Edexcel A-Level Chemistry Paper 2, June 2017: Question 9

18 marks · Hard difficulty · Calculations

Calculate reaction orders from initial rates, determine rate constants, find overall order from a rate equation, and calculate activation energy from Arrhenius graph data.

Practise this question

Question

Exam question with four parts about chemical kinetics. Part (a) asks to calculate the time for a first-order decomposition of compound A. Part (b) provides a table of initial rates and concentrations for reactants X, Y, and Z, asking to calculate orders and the rate constant. Part (c) gives a rate equation for the bromine clock reaction, asking for overall order and a concentration calculation. Part (d) provides a table of temperatures, inverse temperatures, rate constants, and natural logarithms of rate constants, requiring table completion, plotting a graph of ln k against 1/T on a grid provided, and calculating activation energy.
Question text

9 This question is about reaction kinetics.

(a) Compound A decomposes in a first order reaction.

Calculate the time it takes for the mass of A to decrease from 600 g to 37.5 g if the

decomposition has a constant half-life of 14 minutes.

(1)

(b) The ‘initial rates’ method was used to investigate the orders of reaction with

respect to reactants X, Y and Z. The table shows the results obtained.

Initial concentration / mol dm−3

Initial rate

Run −3 −1

X Y Z / mol dm s

1 0.00100 0.00300 0.00600 2.17 × 10−6

2 0.00100 0.00600 0.00600 8.68 × 10−6

3 0.00050 0.00600 0.00600 4.34 × 10−6

4 0.00300 0.00300 0.00300 6.51 × 10−6

(i) Calculate the orders with respect to X, Y and Z.

(3)

X …

Y …

Z …

(ii) Give the rate equation for the reaction and hence calculate the rate constant, k,

to an appropriate number of significant figures. Include units in your answer.

(4)

(c) The kinetics of the ‘bromine clock’ were investigated and the rate equation was

found to be

Rate = k[BrO−][Br−][H+]2

(i) What is the overall reaction order?

(1)

A First

B Second

C Third

D Fourth

(ii) Calculate the concentration of bromide ions required to produce a

18 reaction rate of 4.08 × 10−3 mol dm−3 s−1 at 298 K given that

k = 8.00 dm*P48059A01824*9mol−3s−1

[BrO−] = 0.200 mol dm−3

[H+] = 0.100 mol dm−3

(2)

(d) The rate constant for the reaction between bromoethane and

aqueous hydroxide ions was determined at five different temperatures.

The results are given in the table.

Temperature (T) 1 / Temperature (1 / T) Rate constant, k

−1 3 –1 –1 ln k

/ K / K / dm mol s

293 3.41 × 10−3 5.83 × 10−5 −9.75

303 1.67 × 10−4

313 5.26 × 10−4

323 1.36 × 10−3

333 3.00 × 10−3 3.77 × 10−3 −5.58

Complete the data in the table and use them to plot a graph of ln k against 1 / T

and hence determine the activation energy, E , in kJ mol−1.

a

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

The Arrhenius equation can be expressed as

Ea 1

lnk = – × + constant

R T (7)

*P48059A01924*

*P48059A02024*

(Total for Question 9 = 18 marks)

Mark scheme

Show the mark scheme Mark scheme corresponding to the kinetics question. Shows acceptable answers, mark allocations, example calculations, and a completed grid showing the correct line of best fit and gradient calculation for the Arrhenius plot.

Question

Answer Additional Guidance Mark

Number

Example of calculation: (1)

9(a) Correct answer to 2 SF

(Four half-lives to decrease 600 g to 37.5 g

so 4 x 14 mins)

= 56 (mins)

Penalise wrong units, e.g. “m”

Question

Answer Additional Guidance Mark

Number

Reaction Orders: (3)

9(b)(i)

X First/1 (1)

Y Second/2 (1)

Z Zero/0 (1) Allow “none”/”no order”

Question

Answer Additional Guidance Mark

Number

Reactants can be in any order (4)

9(b)(ii) Z does not have to be included in

Marking point 1 the rate equation

Rate = k[X][Y]2[Z]0 (1) TE from (b)(i) which will apply for all

four marking points

Marking point 2

rearrangement of rate expression (1) Example of calculation:

k = rate / [X][Y]2

Marking point 3

calculation of value for k to 2/3 SF (1) k = 2.17 x 10−6

0.00100 x 0.003002

= 241.11

= 241/240

Any ‘run’ can be used

No TE on incorrect rearrangement

Marking point 4

units dm6 mol-2 s−1 (1) Allow units in any order

Correct answer without working and

with correct units to 2/3 SF scores

marking points 2, 3 and 4

Question

Answer Mark

Number

The only correct answer is D (Fourth) (1)

A is not correct because this is the individual reaction order with respect to bromate(V) ions and

with respect to bromide ions but is not the overall reaction order

9(c)(i) B is not correct because this is the reaction order with respect to hydrogen ions but is not the

overall reaction order

C is not correct because this is the number of species in the rate equation but is not the overall

reaction order

Question

Answer Additional Guidance Mark

Number

Example of calculation: (2)

9(c)(ii)

rearrange rate equation so [Br−] = (1) [Br−]= rate

k [BrO −][H+]2

calculation of value to 2/3 SF (1) = 0.255/0.26 (mol dm−3)

Correct answer without working to

2/3 SF scores 2 marks

If units given then must be correct

No TE on incorrect rearrangement

Question

Answer Additional Guidance Mark

Number

Example of suitable graph (7)

