OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 17

11 marks · Hard difficulty · Calculations

Determine the rate constant, reaction mechanism, activation energy, and pre-exponential factor from experimental initial rates and an Arrhenius plot for the reaction between iron(III) and iodide ions.

Practise this question

Question

Question 17 presents data for the reaction between Fe3+ and I- ions, including a table of initial rates for three experiments, followed by part (a) asking to determine the rate constant and a two-step mechanism. Part (b) provides an Arrhenius plot of ln k against 1/T with data points ranging from 1/T = 2.50 to 4.50 × 10^-3 K^-1, and asks students in (i) to calculate the activation energy and in (ii) to calculate the pre-exponential factor A.
Question text

17 This question is about reaction rates.

Aqueous iron(III) ions, Fe3+(aq), react with aqueous iodide ions, I–(aq), as shown below.

2Fe3+(aq) + 2I–(aq) 2Fe2+(aq) + I (aq)

A student carries out three experiments to investigate how different concentrations of Fe3+(aq)

and I–(aq) affect the initial rate of this reaction. The results are shown below.

[Fe3+(aq)] [I–(aq)] Initial rate

Experiment –3 –3 –3 –1

/ mol dm / mol dm / mol dm s

14.00 × 10–2 3.00 × 10–2 8.10 × 10–4

28.00 × 10–2 3.00 × 10–2 1.62 × 10–3

34.00 × 10–2 6.00 × 10–2 3.24 × 10–3

(a)* Determine the rate constant and a possible two-step mechanism for this reaction that are

consistent with these results. [6]

Additional answer space if required

(b) A student carries out an investigation to find the activation energy, Ea, and the pre-exponential

factor, A, of a reaction.

The student determines the rate constant, k, at different temperatures, T.

The student then plots a graph of lnk against 1 /T as shown below.

33.00

32.00

31.00

30.00

In k

29.00

28.00

27.00

26.00

0.00 1.00 2.00 3.00 4.00 5.00

1/T

/10–3 K–1

(i) Draw a best-fit straight line and calculate the activation energy, in J mol–1.

Give your answer to three significant figures.

Show your working.

activation energy, E = + … J mol–1 [3]

a

(ii) Use the graph to calculate the value of the pre-exponential factor, A.

Show your working.

pre-exponential factor, A = … [2]

Mark scheme

Show the mark scheme The mark scheme provides a level-based response grid for part (a) detailing criteria for orders, rate constant calculation, and mechanism steps. For part (b)(i), it shows the expected gradient calculation from the line of best fit and activation energy to 3 significant figures. For part (b)(ii), it shows the intercept determination and calculation of the pre-exponential factor A using e^(y-intercept).

Question Answer Marks Guidance

17 (a)* Please refer to the marking instructions on page 5 of this mark 6 Indicative scientific points may include:

scheme for guidance on how to mark this question. Orders and rate equation

Fe3+ 1st order AND I– 2nd order

Level 3 (5–6 marks) OR rate = k[Fe3+] [I–]2

A comprehensive conclusion which uses quantitative results for Supported by experimental results

determination of the reaction orders.

AND Calculation of k, including units

Determines k from correct rate equation. k correctly calculated AND correct units,

AND e.g.

Proposes the two-step mechanism which adds up to overall 8.10 10 4

equation with no intermediate electrons. k = 2 2 2 = 22.5

(4.00 10 ) (3.00 10 )

dm6 mol–2 s–1 OR mol–2 dm6 s–1

There is a well-developed line of reasoning which is clear and

logically structured. The information presented is relevant and Two-step mechanism

substantiated. The working for the scientific content is clearly

Two steps add up to give overall equation

linked to the experimental evidence.

Slow step/ rate-determining step matches

Level 2 (3–4 marks) stoichiometry of rate equation.

Reaches a sound, but not comprehensive, conclusion based on Each step balances by species and charge

the quantitative results. e.g.

