Rate Equations (A Level only) (AQA A Level Chemistry)

Exam Questions

3 hours30 questions
1a1 mark

Define the term rate of reaction.

1b4 marks

Substances A and B react in solution according to the equation.

A + 2B → C + 2D

The rate equation for the formation of C and D from A and B is:

rate = k [B]

The data in Table 1 was obtained in a series of experiments on the rate of the reaction between 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

0.15

0.15

3.5 x 10-4

2

0.30

0.15

 

3

0.30

0.30

 

i)
State the order of reaction with respect to A

ii)
State the initial rate for experiment 2.

iii)
The reaction is first order with respect to B. Calculate the initial rate for experiment 3.

iv)
State the overall order of the reaction.
1c3 marks

Substances E and F react in solution according to the equation.

2E + 2F → 2G + H

Overall, the reaction is second order.

Suggest the three combinations for the orders with respect to E and F.

1d5 marks

The data in Table 2 was obtained in a series of experiments on the rate of the reaction between E and F at a constant temperature.

Table 2

Experiment

Initial concentration of E / mol dm–3

Initial concentration of F / mol dm–3

Initial rate / mol dm–3 s–1

1

0.20

0.25

2.4 x 10-3

2

0.40

0.25

2.4 x 10-3

3

0.40

0.50

9.6 x 10-3

i)
Use experiments 1 and 2 to explain the order of reaction with respect to E

ii)
Using experiments 2 and 3, describe the effect of doubling the concentration of F on the initial rate.

iii)
Deduce the order of reaction with respect to F.

iv)
Write the rate equation for the reaction between E and F.

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2a1 mark

For a general reaction between substances A and B, what is meant by order of reaction with respect to A?

2b1 mark

Nitrogen reacts with hydrogen according to the equation.

N2 (g) + 3H2 (g) → 2NH3 (g)

Explain why the rate equation for this reaction is not rate = k [N2] [H2]3.

2c3 marks

Nitrogen monoxide is oxidised to nitrogen dioxide according to the following equation.

2NO (g) + O2 (g) → 2NO2 (g)

The data in Table 1 shows the results of experiments completed to study the reaction kinetics of the oxidation of nitrogen monoxide, at a constant temperature.

Table 1

Experiment

1

2

3

Initial concentration of NO (g) / mol dm–3

3.20 x 10-3

6.40 x 10-3

6.40 x 10-3

Initial concentration of O2 (g) / mol dm–3

1.20 x 10-3

1.20 x 10-3

3.60 x 10-3

Initial rate / mol dm–3 s–1

5.60 x 10-4

2.24 x 10-3

6.72 x 10-3

i)
Use experiments 1 and 2 to deduce the order of reaction with respect to NO. 

ii)
Use experiments 2 and 3 to deduce the order of reaction with respect to O2.

iii)
Write the rate equation for the reaction of nitrogen monoxide with oxygen.
2d2 marks

Nitrogen monoxide reacts with hydrogen according to the equation.

2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)

The rate equation for this reaction is:

rate = k [NO]2 [H2]

i)
Rearrange the rate equation to write an expression for the rate constant, k.

ii)
Deduce the units of the rate constant, k.

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3a5 marks

The Arrhenius equation can be written as:

                                straight k space equals space Ae to the power of negative straight E subscript straight a divided by RT end exponent

State what each of the following terms represents, including units where applicable.

  • A
  • Ea
  • R
  • T
3b1 mark

Rearrange the Arrhenius equation given in part (a) to make A the subject.

3c3 marks

The Arrhenius equation can also be written in natural logarithmic forms.

ln k = ln A - straight E subscript straight a over RT

A plot of ln k against begin mathsize 16px style 1 over straight T end style gives a straight line graph of the type y = mx + c.

Complete Table 1 which relates the terms from the natural logarithmic Arrhenius equation to the equation of a straight line.

Table 1

Straight line term

Arrhenius term

y

ln k

m

 

x

 

c

 

3d3 marks

A graph of ln k against begin mathsize 16px style 1 over straight T end style is shown in Figure 1.

Figure 1

mZZmWz2R_1

i)
Calculate the gradient of the straight line.

ii)

Calculate the activation energy, Ea.
The gas constant R = 8.31 J K-1 mol-1

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4a1 mark

Bromide ions and bromate (V) ions react in acidic conditions according to the following equation.

