OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 20

20 marks · Hard difficulty · Structured Questions

Analyze halogen boiling points, complete an enthalpy profile diagram for hydrogen iodide decomposition, determine the molecular formula of a chlorine oxide using gas laws, and calculate the formula of a Group 1 iodate(V) using titration data.

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Question text

20 This question is about the halogen group of elements and some of their compounds.

(a) The halogens show trends in their properties down the group.

The boiling points of three halogens are shown below.

Halogen Boiling point / °C

Chlorine –35

Bromine 59

Iodine 184

Explain why the halogens show this trend in boiling points.

… [3]

(b) Hydrogen iodide, HI, is decomposed by heat into its elements:

2HI(g) H (g) + I (g) ∆H = +9.5 kJ mol–1

The decomposition is much faster in the presence of a platinum catalyst.

Complete the enthalpy profile diagram for this reaction using formulae for the reactants and

products.

• Use Ea to label the activation energy without a catalyst.

• Use Ec to label the activation energy with a catalyst.

• Use ∆H to label the enthalpy change of reaction.

enthalpy

progress of reaction

[3]

has a boiling point of 83 °C.

When 0.4485 g of A is heated to 100 °C at 1.00 × 105 Pa, 76.0 cm3 of gas is produced.

Determine the molecular formula of compound A.

Show all your working.

molecular formula of A = … [4]

(d) Compound B is an iodate(V) salt of a Group 1 metal.

The iodate(V) ion has the formula IO –.

A student carries out a titration to find the formula of compound B.

Step 1: The student dissolves 1.55 g of B in water and makes up the solution to 250.0 cm3 in

a volumetric flask.

Step 2: The student pipettes 25.00 cm3 of the solution of B into a conical flask, followed by

10 cm3 of dilute sulfuric acid and an excess of KI(aq).

The iodate(V) ions are reduced to iodine, as shown below.

IO –(aq) + 6H+(aq) + 5I–(aq) 3I (aq) + 3H O(l)

32 2

Step 3: The resulting mixture is titrated with 0.150 mol dm–3 Na S O (aq).

22 3

2S O 2–(aq) + I (aq) S O 2–(aq) + 2I–(aq)

23 2 4 6

The student repeats step 2 and step 3 until concordant titres are obtained.

Titration readings

Titration Trial 1 2 3

Final burette reading / cm3 24.00 47.40 23.75 47.05

Initial burette reading / cm3 0.00 24.00 0.00 23.20

Titre / cm3

Table 20.1

(i) Complete Table 20.1 and calculate the mean titre that the student should use for

analysing the results.

mean titre = … cm3 [2]

(ii) The uncertainty in each burette reading is ±0.05 cm3.

Calculate the percentage uncertainty in the titre obtained from titration 1.

Give your answer to two decimal places.

percentage uncertainty = … % [1]

(iii) Describe and explain how the student should determine the end point of this titration

accurately.

… [2]

(iv) Determine the relative formula mass and formula of the Group 1 iodate(V), B.

Show your working.

relative formula mass of B = …

formula of B = … [5]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

20 (a) ASSUME trend is down the group 3 FULL ANNOTATIONS MUST BE USED

(unless stated otherwise) ----------------------------------------------------------- ALLOW

reverse argument throughout

Forces

London forces increase IGNORE van der Waals’/vdW forces

OR induced dipole(–dipole) interactions increase DO NOT ALLOW hydrogen bonds OR permanent dipole(-

dipole) interactions for first and third marking points

Reason

(Number of) electrons increases ALLOW more (electron) shells

Link to energy and particles DO NOT ALLOW covalent bonds break

More energy to break intermolecular forces

to break London forces

to break induced dipole(–dipole) interactions

(b) Ea: without catalyst 3 FULL ANNOTATIONS MUST BE USED

E : with catalyst Mark each point independently

IGNORE state symbols.

H2(g) + I2(g)

Ec Ea 19 ∆H: DO NOT ALLOW –∆H.

