Syllabus Edition

First teaching 2014

Last exams 2024

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Activation Energy (DP IB Chemistry: HL)

Exam Questions

2 hours26 questions
1a5 marks

The Arrhenius equation can be written as:

         begin mathsize 14px style k space equals A e to the power of fraction numerator negative E subscript a over denominator R T end fraction end exponent end style

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

  • A
  • Ea
  • R
  • T
1b1 mark

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

1c1 mark

State how the rate constant, k varies with temperature, T.

1d1 mark

State how the activation energy, Ea, varies with rate constant, k.

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2a3 marks

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

               ln k space equals space ln A space minus space fraction numerator E a over denominator R T end fraction

A plot of ln k against begin mathsize 14px style 1 over T end style gives a straight-line graph of the type y = mx + c.

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

Straight-line term

Arrhenius term

y

ln k 

m

 

x

 

c

 

2b
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2 marks

A graph of ln k against 1 over T is shown below.

arrhenius-graph

 Calculate the gradient of the straight line.

2c
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1 mark

Using section 2 of the data booklet, calculate the activation energy, Efor the graph in part b).

2d
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2 marks

Calculate the frequency factor, A, for the graph in part b) to 2 decimal places.

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3a1 mark

Arrhenius plots for two reactions with different activation energies are shown below.

arrhenius-graph-question-

State which plot shows the reaction with the greatest activation energy.

3b1 mark

The temperature of both reactions from part a) is increased from 20° to 45°.

Using section 1 of the data booklet, determine which of the reactions will experience the largest change in the rate of reaction.

3c
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2 marks

The decomposition of hydrogen peroxide into water and oxygen occurs at a slow rate with a rate constant of k = 6.42 x 10-4 mol dm-3 s-1 and at a temperature of 290 K.

When the temperature is increased to 340 K the rate constant k = 6.47 x 10-2 mol dm-3 s-1.

Using sections 1 and 2 of the data booklet, calculate the activation energy for this reaction.

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1a3 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. 

Using Sections 1 and 2 of the Data Booklet, calculate the activation energy, Ea, correct to three significant figures and state its units.

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

1b2 marks

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

2H2O2 (aq) → 2H2O (l) + O2 (g) 

The table below shows the value of the rate constant at different temperatures for a reaction.

Rate constant k / s-1

ln k

Temperature / K

 italic 1 over italic 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 14px style italic 1 over italic T end style at each temperature.

1c4 marks

The results of the experiment can be used to calculate the activation energy, Ea. Use the results table to plot a graph of ln k against begin mathsize 14px style italic 1 over italic T end style.

q1c_16-2_ib_hl_medium_sq

1d4 marks

Using Sections 1 and 2 of the Data Booklet and your graph, calculate a value for the activation energy, Ea, for this reaction. To gain full marks you must show all of your working.

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2a3 marks

The Arrhenius equation can be represented as k = Aitalic e to the power of negative sign italic E italic a forward slash italic R italic T end exponent in its exponential form.

 State the effect on k of an increase in; 

i)     The constant, A, (frequency factor) 

ii)     Activation energy, Ea 

iii)    Temperature, T

2b2 marks

Using Sections 1 and 2 of the Data Booklet, calculate the activation energy, Ea, of a reaction at 57℃ and a rate constant of 1.30 x 10-4  mol dm-3 s-1. The constant A = 4.55 × 1013.

2c3 marks

The table below shows how temperature affects the rate of reaction.

Rate constant k/s-1

ln k

Temperature / K

 italic 1 over T

2.0 x 10-5

-10.8

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

 

Use the results to plot a labelled graph of ln k against begin mathsize 14px style italic 1 over italic T end style.q2c_16-2_ib_hl_medium_sq

2d4 marks

Using Sections 1 and 2 of the Data Booklet and your graph, calculate a value for the activation energy, Ea, for this reaction.

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3a1 mark

Nitrogen dioxide and ozone react according to the following equation. 

            2NO2 (g) + O3(g) → N2O5 (g) + O2 (g) 

Experimental data shows the reaction is first order with respect to NO2 and first order with respect to O3

State the rate expression for the reaction.

3b3 marks

At 30°C, 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.

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

 

3c4 marks

Using Sections 1 and 2 of the Data Booklet and your answer from part (b), calculate a value for the activation energy of this reaction at 30 °C.

For this reaction ln A = 15.8 mol−1 dm3.