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

A series of experiments was carried out at a given temperature to deduce the rate equation for the reaction.

Figure 1 shows a graph of the results from experiments using different hydrogen ion concentrations.

Figure 1

RCK~d3_4_3

Use the information in Figure 1 to deduce the order of reaction with respect to hydrogen ions.

4b1 mark

Figure 2 shows a graph of the results from experiments using different bromide ion concentrations.

Figure 2

NKb1xPu~_4

Use the information in Figure 2 to deduce the order of reaction with respect to bromide ions.

4c2 marks

Overall, the reaction between bromide ions and bromate (V) ions in acidic conditions is fourth order.

i)
Deduce the order with respect to bromate (V) ions.

ii)
Using your answers to parts (a) and (b), write the rate equation for the reaction.
4d2 marks

Bromide ions and bromate (V) ions react in acidic conditions according to the following equation.

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

i)

Using your answer to part (c), write an expression for the rate constant for the reaction between bromide ions and bromate (V) ions in acidic conditions.

ii)

Suggest suitable units for the rate constant.
(If you did not get an answer for (c), you may assume that rate = k [BrO3-]2 [H+]. This is not the correct answer)

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5a2 marks

Substances A and B react in solution according to the equation.

A (aq) + 2B (aq) → C (aq) + 2D (aq)

In an experiment, the volume of C produced was measured every 10 seconds. The results of the experiment are shown in Table 1.

Table 1

Time / s

0

10

20

30

40

50

60

70

80

90

100

Volume of C produced / cm3

0

12

24

35

40

46

49

50

51

52

52


Plot these results on Figure 1 and draw a curve of best fit.

Figure 1

Hs96751R_5

5b2 marks

Concentration-time graphs are used to calculate initial rates of reaction.

i)
On your Figure 1 graph from part (a), draw a tangent to the curve at t = 0.

ii)

Use this tangent to deduce the initial rate of the reaction.
Initial rate _____________________ mol dm−3 s−1

5c1 mark

Suggest why initial rates of reaction are used to determine orders, rather than rates of reaction at other times during the reaction.

5d5 marks

The general equation for a reaction as described in part (a) is shown.

A (aq) + 2B (aq) → C (aq) + 2D (aq)

Compounds A and B are colourless. Compound C is dark blue.

A series of initial rate experiments is planned to determine the order of this reaction with respect to A

i)
State the independent variable.

ii)
State three control variables.

iii)
State the dependent variable.

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1a4 marks

Two compounds, X and Y, were reacted together.

The initial rate of reaction when compound X and compound Y were reacted together was measured in a series of experiments.

The temperature was kept constant and the results of the experiments are shown in Table 1 below.

Table 1

Expt.

Initial [X] / mol dm-3

Initial [Y] / mol dm-3

Initial rate / mol dm-3 s-1

1

0.030

0.040

4.0 x 10-4

2

0.045

0.040

6.0 x 10-4

3

0.045

0.060

9.0 x 10-4

4

0.060

0.120

2.4 × 10-3

i)
Use the data in Table 1 to deduce the order of reaction with respect to X.

ii)
State the order of the reaction with respect to Y.

iii)
Determine the overall order of the reaction.

iv)
Write the rate equation for the reaction.
1b5 marks

Table 2 shows the results from three different experiments carried out at a constant temperature, to investigate the rate of reaction between compounds A and B in a different chemical reaction.

A + B → Products

Table 2

Expt.

Initial concentration of A / mol dm-3

Initial concentration of B / mol dm-3

Initial rate / mol dm-3 s-1

1

0.50

0.30

7.6 x 10-4

2

0.25

0.30

1.9 x 10-4

3

0.25

0.60

3.8 x 10-4


Use the data from Table 2 to calculate a value for the rate constant, k.

1c2 marks

An overall rate equation is given below.

Rate = k[P][Q]2

i)
State what the units would be of the rate constant, k, in this reaction.

ii)
State the overall order of the reaction above.
1d3 marks

Chemists measured the rate of a chemical reaction in a series of experiments between compounds C and D at a fixed temperature as shown in Table 3 below.

Table 3

Experiment

Initial concentration of C / mol dm-3

Initial concentration of D / mol dm-3

Initial rate / mol dm-3 s-1

1

0.13

0.12

0.32 x 10-3

2

0.39

0.12

2.88 x 10-3

3

0.78

0.24

11.52 x 10-3

i)
Deduce the order of the reaction with respect to C.

ii)
Deduce the order of the reaction with respect to D.

iii)
Write the overall rate equation for this reaction.