∆H ALLOW ∆H arrow even with a gap at the top and

2HI(g) bottom, i.e. does not quite reach reactant or product

line

Progress of reaction Ea: ALLOW no arrowhead or arrowheads at both ends

of Ea line

2HI(g) on LHS AND H2(g) + I2(g) on RHS Ea line must reach (near or not too far beyond)

maximums regardless of position

ΔH labelled with product above reactant

AND arrow upwards ALLOW AE or EA for Ea

Ea AND Ec correctly labelled with Ec below Ea Exothermic diagram can access the first and third marks

IF M = 183 AND Formula = Cl2O7 award 4 marks is any ECF credit possible using working below

IF M = 183 award 3 marks

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

Use of data and unit conversions

(R = 8.314)

T in K: 373K

V in m3: 76.0 10-6

(p in Pa: 1.00 105)

Calculation of n

(1.00 105) (76.0 10–6)

n =

8.314 373

Correct value of n subsumes first mark

n = 2.45 10–3 (mol)

ALLOW ECF from incorrectly calculated n

Molar mass

m 0.4485 –1

M = n = –3 = 183 (g mol )

2.45 10

ALLOW ECF from incorrect M if formula of ClxOy is the

Molecular formula closest to the with calculated value of M

Cl2O7

IGNORE use of 24 000 cm3 for calculation of n

BUT then Mark molar mass and Molecular formula by

ECF for two marks maximum.

76.0 –3

n = 24000 = 3.17 10 (mol)

0.4485 –1

M = –3 = 141.6/141.5 (g mol )

3.17 10

Molecular formula = Cl3O2

(d) (i) Titres correct and ALL recorded to 2 decimal places 2

Titre: 24.00 23.40 23.75 23.85

mean titre = 23.80 (cm3) ALLOW 23.8 cm3

(d) (ii) 0.05 × 2 1 ALLOW ECF from incorrect subtraction in (i) or incorrect

Percentage uncertainty = 23.40 100 = 0.43 (%) mean

ALLOW 0.42% from titre values 2, 3 or 4 or mean titre or

trial titre.

2 DP required

(d) (iii) Add starch (near the end point) 2

ALLOW blue/black OR black OR purple for colour of

Blue to colourless mixture

ALLOW blue colour disappears (to colourless)

IGNORE ‘clear’

IGNORE ‘colorimetry’

(d) (iv) FIRST CHECK THE ANSWER ON THE ANSWER LINE22 5

IF B = RbIO3 AND relative formula mass = 260.5 award 5

marks

IF relative formula mass = 260.5 award 4 marks

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

n(S O 2–) in titration ALLOW ECF from incorrect mean titre in (a)(i)

0.150 × 23.80 –3

= 1000 = 3.57 × 10 (mol)

– ECF from n(S O 2–) in titration

n(IO3 ) in titration 2 3

3.57 × 10–3 ALLOW a two-step calculation

= = 5.95 10–4 (mol)

n(I ) = n(S O 2–) ÷2 and n(IO –) = n(I ) ÷3

62 2 3 3 2

– 3 ECF from n(IO –) in titration

n(IO3 ) in original 250 cm 3

= 10 × 5.95 10–4 = 5.95 10–3 (mol)

ECF from n(IO –) in original 250 cm3

Relative formula mass of B 3

1.55 –1 IF scaling 10 is omitted,

= –3 = 260.5 (g mol ) ALLOW ECF from n(IO –) in titration

5.95 10 3

Formula of B (must be derived from relative formula

mass) ALLOW ECF from incorrect RFM of B provided metal is

Iodate of Group 1 metal that most closely matches from Group 1

ALLOW RbIO –

calculated molar mass of B 3

DO NOT ALLOW RbIO3 without relative formula mass

Formula from 260.5 = RbIO3 value.

DO NOT ALLOW 260.4 (without working) and RbIO3

IF B = RbIO3 AND relative formula mass = 261 award 5

marks

Total 20

How to answer it

Halogen Chemistry & Quantitative Analysis Study Guide

📋 What this question tests

This comprehensive multi-part question tests your knowledge of Group 7 physical trends (boiling points), energetic profiles for catalyzed reactions, ideal gas calculations involving moles and molar mass, redox titrations (iodometry/thiosulfate titrations), percentage uncertainty calculations, practical end-point identification, and complex multi-step stoichiometry to determine the formula of an unknown Group 1 iodate(V) salt.

Part (a): Trends in Halogen Boiling Points

💡 Key Knowledge

Halogens are simple molecular substances held together by weak London forces (induced dipole-dipole interactions).
Down Group 7, molecules become larger with more electrons.
Larger electron clouds set up larger temporary dipoles, increasing London forces.

✅ Correct Answer & Marking Points

Mark 1: Mention London forces (or induced dipole-dipole interactions) increase.
Mark 2: State that molecules have more electrons down the group.
Mark 3: Explain that more energy is needed to overcome these stronger intermolecular forces.

❌ Common Errors

Describing covalent bonds breaking during boiling (covalent bonds stay intact; only intermolecular forces break).
Using forbidden terminology such as "van der Waals forces" (insufficiently specific for OCR) or "hydrogen bonds".

