OCR A-Level Chemistry Unified chemistry (03), November 2021: Question 5

20 marks · Hard difficulty · Structured Questions

Calculate the mass of sodium azide decomposed using the ideal gas equation, determine the pH of a hydrazoic acid solution, write equations for acid-base and organic reactions involving hydrazoic acid, and identify unknown substances E-J from reaction data and an IR spectrum.

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Question

A multi-part chemistry question about nitrogen compounds, sodium azide decomposition, hydrazoic acid equilibria, and reactions of ammonia including an IR spectrum for compound I.
Question text

5 This question is about nitrogen and its compounds.

(a) Sodium azide, NaN3, has been used in car airbags.

The airbag inflates when the NaN3 decomposes to form nitrogen gas:

2NaN3(s) 2Na(s) + 3N2(g)

(i) This is a redox reaction.

Write half-equations for the reduction and oxidation processes that take place.

Reduction …

Oxidation … [2]

(ii) A 16.0 dm3 airbag is inflated at 17.0 ºC.

The pressure in the inflated airbag is 1.20 × 105 Pa.

Calculate the mass of NaN3 that has decomposed.

Give your answer to 3 significant figures.

mass of NaN3 = … g [5]

(b) Hydrazoic acid, HN , is a weak acid (K = 2.51 × 10–5 mol dm–3).

3 a

(i) Calculate the pH of 0.125 mol dm–3 hydrazoic acid.

Give your answer to 2 decimal places.

pH = … [2]

(ii) When added to water, hydrazoic acid forms an equilibrium mixture containing conjugate

acid–base pairs.

Complete the equation for this equilibrium and label the conjugate acid–base pairs as:

A1, B1 and A2, B2.

Equation HN3 + … + …

Acid-base pairs …

[2]

(iii) In the Schmidt reaction, hydrazoic acid, HN3, reacts with carboxylic acids to form primary

amines.

For example, HN3 reacts with RCOOH to form RNH2 and two gases that are found in the

atmosphere.

Write the equation for the reaction of HN3 with 2-methylbutanoic acid.

Show structures for organic compounds.

[3]

(c)* This question is about two reactions of ammonia.

Reaction 1

Excess ammonia is reacted with 4.77 g of copper(II) oxide. The reaction produces 3.81 g of

solid E, liquid F and 0.560 g of gas G, which has a volume of 480 cm3 at RTP.

Reaction 2

Ammonia reacts with compound H to form compound I, C2H5NO, and chloride salt J.

The IR spectrum of I is shown below.

transmittance 50

(%)

4000 3000 2000 1500 1000 500

wavenumber / cm–1

Identify E, F, G, H, I and J, and write equations for the two reactions.

Show your reasoning. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme Mark scheme providing answers and guidance for the sodium azide, hydrazoic acid, and ammonia reaction questions, including the IR spectrum analysis and identification of substances E to J.

AO

Question Answer Marks Guidance

element

5 (a) (i) Reduction: Na+ + e– → Na 2 AO1.2 ALLOW multiples

e.g. 2Na+ + 2e– → 2Na

Oxidation: 2N – → 3N + 2e–

ALLOW 1 mark for 2 correct equations but wrong way round IGNORE state symbols

(ii) FIRST CHECK ANSWER ON ANSWER LINE 5 TAKE CARE as value written down may be

IF mass = 34.5 (g) AND working using ideal gas equation truncated value stored in calculator.