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2a5 marks

Chemists can use kinetic studies to help suggest mechanisms for reactions.

The following data in Table 1 was obtained by reacting compounds A and B together in a series of experiments at a constant temperature. 

Table 1

Experiment

Initial [A]  / mol dm-3

Initial [B] / mol dm-3

Initial rate / mol dm-3 s-1

1

0.14

0.22

0.26 x 10-3

2

0.42

0.22

2.34 x 10-3

3

0.84

0.44

9.36 x 10-3

i)
Show how the data in the table can be used to deduce that the reaction is second-order with respect to A.

ii)
Deduce the order with respect to B.

iii)
Calculate the value for the rate constant, k. State units in your answer.
2b4 marks

The set of data below in Table 2 was obtained from a series of experiments reacting compounds X and Y together at a fixed temperature. The rate equation for this reaction is:

Rate = k[X]2[Y]

Table 2

Experiment

Initial concentration of X / mol dm-3

Initial concentration of Y / moldm-3

Initial rate / mol dm-3 s-1

1

0.85

1.70

9.30 x 10-5

2

0.30

0.15

To be calculated

i)
Use the data from experiment 1 in Table 2, calculate the value for the rate constant, k and state the units.

ii)

Using the rate constant k from part (i), calculate the initial rate of the reaction for experiment 2.

(If you have been unable to calculate and answer for (i), you may assume a value of 6.10 x 10-5. This is not the correct answer.)

2c4 marks

The rate of reaction data between nitrogen monoxide (NO) and oxygen (O2) was obtained across a series of experiments at a constant temperature, as shown in Table 3 below.

The rate equation for this reaction is:

Rate = k [NO]2 [O2]

Table 3

Experiment

Initial concentration of O2 / mol dm-3

Initial concentration of NO / mol dm-3

Initial rate / mol dm-3 s-1

1

1.0 x 10-2

5.0 x 10-2

6.5 x 10-4

2

3.4 x 10-2

6.4 x 10-2

To be calculated

i)

Use the data for Experiment 1, calculate the value for the rate constant k at this temperature and state its units.

ii)
Calculate the value for the initial rate in Experiment 2.
2d5 marks

Nitrogen dioxide reacts with carbon monoxide at 100 degreeC to form nitrogen monoxide and carbon dioxide, as shown in the equation below;

2NO2 (g) + CO (g) → NO (g) + CO2 (g)

The initial rate of reaction was measured in a series of experiments at a constant pressure. The rate equation below was determined;

Rate = k[NO2]2

An incomplete table of data for the reaction between NO2 and CO is shown in Table 4.

Table 4

Experiment

Initial concentration of NO2 / mol dm-3

Initial concentration of CO / mol dm-3

Initial rate / mol dm-3 s-1

1

4.1 x 10-2

2.8  x 10-3

3.3  x 10-5

2

7.8  x 10-3

2.8  x 10-3

 

3

 

5.6  x 10-3

1.8  x 10-4

i)

Use the data from Experiment 1 to calculate a value for the rate constant, k, at this temperature. State its units.

ii)

Use your value of k from (i) to complete the table for the reaction between NO2 and CO.
(If you have been unable to calculate and answer for (i), you may assume a value of 2.3. This is not the correct answer.)

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3a2 marks

For this question look at Figure 1 below.

Figure 1

KBCMhSeW_8

i)

Identify which of the graphs A, B, C or D shows how the rate constant, k, varies with temperature.

ii)

State the effect, if any, increasing the concentration of a reactant would have on the value of the rate constant, k.

3b3 marks

Compound X and Y react together in the following equation.

X + 4Y → XY4

The rate equation for this reaction is below.

rate = k[X][Y]2

A possible mechanism for this reaction is

Step 1              X + 2Y → XY2

Step 2              XY2 + Y → XY3

Step 3              XY3 + Y → XY4

i)
State the overall order for this reaction.

ii)

Deduce which one of the three steps is the rate determining step. Give reasons for your answer.

3c5 marks

Chemists studied the rate of reaction of ethanal dimerises in dilute alkaline solution to form 3-hydroxybutanal in the following equation at a constant temperature of 298K.

