Edexcel A-Level Chemistry Paper 3, June 2018: Question 8

19 marks · Hard difficulty · Calculations

Calculate the equilibrium constant Kc for the esterification reaction between ethanoic acid and propan-1-ol based on titration data and equilibrium amounts.

Practise this question

Question

An examination question about determining the equilibrium constant Kc for the esterification of ethanoic acid and propan-1-ol, broken down into parts (a) through (g) with a total of 19 marks.
Question text

8 This question is about an experiment to determine the equilibrium constant, Kc, for

an esterification reaction producing propyl ethanoate.

The equation for the reaction is

CH3COOH(l) + CH3CH2CH2OH(l) CH3COOCH2CH2CH3(l) + H2O(l)

ethanoic acid propan-1-ol propyl ethanoate

In an experiment to determine the equilibrium constant, Kc, the following steps were

carried out.

• 6.0 cm3 of ethanoic acid (0.105 mol), 6.0 cm3 of propan-1-ol (0.080 mol) and

2.0 cm3 of dilute hydrochloric acid were mixed together in a sealed boiling tube.

In this pre-equilibrium mixture, there is 0.111mol of water

• The mixture was left for one week, at room temperature and pressure,

to reach equilibrium

• The equilibrium mixture and washings were transferred to a volumetric flask and

the solution made up to exactly 250.0 cm3 using distilled water

• 25.0 cm3 samples of the diluted equilibrium mixture were titrated with a solution

of sodium hydroxide, concentration 0.200 mol dm−3, using phenolphthalein as the

indicator

• The mean titre was 23.60 cm3 of 0.200 mol dm−3 sodium hydroxide solution.

(a) State the role of the hydrochloric acid in the esterification reaction.

(1)

(b) (i) Calculate the total amount, in moles, of acid present in the volumetric flask

in the equilibrium mixture.

(2)

(ii) The 2.0 cm3 of dilute hydrochloric acid contained 0.00400 mol of H+(aq) ions.

Use this and your answer to part (b)(i) to calculate the amount, in moles, of

ethanoic acid present in the equilibrium mixture.

(1)

(c) (i) The initial mixture in the boiling tube contained 0.105mol of ethanoic acid.

Use your answer to (b)(ii) to calculate the amount, in moles, of ethanoic acid

that reacted to form the ester in the equilibrium mixture.

(1)

*P52304A02632*

(ii) Use information given in the method, and your answer to (c)(i), to calculate

the amounts, in moles, of propan-1-ol, propyl ethanoate and water that are

present in the equilibrium mixture.

(3)

Moles of propan-1-ol at equilibrium …

Moles of propyl ethanoate at equilibrium …

Moles of water at equilibrium …

(d) (i) Write the expression for the equilibrium constant, Kc, for this reaction.

CH3COOH(l) + CH3CH2CH2OH(l) CH3COOCH2CH2CH3(l) + H2O(l)

(1) 27

*P52304A02732*

(ii) Explain why it is possible, in this case, to calculate Kc using equilibrium

amounts in moles, rather than equilibrium concentrations.

(2)

(iii) Calculate the value of Kc.

Give your answer to an appropriate number of significant figures.

(2)

(e) The pink colour of the phenolphthalein fades after the end-point of the titration

has been reached.

Give a possible explanation for this observation.

(2)

(f ) Explain what you could do to confirm that one week is sufficient time for the

mixture to reach equilibrium.

(2)

(g) A student repeated the experiment, but left the mixture in a water bath at 40°C

until equilibrium was reached.*P52304A02832*

CH COOH(l) + CH CH CH OH(l) CH COOCH CH CH (l) + H O(l) ∆ H = +21.4 kJ mol−1

33 2 2 3 2 2 3 2 r

Deduce the effect, if any, on this student’s value for Kc compared with that

obtained in part (d)(iii).

(2)

(Total for Question 8 = 19 marks)

Mark scheme

Show the mark scheme The official mark scheme showing acceptable answers, marking points, and calculations for parts (a) through (g) of the esterification equilibrium question.

Question

Acceptable Answers Additional Guidance Mark

Number

8(a) Any one from: Ignore any mention of protonation or (1)

mechanism for catalysis

Catalyst / speeds up reaction / increases rate / increases Do not award additional incorrect types of

rate of attainment of equilibrium / lowers activation energy reaction

Question

Acceptable Answers Additional Guidance Mark

Number

8(b)(i) Ignore SF throughout 8(b)(i) to 8(c)(ii) except 1 SF, (2)

which should be penalised once only

Example of calculation:

calculation of moles of H+ in 25.0 cm3 (1) (moles NaOH = 0.200 x 23.60 )

1000

= 0.00472 (mol) (= mol H+ in 25.0 cm3)

calculation of moles of H+ in 250 cm3 flask (1) (= 10 x 0.00472) = 0.0472 (mol) (in 250 cm3)

