = 1 OR first OR [D] OR [D]1
Ignore working
1
Order wrt E = 2 OR second OR [E]2
1

(b)     (At time zero/start) the concentrations are known
1

(c)     M1     (Calculate) gradient (of tangent/curve/graph)
Allow description of gradient calculation:
Change in conc / time
1
M2     at t=0 or at start of graph/curve
M2 scored only if M1 gained
Ignore the word initial
1
[5]

M17.          (a)     3-hydroxybutanal
ignore number 1   i.e. allow 3-hydroxybutan-1-al
not hydroxyl
1

(b)     
1
= 1.1
1
mol–1 dm3 s–1
1

(c)     planar or flat C=O or molecule
allow planar molecule
1
equal probability of attack from above or below
must be equal; not attack of OH–
1

(d)     (i)      Step 1 if wrong – no mark for explanation.
1
involves ethanal and OH– or species/ “molecules”
in rate equation
1

(ii)     (B-L) acid or proton donor
not Lewis acid
1

(iii)     nucleophilic addition
QOL
1

(iv)

not allow M2 before M1, but allow M1 attack on C+ after non-scoring carbonyl arrow
ignore error in product
2

(e)     

1
[13]

 

M18.(a)    Exp 2   14.(4) ×10–3   OR   1.4(4) ×10–2   or 0.014
Allow 2sf
1

Exp 3   0.1(0)
1

Exp 4   0.3(0)
If three wrong answers, check their value of k in 1(b).
They can score all 3 if they have used their (incorrect) value of k. see below.
Exp 2   rate = 0.096 × k
Exp 3   [Q] = 0.015/k
Exp 4   [P] = 0.116/√k
1

(b)      
mark is for insertion of numbers into a correctly rearranged rate equ , k = etc
1

= 0.15 (min 2sfs)                   (allow )
if upside down, score only units mark
AE (–1) for copying numbers wrongly or swapping two numbers
1

mol–2 dm+6 s–1
Any order
If k calculation wrong, allow units conseq to their k
1

(c)     G
1
[7]

M19.          (a)     (i)      
mark is for insertion of numbers into a correctly
rearranged rate equ, k = etc
AE (-1) for copying numbers wrongly or swapping two numbers
1
= 0.32 (min 2sfs)
1
mol–2 dm6 s–1 Units must be conseq to their k
Any order
If k calculation wrong, allow units conseq to their k
1

(ii)     4.95 × 10–5 to 4.97 × 10–5 or 5.0 × 10–5 (min 2 sfs)
(ignore units)
rate = their k × 1.547 × 10–4
1

(b)     Step 2
If wrong no further mark
1
One H2 (and two NO) (appear in rate equation)
or species (in step 2) in ratio/proportion as in the rate equation
1
[6]

 

M20.          (a)     (i)      propyl methanoate
must be correct spelling
1

(ii)     rate = k[X][OH–]
allow HCOOCH2CH2CH3 (or close) for X
allow ( ) but penalise missing minus
1

(iii)     k = 
In (a)(iii), if wrong orders allow
mark is for insertion of numbers in correct expression for k
If expression for k is upside down, only score units conseq to their expression
1
= 0.10(12)      2sf minimum
1 for conseq answer
1
mol–1 dm3 s–1
1 for conseq units
any order
1

(iv)    2.1(3) × 10–5
or 2.1(2) × 10–5          ignore units
allow 2 sf
NB If wrong check the orders in part (a)(iii) and allow (a)(iv) if conseq to wrong k
See * below
1

(v)     1.3 ×10–4 (1.28 ×10–4)
allow (1.26 × 10–4) to (1.3 × 10–4)        ignore units
allow 2 sf
NB If wrong check the orders in part (a)(iii) and allow (a)(iv) if conseq to wrong k
See ** below
1

For example, if orders given are 1st in X and second in OH–
[The mark in a(ii) and also first mark in a(iii) have already been lost]
So allow mark   * in (iv) for rate = their k × (0.012)(0.0175)2 = their k ×(3.7 × 10–6)
                           (allow answer to 2sf)
** in (v) for rate = their k × (0.012)(0.105)2 = their k ×(1.32 × 10–4)
                           (allow answer to 2sf)
The numbers will of course vary for different orders.

