OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 21

✅ Correct Answer

Compare Experiments 1 and 2: When [S₂O₃²⁻] halves (from 0.16 to 0.08 mol dm⁻³) while [H⁺] remains constant (1.00 mol dm⁻³), the initial rate halves (from 0.0120 to 0.0060 mol dm⁻³ s⁻¹). This proves the reaction is first order with respect to [S₂O₃²⁻] .

*(Alternatively, comparing Experiments 2 and 3 shows zero order with respect to [H⁺] ).*

💡 Key Knowledge

To justify a rate order from data tables, you must explicitly state:

• What happens to the concentration of the chosen reactant.
• What happens to the concentration of other reactants (confirming they are kept constant or accounting for them).
• The corresponding effect on the initial rate of reaction.

✅ Correct Answer

Step 1 (RDS): S₂O₃²⁻ → S + SO₃²⁻

Step 2: SO₃²⁻ + 2H⁺ → SO₂ + H₂O

(Alternative valid pair: Step 1: S₂O₃²⁻ → SO₂ + SO²⁻ followed by Step 2: SO²⁻ + 2H⁺ → S + H₂O )

🧠 Exam Technique

The rate equation is rate = k[S₂O₃²⁻] . This tells you that only one S₂O₃²⁻ ion is involved in the rate-determining step (Step 1), and no H⁺ ions appear in Step 1. All other species from the overall equation must be balanced out across the two steps.

📐 Step-by-Step Calculation

Step 1: Determine the gradient.
Pick two points on the line of best fit. Ensure your triangle is large.
Gradient = Δ(ln k) / Δ(1/T).
Using points from the scheme, gradient is typically around -5900 (acceptable range: -5700 to -6100).

Step 2: Relate gradient to E_a.
The Arrhenius equation in linear form is: ln k = (-E_a / R) × (1/T) + ln A .
Therefore, Gradient = -E_a / R, meaning E_a = -gradient × R (where R = 8.314 J mol⁻¹ K⁻¹).

Step 3: Calculate and convert units.
E_a = -(-5900) × 8.314 = 49052.6 J mol⁻¹.
Convert to kJ mol⁻¹ by dividing by 1000: 49.1 kJ mol⁻¹ (to 3 sig figs).

❌ Common Errors & Traps

• Power of 10 omission: The x-axis is labelled 1/T / 10⁻³ K⁻¹ . Students frequently forget to multiply their temperature axis readings by 10⁻³ when finding the gradient, leading to an activation energy that is out by a factor of 1000!
• Negative E_a: Activation energy can never be negative. If your gradient is negative (as expected), remember to multiply by -1.

✅ Correct Answer

The student assumed the y-intercept ( ln A ) occurs when the x-axis value is zero. However, the x-axis scale ( 1/T ) starts at 3.00 × 10⁻³ K⁻¹, not zero. The true y-intercept is off the grid to the left where 1/T = 0 .

💡 Key Knowledge

In Arrhenius plots ( ln k vs 1/T ), the y-axis intercept represents ln A only if the graph axes are plotted with a true origin where 1/T = 0 . Truncated axes require extrapolating the line back to zero to find ln A .

📐 Step-by-Step Calculation

Step 1: Calculate ln k.
Given k = 0.075 s⁻¹ .
ln(0.075) = -2.59.

Step 2: Read 1/T from the graph.
Locate -2.59 on the y-axis (ln k) and move across to the line of best fit, then read down to the x-axis ( 1/T ).
1/T = 3.10 × 10⁻³ K⁻¹ (Acceptable range: 3.09 to 3.11 × 10⁻³).

Step 3: Convert 1/T to temperature in °C.
T (in Kelvin) = 1 / (3.10 × 10⁻³) = 322.58 K.
Convert to Celsius by subtracting 273.15:
322.58 - 273.15 = 49.43 °C → 49.6 °C (Acceptable range: 48.5 °C to 50.6 °C).

🧠 Exam Technique

Always double-check your temperature conversions! Examiners frequently see students calculate the Kelvin temperature correctly and then forget to subtract 273 to convert it into degrees Celsius as requested by the question prompt.