Allow TE for M2 on moles of NaOH

Correct answer with or without working scores 2 marks

Question

Acceptable Answers Additional Guidance Mark

Number

8(b)(ii) Example of calculation: (1)

subtracts moles of H+ in HCl from answer to (b)(i) 0.0472 – 0.00400 = 0.0432 (mol)

Allow TE on answer to part (b)(i)

Question

Acceptable Answers Additional Guidance Mark

Number

8(c)(i) Example of calculation: (1)

calculation of moles of CH3COOH that have reacted (0.105 — 0.0432) = 0.0618

Allow TE on part (b)(ii) unless negative value

Question

Acceptable Answers Additional Guidance Mark

Number

8(c)(ii) Example of calculation: (3)

calculation of equilibrium moles of CH3CH2CH2OH

(1) 0.0800 — 0.0618 = 0.0182

calculation of equilibrium moles of

CH3COOCH2CH2CH3 (1) 0.0618

calculation of equilibrium moles of H2O (1) 0.111 + 0.0618 = 0.1728

Allow TE on answer to part (c)(i) unless negative

value

Question

Acceptable Answers Additional Guidance Mark

Number

8(d)(i) (1)

(Kc =) [CH3COOCH2CH2CH3][H2O] IGNORE state symbols even if incorrect

[CH3COOH][CH3CH2CH2OH] Do not award round brackets

Question

Acceptable Answers Additional Guidance Mark

Number

8(d)(ii) An explanation that makes reference to the following 2 marks could be scored by a correct (2)

points: mathematical expression showing V or dm3

cancel

Same number of moles/molecules on both sides of

the equation

(1) Allow same number of terms on top and

bottom of Kc expression

(so) volume / V cancels in Kc expression (1) Allow units cancel out

Allow “all divided by the same volume”

Question

Acceptable Answers Additional Guidance Mark

Number

8(d)(iii) Example of calculation 2

calculates value of Kc (1) Kc = (0.0618) x (0.1728) = 13.58241758

(0.0432) x (0.0182)

final value of Kc quoted to 2 or 3 SF (1) = 14 / 13.6 (no units)

Correct answer with no working gains full

marks

Ignore units

No TE on wrong Kc expression

Question

Acceptable Answers Additional Guidance Mark

Number

8(e) An explanation that makes reference to the following points: Mark independently (2)

the equilibrium shifts to the left

or

the mixture absorbs carbon dioxide from the atmosphere

(1)

so the mixture is (becoming more) acidic / the acid Allow no longer alkaline

reforms (1) Do not award just “pH decreases”

Question

Acceptable Answers Additional Guidance Mark

Number

8(f) An explanation that makes reference to the following Ignore pH probes / checking pH (2)

points:

carry out / repeat experiment and leave for longer Allow repeat experiment and check titres

than a week (1) within first week

the titre value / Kc value will remain unchanged (if Allow moles / concentration are unchanged

equilibrium has been established) (1) Ignore just “results unchanged”

Question

Acceptable Answers Additional Guidance Mark

Number

8(g) An answer that makes reference to the following points: M2 depends on M1 (2)

Kc value will be greater than that calculated in (d)(iii) (1)

because the (forward) reaction is endothermic Ignore

or References to the equilibrium position shifting

backward / reverse reaction is exothermic (1) to the right (with increasing temperature)

(Total for Question 8 = 19 marks)

TOTAL FOR PAPER = 120 MARKS

How to answer it

Determining Equilibrium Constant (Kc) for Esterification

Edexcel A-Level Chemistry • Core Practical & Calculations

📌 What this question tests

This multi-step synoptic question tests your understanding of homogeneous chemical equilibria, esterification kinetics, acid-base titrations involving strong acid catalysts, mole calculations using ICE tables (Initial, Change, Equilibrium), Kc expressions, and Le Chatelier's Principle regarding temperature and time.

Part (a): Role of Hydrochloric Acid

Question Analysis & Mark Scheme

✅ Correct Answer

Catalyst / speeds up the reaction / increases the rate of attainment of equilibrium / lowers activation energy.

💡 Key Knowledge

In esterification, carboxylic acids and alcohols react extremely slowly on their own. A strong acid catalyst like HCl is added to speed up the rate at which equilibrium is reached without altering the position of equilibrium or the final Kc value.

🎯 Mark: 1 mark. (Ignore any mention of protonation or mechanism).

Parts (b)(i) & (b)(ii): Titration & Acid Calculations

Step-by-Step Calculation Guide

📐 Step-by-Step Calculation for (b)(i)

  1. Moles of NaOH in titration: 0.200 × (23.60 / 1000) = 0.00472 mol
  2. Moles of acid in 25.0 cm³ sample: Since stoichiometry is 1:1, moles of total acid in the 25.0 cm³ aliquot = 0.00472 mol
  3. Scale up to 250 cm³ volumetric flask: 0.00472 × 10 = 0.0472 mol total acid in the flask.

