AQA A-Level Chemistry Paper 2, 2023: Question 9

12 marks · Medium difficulty · State/Explain/Numerical

Decomposition of Ethanoic Anhydride This question involves deducing the general formula for the ketene homologous series, completing a reaction mechanism with curly arrows, using the Arrhenius equation to calculate 𝑇 1 T 1 ​ , and sketching the Maxwell–Boltzmann distribution to explain why the rate of reaction increases at a higher temperature.

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AQA A-Level Chemistry Paper 2, 2023: Question 9
Question text

09 This question is about ethanoic anhydride.

In the gas phase, ethanoic anhydride (CH3CO)2O decomposes to form ethenone.

The equation is

(CH3CO)2O → H2C=C=O + CH3COOH

Ethenone

09.1 Ethenone is the simplest member of the ketene homologous series.

Ketenes all contain one C=C double bond and one C=O double bond.

Deduce the general formula for the ketene homologous series.

[1 mark]

09.2 Figure 11 shows an incomplete suggested mechanism for the decomposition of

ethanoic anhydride.

Figure 11

Complete the mechanism in Figure 11 by adding three curly arrows and any relevant

lone pairs of electrons.

[3 marks]

09.3 For a chemical reaction the relationship between the rate constant, k, and the

temperature, T, is shown by the Arrhenius equation.

−Ea

k = Ae RT

For the decomposition of gaseous ethanoic anhydride

the activation energy, E = 34.5 kJ mol–1

a

the Arrhenius constant, A = 1.00 × 1012 s–1

At temperature T the rate constant, k = 2.48 × 108 s–1

Calculate T1

*26* –1 –1

The gas constant, R = 8.31 J K mol

[3 marks]

T1 K

09.4 Sketch the Maxwell–Boltzmann distribution of molecular energies for

gaseous ethanoic anhydride at temperature T1 and at a higher temperature T2

Include a label for each axis, and mark on the appropriate axis a typical position for

the activation energy.

Explain why the rate of reaction is faster at T2

[5 marks]

*27* Explanation

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2023: Question 9

Question Answers Additional Comments/Guidelines Mark

CnH2n-2O Allow CnH2nCO or (CH2)nCO or CnH2(n-1)O 1

09.1

(AO2)

Allow other C-O bond breaking for M1

09.2

(3 x AO2)

k –Ea/RT Ea

M1 = e OR via ln k = ln A – or shown with numbers

A RT

34500 3

09.3 M2 8.302 =

8.31 x T (3 x AO2)

M3 T = 500 K

M1 x axis labelled correctly (kinetic not required)

AND y axis labelled correctly allow particles

M2 Ea labelled on x axis

M3 Distribution correct shape for T1

M4 Peak at T lower with max shifted right and 5

09.4 2

only crosses once (5 x AO1)

M5 At T2 (many) more particles have E≥Ea

How to answer it

Step-by-Step Guide: Decomposition of Ethanoyl Anhydride

Question (a): General Formula of Ketene Homologous Series

Step 1: Identify the Pattern of Ketenes

Ketenes contain:

  • One C=C double bond
  • One C=O double bond

Step 2: Deduce the General Formula

For ketenes, the general formula is:

CnH2n-2O

Alternative Answers: CnH2nCO or (CH2)nCO

Question (b): Completing the Mechanism

Step 1: Add Curly Arrows

Complete the mechanism by adding:

  • A curly arrow from the C-H bond to the C=C bond in Step 1.
  • Another curly arrow from the C-O bond to break and form the double bond.

Step 2: Lone Pairs

Add a lone pair on the oxygen atom where the bond breaks.

Tip: Ensure arrows start from bonds or lone pairs and point to where electrons are transferred.

Question (c): Calculating T1 Using the Arrhenius Equation

Step 1: Rearrange the Arrhenius Equation

Start with:

k = Ae-Ea / RT

Take natural logs of both sides:

ln k = ln A - (Ea / RT)

Step 2: Substitute Values

  • k = 2.48 × 108 s-1
  • A = 1.00 × 1012 s-1
  • Ea = 34.5 kJ mol-1 = 34,500 J mol-1
  • R = 8.31 J K-1 mol-1

Substitute into the rearranged equation:

ln(2.48 × 108) = ln(1.00 × 1012) - (34,500 / (8.31 × T1))

Step 3: Solve for T1

Rearranging and solving gives:

T1 = 500 K

Question (d): Maxwell-Boltzmann Distribution

Step 1: Draw the Distribution

Your graph should include:

  • X-axis: Energy
  • Y-axis: Number of molecules
  • Two curves: T1 (lower temperature) and T2 (higher temperature).

Key Features:

  • The curve at T2 is lower and shifts to the right.
  • Label the activation energy (Ea) on the x-axis.

Step 2: Explain Why the Rate is Faster

At a higher temperature T2 :

  • More molecules have energy E ≥ Ea (activation energy).
  • This leads to more successful collisions per unit time.

Conclusion: The rate of reaction increases at T2 because a greater proportion of molecules have sufficient energy to overcome the activation barrier.

Topics

Physical Chemistry · Organic Chemistry · 3.1.5 Kinetics · 3.1.9 Rate Equations · 3.3.1 Introduction to Organic Chemistry

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.