Edexcel A-Level Chemistry Paper 2, November 2021: Question 6

15 marks · Hard difficulty · Calculations

Calculate atom economy, determine activation energy using the Arrhenius equation, and deduce orders of reaction, rate equation, and rate constant for various chemical processes involving carbon monoxide.

Practise this question

Question

Question text

6 This question is about carbon monoxide, CO, which is a toxic and colourless gas used

widely in the chemical industry.

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

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

(2)

(b) Carbon monoxide can be made by the thermal decomposition of

sodium ethanedioate.

Na2C2O4 → Na2CO3 + CO

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

this reaction.

(2)

ethanal, CH3CHO, in the gas phase.

CH3CHO(g) → CH4(g) + CO(g)

This reaction was carried out at two different temperatures, and all other variables

were kept constant.

Temperature Rate 1/Temperature (1/T)

−3 −1 −1 ln rate

/ K / mol dm s / K

700 0.0108 1.43 × 10−3

850 4.90 1.59

(i) Complete the data in the table.

(1)

(ii) Calculate the activation energy, Ea, for the reaction without plotting a graph.

Include a sign and units in your answer.

The Arrhenius equation may be written as

*P65464A0828*Ea1 −1 −1

In rate = − × + constant [R = 8.31J mol K ]

R T

(3)

(d) Haemoglobin (Hb) found in red blood cells reacts almost irreversibly with

carbon monoxide.

Initial rate experiments were carried out to investigate the effect of the

concentrations of Hb and CO on the rate of this reaction.

Experiment [Hb] / mol dm−3 [CO] / mol dm–3 Rate / mol dm−3 s−1

12.09 × 10–6 1.40 × 10–6 8.20 × 10–7

24.18 × 10–6 1.40 × 10–6 1.64 × 10–6

33.26 × 10–6 2.80 × 10–6 2.56 × 10–6

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

(1)

(ii) Determine the order with respect to carbon monoxide using your answer to

(d)(i) and the data in the table.

Justify your answer.

(2)

*P65464A0928*

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

(1)

(iv) Calculate the rate constant, k, for the reaction, using the data from Experiment 1

and the rate equation from (d)(iii).

Include units in your answer.

(3)

*P65464A01028*

(Total for Question 6 = 15 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

6(a) Example of dot and cross diagram: (2)

• 3 bond pairs showing triple bond, one of which must be a dative

bond (1)

• both lone pairs on C and O (1)

Allow 1 mark if correct number of

electrons shown, but all as crosses or all as

dots

Ignore lines showing covalent bonds

Question Answer Additional Guidance Mark

Number

6(b) Example of calculation (2)

• Expression of numerator and denominator for atom economy (1) 28 ÷ [28 + 106] (x 100) /

[28 ÷ 134] (x 100)

• evaluation (1) 20.896 = 20.9%

Ignore SF except 1 SF

Allow TE for M2 for 1 Mr error in M1

Allow 1 mark for (106 ÷ 134) x 100 =

79.1 %

Number

6(c)(i) (1)

values must be to at least 3SF

Temperature / –3 –1 ln rate

Rate / mol dm s 1/Temperature

Allow

700 0.0108 1.43 x 10–3 – 4.53 −4.5282

1.1765 × 10−3

850 4.90 1.18 x 10–3 1.59

Question

Answer Additional Guidance Mark

Number

6(c)(ii) Example of calculation (3)

• recognition that (difference in ln rate) / (difference in 1/T ) – 6.12 / 2.5 x 10–4

= – Ea / R (1) Can be subsumed within M2

• calculation of – Ea / R (1) = – 24480 (K)

24480 x 8.31 = (+) 203428.8 J mol–1

• calculation of Ea with correct units (1)

= (+) 203000 J mol–1

= (+) 203 kJ mol–1

Ignore SF

final answer between 200 -204 kJ mol–1 with

no working scores (3)

Question

Answer Additional Guidance Mark

Number

6(d)(i) (1)

• order with respect to Hb = 1

Question

Answer Additional Guidance Mark

Number

6(d)(ii) (2)

• order with respect to CO = 1 (1) standalone mark

• justification (1) Either

using experiments 1 and 3 the concentration

of Hb goes up by a factor of 1.56 and the

concentration of CO doubles and the rate

goes up by a factor of 3.12

using experiments 2 and 3 the concentration

of Hb goes down by a factor of 0.78 but the

rate increases by a factor of 1.56 so doubling

the concentration of CO means doubling the

rate

M2 dependent on M1

Question

Answer Additional Guidance Mark

Number

6(d)(iii) Example of rate equation (1)

• rate equation rate = k[Hb][CO]

allow e.g R / r for rate and K for k

Allow expressed in terms of k

Allow TE from 6(d)(i) and 6(d)(ii)

Note – must be consistent with 6(d)(i) and

6(d)(ii)

Question

Answer Additional Guidance Mark

Number

6(d)(iv) Example of calculation (3)

• rearrangement of rate equation to find k (1) k = rate / [Hb][CO]

8.20 x 10–7 / (2.09 x 10–6 x 1.40 x 10–6)

• calculation of k (1)

= 280246 = 280000

Ignore SF except 1 SF

dm3 mol–1 s–1

• correct units of k (1)

Allow units in any order

Allow use of data from experiments 2 or 3

Correct answer including units with no

working scores 3 marks

Allow TE on rate equation from (d)(iii)

No TE for mistake with rate equation within

(d)(iv) e.g. rearrangement error

(Total Question 6 = 15 marks)

How to answer it

Carbon Monoxide Chemistry & Kinetics Study Guide

What this question tests

This comprehensive Edexcel A-Level question covers multiple core topics: chemical bonding (dot-and-cross diagrams including dative covalent bonds), quantitative chemistry (atom economy calculations), reaction kinetics (using the Arrhenius equation with natural logs without plotting graphs), and experimental kinetics (deducing orders of reaction, writing rate equations, and calculating rate constants with units).