Award 5 marks for calculation

-------------------------------------------------------------------------- pV

Rearranging ideal gas equation IF n = RT is omitted, ALLOW when values are

pV substituted into rearranged ideal gas equation.

n = RT AO2.4

pV ×5

Unit conversion AND substitution into n = RT : Calculator: 0.7963302448

• R = 8.314 OR 8.31

• V = 16(.0) × 10–3

1.20 × 105 × 16.0 × 10–3 From unrounded 0.7963302448,

• T in K: 290 K e.g. n(NaN3) = 0.5308868299

8.314 × 290

Calculation of n

n = 0.796 (mol) mass = 0.5308868299 × 65 = 34.50764394

→ 34.5 to 3 SF

Calculation of mass

2 COMMON ERROR

n(NaN3) = × 0.796 = 0.531 (mol) 51.7 OR 51.8 → 4 marks (2/3 omitted

depending on intermediate rounding

mass NaN3 = 0.531 × 65 = 34.5 (g) 0.796 × 65 = 51.7 OR 51.8

3 SF required 54.4 → 4 marks (inverted gas equation)

RT

n = pV →1.255760417→ 0.8371736111

→ 54.4 (g) CARE with intermediate rounding

81.6 OR 81.7 → 3 mks (as above but no 2/3)

16 AO

element

(b) (i) FIRST CHECK THE ANSWER ON ANSWER LINE 2 AO2.2 ALLOW ECF throughout

If answer = 2.75 award 2 marks ×2

-------------------------------------------------------------------------- IGNORE error with HN3 shown as NH3

[H+]2 = K × [HN ]) = 2.51 × 10–5 × 0.125

a 3

[H+] = √(K × [HN ])

a 3

[H+]2 = 2.51 × 10–5 × 0.125

OR [H+] = √ (2.51 × 10–5 × 0.125)

OR [H+] =1.77…. × 10–3 (mol dm–3)

pH = –log 1.77…. × 10–3 = 2.75 (Must be to 2DP) ALLOW pH mark by ECF

ONLY if 2.51 × 10–5 × 0.125 used AND pH <7

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

Common errors (Must be to 2 DP)

pH = 5.50 → 1 mark (No square root)

[H+] = 6.26 × 10–4 from √ (2.51 × 10–5 ) × 0.125

pH = 3.20 → 1 mark

[H+] = 8.87 × 10–6 from √ (0.125) × 2.51 × 10–5

pH = 5.05 → 1 mark

(ii) • Correct equation 2 AO1.2

• Correct acid–base pair labels for correct equation ×2 ALLOW 1 mark for one correct acid–base pair

WITH correct labels

HN + H O N – + H O+ e.g. H O H O+

32 3 3 2 3

A1 B2 B1 A2 WITH B1 A1

OR OR B2 A2

A2 B1 B2 A1

AO

Question Answer 17 Marks Guidance

element

(iii) Structure of 2-methylbutanoic acid 3 AO3.2 ALLOW correct structural OR skeletal

×2 OR displayed formula OR mixture of the

Structure of organic product (primary amine) above as long as non-ambiguous

CO2 AND N2 as products AO2.6 Common error

With NH3, → CO2 + H2

ALLOW ECF for equation using a different

amine isomer of the organic product

e.g. (CH3)2CHCH2NH2

DO NOT ALLOW ECF from unbranched

species, e.g. CH3CH2CH2NH2

IGNORE HN3 in equation, even if missing

IGNORE poor connectivity to all groups

18 AO

element

(c)* Please refer to the marking instructions on page 6 of this 6 AO3.1 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. ×2

Identify of E, F, G, H, I and J

Level 3 (5–6 marks) AO3.2 • E Cu/copper

Reaches a comprehensive conclusion to determine the ×4

• F: H2O/water

correct formulae of almost all of E, F, G, H, I and J

• G: N2/nitrogen

There is a well-developed line of reasoning which is clear and • H: CH3COCl OR ClCH2CHO OR C2H3OCl

logically structured. • I: CH CONH OR H NCH CHO

32 2 2

The information presented is relevant and substantiated.