2CH3CHO → CH3CH(OH)CH2CHO

Rate = k[CH3CHO][:OH]

Table 1

Initial concentration of ethanal / mol dm-3

Initial concentration of sodium hydroxide / mol dm-3

Initial rate / mol dm-3 s-1

0.15

0.025

2.7 x 10-3

 

i)

Use the data from Table 1 to calculate a value for the rate constant k and state its units.

A three step mechanism for this reaction has been suggested and is shown in Figure 2. 

Figure 2

Step 1:

CH3CHO + :OH- → :CH2CHO + H2O

Step 2:

CH3CHO + :CH2CHO → CH3CH(O:-)CH2CHO

Step 3:

CH3CH(O:-)CH2CHO + H2O → CH3CH(OH)CH2CHO + :OH-



ii)

Using the rate equation, predict which of these steps is the rate-determining step and explain why.

3d3 marks

The overall rate equation for a reaction that occurred is below.

Rate = k[A]

Explain why doubling the temperature of this reaction would have a much greater effect on the rate of the reaction, compared to doubling the concentration of A.

You must refer to kinetic theory in your answer.

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4a2 marks

The decomposition of hydrogen peroxide into water and oxygen occurs at a slow rate with a rate constant of k = 6.62 x 10-3 mol dm-3 s-1 and at a temperature of 290 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 gas constant R = 8.31 J K−1 mol−1.

The constant A = 3.18 × 1011 mol−1 dm3.

4b3 marks

The Arrhenius equation can be represented as k = A begin mathsize 16px style straight e to the power of fraction numerator negative straight E subscript straight a over denominator RT end fraction end exponent end style in its exponential form.

State the effect on k of an increase in;

i)
The Arrhenius constant, A, (frequency factor)

ii)
Activation energy, Ea

iii)
Temperature, T
4c2 marks

Calculate the activation energy, Ea, of a reaction at 57 degreeC and a rate constant of 1.30 x 10-4  mol dm-3 s-1.

The gas constant R = 8.31 J K−1 mol−1.

The constant A = 4.55 × 1013.

4d7 marks

A series of experiments was carried out with nitrogen dioxide and ozone as shown in the equation below;

2NO2 (g) + O 3(g)  →  N2O5 (g) + O2 (g)

The rate equation for the reaction between the two compounds is below.

Rate = k [NO2] [O3]

In one experiment at 30 degreeC, the initial rate of reaction is 3.46 × 10−3 mol dm−3 s−1 when the initial concentration of NO2 is 0.50 mol dm−3 and the initial concentration of O3 is 0.21 mol dm−3.

i)
Calculate a value for the rate constant k at this temperature and state its units.

ii)

An equation that relates the rate constant, k, to the temperature, T and activation energy, Ea, is

ln k = ln A - straight E subscript straight a over RT

Use this equation and your answer from part (i) to calculate a value, in kJ mol−1, for the activation energy of this reaction at 30 degreeC.
For this reaction ln A = 15.8 mol−1 dm3.
The gas constant R = 8.31 J K−1 mol−1.
(If you were unable to complete part (i) you should use the value of 3.6 × 10−3 for the rate constant. This is not the correct value.)

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5a4 marks

Manganate (VII) ions, MnO4-, oxidise hydrogen peroxide, H2O2, to oxygen gas in acidic conditions.

2MnO4- + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2

The results of the reaction mixture at a fixed temperature are shown in Table 1.

Table 1

Time /s

0

100

200

400

600

800

1000

1200

[H]+ / mol dm-3

0.50

0.44

0.39

0.30

0.23

0.18

0.14

0.11

i)

Plot these results on the grid in Figure 1 and draw a line of best fit. The first three points have been plotted.

Figure 1

mtjdmP8G_10

ii)
Calculate the rate of reaction when [H+] = 0.35 mol dm–3. Show your working using a suitable construction on the graph in Figure 1.
5b2 marks

Hydrogen peroxide decomposes to form water and oxygen as shown in the equation below.

2H2O2 (aq) → 2H2O (l) + O2 (g)

Table 2 below shows the value of the rate constant at different temperatures for a reaction.

Table 2

Rate constant k / s-1

ln k

Temperature / K

bold 1 over bold T

0.000493

 

295

 

0.000656

 

298

 

0.001400

 

305

 

0.002360

 

310

 

0.006120

 

320

 


Complete the table by calculating the values of ln k and begin mathsize 16px style 1 over straight T end style
 at each temperature.          

