Edexcel A-Level Chemistry Paper 1, November 2021: Question 5

13 marks · Medium difficulty · Calculations

Calculate the enthalpy change for the reaction between chlorine and phosphorus, explain why bond enthalpies are positive, show disproportionation using oxidation numbers, balance a chlorine-alkali equation, identify chlorination/halogen conditions and colours, write an ionic equation for halide precipitation, and identify a halide ion by mass calculation.

Practise this question

Question

Question text

5 The halogens are elements in Group 7 of the Periodic Table.

(a) Chlorine compounds have many uses, including water treatment.

(i) Chlorine and phosphorus (P4) can react to form phosphorus(V) chloride.

The structure of a molecule of phosphorus is

P P

Some mean bond enthalpy values are shown in the table.

Bond Mean bond enthalpy / kJ mol–1

P P +198

Cl Cl +243

P Cl +326

Calculate the enthalpy change for the reaction between chlorine and

phosphorus to form phosphorus(V) chloride.

10Cl2 + P4 → 4PCl5

(3)

(ii) Give a reason why bond enthalpy values are always positive.

(1)

(b) Sodium chlorate(I) is a bleaching agent.

(i) Sodium chlorate(I) can be made by the reaction of chlorine with

sodium hydroxide.

Show, by using oxidation numbers, that this reaction is disproportionation.

2NaOH + Cl2 → NaClO + NaCl + H2O

(2)

… 11

… *P65463A01128*

(ii) A different bleaching agent can be made by the reaction of chlorine with

sodium hydroxide under different conditions.

Balance this equation.

… NaOH + … Cl2 → NaClO3 + … NaCl + … H2O

(1)

(iii) What conditions are required for the reaction in (b)(ii)?

(1)

A cold and dilute alkali

B cold and concentrated alkali

C hot alkali

D excess chlorine

hexane or cyclohexane.

Which sequence of colours is correct for chlorine, bromine and iodine dissolved in

an organic solvent?

(1)

Chlorine Bromine Iodine

A orange red-brown black

B pale green *P65463A01228*orangeblack

C orange red-brown purple

D pale green orange purple

(d) Halide ions can be identified by their reaction with silver nitrate.

(i) Write the ionic equation for the reaction between aqueous solutions of

sodium iodide and silver nitrate.

Include state symbols.

(2)

(ii) A solution containing 0.010 mol of a halide ion was reacted with excess

silver nitrate and produced 1.88g of precipitate.

Identify the halide ion.

Justify your answer.

(2)

(Total for Question 5 = 13 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

5(a)(i) Example of calculation (3)

• sum of bonds broken (1) bonds broken = (6 x 198) + (10 x 243)

= 3618 (kJ mol−1)

bonds made = (20 x 326) = (−)6520 (kJ mol−1)

• and sum of bonds made (1)

• answer and with negative sign (1) enthalpy change

= Bonds broken – bonds made

= (3618 − 6520)

=−2902 (kJ mol−1)

Correct answer with no working scores (3)

TE on bonds broken and made

Question

Answer Additional Guidance Mark

Number

5(a)(ii) (1)

• bond breaking requires energy ALLOW bond breaking is endothermic

or ALLOW bond making is exothermic

by convention bond enthalpies refer to dissociation and

so are endothermic Ignore just ‘bonds are broken’ / ‘it is endothermic’

Question

Answer Additional Guidance Mark

Number

5(b)(i) Check the equation (2)

• chlorine is oxidised and from 0 to +1 (in NaClO) (1)

• chlorine is reduced and from 0 to -1 (in NaCl) (1)

Allow (1) for three correct oxidation numbers if

no other mark is awarded.

Allow (1) max for general definition of

disproportionation

Question

Answer Additional Guidance Mark

Number

5(b)(ii) (1)

6 NaOH + 3 Cl2 → NaClO3 + 5 NaCl + 3 H2O

• equation

Allow multiples

Question

Answer Mark

Number

5(b)(iii) The only correct answer is C (hot alkali) (1)

A is not correct because high temperature is required

B is not correct because high temperature is required

D is not correct because high temperature and not excess chlorine is required

Question

Answer Mark

Number

5(c) The only correct answer is D (pale green – orange - purple) (1)

A is not correct because chlorine is not orange and the colour stated for bromine is for the pure liquid state and

solid iodine can appear black but not in an organic solvent

B is not correct because solid iodine can appear black but not in an organic solvent

C is not correct because chlorine is not orange and the colour stated for bromine is in the pure liquid state

Question

Answer Additional Guidance Mark

Number

5(d)(i) Example of equation (2)

• ionic equation (1) Ag+(aq) + I−(aq) → AgI(s)

Allow multiples

• state symbols (1) M2 dependent on M1 or near miss

Question

Answer Additional Guidance Mark

Number

5(d)(ii) An answer that includes Incorrect halide scores (0) (2)

• halide ion with some justification attempt (1) Bromide (ion)/Br−

Do not award ‘bromine (ion)’

• calculation of expected mass of silver halides (1) 0.01 mol of AgCl = 1.43 (g)

AgBr = 1.88 (g) AgI = 2.35 (g)

Mass of 1.0 mol is 188 g so subtraction of 107.9

for Ag means X = 80.1 so closest is Br

TE on incorrect formula silver halide in d(i)

(Total Question 5 = 13 marks)

How to answer it

Edexcel A-Level Chemistry: Group 7 (The Halogens) & Energetics

What this question tests

This comprehensive question assesses your knowledge of Group 7 chemistry, including mean bond enthalpy calculations, disproportionation reactions of chlorine, balancing equations under different temperature conditions, observations of halogens in organic solvents, and qualitative identification of halide ions using silver nitrate coupled with quantitative mass calculations.