• J: NH4Cl/ammonium chloride

Level 2 (3–4 marks)

Reaches a sound conclusion to determine the correct Examples of reasoning

formulae of at least half of E, F, G, H, I and J Working

4.77

n(CuO) = (63.5 + 16) = 0.06 (mol)

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by some M(E) = 3.81 ÷ 0.06 = 63.5

evidence. 480

n(G) = = 0.02

24000

Level 1 (1–2 marks) 0.560 –1

Reaches a simple conclusion to determine the correct M(G) = = 28 (g mol )

0.02

formulae of some of E, F, G, H, I and J Infrared spectrum

I contains

There is an attempt at a logical structure with a line of reasoning. –1

The information is in the most part relevant. • C=O ( ~1700 cm )

• NH ( ~3200–3400 cm–1)

0 marks No response or no response worthy of credit.

Equations

3CuO + 2NH3 → 3Cu + 3H2O + N2

CH3COCl + 2NH3→ CH3CONH2 + NH4Cl

OR

ClCH2CHO + 2NH3→ H2NCH2CHO + NH4Cl

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How to answer it

Nitrogen and its Compounds Study Guide

What this question tests

This comprehensive OCR A-Level Chemistry question tests your mastery across multiple core physical, inorganic, and organic topics: constructing redox half-equations, applying the ideal gas equation, calculating weak acid pH using Ka approximations, identifying conjugate acid-base pairs, predicting products of nitrogen-containing organic reactions (Schmidt reaction), and executing a multi-step stoichiometry and structural identification problem using quantitative mass/volume data and infrared (IR) spectroscopy.

Question 5(a)(i)

Redox Half-Equations in Sodium Azide Decomposition

✅ Correct Answer

Reduction: Na⁺ + e⁻ → Na
*(Multiples like 2Na⁺ + 2e⁻ → 2Na are fully accepted)*

Oxidation: 2N₃⁻ → 3N₂ + 2e⁻

💡 Key Knowledge

Sodium azide contains the azide ion ( N₃⁻ ) where nitrogen has an oxidation state of -1/3, which is oxidized to elemental nitrogen ( N₂ , oxidation state 0). Sodium ions ( Na⁺ ) are reduced to metallic sodium ( Na ).

❌ Common Errors

Students frequently write half-equations with incorrect balancing for electrons or atomic species, or write equations in reverse. Ensure state symbols are handled carefully and species formulas match ionic components.

Question 5(a)(ii)

Ideal Gas Calculation & Stoichiometry

✅ Correct Answer

Mass of NaN₃ = 34.5 g (to 3 significant figures)

📐 Step-by-Step Calculation

  1. Convert units:
    Pressure P = 1.20 × 10⁵ Pa
    Volume V = 16.0 dm³ = 0.0160 m³ (or keep in cm³ with appropriate R)
    Temperature T = 17.0 °C + 273.15 = 290.15 K (or 290 K)
  2. Rearrange ideal gas equation: n = PV / RT
  3. Calculate moles of N₂ gas:
    n(N₂) = (1.20 × 10⁵ × 0.0160) / (8.314 × 290) = 0.7963 mol
  4. Use stoichiometry from equation ( 2NaN₃ → 2Na + 3N₂ ):
    n(NaN₃) = (2 / 3) × n(N₂) = (2 / 3) × 0.7963 = 0.5309 mol
  5. Calculate mass:
    M(NaN₃) = 22.99 + (3 × 14.01) = 65.0 g mol⁻¹
    Mass = 0.5309 × 65.0 = 34.507 g → 34.5 g (3 SF)

🧠 Exam Technique & Traps

Top Tip: Always convert temperature to Kelvin immediately ( +273 ). Watch out for rounding too early in your calculator—keep unrounded intermediate values and only round to 3 significant figures at the very end.
Question 5(b)(i)

Weak Acid pH Calculation

✅ Correct Answer

pH = 2.75 (Must be to exactly 2 decimal places)

📐 Step-by-Step Calculation

  1. Write Ka expression: Ka = [H⁺][HN₃⁻] / [HN₃]
  2. Apply weak acid approximations: [H⁺] = [HN₃⁻] and [HN₃] ≈ [HN₃]initial , giving [H⁺]² = Ka × [HN₃]
  3. Substitute values:
    [H⁺]² = 2.51 × 10⁻⁵ × 0.125 = 3.1375 × 10⁻⁶
    [H⁺] = √(3.1375 × 10⁻⁶) = 1.7713 × 10⁻³ mol dm⁻³
  4. Calculate pH:
    pH = -log(1.7713 × 10⁻³) = 2.7518 → 2.75

❌ Common Errors

Failing to give the final pH to 2 decimal places loses accuracy marks. Remember: the number of decimal places in a pH value corresponds to the number of significant figures in the concentration/H⁺ ion value.

