OCR A-Level Chemistry Unified chemistry (03), June 2018: Question 5

11 marks · Medium difficulty · Calculations

Calculate the enthalpy change of neutralisation and determine the equilibrium constant Kc for the esterification of succinic acid with ethanol.

Practise this question

Question

A two-part chemistry exam question about succinic acid. Part (a) asks to calculate the enthalpy change of neutralisation using temperature data from a polystyrene cup experiment provided in a table. Part (b) involves an esterification equilibrium reaction with ethanol, featuring an equilibrium expression (Expression 5.1) and five sub-questions covering analytical techniques, writing equations, drawing skeletal structures, explaining why Kc can be calculated using moles, and calculating the value of Kc.
Question text

5 A student carries out two experiments in the laboratory based on succinic acid (butanedioic acid),

(CH2COOH)2.

(a) Aqueous succinic acid can be neutralised by aqueous sodium hydroxide, NaOH(aq):

(CH2COOH)2(aq) + 2NaOH(aq) → (CH2COONa)2(aq) + 2H2O(l)

This reaction can be used to determine a value for the enthalpy change of neutralisation,

∆neutH.

The student follows this method:

• Add 50.0 cm3 of 0.400 mol dm–3 succinic acid to a polystyrene cup.

• Measure out 50.0 cm3 of 1.00 mol dm–3 NaOH(aq), which is in excess.

• Measure the temperature of both solutions.

• Add the NaOH(aq) to the aqueous succinic acid in the polystyrene cup, stir the mixture,

and record the maximum temperature.

Temperature readings

Maximum temperature of mixture/°C 26.5

Initial temperature of both solutions/°C 21.5

Calculate a value for the enthalpy change of neutralisation, ∆ H, in kJ mol–1.

neut

Assume that the density of all solutions and the specific heat capacity, c, of the reaction

mixture are the same as for water.

Δ H = … kJ mol–1 [4]

neut

(b) Succinic acid is esterified by ethanol, C2H5OH, in the presence of an acid catalyst to form an

equilibrium mixture.

The equilibrium constant, Kc, for this equilibrium can be calculated using the amounts, in

moles, of the components in the equilibrium mixture, using expression 5.1.

n( (CH COOC H ) ) × n( H O )2

22 5 2 2

Kc = 2 Expression 5.1

n( (CH2COOH)2) × n( C2H5OH )

A student carries out an experiment to determine the value of Kc for this equilibrium.

• The student mixes together 0.0500 mol of succinic acid and 0.150 mol of ethanol,

with a small amount of an acid catalyst.

• The mixture is allowed to reach equilibrium.

• The student determines that 0.0200 mol of succinic acid are present in the

equilibrium mixture.

(i) Which technique could be used to determine the equilibrium amount of succinic acid?

… [1]

(ii) Write the equation for the equilibrium reaction that takes place.

… [1]

(iii) Draw the skeletal formula of the ester present in the equilibrium mixture.

[1]

(iv) Kc is the equilibrium constant in terms of equilibrium concentrations.

Why can expression 5.1 be used to calculate Kc for this equilibrium?

… [1]

(v) Calculate the value of Kc for this reaction.

Show your working.

Kc = … [3]

Mark scheme

Show the mark scheme The mark scheme providing detailed answers and marking points for all parts of question 5, including calculation steps for enthalpy change, acceptable answers for experimental techniques and equations, the skeletal structure of the ester, reasons why volume cancels in Kc expressions, and equilibrium moles calculations.

Question Answer Marks Guidance

5 (a) TAKE CARE: Correct final answer of –52.3 OR –52.25 can be 4

obtained from two cancelling errors:

Use of 50 for energy released

(no 2 of 50 for two solutions mixed)

No ÷ 2 in final step

–52.3 OR –52.25 would then be awarded 2 marks out of 4

------------------------------------------------------------------------------

Correctly calculates n(succinic acid) ALLOW ECF throughout

50.0

= 0.400 1000 = 0.02(00) (mol)

Energy released in J OR kJ

= 100.00 4.18 5.0 = 2090 (J) OR 2.090 (kJ) DO NOT ALLOW less than 3 SF

IGNORE units

Energy released, in kJ or J, for formation of 2 mol H2O

2090 -----------------------------------------------------------------

± 0.0200 = ± 104500 (J) ALTERNATIVE METHOD

OR n(succinic acid) = 0.02(00) (mol)

2.090

± 0.0200 = ± 104.5 OR ± 105 (kJ) Energy released = 2090 (J) OR 2.090 (kJ)

∆ H to 3 or more SF AND correct – sign n(H2O) formed = 2 0.02(00) = 0.04(00) (mol)

neut

104.5 2.090 –1

= – = – 52.3 OR – 52.25 kJ mol–1 ∆neutH = – = – 52.3 OR – 52.25 kJ mol

2 0.0400

(b) (i) Titration 1 IGNORE type of titration

(ii) (CH2COOH)2 + 2C2H5OH ⇌ (CH2COOC2H5)2 + 2H2O 1 ALLOW instead of ⇌ sign

ALLOW molecular formulae or hybrid formulae

Structures provided on QP

e.g. C4H6O4 + 2C2H6O ⇌ C8H14O4 + 2H2O

(iii) O 1 IGNORE displayed formulae

O 16

O

O

(iv) Volume cancels 1 ALLOW units cancel

OR

Same number of moles on each side of equation ALLOW (sum of) balancing numbers/coefficients on

each side of equation are the same

OR same number of (moles of) reactants and

products

IGNORE volume is the same; Kc has no units

(v) Moles of equilibrium products 1 mark 3

n((CH2COOC2H5)2) = 0.0300 (mol)

AND

n(H2O) = 0.0600 (mol)

Moles of C2H5OH 1 mark

n(C2H5OH) = 0.150 – 0.060 = 0.0900 (mol)

Kc calculated 1 mark

0.03 0.062 ALLOW ECF

= 2 = 0.667 OR 0.67

0.02 0.09

NOTE: 0.02 must be used for n(succinic acid) ALLOW 0.66, 0.666, etc. (2 SF and more)

Treated as meaning 0.6 recurring

ALLOW 2/3

IGNORE any units

Total 11

How to answer it

Enthalpy of Neutralisation & Chemical Equilibrium Study Guide

What this question tests

This multi-part question assesses core physical and organic chemistry topics: calculating enthalpy changes of neutralisation from calorimetric temperature data, understanding experimental techniques for tracking equilibria (titrations), writing balanced chemical equations, interpreting equilibrium expressions, and calculating equilibrium constants (Kc) using ICE tables.