Question 5 (a)

Energetics and Bond Enthalpies

✅ Correct Answer (5a.i)

Enthalpy change = -2902 kJ mol⁻¹

Mark breakdown: (1) Sum of bonds broken, (1) Sum of bonds made, (1) Correct final value with negative sign.

📐 Step-by-Step Calculation

Analyze Equation: 10Cl₂ + P₄ → 4PCl₅ . Note that PCl₅ has 5 P-Cl bonds per molecule, so 4 molecules = 20 P-Cl bonds. In P₄, each P is bonded to 3 others, giving 6 P-P bonds. Cl₂ has 10 Cl-Cl bonds.
Bonds Broken (Reactants): (6 × 198) + (10 × 243) = 1188 + 2430 = +3618 kJ mol⁻¹
Bonds Made (Products): 20 × 326 = -6520 kJ mol⁻¹
Calculate Enthalpy Change: Σ(Bonds broken) - Σ(Bonds made) = 3618 - 6520 = -2902 kJ mol⁻¹

❌ Common Calculation Traps

Forgetting to multiply individual bond enthalpies by the stoichiometric balancing numbers (e.g., missing the 10 for Cl-Cl or 20 total P-Cl bonds).
Omitting the negative sign in the final enthalpy change.

✅ Correct Answer (5a.ii)

Bond breaking requires energy (or: by convention, bond enthalpies refer to bond dissociation and are therefore endothermic).

Mark breakdown: (1) mark.

Question 5 (b)

Disproportionation of Chlorine & Conditions

✅ Correct Answer (5b.i)

Chlorine is simultaneously oxidised and reduced in the same reaction.

Oxidation state of Cl changes from 0 in Cl₂ to +1 in NaClO.
Oxidation state of Cl changes from 0 in Cl₂ to -1 in NaCl.

Mark breakdown: (1) for oxidation state change/identification, (1) for reduction state change/identification.

✅ Correct Answer (5b.ii & 5b.iii)

Balanced Equation (5b.ii): 6NaOH + 3Cl₂ → NaClO₃ + 5NaCl + 3H₂O

Required Conditions (5b.iii): C - hot alkali

Mark breakdown: 1 mark for equation, 1 mark for multiple choice selection.

🧠 Exam Technique

When balancing equations involving disproportionation to form chlorates(V), remember that chlorine forms both NaClO₃ and NaCl in a 1:5 ratio under hot conditions, requiring 6 moles of NaOH.

Question 5 (c)

Halogen Colours in Organic Solvents

✅ Correct Answer

D - pale green – orange – purple

Mark breakdown: (1) mark for choosing option D.

💡 Key Knowledge

Chlorine: Pale green (in both water and organic solvents).
Bromine: Orange / red-brown in organic solvents (hexane/cyclohexane).
Iodine: Purple in organic solvents (though solid iodine can appear black, its solution in non-polar organic solvents is characteristically violet/purple).

Question 5 (d)

Halide Identification & Quantitative Precipitation

✅ Correct Answer (5d.i)

Ag⁺(aq) + I⁻(aq) → AgI(s)

Mark breakdown: (1) for correct ionic species and balancing, (1) for correct state symbols. M2 is dependent on M1.

📐 Step-by-Step Calculation (5d.ii)

Find moles of halide reacted: 0.010 mol. Since the reaction is 1:1, moles of precipitate formed = 0.010 mol .
Calculate molar mass of precipitate: Molar Mass = Mass / Moles = 1.88 g / 0.010 mol = 188 g mol⁻¹ .
Identify the silver halide: Subtract the atomic mass of Silver (Ag = 107.9 g mol⁻¹): 188 - 107.9 = 80.1 g mol⁻¹ , which corresponds to the Bromide ion ( Br⁻ ).
Justification: State clearly that the calculated Mr matches AgBr , therefore the halide ion is bromide.

Mark breakdown: (1) Identification of bromide ion with justification, (1) Calculation of expected mass or molar mass.

❌ Common Errors

Students frequently write "bromine (atom/molecule)" instead of specifying the bromide ion (Br⁻) . Ensure you name ions correctly when asked to identify a species from solution tests.

Topics

Physical Chemistry · Inorganic Chemistry · Organic Chemistry · Topic 8: Energetics I · Topic 3: Redox I · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.