Fe3+(aq) + 2I–(aq) [FeI ]+ SLOW

AND 2

Fe3+(aq) + [FeI ]+ 2Fe2+(aq) + I (aq) FAST

Correctly identifies the orders and rate equation. 2 2

AND 3+ – 2+ –

Fe (aq) + 2I (aq) Fe (aq) + I2 (aq) SLOW

Fe3+(aq) + I –(aq) Fe2+(aq) + I (aq) FAST

Calculates the rate constant 2 2

OR

Fe3+(aq) + 2I–(aq) Fe+ + I SLOW

Proposes the two-step mechanism with reactants of first step 2

matching rate equation or matches orders Fe3+(aq) + Fe+ 2Fe2+(aq) FAST

There is a line of reasoning presented with some structure. The There may be other feasible possibilities

information presented is relevant and supported by some

evidence. The working for the scientific content is clearly linked

to the experimental evidence.

Level 1 (1–2 marks) 10

Attempts to reach a simple conclusion for orders

AND

Attempts a relevant rate equation.

There is an attempt at a logical structure with a line of

reasoning. The information is in the most part relevant The

working for the scientific content is clearly linked to the

experimental evidence.

0 marks

No response or no response worthy of credit.

(b) (i) 3

Gradient

Correct gradient calculated from best-fit straight line ALLOW lines which do not intercept y-axis

drawn within the range ±800 ±1040

ALLOW mark for gradient if correct working shown

within Ea calculation without gradient being calculated

Ea calculation separately

Ea = (–) gradient 8.314 ALLOW ±0.8(00) ±1.04(0)

–1 (omission of 10–3)

e.g. from ±820, Ea = (+)6817.48 (J mol )

ALLOW ECF for calculated gradient x 8.314

E to 3 SF AND use of 10–3 for gradient If value of gradient not shown separately,

a

e.g. from ±820, E = (+)6820 (J mol–1) ALLOW E in range: 6650 8650

a a

OR 6.65 8.65 (omission of 10–3)

This mark subsumes gradient mark

NOTE: Omission of 10–3 can get 1st 2 marks

(ii) Intercept shown on graph 2 ALLOW y = 31.4

could be by extrapolation of line, or label on y axis

AND ln A linked to intercept value ALLOW substitution of correct values of ln k and 1/T

e.g. ln A = 31.4 into ln k = –Ea/R x 1/T + ln A to give a value of ln A

which approximately matches the intercept if given

ln A = ln k + (Ea/R x 1/T)

Calculation of A = elnA

OR

Calculation of A = eintercept eln k+ ( Ea/R x 1/T)

e.g. A = e31.4 = 4.33 1013

ALLOW ECF from incorrect ln A

e31.2 = 3.55 1013

e31.3 = 3.92 1013

e31.35 = 4.12 1013

e31.45 = 4.56 1013

e31.5 = 4.79 1013

e31.6 = 5.29 1013

e31.7 = 5.85 1013

e31.8 = 6.46 1013

e31.9 = 7.14 1013

e32.0 = 7.9(0) 1013

e32.1 = 8.73 1013

IF 2 DP answer given, check rounding from calculator

value, not 3 DP values given

Eg e31.7 = 5.8497 1013 and = 5.8 1013(2SF)

Total 11

How to answer it

Kinetics, Rate Equations and Arrhenius Calculations

What this question tests

This multi-part synoptic question assesses core chemical kinetics skills: deducing orders of reaction from experimental rate data, calculating rate constants with correct units, proposing multi-step reaction mechanisms consistent with rate laws, extracting activation energy (Eₐ) from Arrhenius graphical data (ln k vs 1/T), and using intercepts to calculate the pre-exponential factor (A).

Part (a) — 6 Marks

Orders, Rate Constant, and Reaction Mechanisms

💡 Key Knowledge

  • Initial Rates Method: Compare experiments where one reactant concentration changes while others remain constant to find individual orders.
  • Rate Equation: rate = k[Fe³⁺][I⁻]²
  • Mechanism Rules: The slow step (rate-determining step) must match the stoichiometry of the rate equation. All steps must add up to the overall balanced equation.