5c6 marks

The Arrhenius equation when expressed as a logarithmic relationship is as follows:

ln k = begin mathsize 16px style fraction numerator negative Ea over denominator RT end fraction end style + ln A

i)
Using your results from part (b) plot a graph of ln k against begin mathsize 16px style 1 over straight T end style

ii)

Use your graph to calculate a value for the activation energy, in kJ mol−1, for this reaction. To gain full marks you must show all of your working.
The gas constant R = 8.31 J K−1 mol−1.

6NyUdP3s_11

5d7 marks

Table 3 below shows a series of experiments carried out to investigate how temperature affects the rate of reaction.

Table 3

Rate constant k/s-1

ln k

Temperature / K

begin mathsize 16px style bold 1 over bold T end style

3.1 x 10-5

-10.4

278

0.00360

4.7 x 10-4

-7.7

298

0.00336

1.7 x 10-3

-6.4

308

0.00325

5.2 x 10-3

-5.3

318

0.00314


The logarithmic form of the Arrhenius equation is; ln k = fraction numerator negative straight E subscript straight a over denominator RT end fraction
+ ln A

i)
Use the completed set of results above, to plot a labelled graph of ln k against begin mathsize 16px style 1 over straight T end style.

ii)
Use your graph to calculate a value for the activation energy, in kJ mol−1, for this reaction. The gas constant R = 8.31 J K−1 mol−1.
WEU_k17J_12

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1a3 marks

For the general reaction V + W + X Y + Z, the rate equation is rate = k [V] [X]2.

State the effect of doubling the concentration of each reactant individually on the rate of reaction. Justify your statements.

1b3 marks

For the general reaction Q + R + ST + U, the rate equation is rate = k [Q] [R]2 [S].

Explain the overall effect of tripling the concentration of each reactant on the rate of reaction.

1c3 marks

Solutions of two compounds, L and M, react together according to the following equation:

2L + MP

When the concentrations of L and M are doubled, the rate of reaction increases by a factor of four.

The rate equation for this reaction is rate = k [M] [N].

State and explain the role of the chemical, N.

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2a3 marks

The results of three experiments carried out at a temperature, T1,  to investigate the rate of the reaction between compounds A and B are shown in Table 1.

Table 1

 

Experiment 1

Experiment 2

Experiment 3

Initial concentration of A / mol dm-3

0.12

0.27

0.12

Initial concentration of B / mol dm-3

0.80

0.80

1.16

Initial rate / mol dm-3 s-1

2.842 x 10-5

6.394 x 10-5

5.974 x 10-5


Use the data in
Table 1 to deduce the rate equation for the reaction between compounds A and B.

2b2 marks

A separate experiment is performed at a different temperature, T2.

When the initial concentration of A is 0.27 mol dm–3 and B is 0.80 mol dm–3, the value of the rate constant is 1.894 x 10-4 mol–2 dm6 s–1.

Use the information from part (a), to deduce whether T1 or T2 is the higher temperature.

(If you did not get an answer for (a), you may assume that rate = k [A]2 [B]2. This is not the correct answer.)

2c3 marks

Use your answer to part (b) to sketch a labelled Maxwell-Boltzmann energy distribution on Figure 1 below, to show the reaction between compounds A and B at temperatures T1 and T2.

Figure 1

24

2d4 marks

Use your knowledge of kinetic theory to explain whether increasing temperature or increasing concentration will have a greater effect on the rate of reaction.

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3a1 mark

Compounds C and D react together in three different, third order experiments at the same fixed temperature, T1. The results of these experiments are shown in Table 1 .

Table 1

 

Experiment 1

Experiment 2

Experiment 3

[C] / mol dm-3

2.35 x 10-2

7.05 x 10-2

1.41 x 10-1

[D] / mol dm-3

5.92 x 10-2

5.92 x 10-2

1.18 x 10-1

Initial rate / mol dm-3 s-1

2.851 x 10-2

2.566 x 10-1

To be calculated


State the assumption made about [C] and [D].

3b4 marks

Using the information in part (a), deduce the rate equation for the reaction between compounds C and D.

3c2 marks

Using your answer to part (b), calculate the missing value in Table 1.