📐 Calculation for (b)(ii)

  1. The total acid in the flask is a mixture of unreacted ethanoic acid AND the HCl catalyst added at the start.
  2. Moles of H⁺(aq) from HCl = 0.00400 mol .
  3. Subtract catalyst moles: 0.0472 − 0.00400 = 0.0432 mol of ethanoic acid at equilibrium.

❌ Common Errors & Exam Technique

Students often forget to multiply by 10 when scaling up from the 25 cm³ titre aliquot to the full 250 cm³ volumetric flask. Always check your volume scale factors!

🎯 Marks: (b)(i) = 2 marks, (b)(ii) = 1 mark. (Allow Error Carried Forward [TE] from (b)(i) to (b)(ii)).

Parts (c)(i) & (c)(ii): ICE Calculations for Equilibrium Moles

Deducing Equilibrium Quantities

📐 Part (c)(i): Ethanoic Acid Reacted

Initial moles of ethanoic acid = 0.105 mol

Equilibrium moles (from b(ii)) = 0.0432 mol

Reacted: 0.105 − 0.0432 = 0.0618 mol

📐 Part (c)(ii): Equilibrium Moles Table

Using 1:1 stoichiometry from the equation:

  • Propan-1-ol: 0.0800 − 0.0618 = 0.0182 mol
  • Propyl ethanoate: 0 + 0.0618 = 0.0618 mol
  • Water: 0.111 (pre-existing) + 0.0618 (formed) = 0.1728 mol

❌ Common Errors

Do not forget that water was already present in the pre-equilibrium mixture ( 0.111 mol ). You must add the newly formed water moles to this initial amount.

🎯 Marks: (c)(i) = 1 mark, (c)(ii) = 3 marks.

Parts (d)(i), (d)(ii) & (d)(iii): Kc Expression & Calculation

Equilibrium Law & Mathematics

✅ (d)(i) Kc Expression

Kc = ([CH₃COOCH₂CH₂CH₃][H₂O]) / ([CH₃COOH][CH₃CH₂CH₂OH])

Note: Square brackets indicate concentrations. Do not use round brackets.

💡 (d)(ii) Why Moles can be used

There are an equal number of moles/molecules on both sides of the stoichiometric equation (2 moles reactants give 2 moles products). Therefore, the volume terms ( V or dm³ ) in the Kc expression completely cancel out.

📐 (d)(iii) Step-by-Step Kc Calculation

  1. Substitute equilibrium values into Kc expression:
    Kc = (0.0618 × 0.1728) / (0.0432 × 0.0182)
  2. Evaluate numerator and denominator:
    Kc = 0.010679 / 0.00078624 = 13.5824...
  3. Apply appropriate significant figures:
    Final Answer: 14 or 13.6 (quoted to 2 or 3 SF, matching data given in question). No units required.
🎯 Marks: (d)(i) = 1 mark, (d)(ii) = 2 marks, (d)(iii) = 2 marks.

Part (e): Phenolphthalein Fading Observation

Examiner Insight & Explanation

✅ Correct Answer

1. The equilibrium shifts to the left OR the mixture absorbs carbon dioxide from the atmosphere.
2. So the mixture becomes more acidic / the ethanoic acid reforms.

🧠 Exam Technique

At the end-point, the solution is very weakly alkaline due to the slight excess of NaOH indicator transition. Atmospheric CO₂ dissolves into the aqueous titration mixture, forming carbonic acid ( H₂CO₃ ), which neutralises the alkali and shifts the esterification equilibrium back toward the reactants, re-introducing acid and lowering the pH.

🎯 Marks: 2 marks (marked independently).

Part (f): Confirming Equilibrium

Experimental Design

✅ Correct Answer

Carry out / repeat the experiment and leave for longer than a week. The titre value / Kc value will remain unchanged if equilibrium has been established.

💡 Key Knowledge

To prove a system has reached dynamic equilibrium, you must demonstrate that macroscopic properties (such as concentrations or titration titre values) no longer change over time.

🎯 Marks: 2 marks.

Part (g): Temperature Change Effect on Kc

Le Chatelier's Principle & Thermodynamics

✅ Correct Answer

The Kc value will be greater than calculated in (d)(iii) because the forward reaction is endothermic (or the reverse reaction is exothermic).

💡 Key Knowledge & Analysis

Given ΔH = +21.4 kJ mol⁻¹ , the forward reaction is endothermic. When temperature is increased (from room temp to 40 °C), Le Chatelier's principle states the equilibrium shifts in the endothermic (forward) direction to oppose the increase in temperature. This increases the yield of products relative to reactants, resulting in a larger Kc value.

🎯 Marks: 2 marks. (M2 depends on M1 being correct).

Topics

Physical Chemistry · Topic 10: Equilibrium I · Topic 11: Equilibrium II · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.