Question 5(b)(ii)

Conjugate Acid-Base Pairs

✅ Correct Answer

Equation: HN₃ + H₂O &rightleftharpoons; N₃⁻ + H₃O⁺

Pairs Labelling:
Pair 1: HN₃ (A1) and N₃⁻ (B1)
Pair 2: H₂O (B2) and H₃O⁺ (A2) *(Note: A1/B1 and A2/B2 designations can be swapped as long as they match correctly)*

💡 Key Knowledge

An acid is a proton ( H⁺ ) donor and a base is a proton acceptor. Conjugate acid-base pairs differ by a single proton ( H⁺ ).

Question 5(b)(iii)

Schmidt Reaction Organic Mechanism & Products

✅ Correct Answer

Reactant structure: 2-methylbutanoic acid ( CH₃CH₂CH(CH₃)COOH )

Organic Product: Primary amine where the -COOH carbon is lost as CO₂ , yielding sec-butylamine or the corresponding amine structure: CH₃CH₂CH(CH₃)NH₂ (butan-2-amine)

Inorganic Gases: CO₂ and N₂

🧠 Exam Technique

Ensure structural, displayed, or skeletal formulas clearly show correct connectivity. For instance, the amine group must attach correctly to the carbon chain where the carboxylic acid carbon was eliminated.

Question 5(c)* - 6 Mark Synthesis Problem

Reaction of Ammonia with Copper(II) Oxide & Identification Task

✅ Correct Identification of Species (E to J)

  • E (solid): Copper / Cu
  • F (liquid): Water / H₂O
  • G (gas): Nitrogen / N₂
  • H (compound): Ethanoyl chloride ( CH₃COCl ) or chloroacetaldehyde ( ClCH₂CHO ) / acyl chloride formula matching data
  • I (compound): Amide / organic product, e.g., ethanamide ( CH₃CONH₂ )
  • J (salt): Ammonium chloride ( NH₄Cl )

📐 Step-by-Step Quantitative Reasoning (Reaction 1)

  1. Moles of CuO: Mr(CuO) = 63.5 + 16.0 = 79.5 g mol⁻¹ .
    Moles = 4.77 g / 79.5 = 0.060 mol
  2. Identify solid E (Cu): Mass = 3.81 g. Moles of Cu = 3.81 / 63.5 = 0.060 mol (1:1 ratio with CuO). Therefore E is Cu.
  3. Identify gas G (N₂): Volume = 480 cm³ at RTP. Moles = 480 / 24000 = 0.020 mol .
  4. Identify liquid F (H₂O): By mass difference or stoichiometry, remaining mass/moles give water, so F is H₂O.

💡 Infrared (IR) Spectroscopy Clues for Compound I

The IR spectrum of I shows characteristic peaks:

  • Strong sharp absorption around 1700 cm⁻¹ corresponding to C=O (carbonyl stretch).
  • Peaks in the 3200–3400 cm⁻¹ region corresponding to N-H stretches (primary amide).

🧠 Top-Level Exam Strategy (Level 3 Response)

To achieve full Level 3 marks (5-6 marks), your answer must present a logically structured, coherent narrative that connects quantitative mole calculations from Reaction 1 with structural deductions from IR data in Reaction 2, supported by balanced chemical equations for both reactions.

Topics

Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 5.1 Rates, equilibrium and pH · 2.1 Atoms and reactions · 6.2 Nitrogen compounds, polymers and synthesis · 6.3 Analysis

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.