Table 1

 

Experiment 1

Experiment 2

Experiment 3

[C] / mol dm-3

2.35 x 10-2

7.05 x 10-2

1.41 x 10-1

[D] / mol dm-3

5.92 x 10-2

5.92 x 10-2

1.18 x 10-1

Initial rate / mol dm-3 s-1

2.851 x 10-2

2.565 x 10-1

To be calculated

3d2 marks

In a further experiment at temperature, T1, the initial rate of reaction was found to be 3.74 x 10-2 mol dm-3 s-1.

The initial concentrations of C and D were both y mol dm-3.

Using your answers from parts (b) and (c), calculate the concentration of C and D.

(If you did not get answers for (b) and (c), you may assume that rate = k [C] [D] and that the rate constant k = 52.7 mol-1 dm3 s-1. These are not the correct answers.)

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4a2 marks

Bromate (V) ions and bromide ions react in acidic conditions to form bromine and water.

Write the ionic half equations and overall balanced symbol equation, including state symbols, for this reaction.

4b3 marks

A series of experiments were carried out to investigate how the rate of the reaction of bromate and bromide in acidic conditions varies with temperature.

The time taken, t, was measured for a fixed amount of bromine to form at different temperatures. The results are shown in Table 1.

Table 1

Temperature (T) / K

begin mathsize 16px style 1 over straight T end style x 10-3 / K-1

Time (t) / s

begin mathsize 16px style 1 over straight t end style / s-1

ln begin mathsize 16px style 1 over straight t end style

408

2.451

21.14

0.0473

-3.051

428

2.336

10.57

 

 

448

 

5.54

0.1805

-1.712

468

2.137

3.02

0.331

-1.106

488

2.049

 

 

-0.536


Complete Table 1.

4c4 marks

The Arrhenius equation relates the rate constant, k, to the activation energy, Ea, and temperature, T.

ln k = ln Afraction numerator negative straight E subscript straight a over denominator RT end fraction

In this experiment, the rate constant, k, is directly proportional to begin mathsize 16px style 1 over straight t end style. Therefore,

ln begin mathsize 16px style 1 over straight t end style = ln Afraction numerator negative straight E subscript straight a over denominator RT end fraction

Use your answers from part (b) to plot a graph of ln begin mathsize 16px style 1 over straight t end style against begin mathsize 16px style 1 over straight T end style x 10-3 on Figure 1 below.

Figure 1

dDDO8CAX_20

4d4 marks

Use your graph and information from part (c) to calculate a value for the activation energy, in kJ mol–1, for this reaction. To gain full marks you must show all of your working.

The value of the gas constant, R = 8.31 J K–1 mol–1.

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5a3 marks

Three experiments were carried out at a temperature, T1,  to investigate the rate of the reaction between compounds F and G. The results are shown in Table 1.

Table 1

 

Experiment 1

Experiment 2

Experiment 3

Initial concentration of F / mol dm-3

1.71 x 10-2

5.34 x 10-2

7.62 x 10-2

Initial concentration of G / mol dm-3

3.95 x 10-2

6.24 x 10-2

3.95 x 10-2

Initial rate / mol dm-3 s-1

3.76 x 10-3

1.85 x 10-2

1.68 x 10-2


Use the data in
Table 1 to deduce the rate equation for the reaction between c
ompounds F and G.

5b2 marks

F and G react together to form FG2 according to the following two step mechanism.

Step 1:            F + G → FG

Step 2:            FG + G → FG2

Using your answer to part (a), explain which step is the rate-determining step.

5c2 marks

Use the information in Table 1 to calculate a value for the rate constant, k, for this reaction between 0.0534 mol dm-3 F and 0.0624 mol dm-3 G.

Give your answer to the appropriate number of significant figures.

State the units for k.

(If you did not get an answer for (a), you may assume that rate = k [F]2 [G]2. This is not the correct answer.)

5d3 marks

The Arrhenius equation shows how the rate constant, k, for a reaction varies with temperature, T.

                                                                      k equals A e to the power of negative E subscript a divided by R T end exponent

For the reaction between 0.0534 mol dm-3 F and 0.0624 mol dm-3 G at 25 degreeC, the activation energy, Ea, is 16.7 kJ mol–1.

Use your answer to part (c) to calculate a value for the Arrhenius constant, A, for this reaction.

The gas constant R = 8.31 J K–1 mol–1.

Give your answer to the appropriate number of significant figures.

(If you did not get an answer for (a), you may assume that k has a value of 4.97. This is not the correct answer.)

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