OCR A-Level Chemistry Unified chemistry (03), November 2021: Question 1

18 marks · Hard difficulty · Structured Questions

Answer a series of short chemistry questions covering partial pressure, equilibrium constants, boiling points and intermolecular forces, percentage composition and stoichiometry, reactions of carboxylic acids, and organic reaction mechanisms.

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Question text

1 These short questions are from different areas of chemistry.

(a) A flask contains 110 g CO2 and 120 g O2 at a pressure of 50.0 atm.

What is the partial pressure of the CO2?

partial pressure of CO2 = … atm [2]

(b) Pure PCl5 is placed in a sealed container which is heated to 200 ºC.

An equilibrium is set up: PCl5(g) PCl3(g) + Cl2(g)

Under these conditions:

• K = 8.00 × 10–3 mol dm–3

• equilibrium concentration of PCl = 0.0500 mol dm–3.

Calculate the concentrations of PCl3 and Cl2 at equilibrium.

concentration of PCl = … mol dm–3

concentration of Cl = … mol dm–3

[2]

hydrogen fluoride are shown below.

Electronegativity H: 2.1 C: 2.5 N: 3.0 F: 4.0

Compound CH4 NH3 HF

Boiling point / ºC –161.5 –33.3 19.5

Explain the difference in the boiling points of CH4, NH3 and HF.

… [3]

(d) Compound A has the following percentage composition by mass:

Ca, 81.10%; N, 18.90%.

A student reacts compound A with water to form alkaline gas B and alkali C.

Identify A, B and C and write the equation for the reaction of compound A with water.

… [4]

(e) The structure of lactic acid is shown below.

lactic acid

Complete and balance the equations for two reactions of lactic acid.

Reaction with sodium carbonate

CH3CH(OH)COOH + …

Reaction with aluminium

CH3CH(OH)COOH + …

[4]

(f) The ester, methyl ethanoate, can be synthesised by reacting a haloalkane with a carboxylate

ion.

The mechanism is nucleophilic substitution.

Outline the mechanism for this reaction.

[3]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

1 (a) FIRST CHECK THE ANSWER ON ANSWER LINE 2 AO1.2

If answer = 20 award 2 marks × 2

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

n(CO2) = OR 2.5 (mol)

AND

n(O2) = OR 3.75 (mol)

2.5

p(CO2) = × 50.0 OR 0.4 × 50.0 = 20(.0) (atm) ALLOW ECF

6.25

from incorrect Σ ( n(CO2) + n(O2) ) ONLY

(b) FIRST CHECK THE ANSWER ON ANSWER LINES 2

If [PCl3] = [Cl2] = 0.02(00) award 2 marks

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

[PCl3] [Cl2] [PCl3] [Cl2] AO1.1 Square brackets required

Kc = OR with number(s), e.g. Kc =

[PCl5] 0.05(00)

[PCl3] = [Cl2] = √(Kc × [PCl5] ) ------------------------------------------------------------

= √(8.00 × 10–3 × 0.0500 ) Common errors

= √(4.00 × 10–4) [PCl3] [Cl2]

2.00 × 10–4 from K = 1 mark

c [PCl ]

= 2.00 × 10–2 (mol dm–3) AO2.2 ÷2 instead of √

[PCl5]

2.5 from Kc = [PCl 1 mark

3] [Cl2]

Inverse Kc expression

Question Answer 4 Marks Guidance

element

3 ANNOTATE WITH TICKS AND CROSSES

ALLOW ORA throughout

Electronegativity and boiling point -------------------------------------------------------------------

Boiling point/Energy increases with increased ORA

electronegativity (difference) AO1.1

Type of intermolecular force

HF AND NH3 have hydrogen bonding IGNORE permanent dipole interactions

AND IGNORE IDID

CH4 has London forces/induced (dipole–)dipole interactions AO1.2 IGNORE HF and NH3 are polar/CH4 is non-polar

Comparison between strength of intermolecular forces

HF has stronger hydrogen bonding than NH3 AO2.1 IGNORE strength of ionic and covalent bonds

OR hydrogen bonding is stronger than London forces

(d) 4 IGNORE working

A: Ca3N2 (formula required) AO1.1

If B and C labels are the wrong way round OR

B: NH3 OR ammonia AO2.7 missing, award 1/2 for B and C labels,

×2 i.e. for B Ca(OH)2 C NH3 1/2 marks

C: Ca(OH)2 OR calcium hydroxide

AO2.6 ALLOW CaO2H2

Equation:

Ca3N2 + 6H2O → 2NH3 + 3Ca(OH)2 ALLOW multiples for equation

IF C = CaO, ALLOW ECF for:

Ca3N2 + 3H2O → 2NH3 + 3CaO

5 AO

element

(e) 4 AO2.6 ALLOW multiples

×4 IGNORE state symbols

2CH3CH(OH)COOH + Na2CO3 →

2CH3CH(OH)COONa + CO2 + ALLOW ions shown separately

H2O

For CO2 AND H2O, ALLOW H2CO3

CO2 and H2O OR CH3CH(OH)COONa as product(s)

ALLOW ….COONa+ (i.e. one of charges missing)

Balanced equation correct

ALLOW ….COO) Al3+ (i.e. one of charges missing)

3CH3CH(OH)COOH + Al → (CH3CH(OH)COO)3Al + 1½ H2

H2 OR (CH3CH(OH)COO)3Al as product

Balanced equation correct

6 element

(f) Mechanism: 3 ANNOTATE ANSWER TICKS AND CROSSES

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

NOTE: Curly arrows can be straight, snake-like, etc.

but NOT double headed or half headed arrows

1st curly arrow must start from, OR be traced back

to, any part of C–Cl bond and go to Cl

NOTE: Can be any C–X bond, e.g. C–Cl, C–Br, C–I but

must be consistent.

2nd curly arrow must

Curly arrow on C–X • go to the C of C–Cl

δ+ δ– AND

Dipole shown on C–X bond of CH3X, C and X

AND • start from, OR be traced back to any point

across width of lone pair on O of CH COO–

curly arrow from C−X bond to X atom 3

AO2.5

Curly arrow from CH COO–

Curly arrow from CH COO– to C atom of C–X bond

3 AO1.2

• OR start from ‘–‘ on O of CH COO– ion

(Lone pair NOT needed

if curly arrow from O–)

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

Products AO2.5 If CH COOH used instead of CH COO–,

Correct organic product AND X– –

ALLOW X OR HX as 2nd product

7 AO

element

ALLOW SN1 mechanism

First mark

Dipole shown on C–Cl bond, Cδ+ and Clδ−,

AND curly arrow from C−Cl bond to Cl atom

Second mark

Correct carbocation AND curly arrow from CH COO–

to carbocation

Curly arrow must be from lone pair on O of

CH COO–

OR from minus on O of CH COO– ion (no need to

show lone pair if curly came from – charge)

Third mark

Correct organic product AND Cl–

How to answer it

OCR A-Level Chemistry Synoptic Multi-Topic Revision Guide

What this question tests

This comprehensive multi-topic question assesses core concepts across physical, inorganic, and organic chemistry. You will be tested on partial pressures and mole fractions, equilibrium constants (Kc calculations), intermolecular forces and electronegativity trends, quantitative formula analysis combined with inorganic synthesis, carboxylic acid reactions with metals and carbonates, and halogenoalkane nucleophilic substitution mechanisms.

Part (a) — Partial Pressures

Calculating Partial Pressure from Mass Data

✅ Correct Answer

partial pressure of CO₂ = 20 (atm)

📐 Calculation Steps

Find moles of CO₂: 110 g / 44.0 g mol⁻¹ = 2.50 mol
Find moles of O₂: 120 g / 32.0 g mol⁻¹ = 3.75 mol
Calculate total moles: 2.50 + 3.75 = 6.25 mol
Apply mole fraction formula: (2.50 / 6.25) × 50.0 atm = 20 atm

❌ Common Errors

Forgetting to calculate the total moles of all gases in the mixture before finding the mole fraction.
Using atomic masses instead of molecular masses (e.g., using 16 instead of 32 for O₂).

Mark breakdown: 2 marks total. 1 mark for calculating moles of CO₂ and O₂; 1 mark for the correct partial pressure.

Part (b) — Equilibrium Constants (Kc)

Determining Equilibrium Concentrations

✅ Correct Answer

[PCl₃] = 2.00 × 10⁻² mol dm⁻³

[Cl₂] = 2.00 × 10⁻² mol dm⁻³

💡 Key Knowledge

For the dissociation equilibrium PCl₅(g) &rightleftharpoons; PCl₃(g) + Cl₂(g) , since 1 mole of PCl₅ produces 1 mole of PCl₃ and 1 mole of Cl₂, their equilibrium concentrations are equal: [PCl₃] = [Cl₂] .

