AQA A-Level Chemistry Paper 2, June 2025: Question 9

✅ Correct Answer

The product (silver bromide / silver halide) is a precipitate / is insoluble.

💡 Key Knowledge

Hydrolysis of a bromoalkane releases bromide ions into solution:

R-Br + H₂O → R-OH + H⁺ + Br⁻

These immediately react with aqueous Ag⁺ ions to form an insoluble cream precipitate:

Ag⁺(aq) + Br⁻(aq) → AgBr(s)

❌ Common Errors

• Saying simply that "a colour change occurs" or "it turns yellow" without explicitly mentioning a precipitate or insoluble solid.
• Confusing the reaction with the thiosulfate/acid reaction and stating that "sulfur is formed".

✅ Correct Answer

Any one of the following:

• Start timer at the same point of addition
• View the cross from the same distance / angle
• Constant light level in the room / background lighting
• Same swirling / agitation / stirring rate
• Same person observing the disappearance of the cross

🧠 Exam Technique

The stem specifies that "volume and concentration of each substance used at each temperature is kept constant". Therefore, do not write volume, concentration, or temperature as your answer. Always look at the human observation and practical handling aspects of visual endpoint experiments.

📐 Step-by-Step Calculation

1. Convert Celsius to Kelvin:
   T = 52.2 + 273(.15) = 325.2 K (or 325.35 K)
2. Calculate reciprocal temperature (1/T):
   1 / 325.2 = 0.003075 K⁻¹
3. Express to consistent standard form (matching table format):
   3.08 × 10⁻³ K⁻¹

✅ Accepted Values

• 3.08 × 10⁻³ (best practice, matches the 3 s.f. format in Table 2)
• Also allowed: 3.075 × 10⁻³, 0.003075, or 0.00308

❌ Common Errors

• Forgetting to convert to Kelvin: calculating 1 / 52.2 = 0.0192 (scores 0).
• Rounding prematurely to 3.1 × 10⁻³ (must match the precision in the table, at least 3 sig figs).

🧠 Graph Plotting Rules

• M1 (Scale): The x-axis (1/T) scale must be linear, sensible (e.g. 2.90 × 10⁻³ to 3.20 × 10⁻³), and span at least half of the available horizontal grid. Do not start at 0! Use a broken axis / start directly near 2.90 × 10⁻³.
• M2 (Plotting & Best Fit): All 5 points plotted accurately within ±0.5 small square. Draw a clean, continuous straight line of best fit using a ruler.

📐 Step-by-Step E_a Determination

1. Determine the gradient: Draw a large gradient triangle (change in y / change in x).
   Gradient = Δ(ln k) / Δ(1/T)
   From the mark scheme line of best fit:
   Gradient = - (2.38) / (2.25 × 10⁻⁴) = -10578 K
   (Accepted range: -10469 to -10830) → [M3]
2. Equate gradient to Arrhenius relation:
   Gradient = - E_a / R
   E_a = - Gradient × R
   E_a = -(-10578) × 8.31 = +87903 J mol⁻¹ → [M4]
3. Convert J mol⁻¹ to kJ mol⁻¹:
   E_a = 87903 / 1000 = +87.9 kJ mol⁻¹
   (Accepted range: +87 to +90 kJ mol⁻¹) → [M5]

❌ Common Errors & Pitfalls

• Forgetting the 10⁻³ multiplier: Reading Δ(1/T) as 0.225 instead of 0.225 × 10⁻³ gives an answer 1000 times too small.
• Not converting J mol⁻¹ to kJ mol⁻¹: Giving 87900 instead of 87.9 (the unit on the answer line is kJ mol⁻¹).
• Sign errors: Leaving E_a as a negative value. Activation energy is always positive!
• Small gradient triangles: Using points too close together introduces large rounding errors. Use at least half the length of your drawn line.

✅ Correct Answer: Option B

Option B is the correct box to tick.

💡 Connecting Rate Equations to Mechanisms

• The given rate equation is: rate = k[R₂CHBr]
• Notice that [KOH] / [OH⁻] does NOT appear in the rate equation (it is zero order with respect to OH⁻).
• This means OH⁻ cannot be involved in or before the Rate-Determining Step (RDS).
• Hence, the slow step must involve only R₂CHBr breaking down to form a carbocation intermediate (an S_N1 mechanism).
• Option A is a one-step S_N2 mechanism (which would have rate = k[R₂CHBr][OH⁻] ). Option C has the attack of OH⁻ as the slow step, which would also make OH⁻ first order.

✅ Mark Scheme Breakdown

• Mark 1: Bromine is more electronegative than carbon, making the Br atom partially negative (δ-) and the C atom partially positive (δ+).
• Mark 2: The lone pair of electrons on the hydroxide ion (OH⁻) attacks / is attracted to the partially positive carbon (C^δ+) [OR to the carbocation C⁺ formed after C–Br bond cleavage].

🧠 What Examiners Want to See

Make sure you clearly state both charges:

Br is δ- and C is δ+

Then clearly identify the nucleophile's active site:

The lone pair on the oxygen of HO⁻ attacks the δ+ carbon.

❌ Common Errors

• Stating bromine is "more electropositive" or confusing which atom is δ+ and δ-.
• Failing to mention the lone pair on the hydroxide ion when describing the attack.
• Omitting the reason for the polarity (electronegativity difference between C and Br).