Question Part (a)

Enthalpy of Neutralisation Calculation

✅ Correct Answer

ΔneutH = -52.3 kJ mol⁻¹ (or -52.5 kJ mol⁻¹ depending on rounding path)

💡 Key Knowledge

  • Use q = m × c × ΔT to find energy transferred. Total mass m = 50.0 + 50.0 = 100.0 g .
  • Enthalpy change of neutralisation is always defined per mole of H₂O formed (or per mole limiting reagent reacting according to stoichiometry).
  • Always include the negative sign for exothermic neutralisation reactions.

📐 Step-by-Step Calculation

  1. Moles of succinic acid: 0.400 × (50.0 / 1000) = 0.0200 mol . (Note: NaOH is in excess).
  2. Temperature change (ΔT): 26.5 - 21.5 = 5.0 °C .
  3. Energy released (q): 100.0 × 4.18 × 5.0 = 2090 J = 2.090 kJ .
  4. Per mole of reaction / H₂O: Since 2 moles of NaOH react with 1 mole of succinic acid to produce 2 moles of H₂O, divide energy by 0.0200 (or calculate moles of water formed = 0.0400 mol ).
  5. Final ΔneutH: -2.090 / 0.0400 = -52.3 kJ mol⁻¹ .

❌ Common Errors & Examiner Pitfalls

  • Mass trap: Using only 50 g instead of 100 g for total solution mass.
  • Stoichiometry trap: Forgetting to divide by the moles of water formed or failing to account for the 1:2 stoichiometric ratio, leading to double the actual enthalpy value.
  • Significant figures: OCR strictly penalises answers given to fewer than 3 significant figures.
🎯 Marks: 4 marks available. Awarded for correct moles, correct energy calculation, correct scaling to 1 mole of water, and correct final value with negative sign and valid SF.
Question Part (b)(i)

Technique to Determine Equilibrium Amount

✅ Correct Answer

Titration (using a standard solution of sodium hydroxide to find unreacted acid).

🧠 Exam Technique

Keep answers concise. Examiners accept "Titration" directly without needing extra descriptors about acid-base indicators, though mentioning quenching/titrating against NaOH is safe.

🎯 Mark: 1 mark.
Question Part (b)(ii)

Equation for the Equilibrium Reaction

✅ Correct Answer

(CH₂COOH)₂ + 2C₂H₅OH ⇌ (CH₂COOC₂H₅)₂ + 2H₂O

💡 Key Knowledge

Make sure to use the equilibrium arrow ( ⇌ ) instead of a straight reaction arrow, and ensure stoichiometry matches the esterification of a dicarboxylic acid (needs 2 moles of alcohol).

🎯 Mark: 1 mark.
Question Part (b)(iii)

Skeletal Formula of the Ester

✅ Correct Answer

Draw the skeletal formula for diethyl succinate: CH₃CH₂OOC-CH₂-CH₂-COOCH₂CH₃ (drawn with zig-zag carbon chains attached to oxygen atoms on both carbonyl groups).

🧠 Exam Technique

Check that terminal bonds connect correctly from oxygen to the ethyl group carbons. Examiners accept displayed, structural, or skeletal formulas as long as connectivity is completely unambiguous.

🎯 Mark: 1 mark.
Question Part (b)(iv)

Validity of Equilibrium Expression

✅ Correct Answer

Volumes cancel out OR There is the same number of moles of gaseous/aqueous species on both sides of the equation (2 moles of reactants vs 2 moles of products in terms of volume terms).

❌ Common Errors

Stating simply "units cancel" without explaining why (volumes cancel out from top and bottom of the Kc expression because concentration ratios have equal powers).

🎯 Mark: 1 mark.
Question Part (b)(v)

Calculating Kc

✅ Correct Answer

Kc = 0.667 (or 0.67 / 0.66)

📐 Step-by-Step ICE Table Calculation

  1. Initial Moles: Succinic acid = 0.0500 , Ethanol = 0.150 , Ester = 0 , Water = 0 .
  2. Equilibrium Moles of Acid: Given as 0.0200 mol .
  3. Change in Moles: 0.0500 - 0.0200 = 0.0300 mol of succinic acid reacted.
  4. Equilibrium Products: Ester formed = 0.0300 mol ; Water formed = 2 × 0.0300 = 0.0600 mol .
  5. Equilibrium Ethanol: 0.150 - (2 × 0.0300) = 0.0900 mol .
  6. Substitute into Kc Expression: Kc = (0.0300 × 0.0600²) / (0.0200 × 0.0900²) = 0.667 . (Volumes cancel, so mole values can be used directly).
🎯 Marks: 3 marks available. Awarded for correct equilibrium moles of products, correct equilibrium moles of ethanol, and final correct Kc calculation. ECF applies.

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · Practical Activity Groups · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · 6.1 Aromatic compounds, carbonyls and acids · PAG 3: Enthalpy determination

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.