📐 Step-by-Step Calculation

Write the Kc expression: Kc = ([PCl₃][Cl₂]) / [PCl₅]
Rearrange for unknown concentration squared: [PCl₃]² = Kc × [PCl₅]
Substitute values: [PCl₃]² = (8.00 × 10⁻³) × (0.0500) = 4.00 × 10⁻⁴
Take the square root: [PCl₃] = √(4.00 × 10⁻⁴) = 2.00 × 10⁻² mol dm⁻³

❌ Common Errors

Failing to take the square root of the combined value, leaving the answer as 4.00 × 10⁻⁴ (loses 1 mark). Inverting the Kc expression is also a frequent mistake.

Mark breakdown: 2 marks total. 1 mark for setting up/rearranging the expression; 1 mark for correct evaluation with proper units/significant figures.

Part (c) — Intermolecular Forces

Explaining Trends in Boiling Points

💡 Key Knowledge

Electronegativity trend: Electronegativity increases across the period/group, creating greater bond polarity.
Intermolecular types: CH₄ has London forces only. NH₃ and HF possess permanent dipoles and hydrogen bonding (due to H bonded to N, F, or O).
Comparison: HF has a higher boiling point than NH₃ because fluorine is more electronegative than nitrogen, resulting in stronger hydrogen bonds.

🧠 Exam Technique

To secure all 3 marks, structure your answer logically: (1) State the types of intermolecular forces present in each molecule; (2) Explicitly mention hydrogen bonding for HF and NH₃, and London forces for CH₄; (3) Compare the relative strengths to explain the boiling point order.

Mark breakdown: 3 marks total. 1 mark for linking electronegativity/boiling point trend; 1 mark for identifying types of intermolecular forces; 1 mark for comparing hydrogen bond strengths.

Part (d) — Quantitative Analysis & Inorganic Synthesis

Identifying Unknown Compounds A, B, and C

✅ Correct Answers

Compound A: Ca₃N₂ (Calcium nitride)
Compound B: NH₃ (Ammonia gas)
Compound C: Ca(OH)₂ (Calcium hydroxide / alkali)

📐 Empirical Formula Calculation for A

Ca: 81.10 / 40.1 = 2.022 mol → 3
N: 18.90 / 14.0 = 1.350 mol → 2
Formula = Ca₃N₂

💡 Balanced Equation

Ca₃N₂ + 6H₂O → 2NH₃ + 3Ca(OH)₂

A nitride reacting with water produces ammonia gas (alkaline gas B) and a metal hydroxide solution (alkali C).

Mark breakdown: 4 marks total. 1 mark for formula of A; 1 mark for identity of B; 1 mark for identity of C; 1 mark for the balanced equation.

Part (e) — Reactions of Carboxylic Acids

Completing Equations for Lactic Acid

✅ Correct Answers & Equations

1. Reaction with Sodium Carbonate:

2CH₃CH(OH)COOH + Na₂CO₃ → 2CH₃CH(OH)COONa + CO₂ + H₂O

2. Reaction with Aluminium:

3CH₃CH(OH)COOH + Al → (CH₃CH(OH)COO)₃Al + 1½ H₂

❌ Common Errors

Balancing errors with stoichiometric coefficients (especially forgetting the multiplier for water with carbonates, or fractional coefficients for hydrogen gas with aluminium). Ensure carboxylate salt formulas enclose the ion correctly with subscripts.

Mark breakdown: 4 marks total (2 marks per fully balanced equation, including correct species and balancing).

Part (f) — Organic Mechanisms

Nucleophilic Substitution: Synthesis of Methyl Ethanoate

💡 Mechanism Requirements

Dipole: Must show partial charges (δ+ on carbon, δ- on halogen) across the C-X bond.
First Curly Arrow: Starts from the polar C-X bond (or bond pair) and goes to the halogen atom, showing heterolytic fission.
Second Curly Arrow: Starts from a lone pair (or negative charge) on the oxygen of the carboxylate ion ( CH₃COO⁻ ) and points to the electron-deficient carbon atom.
Products: Correct ester ( CH₃COOCH₃ ) and halide leaving ion ( X⁻ ).

🧠 Examiner Tip on Curly Arrows

Make sure your arrows originate precisely where required. The arrow for the nucleophile must start from the lone pair or negative charge on the oxygen atom. Loose arrowheads that float in empty space will lose marks!

Mark breakdown: 3 marks total. 1 mark for dipole and first curly arrow; 1 mark for nucleophile curly arrow from carboxylate ion; 1 mark for correct organic product and leaving group.

Topics

Module 2: Foundations in chemistry · Module 3: Periodic table and energy · Module 4: Core organic chemistry · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 5.1 Rates, equilibrium and pH · 3.1 The periodic table · 4.2 Alcohols, haloalkanes and analysis · 2.1 Atoms and reactions · 6.2 Nitrogen compounds, polymers and synthesis

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.