OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 16
10 marks · Hard difficulty · Structured Questions
Explain the effect of temperature on the rate of reaction using a Boltzmann distribution, determine the orders, rate equation, rate constant and units from experimental data, and suggest a two-step mechanism for a reaction.
Practise this questionQuestion
Question text
16 A catalytic converter in a car removes nitrogen monoxide, NO, and carbon monoxide, CO, from
the exhaust gases.
(a) One reaction that happens in a catalytic converter is shown below.
2CO(g) + 2NO(g) N2(g) + 2CO2(g) Reaction 16.1
(i) Explain how increasing the temperature increases the rate of Reaction 16.1.
Include a labelled sketch, using Boltzmann distributions, on the grid below.
Label the axes.
… [3]
(ii) The rate of Reaction 16.1 is investigated by carrying out three experiments at the same
temperature. The results are shown below.
[NO(g)] [CO(g)] Initial rate
Experiment –3 –3 –3 –1
/ mol dm / mol dm / mol dm s
12.75 × 10–4 7.25 × 10–4 1.85 × 10–4
25.50 × 10–4 7.25 × 10–4 7.40 × 10–4
31.10 × 10–3 2.90 × 10–3 1.18 × 10–2
Determine the orders with respect to NO and CO, the rate equation, and the rate
constant, k, including units.
Explain your reasoning.
k = … units … [5]
(b) Carbon monoxide also reacts with nitrogen dioxide as shown in Reaction 16.2.
CO(g) + NO2(g) NO(g) + CO2(g) Reaction 16.2
The rate equation for Reaction 16.2 is shown below:
rate = k [NO (g)]2
Suggest a possible two-step mechanism for Reaction 16.2.
The first step is much slower than the second step.
step 1 …
step 2 …
[2]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
16 (a) (i) 3 AO1.1
T1 ×3
T2
Axes labelled (number of) molecules ALLOW particles on the y-axis
AND (kinetic) energy DO NOT ALLOW atoms on y-axis
AND correct drawing of a Boltzmann distribution DO NOT ALLOW enthalpy on x-axis
i.e. curve must start within the first small square nearest DO NOT ALLOW an increase of more than one
to the origin small square at the high energy end of the curve
AND must not touch the x-axis at high energy i.e. allow a small inflection
Drawing of correct Boltzmann distributions at two
different temperatures with one termperature identified. ALLOW T2 as ‘higher termperature’
Maximum of curve for higher temperature must be
to the right AND lower than the maximum of the
curve for lower temperature
Lines can only cross once
(At higher temperature) more molecules/particles have
energy above activation energy ALLOW ORA if states the effect when the
temperature is lower
ALLOW has enough energy to react
ALLOW Ea shown on graph AND greater area
under the curve to the right of Ea
DO NOT ALLOW lowers Ea
DO NOT ALLOW atoms for molecules
IGNORE (more) successful collisions
AO
10 element
(a) (ii) Orders 5 ALLOW ORA throughout
(Expt 1+2) e.g. expt 2+1 [NO] halves, rate quarters etc.
When [NO] × 2, rate × 4 AO3.1
AND 2nd order with respect to NO IGNORE [CO] constant
(Expt 2+3) ALLOW if working shown with the table.
When [NO] × 2 AND [CO] × 4, rate × 16 AO3.2 ALLOW if seen in 2 steps i.e.
AND 1st order with respect to CO When [NO] × 2, rate x 4 AND
[CO] × 4, intermediate rate × 4.
AO2.6
ALLOW comparing Expt 1+3
AO1.2 When [NO] × 4 AND [CO] × 4, rate × 64
×2 AND 1st order with respect to CO
Rate Equation ALLOW ECF from incorrect orders
rate = k [NO]2[CO] ALLOW rate = k [NO]2[CO]1
ALLOW rate equation with correct numbers
Value of k substituted
1.85 × 10–4
(k = –4 2 –4)
(2.75 × 10 ) × 7.25 × 10
6 ALLOW 3.36 x 106 from the use of Expt 3
= 3.37 × 10
IGNORE errors in working out – the mark is for the
value
ALLOW 3 SF upto the calculator value
3374180.678 OR 3.374180678 x106
IGNORE rounding errors past 3SF
Units of k
6 –2 –1 ALLOW units in any order e.g. mol–2 dm6 s–1
dm mol s
ALLOW ECF from incorrect rate equation.