 

3d1 mark

The relationship between the rate constant and temperature is given by the Arrhenius equation, k = Ae to the power of italic minus fraction numerator E a over denominator R T end fraction end exponent 

State how temperature affects activation energy.

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4a1 mark

A common relationship exists between temperature and rate. 

What temperature change is associated with a doubling of rate? 

4b2 marks

An Arrhenius plot of ln k against italic 1 over T for the reaction between A (g) and B (g) at different temperatures is shown in Figure 1 below.

q4b_16-2_ib_hl_medium_sq

The equation of the line of best fit was found to be: 

            ln k = -6154begin mathsize 14px style begin italic style stretchy left parenthesis 1 over italic T stretchy right parenthesis end style end style - 8.2 

Calculate the activation energy, Ea, for the reaction between A (g) and B (g).

4c2 marks

Define the Arrhenius constant, A.

4d2 marks

Using the Arrhenius plot, calculate an approximate value for the constant, A.

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5a1 mark

The graph of ln k against begin mathsize 14px style italic 1 over italic T end style for a general reaction is shown.q5a_16-2_ib_hl_medium_sq



Sketch the expected line for a different reaction with a higher activation energy.

5b2 marks

A graph of ln k against begin mathsize 14px style italic 1 over italic T end style for another general reaction is shown.q5b_16-2_ib_hl_medium_sq

 

Sketch the expected line for the same reaction with an added catalyst.

5c3 marks

Rate constant data for the reaction of hydrogen and iodine at two different temperatures is shown in the table below. 

H2 (g) + I2 (g) → 2HI (g) 

Table 1

Experiment

Temperature / K

Rate constant, k / mol dm-3 s-1

1

599

5.40 x 10-4

2

683

2.80 x 10-2

            

Using Sections 1 and 2 of the Data Booklet, calculate the activation energy, in kJ mol-1, for the reaction.

5d2 marks

Using the data from experiment 1 and Sections 1 and 2 in the Data Booklet, calculate a value for the constant, A.

Table 2

Experiment

Temperature / K

Rate constant, k / mol dm-3 s-1

1

599

5.40 x 10-4

2

683

2.80 x 10-2

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1a
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3 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 below. 

Temperature (T) / K

1 over Tx 10-3 / K-1

Time (t) / s

1 over t/ s-1

ln1 over t

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

0.3311

-1.106

488

2.049

   

-0.536

 

Complete the table above.

1b
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4 marks

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

ln k = ln A + fraction numerator negative E subscript a over denominator R T end fraction

In this experiment, the rate constant, k, is directly proportional to 1 over t. Therefore, 

ln 1 over t= ln A + fraction numerator negative E subscript a over denominator R T end fraction

Use your answers from part (a) to plot a graph of ln 1 over t against 1 over Tx 10-3 on the graph below.

graph

1c
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4 marks

Use section 2 of the data booklet along with your graph and information from part (b) 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.

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2a
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3 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 the table below:

 

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 the table to deduce the rate equation for the reaction between compounds F and G.

2b
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2 marks

Use the information in the table in part (a) 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)

2c
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2 marks

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

k space equals space A e to the power of fraction numerator negative E subscript a over denominator R T end fraction end exponent

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

Use section 2 of the data booklet and your answer to part (b) to calculate a value for the Arrhenius constant, A, for this reaction. 

Give your answer to the appropriate number of significant figures.

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

2d
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2 marks

The temperature of the reaction is increased to twice the original temperature, T1.
The value of k increases to 0.28 mol-1 dm3 s-1 at this new temperature.
 

Using sections 1 and 2 of the data booklet and your answer to part (b), determine the original temperature, T1.

(If you did not get an answer for (b), you may assume that k = 16700 mol-1 dm3 s-1 This is not the correct answer)

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3a
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2 marks

The rate constant for a reaction doubles when the temperature is increased from 25.0 °C to 35 °C.

Calculate the activation energy, Ea, in kJ mol−1 for the reaction using section 1 and 2 of the data booklet.

3b
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2 marks

The rate constant is 6.2 x 103 s-1 when the temperature is reduced by a factor of a fifth from the original starting temperature, 25 °C.

Calculate the rate constant, in min-1, using sections 1 and 2 of the data booklet.

3c2 marks

A different reaction route is used which reduces the activation energy of the reaction. 

Explain how the rate constant calculated in part(b) would differ.

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