9(d)

calculation of all three 1/T values x 10−3 (1) (3.41), 3.30, 3.19, 3.10, (3.00)

calculation of all three ln k values (1) (−9.75), −8.70, −7.55, −6.60,

(−5.58)

Allow omission of end zero

Penalise more than 3SF once only

axes: correct way round, labelled, suitable scale (1) Plotted points must cover at least ½

the graph paper on each axis

Do not award 1/t

all points plotted correctly, with best-fit straight line(1) Allow ±½ square

calculation of gradient with sign (1) Gradient = − 10200 Allow ±500

Allow this mark if the value is seen in

the Ea calculation

units of gradient (1) K

use of gradient to calculate activation energy (1) Ea = 10200 x 8.31 / 1000

= (+) 84.8 (kJ mol−1)

Final answer must be positive and in

the range (+) 80.6 – 88.9 (kJ mol−1)

Allow value given in J mol−1 but then

these units are essential

Ignore SF for gradient and activation

energy values

(Total for Question 9 = 18 marks)

TOTAL FOR PAPER = 90 MARKS

How to answer it

Reaction Kinetics & Arrhenius Equations

Edexcel A-Level Chemistry • Comprehensive Study Guide

What this question tests

This multi-part question assesses core kinetics principles: calculating half-lives for first-order reactions, determining orders of reaction using initial rates data, formulating rate equations, calculating rate constants with correct units, identifying overall reaction orders, rearranging rate expressions for unknown concentrations, processing Arrhenius data, plotting graphical logarithmic relationships, and calculating activation energy (Eₐ).

Part (a): First-Order Half-Life Calculation

✅ Correct Answer

56 minutes (to 2 Significant Figures)

📐 Step-by-Step Calculation

  1. Find mass fractions: 600 g → 300 g → 150 g → 75 g → 37.5 g.
  2. Count half-lives elapsed: Exactly 4 half-lives.
  3. Multiply by half-life duration: 4 × 14 mins = 56 minutes.

❌ Common Errors

Writing incomplete units or mathematical symbols like "m" instead of minutes. Ensure answers match the requested significant figures.

🎯 Mark: 1 mark

Part (b): Initial Rates Method & Rate Constants

(i) Determining Orders of Reaction

✅ Correct Answers

  • X: First order (1)
  • Y: Second order (2)
  • Z: Zero order (0)

🧠 Exam Technique

Compare runs where two concentrations are held constant while one changes. Match concentration factor changes to rate factor changes to deduce individual orders.

(ii) Rate Equation & Rate Constant Calculation

✅ Correct Answers

Rate = k[X][Y]²

k = 240 (or 241)

Units: dm⁶ mol⁻² s⁻¹

📐 Calculation Steps

  1. Rearrange rate equation: k = Rate / ([X][Y]²)
  2. Substitute values from Run 1: k = (2.17 × 10⁻⁶) / ((0.00100) × (0.00300)²)
  3. Calculate raw value: 241.11...
  4. Round to 2/3 SF: 240 or 241.

❌ Common Errors

Incorrect unit derivation. Remember that k = Rate / (concentration)³ , leading to (mol dm⁻³ s⁻¹) / (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹ .

🎯 Marks: 3 marks for (i) + 4 marks for (ii)

Part (c): Bromine Clock Kinetics

(i) Overall Reaction Order

✅ Correct Answer

D (Fourth)

💡 Key Knowledge

Overall order is the sum of all individual reactant powers in the rate equation: Rate = k[BrO₃⁻][Br⁻][H+]² . Sum: 1 + 1 + 2 = 4.

(ii) Calculating Reactant Concentration

✅ Correct Answer

0.25 or 0.26 mol dm⁻³

📐 Step-by-Step Calculation

  1. Rearrange for bromide ions: [Br⁻] = Rate / (k [BrO₃⁻][H+]²)
  2. Substitute numbers: [Br⁻] = (4.08 × 10⁻³) / (8.00 × 0.200 × (0.100)²)
  3. Evaluate: 0.255 mol dm⁻³ → round to 0.26 mol dm⁻³ (2 SF).
🎯 Marks: 1 mark for (i) + 2 marks for (ii)

Part (d): Arrhenius Equation & Activation Energy

💡 Key Knowledge & Data Completion

Calculate missing 1 / T values (by dividing 1 / Temperature ) and missing ln k values using natural logarithms of the rate constants provided.

  • T = 303 K → 1/T = 3.30 × 10⁻³ K⁻¹ | ln k = -8.70
  • T = 313 K → 1/T = 3.19 × 10⁻³ K⁻¹ | ln k = -7.55
  • T = 323 K → 1/T = 3.10 × 10⁻³ K⁻¹ | ln k = -6.60

🧠 Graphical & Calculation Technique

  1. Axes: Plot ln k on the y-axis against 1 / T on the x-axis. Use sensible scales covering over half the grid.
  2. Gradient: Calculate Δ(ln k) / Δ(1 / T) . Expected gradient is negative (~ -10200 K).
  3. Activation Energy: Since Gradient = -Eₐ / R , multiply gradient by -R ( R = 8.31 J mol⁻¹ K⁻¹ ) and divide by 1000 to convert from J to kJ mol⁻¹.

✅ Final Values

Gradient: ~ -10200 K (Units: K )

Activation Energy (Eₐ): +84.8 kJ mol⁻¹ (Acceptable range: 80.6 – 88.9 kJ mol⁻¹ )

🎯 Marks: 7 marks total

Topics

Physical Chemistry · Topic 16: Kinetics II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.