Part (b): Enthalpy Profile for Catalyzed Decomposition

🧠 Exam Technique & Diagram Rules

Reactants & Products: Label reactants 2HI(g) on the left and products H₂(g) + I₂(g) on the right.
Endothermic curve: Products must be drawn at a higher energy level than reactants.
Activation Energies: Draw two peaks. The uncatalyzed peak ( Eₐ ) must be higher than the catalyzed peak ( E꜀ ).
Enthalpy Change: Label ΔH with an arrow pointing upwards from the reactant level to the product level.

❌ Common Errors

Drawing the catalyzed peak higher than the uncatalyzed peak.
Placing the arrow for ΔH downwards or omitting formula labels for reactants and products.

Part (c): Determining the Molecular Formula of Oxide A

📐 Step-by-Step Calculation

Convert units:
- Pressure ( p ) = 1.00 × 10⁵ Pa
- Volume ( V ) = 76.0 cm³ = 76.0 × 10⁻⁶ m³
- Temperature ( T ) = 100 °C + 273.15 = 373 K
- Gas Constant ( R ) = 8.314 J mol⁻¹ K⁻¹
Calculate moles of gas ( n ) using Ideal Gas Equation ( pV = nRT ):
n = pV / RT = (1.00 × 10⁵ × 76.0 × 10⁻⁶) / (8.314 × 373) = 2.45 × 10⁻³ mol
Calculate molar mass ( M ):
M = mass / n = 0.4485 / (2.45 × 10⁻³) = 183 g mol⁻¹
Deduce Molecular Formula:
An oxide of chlorine with Mᵣ = 183 is Cl₂O₇ (Cl: 35.5×2 + O: 16×7 = 71 + 112 = 183).

Mark Allocation (4 marks): 1 mark for correct data conversion/rearrangement, 1 mark for moles of gas, 1 mark for molar mass calculation, 1 mark for molecular formula Cl₂O₇ .

Part (d): Titration & Formula of Group 1 Iodate(V) Salt B

📋 Subpart (i): Titre Processing & Mean

Trial: 24.00 cm³ | Titration 1: 47.40 - 24.00 = 23.40 cm³ | Titration 2: 23.75 - 0.00 = 23.75 cm³ | Titration 3: 47.05 - 23.20 = 23.85 cm³

Using concordant titrations 1 and 3 (or 2 and 3 depending on student data selection, here mark scheme accepts averaging concordant values, e.g., using 23.80 cm³ from concordant values).
Mean Titre = 23.80 cm³ (Must be recorded to 2 decimal places).

📋 Subpart (ii): Percentage Uncertainty

Percentage Uncertainty = (± uncertainty / titre value) × 100

= (0.05 × 2 / 23.40) × 100 = 0.43% (Double buret reading accounts for two absolute uncertainties: initial and final).

📋 Subpart (iii): End-Point Procedure

Reagent added: Add starch indicator near the end point (when the iodine colour fades to pale yellow/straw).
Colour change: Solution turns from blue-black to colourless.

📐 Subpart (iv): Step-by-Step Calculation for Formula of B

Moles of Na₂S₂O₃ in titration:
n = (0.150 × 23.80) / 1000 = 3.57 × 10⁻³ mol
Moles of IO₃⁻ in the 25 cm³ pipette sample:
From equation, 1 mol IO₃⁻ reacts with 6 mol S₂O₃²⁻ .
n(IO₃⁻) = (3.57 × 10⁻³) / 6 = 5.95 × 10⁻⁴ mol
Moles of IO₃⁻ in original 250 cm³ volumetric flask:
Scale up by factor of 10 ( 250 / 25 ):
n = 5.95 × 10⁻⁴ × 10 = 5.95 × 10⁻³ mol
Calculate Relative Formula Mass ( Mᵣ ) of B:
Mᵣ = mass / moles = 1.55 g / (5.95 × 10⁻³ mol) = 260.5 g mol⁻¹
Identify Group 1 Metal and Formula:
Let Group 1 metal be M . Formula is MIO₃ .
Mᵣ(IO₃) = 126.9 + (16.0 × 3) = 174.9
Ar(M) = 260.5 - 174.9 = 85.6
The closest Group 1 metal is Rubidium ( Rb , Aᵣ = 85.5 ).
Formula of B: RbIO₃

Mark Allocation for Part (d)(iv) (5 marks): Correct titration moles, stoichiometric scaling to find iodine/iodate, scaling up to the 250 cm³ flask, calculating molar mass, and correctly identifying the salt as RbIO₃ .

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Practical Activity Groups · PAG 2: Acid-base titration · 3.1 The periodic table · 3.2 Physical chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.