AO
11 element
Common errors
4 marks (including units)
4.65 x 109 mol -3 dm9 s-1 (use of 2nd order
with respect to CO)
2446 mol-1 dm3 s-1 (use of zero order
wrt CO)
(b) 2NO2 only on LHS of step 1 2 AO3.1 M2 dependent on M1
×2
Rest of mechanism Examples:
Step 1 : 2NO2 → NO + NO3
Step 2 : NO3 + CO → NO2 + CO2
OR
Step 1 : 2NO2 → N2O4
Step 2 : N2O4 + CO → NO + NO2 + CO2
OR
Step 1 : 2NO2 → N2 + 2O2
Step 2 : N2 + 2O2 + CO → NO + NO2 + CO2
OR
Step 1 : 2NO2 → 2NO + O2
Step 2 : NO + O2 + CO → NO2 + CO2
Total 10
How to answer it
Catalytic Converters, Boltzmann Distributions & Reaction Kinetics
What this question tests
This question assesses your understanding of chemical kinetics, specifically how temperature affects reaction rates using Boltzmann distributions, determining orders of reaction and rate constants from experimental data, calculating rate constant units, and proposing multi-step reaction mechanisms consistent with a given rate equation.
Effect of Temperature on Rate & Boltzmann Distributions
💡 Key Knowledge: Boltzmann Distribution
- Axes Labels: Y-axis must be Number of molecules (or particles), X-axis must be Energy (or kinetic energy). Do not use "atoms" or "enthalpy".
- Curve Shape: Must start at the origin (0,0) and asymptotically approach (but never touch) the x-axis at high energy.
- Temperature Shift ( T₂ > T₁ ): Higher temperature shifts the peak to the right and lowers the maximum height. Curves should only cross once.
✅ Model Answer & Marking Points (3 Marks)
- Mark 1: Correctly labeled axes (Number of molecules vs Energy) and valid curve starting at the origin without touching the x-axis.
- Mark 2: Drawing a second Boltzmann distribution curve representing a higher temperature ( T₂ ) with a lower peak shifted to the right.
- Mark 3: Explaining that at higher temperatures, a greater number/proportion of molecules have energy equal to or greater than the activation energy ( E ≥ Eₐ ), leading to a higher frequency of successful collisions.
Initial Rates Table, Rate Equation, and Rate Constant ( k )
📐 Step-by-Step Calculation
- Order wrt [NO]: Compare Experiments 1 and 2. [CO] is constant. [NO] doubles (from 2.75 × 10⁻⁴ to 5.50 × 10⁻⁴), and the rate quadruples (from 1.85 × 10⁻⁴ to 7.40 × 10⁻⁴). Therefore, order is 2.
- Order wrt [CO]: Compare Experiments 2 and 3. [NO] doubles (5.50 × 10⁻⁴ to 1.10 × 10⁻³), which would independently multiply the rate by 4. However, the rate increases by 16 times (7.40 × 10⁻⁴ to 1.18 × 10⁻²). Since 4 × 4 = 16, [CO] must also double between exp 2 and 3, meaning the rate is 1st order wrt [CO].
- Rate Equation: rate = k [NO]² [CO]
- Calculate k : Rearrange for k = rate / ([NO]² [CO]) .
Using Exp 1: k = (1.85 × 10⁻⁴) / ((2.75 × 10⁻⁴)² × 7.25 × 10⁻⁴) = 3.37 × 10⁶ - Units of k : dm⁶ mol⁻² s⁻¹
❌ Common Student Errors
- Incorrect Orders: Confusing the effect of doubling [NO] vs [CO] in experiments 2 and 3. Failing to isolate variables properly.
- Significant Figures: Rounding k to 1 or 2 sig fig instead of standard 3 SF (Acceptable values: 3.37 × 10⁶).
- Unit Derivation Mistakes: Forgetting how to cancel out concentration terms, leading to inverted or missing units like mol⁻³ dm⁹ s⁻¹ .
Reaction Mechanisms & Rate-Determining Steps
💡 Key Knowledge: Multi-Step Mechanisms
- Rate Equation Link: The rate equation rate = k[NO₂]² tells you that the rate-determining step (the slow first step) involves collision of two NO₂ molecules.
- Stoichiometry Check: All steps must add up algebraically to give the overall equation: CO(g) + NO₂(g) → NO(g) + CO₂(g) .
✅ Acceptable Mechanisms (2 Marks)
- Mark 1: Step 1 has 2NO₂ on the Left Hand Side (LHS) since it's second order in the rate equation and the slow step.
- Mark 2: Balanced intermediate steps that produce the correct overall products.
Step 1 (slow): 2NO₂ → NO + NO₃
Step 2 (fast): NO₃ + CO → NO₂ + CO₂
Example 2:
Step 1 (slow): 2NO₂ → N₂O₄
Step 2 (fast): N₂O₄ + CO → NO + NO₂ + CO₂
Topics
Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.