(vi)    Lowered
if wrong, no further mark
1
fewer particles/collisions have energy > Ea
OR
fewer have sufficient (activation) energy (to react)
not just fewer successful collisions
1

(b)     Step 2
1
(this step with previous) involves one mol/molecule/particle
A and two Bs
or 1:2 ratio or same amounts (of reactants) as in rate equation
if wrong, no further mark
1
[11]

 

M21.(a)     Exp 2  4.5 ×10–4
Min 2sf
1

Exp 3  4.5 ×10–3
If three wrong answers, check their value of k in (b).
1

Exp 4  0.043 OR 4.3 ×10–2        OR 0.044 OR 4.4 ×10–2
They can score all 3 if they have used their (incorrect) value of k. see below.
Exp 2 rate = k × (1.0125 × 10–4)
Exp 3 [Q] = 0.02/k
Exp 4 [P] = 0.0913/√k
1

(b)     
Mark is for insertion of numbers into a correctly rearranged rate equ , k = etc
If upside down, score only units mark from their k
AE (-1) for copying numbers wrongly or swapping two numbers
1

= 4.4(4) (allow 40/9)
1

mol–2dm+6s–1
Any order
If k calculation wrong, allow units conseq to their k expression
1
[6]

M22.(a)     (i)      2 or two or second or [E]2
1

(ii)     1 or one or first or [F]1 or [F]
1

(b)    (i)      k = 
mark is for insertion of numbers into a correctly rearranged rate equ , k = etc.
AE (−1) for copying numbers wrongly or swapping two numbers.
1

= 22.9 (Allow 22.9 − 24 after correct rounding)
1

mol−2dm+6 s&8722;1
Any order.
1

(ii)     6.8(2) × 10−3 (mol dm&8722;3s−1)
OR if their k is wrong, award the mark consequentially
a quick check can be achieved by using
their answer = 2.9768 × 10−4 Allow 2.9 − 3.1 × 10−4 for the mark
    their k
Allow 6.8 × 10−3 to 6.9 × 10−3
Ignore units.
1
[6]

M23.(a)     (i)  
 
Mark is for insertion of numbers into a correctly rearranged rate equ , k = etc.
If upside down, score only units mark from their k
AE (−1) for copying numbers wrongly or swapping two numbers
1

= 1.8(3)
1

mol−2 dm+6 s−1
Any order
If k calculation wrong, allow units consequential to their k = expression
1

(ii)     5.67 × 10−4 (mol dm−3 s−1)   OR    their k × 3.1 × 10−4
Allow 5.57 × 10−4 to 5.7 × 10−4
1

(b)     (i)      2 or second or [D]2
1

(ii)     0 or zero or [E]0
1

(c)     (i)      Step 1 or equation as shown
Penalise Step 2 but mark on
1

(ii)
Ignore correct partial charges, penalise full / incorrect partial charges
If Step 2 given above, can score the mark here for

allow: OH− (must show lp)
If SN2 mechanism shown then no mark (penalise involvement of :OH− in step 1)
Ignore anything after correct step 1
1
[8]

M24.(a)     k = rate / [A]2 or 
1
= 1.87 or 1.9
Answer scores 2
1.90 scores first mark only (incorrect rounding)
1
mol−1dm3s−1
Any order and independent of calculation
1

(b)     Expt 2 rate = 1.167 × 10−4 − 1.2 × 10−4 (mol dm−3 s−1)
If answers in table are not those given here, check their value of k in part (a) or use of alternative k.
1
Expt 3 [A] = 9.7 × 10−3 − 9.8(1) × 10−3 (mol dm−3)
If their k is incorrect in part (a) mark this part consequentially e.g. if k = 7.9 × 10−3 due to lack of squaring in (a)
Using alternative value for k
expt 2 4.9 × 10−7
Expt 2 rate = 1.4(4) × 10−4 (mol dm−3 s−1)
expt 3 1.5 ×10−1
Expt 3 [A] = 8.85 × 10−3 (mol dm−3)
(expt 2 6.24 × 10−5 × their k)
(expt 3 0.0134 / √k)
1

(c)     Slow step or rds involves only A
OR
B does not appear in the slow step or the rds
OR
B only appears after the slow step or the rds
Not B has no effect on the rate or B is not in the rate equation
Allow “it” for B
1
[6]

M25.(a)     (i)      2
1

(ii)     0
1

(b)    (i)      K = 
 
Correct answer for k with or without working scores 2.
First mark is for insertion of numbers into a correctly rearranged rate equ , k = etc.
1

