WJEC A-Level Chemistry Unit 3, June 2025: Question 8

✅ Correct Answer

State symbols are not strictly required unless requested, but equations must balance for atoms and charge.

❌ Common Errors

• Confusing hot conc. NaOH with cold dilute NaOH (which yields NaClO, sodium chlorate(I): Cl₂ + 2NaOH → NaCl + NaClO + H₂O ).
• Writing incorrect formulae such as NaClO₅ for chlorate(V). The roman numeral refers to the oxidation state of chlorine, not the number of oxygens!

✅ Correct Answer & Mark Scheme

1. Mark 1: State that the same element (chlorine) has been both oxidised and reduced.
2. Mark 2: Provide explicit oxidation states:
   • Oxidised from 0 in Cl₂ to +5 in NaClO₃ (or chlorate(V)).
   • Reduced from 0 in Cl₂ to -1 in NaCl (or chloride).

🧠 Exam Technique

Always state both the species and the numerical oxidation state before and after:

• In Cl₂: Oxidation state = 0
• In NaClO₃: Na = +1, O₃ = -6, so Cl = +5
• In NaCl: Na = +1, so Cl = -1

Do not just say "chlorine is oxidised and reduced"—you must include the numbers to secure both marks.

✅ Correct Answer

Add the sample to ice / ice-cold water (or rapid cooling in an ice bath).

💡 Key Knowledge

Quenching means suddenly stopping or drastically slowing a chemical reaction so that the concentration of species remains fixed at the sampling time.

Diluting with large volumes of ice-cold water achieves this by both rapidly decreasing the temperature (vastly reducing collisions with E ≥ E_a) and lowering concentration.

✅ Correct Answer

The student is incorrect (No).

Reason (either of the following):

• The overall reaction is second order, and second-order reactions do not have a constant half-life.
• As one reactant's concentration halves, the other reactant's concentration also decreases, altering the reaction rate and continuously changing the half-life.

❌ Common Errors

Students often recall that "first-order reactions have constant half-lives" and incorrectly apply this to a multi-reactant system where both change simultaneously.

A constant half-life for a first-order reactant is only observed if all other reactants are in large excess (pseudo-first-order conditions).

📐 Step-by-Step Tangent Method

1. Place a ruler at the origin (t = 0, [Cl⁻] = 0) to draw a tangent to the [Cl⁻] curve.
2. Ensure the tangent accurately follows the curve's slope at the exact moment t = 0.
3. Find the gradient:
   Rate = Δ[Cl⁻] / Δt
4. Using values from a well-drawn tangent (e.g. [Cl⁻] reaches ~2.00 × 10⁻³ mol dm⁻³ at t ≈ 10 min):
   Rate ≈ (2.00 × 10⁻³) / 10 = 2.00 × 10⁻⁴ mol dm⁻³ min⁻¹

🧠 Mark Scheme Criteria

• 2 Marks: A good tangent drawn on the graph at t = 0 giving a calculated rate in the range:
  1.55 × 10⁻⁴ to 2.35 × 10⁻⁴ mol dm⁻³ min⁻¹
• 1 Mark: A reasonable tangent yielding a value in the wider range of:
  1.40 × 10⁻⁴ to 3.00 × 10⁻⁴ mol dm⁻³ min⁻¹

Note: Leaving your tangent visible on the graph is essential to be credited.

📐 Step-by-Step Calculation

1. Identify Initial Concentrations at t = 0 [1 Mark]:
   From graph y-intercepts:
   • [ClO⁻]₀ = 3.50 × 10⁻³ mol dm⁻³
   • [Br⁻]₀ = 4.00 × 10⁻³ mol dm⁻³
2. Write & Rearrange Rate Equation [1 Mark]:
   Rate = k [ClO⁻] [Br⁻]
   k = Rate / ([ClO⁻] [Br⁻])
3. Calculate Numerical Value for k [1 Mark]:

   Using Rate = 2.00 × 10⁻⁴ mol dm⁻³ min⁻¹:

   k = (2.00 × 10⁻⁴) / (3.50 × 10⁻³ × 4.00 × 10⁻³)
   k = 2.00 × 10⁻⁴ / 1.40 × 10⁻⁵ = 14.3

   (Accepts full error-carried-forward from Part I calculated rate)

4. Determine Units [1 Mark]:
   Units = (mol dm⁻³ min⁻¹) / [(mol dm⁻³) × (mol dm⁻³)]
   Units = mol⁻¹ dm³ min⁻¹

❌ Major Traps to Avoid

Trap 1: The Time Unit is Minutes!
The x-axis on the graph is plotted in minutes, not seconds. Writing s⁻¹ instead of min⁻¹ loses the unit mark completely.

Trap 2: Reading the Wrong Reactants
Make sure you identify which curve is [Br⁻] (starts at 4.00 × 10⁻³) and which is [ClO⁻] (starts at 3.50 × 10⁻³). Swapping them numerically gives the same product here, but misreading values loses marks.

✅ Correct Answer

• Difference: No difference (the graphs are identical). [1 Mark]
• Reason: From the stoichiometric equation ( ClO⁻ + Br⁻ → Cl⁻ + BrO⁻ ), the mole ratio of Cl⁻ to BrO⁻ is 1:1. Each time one chloride ion is produced, exactly one bromate(I) ion is produced, so their concentrations are equal at all times. [1 Mark]

🧠 Exam Technique

Always inspect the balanced equation! Since 1 mole of ClO⁻ reacts with 1 mole of Br⁻ to make 1 mole of Cl⁻ and 1 mole of BrO⁻, both products will follow the identical rate of appearance and reach identical concentrations at any point during the reaction.