✅ Correct Answers & Mark Breakdown (Level 3: 5–6 marks)

  • Orders: Fe³⁺ is 1st order; I⁻ is 2nd order. Overall order = 3.
  • Rate Constant (k): 22.5 dm⁶ mol⁻² s⁻¹ (or mol⁻² dm⁶ s⁻¹)
  • Mechanism Example:
    Step 1 (SLOW): Fe³⁺(aq) + 2I⁻(aq) → [FeI₂]⁺(aq)
    Step 2 (FAST): Fe³⁺(aq) + [FeI₂]⁺(aq) → 2Fe²⁺(aq) + I₂(aq)

📐 Calculation Steps for k

  1. Rearrange rate equation: k = rate / ([Fe³⁺][I⁻]²)
  2. Substitute values from Experiment 1:
    k = (8.10 × 10⁻⁴) / ((4.00 × 10⁻️²) × (3.00 × 10⁻²)²)
  3. Calculate value: 22.5
  4. Determine units: s⁻¹ / (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹

❌ Common Errors & Examiner Pitfalls

  • Unit Errors: Forgetting or miscalculating units for a third-order overall rate constant.
  • Mechanism Failures: Proposing steps containing species or intermediates that do not balance out to the overall equation, or having a slow step that doesn't match the rate equation terms.
Mark Scheme Note: Level 3 requires quantitative deduction of orders, a calculated value for k with units, and a chemically valid two-step mechanism.
Part (b)(i) — 3 Marks

Arrhenius Activation Energy Calculation

🧠 Exam Technique & Graph Skills

  • Best-Fit Line: Draw a straight line of best fit through the crosses, ensuring balanced distribution of points above and below the line.
  • Gradient Triangle: Draw a large gradient triangle spanning at least half the length of the plotted line to minimise reading errors.
  • Axis Scale Trap: Notice the axis multiplier / 10⁻³ K⁻¹ on the x-axis! You must multiply your x-coordinate values by 10⁻³ when calculating the gradient.

✅ Correct Answers

  • Gradient: Negative value, typically around -820 to -1040 (accounting for the 10⁻³ factor).
  • Activation Energy (Eₐ): Calculated using Eₐ = -gradient × R (where R = 8.314 J mol⁻¹ K⁻¹).
  • Final Answer to 3 SF: e.g., +6820 J mol⁻¹ (or depending on exact student graph line, ranges like 6650 to 8650 are accepted with ECF).

❌ Common Errors

  • Omitting 10⁻³: Forgetting to account for the axis scale factor is the #1 place students lose marks here.
  • Sign confusion: Arrhenius equation is ln k = (-Eₐ/R)(1/T) + ln A , meaning gradient = -Eₐ/R . Eₐ must end up positive.
Mark Scheme Note: 1 mark for correct gradient calculation, 1 mark for multiplying gradient by R (8.314), 1 mark for correct 3 SF rounding and units.
Part (b)(ii) — 2 Marks

Calculating the Pre-Exponential Factor (A)

💡 Key Knowledge

  • The Arrhenius equation in linear form is ln k = (-Eₐ/R)(1/T) + ln A .
  • The y-intercept of the graph corresponds directly to ln A .
  • To find A from ln A, you must take the inverse natural logarithm: A = e^(ln A) .

✅ Correct Answers

  • Intercept (ln A): Extrapolate the line to the y-axis where 1/T = 0. Value is approximately 31.4 .
  • Pre-exponential Factor (A): 4.33 × 10¹³ (allow range based on student intercept, e.g., values around e³¹.⁴ ).

🧠 Alternative Method (Substitution)

If your line does not comfortably fit on the page to read the y-intercept directly, you can calculate ln A by substituting a coordinate pair ( 1/T, ln k ) and your calculated Eₐ back into the rearranged equation: ln A = ln k + (Eₐ / R)(1/T) .

Mark Scheme Note: 1 mark for identifying/using the y-intercept (ln A), 1 mark for correctly evaluating e^(intercept) to find A. ECF applies from incorrect ln A values.

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 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.