  = 5.05      (range allowed 5.03−5.07)
AE (−1) for copying numbers wrongly or swapping two numbers.
1

mol−2 dm+6 s−1
Mark units separately, ie only these units but can be in any order.
1

(ii)     8.3 × 10−6 (mol dm−3 s−1)
Allow 0.83 × 10−5.
Ignore units.
OR if not 8.3 × 10−6, look at their k in part(i) and if not 5.05
Allow ecf for their (incorrect) k × (1.64 × 10−6)
1
[6]

M26.(a)     Gradient drawn on graph
 
Line must touch the curve at 0.012 but must not cross the curve.
1

(b)     Stage 1: Rate of reaction when concentration = 0.0120 mol dm–3
From the tangent
Change in [butadiene] = –0.0160 – 0 and change in time = 7800 – 0
Extended response
1

Gradient = –(0.0160 – 0) / (7800 – 0) = –2.05 × 10–6
Rate = 2.05 × 10–6 (mol dm–3 s–1)
1

Stage 2: Comparison of rates and concentrations
Initial rate / rate at 0.0120 = (4.57 × 10–6) / (2.05 × 10–6) = 2.23
Marking points in stage 2 can be in either order
1

Inital concentration / concentration at point where tangent drawn = 0.018 / 0.012 = 1.5
1

Stage 3: Deduction of order
If order is 2, rate should increase by factor of (1.5)2 = 2.25 this is approximately equal to 2.23 therefore order is 2nd with respect to butadiene
1
[6]

M27.(a)     Consider experiments 1 and 2: [B constant]
[A] increases × 3: rate increases by 32 therefore 2nd order with respect to A
1

Consider experiments 2 and 3:
[A] increases × 2: rate should increase × 22 but only increases × 2
Therefore, halving [B] halves rate and so 1st order with respect to B
1

Rate equation: rate = k[A]2[B]
1

(b)     rate = k [C]2[D] therefore k = rate / [C]2[D]
1

 
 
Allow consequential marking on incorrect transcription
1

mol–2 dm+6 s–1
Any order
1

(c)     rate = 57.0 × (3.6 × 10–2)2 × 5.4 × 10–2 = 3.99 × 10–3 (mol dm–3 s–1)
OR
Their k × (3.6 × 10–2)2 × 5.4 × 10–2
1

(d)     Reaction occurs when molecules have E>Ea
1

Doubling T by 10 °C causes many more molecules to have this E
1

Whereas doubling [E] only doubles the number with this E
1

(e)     Ea = RT(lnA – lnk) / 1000
Mark is for rearrangement of equation and factor of 1000 used correctly to convert J into kJ
1

Ea = 8.31 × 300 (23.97 – (–5.03)) / 1000 = 72.3 (kJ mol–1)
1
[12]

M28.(a)     Iodine is not involved in (or before) the rate determining / slow(est) / limiting step (in the mechanism)
Ignore, iodine does not appear in the rate equation or iodine concentration does not affect the rate
1

(b)     
Mark for answer
1
mol–1 dm+3 s–1
Mark units separately, i.e. only these units but can be in any order
1

(c)     Rate = k [H+]
If wrong or missing CE = 0
1
(Large excess of propanone) so [CH3COCH3] is (effectively) constant
1
[5]

M29.         (a)     order with respect to P is 2
1
order with respect to Q is 1
1

(b)     (i)      rate = k[R][S]2
(if wrong expression, no further marks)
1
rate = (4.2 × 10–4) × 0.16 × 0.842
1
= 4.7 × 10–5 (mol dm–3 s–1)
ignore units even if wrong
1

(ii)     
1
= 1.1 × 10–4
1

(iii)     T1
 *If calculated value for k > 4.2 × 10–4, then answer to (iii) is T2
1
[8]

 

M30.C
[1]

 

E4.          This question on rate equations was answered well by most candidates even those who scored less well elsewhere in the paper. In part (a) most candidates deduced the overall order correctly, but to deduce that the order with respect to iodine was zero proved more difficult. Part (b) was well answered by most students, but only the better candidates were able to gain both marks in part (c). Many gave a general description of the role of a catalyst rather than the evidence that hydrogen ions appear in the rate equation but not in the overall stoichiometric equation. The use of pH to show a change of hydrogen ion concentration was understood by most candidates although some were unable to convert a pH value into a hydrogen ion concentration. Full marks were available for correct use in part (d) of an incorrect rate constant calculated in part (b).