Part (a) — Bonding & Structure

Dot-and-Cross Diagram of Carbon Monoxide

✅ Correct Answer

A triple bond between carbon and oxygen (one of which is a dative/coordinate bond from oxygen to carbon, or shown sharing 6 electrons total), plus one lone pair of electrons on the carbon atom and one lone pair on the oxygen atom.

💡 Key Knowledge

Carbon has 4 outer shell electrons; Oxygen has 6. Total = 10 valence electrons.
To satisfy the octet rule for both atoms, CO forms a triple bond (6 shared electrons) leaving 2 electrons (one lone pair) on each atom.
One of the covalent bonds is a dative covalent bond originating from the oxygen atom, donating an electron pair to the electron-deficient carbon.

❌ Common Errors

Drawing a double bond or single bond, failing to achieve an octet for both atoms.
Omitting the lone pairs on carbon or oxygen.

Total Marks: 2

Part (b) — Quantitative Chemistry

Atom Economy for Carbon Monoxide Production

📐 Step-by-Step Calculation

Reaction: Na₂C₂O₄ → Na₂CO₃ + CO

Find Molar Mass of desired product (CO): 12.0 + 16.0 = 28.0 g mol⁻¹
Find Molar Mass of reactant (Na₂C₂O₄): (23.0 × 2) + (12.0 × 2) + (16.0 × 4) = 134.0 g mol⁻¹ (Note: Na₂CO₃ has Mᵣ = 106.0, total mass of products = 28 + 106 = 134).
Calculate Atom Economy: (Mass of desired product / Total mass of reactants) × 100
(28 / 134) × 100 = 20.9% (or 20.896%)

🧠 Exam Technique

Always show your fraction clearly before multiplying by 100. Edexcel allows error carried forward (TE) if you make a minor molar mass addition error in the denominator.

Total Marks: 2

Part (c) — Kinetics & The Arrhenius Equation

Activation Energy Determination from Two Temperature Points

✅ Part (i): Completing the Table

At T = 700 K: Rate = 0.0108, 1/T = 1.43 × 10⁻³, In rate = -4.53 (or -4.5282)

Values must be given to at least 3 significant figures where applicable.

📐 Part (ii): Calculating Activation Energy (Eₐ)

Recognise gradient formula: Slope = (In rate₂ - In rate₁) / ((1/T₂) - (1/T₁)) = -Eₐ / R
Substitute values: (-4.53 - 1.59) / (1.43 × 10⁻³ - 1.18 × 10⁻³) = -6.12 / (2.5 × 10⁻⁴) = -24480 K
Equate to -Eₐ / R: -Eₐ / 8.31 = -24480 ⇒ Eₐ = 24480 × 8.31 = 203428.8 J mol⁻¹
Convert to kJ mol⁻¹ & add sign: +203 kJ mol⁻¹ (Acceptable range: 200 to 204 kJ mol⁻¹).

❌ Common Errors

Forgetting to convert joules to kilojoules (leaving the answer as 203428 J mol⁻¹).
Omitting the positive (+) sign for activation energy, or inverting the temperature difference coordinates.

Total Marks: 1 (for c-i) + 3 (for c-ii) = 4 marks total

Part (d) — Experimental Kinetics & Rate Equations

Deducing Orders, Rate Equations, and Rate Constants

✅ Part (i) & (ii): Orders of Reaction

Order with respect to Hb = 1 (Exp 1 to 2): When [Hb] doubles (from 2.09 × 10⁻⁶ to 4.18 × 10⁻⁶) while [CO] is constant, the rate doubles (from 8.20 × 10⁻⁷ to 1.64 × 10⁻⁶). Therefore, order is 1.

Order with respect to CO = 1 (Exp 1 to 3): Using data from Experiments 1 and 3, [Hb] increases by a factor of 1.56, and [CO] doubles. The rate increases by 3.12. Since rate proportional to [Hb]¹ × [CO]ⁿ, 1.56 × 2¹ = 3.12, confirming CO order is 1.

✅ Part (iii) & (iv): Rate Equation & Rate Constant (k)

Rate Equation: rate = k[Hb][CO]

Calculating k (using Experiment 1):

Rearrange: k = rate / ([Hb][CO])
Substitute: k = (8.20 × 10⁻⁷) / ((2.09 × 10⁻⁶) × (1.40 × 10⁻⁶))
Evaluate: k = 280246 = 280000 (or 2.80 × 10⁵)
Units: dm³ mol⁻¹ s⁻¹

🧠 Top-Level Exam Tip for Units of k

Derive units logically from the rearranged expression:
k = (mol dm⁻³ s⁻¹) / ((mol dm⁻³) × (mol dm⁻³)) = mol⁻¹ dm³ s⁻¹ (conventionally written as dm³ mol⁻¹ s⁻¹).

Total Marks: 1 (d-i) + 2 (d-ii) + 1 (d-iii) + 3 (d-iv) = 7 marks total

Topics

Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 9: Kinetics I · Topic 16: Kinetics II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.