E5.          This question on rate equations was answered well by most candidates. Part (a)(i) required that the candidate stated or showed mathematically that the rate doubled when the concentration of A doubled when considering experiments 1 and 2. Part (a)(ii) was usually answered correctly. Problems arose in part (b) where some candidates had difficulty in rearranging the rate equation correctly. However the mark in part (b)(ii) could still be gained for correct working despite using a wrong rate constant calculated in part (b)(i).

E6.          In part (a) few candidates were able to explain correctly what the term ‘order of reaction’ means. However, despite this most were able to use the idea itself and scored highly in the rest of the question. The fact that the rate of reaction increased by a factor of nine, 32 , when the concentration of A increased by three times led some candidates to think that the order with respect to A was three rather than two.

E7.          This question was generally answered well even by weaker candidates. Calculation of the concentration of P in part (a)(i) was the only part found difficult by many. Most answers to part (a)(ii) were correct, including the units. A change in temperature was the usual and correct answer in part (b).

E9.          As is usual in this type of question, high marks were obtained by most candidates. Part (a)(ii) caused the most problems with the answer 2 being common from the weaker candidates. Occasionally in part (b) the value of the concentration of NO was not squared in the calculation and the units were sometimes not given correctly. The value of the initial rate in part (b)(ii) was allowed consequential to an incorrect answer in part (b)(i). 

E10.          This question was generally answered well as is usual in questions about rates of reaction. As expected, the value required in experiment 4 was the least well answered section in part (a)(i). In part (a)(ii), the major problem for some was to rearrange the rate equation to produce k as the subject of an equation. Disappointing numbers of candidates were unable to do this correctly. The units were usually correctly given as was the answer to part (b).

E11.          This question was answered well by most candidates and full marks were often gained. The value of the concentration of NO in experiment 4 was the least well answered part of section (a). Only a few candidates failed to calculate the answer 2000 for the rate constant in part (b). The few errors seen usually involved a failure to square the NO concentration or problems with the powers of 10. The units were usually correctly given.

E12.          The first three parts of this question were well answered. In part (a), most candidates were able to name the ester propyl methanoate correctly although the equation for its reaction with sodium hydroxide was less well answered. The orders and the initial rate in part (b) were correctly deduced by most candidates and many were able to deduce the correct rate equation in part (c) and to use it correctly to calculate the rate constant. Those who gave an incorrect rate equation were awarded a mark for units which were consistent with their rate equation. By comparison, few were able to answer part (c)(iii) correctly by stating that the large excess of sodium hydroxide meant that its concentration was effectively constant. Part (b) distinguished fairly well between those who recognised the alcohol formed by hydrolysis of the ester drawn and were able to deduce that it would distil easily. Either the systematic name, propane-1,2,3-triol, or its common name glycerol, were accepted. The formula of the sodium salt of the acid was accepted in several forms and its use as soap was well known by most, even those who gained no other marks in part (c).

E13.          As usual with questions involving rate equations, this question was answered well, particularly parts (a) and (c). Part (b) however was poorly answered and only the very best candidates gained this mark. Many of the rest only divided the original rate by 2 instead of by 4 as the concentrations of both the ester and of If” ions were halved. Again it was disappointing to see the number of candidates who did not give the pH to two decimal places despite being asked to do so.

E14.          As usual for questions on kinetics, this was answered well. It was also pleasing that, in part 3(d), over half of the candidates gained both marks on a question which required understanding of a new part of the specification. Application in part (e) of their understanding of the use of curly arrows in organic mechanisms was, however, less well done.

E15.In this Section, the graphs seen showed the usual range of mistakes seen in this type of question as mentioned in the marking scheme. The main difficulty was in ensuring that the plotted graph covered at least half the printed grid.
In part (b), most candidates were able to find an appropriate gradient and full marks were common. Allowance was made for consequential errors from the drawing of the graph but in part (c) the order quoted needed to be based on the final answer to part (b).
Candidates had some trouble with part (d), many thinking that a ‘white tile’ would improve the precision of time measurement.

E16.Part (a) was answered well and two thirds of students scored both marks. By contrast, part (b) was a hard question with few correct answers. Many students spoke about the changing concentrations but few mentioned the importance of knowing the initial concentration. The first mark in part (c) for mentioning a gradient was often gained but the location of where to measure the gradient was only given by a third of students.

E17.          This question was generally well done although in part (a), fewer candidates than expected could name compound X correctly. Part (b) was answered very well but in part (c) many candidates did not read the full question and described what a racemic mixture contains rather than how one is formed in this reaction. Part (d) was answered very well except for part (d)(ii) where fewer than 50% of candidates could deduce that ethanal was acting as a proton donor in the reaction. The deduction required in part (e) was also found difficult despite the hint given that the required product contained six carbon atoms.

E18.The kinetics question was, as usual, answered well with only Experiment 4 in part (a) causing much difficulty. However, in part (c) fewer than half of the students identified graph G as that correctly showing how the rate constant k varies with temperature.

 

E19.          Part (a) was well done with many candidates scoring full marks. Although most gave the correct answer to part (a)(ii), many did it by the long route, having failed to see that the answer was simply 8x the initial rate.
Candidates found part (b) more difficult. Many of those who correctly chose step 2 could not offer a full explanation.

E20.          Most parts of this question were answered well. The ester was named correctly by over two thirds of the candidates but a surprising number attempted to write a stoichiometric equation in part (a)(ii) rather than a rate equation. Part (a)(iii) was answered very well but in part (a)(iv) the effect of dilution on the concentration of both reagents was missed by a large majority. However, when the exact concentrations were given in part (a)(v) most candidates were able to calculate the new rate of reaction correctly. The effect on the rate constant of lowering the temperature was well known in part (a)(vi), but few candidates were able to give a correct explanation and many offered a mathematical rather than a kinetic explanation. In part (b), the requirement to use the rate equation to identify the rate determining step proved difficult for the majority. 

E21.          As is usual with kinetics questions, this produced some high marks, with many students scoring full marks. In part (a), the more tricky calculation for Expt 4 was least well done and in part (b) the most common mistake in calculating the value of k was to forget to square the concentration of P.

 

E22.This question was answered well. A few students lost marks through the incorrect use of calculators in the division necessary in part (b)(i) to find the rate constant, k, or by omitting s-1 from the units of the rate calculated in part (b)(ii).

E23.This question was answered well by most students and over 80% scored full marks in part (a). Part (c)(ii) was found slightly harder and, of those who correctly identified Step 1 as the slower step, most gained the mark for the curly arrow. Those who suggested Step 2 could also gain a mark for a correct arrow for that step. However, many gave a full SN2 mechanism in both cases.

E24.The calculation in part (a) was well answered and there were few errors although the units mark was occasionally lost. In part (b) the calculations were again well answered: experiment 3 involving a square root was the least well done. Marks were awarded consequentially on an incorrect value of k. Few students used the value of k of 2.3 supplied.
Here as elsewhere, students should be encouraged to keep their answer in their calculator as they work through a question and then write the final answer to the appropriate number of significant figures (usually 3).
The mark in part (c) proved elusive for many students, who referred to the rate of the reaction and not the mechanism as required by the question. It was well known that the concentration of B would not affect the rate, but how B was involved in the mechanism and the rate determining step was not well known.

E25.Most students scored highly on this question. Only part (b)(ii) proved difficult for some who were unable to calculate the new initial rate in part (b)(ii) either by dividing the given rate by 8 or by using their rate constant to calculate the new rate.

E28.Part (a) was answered correctly by just over half of the students. The majority of incorrect answers were because students didn’t answer the question and simply just repeated what the rate equation told them.
Calculation of a rate constant in part (b) was answered very well as usual but part (c) proved to be very difficult. A large number of students could not deduce the correct rate equation. Those who did gain this mark were not often able to give a reason for their answer and the second mark was rarely seen.

E29.          As is normal, the use of rate equations is well understood and almost all candidates can answer these questions well. Very few failed to score the marks for deducing the orders with respect to P and Q. The calculation in part (b) using the given orders of reaction was well done by all except those candidates who could not rearrange the equation correctly. The same applied in part (b)(ii). In the few cases where the calculated value of the rate constant was incorrect, and greater than 4.2 × 10–4 mol –2 dm 6 s –1 the usual answer of T1 for part (b)(iii) was replaced by the answer T2  consequential on their incorrect value of k.