OCR A-Level Chemistry Unified chemistry (03), November 2021: Question 4

11 marks · Hard difficulty · Structured Questions

Investigate the reaction kinetics between pent-1-ene and iodine, determining reaction orders, initial rates, rate constants, and reaction mechanisms from experimental data and concentration-time graphs.

Practise this question

Question

Question text

4 Pent-1-ene and iodine react as shown in the equation below.

CH3CH2CH2CH=CH2 + I2 CH3CH2CH2CHICH2I

A student investigates the rate of this reaction by monitoring the concentration of iodine over time.

The initial concentrations are shown in the table.

Concentration / mol dm–3

I2 0.0200

CH3CH2CH2CH=CH2 2.0000

In this investigation, the order with respect to pent-1-ene can be assumed to be zero.

The student plots the graph below from the experimental results.

0.0200

0.0150

[I ] 0.0100

/ mol dm–3

0.0050

0.0000

0 1000 2000 3000 4000 5000 6000 7000 8000

time/s

(a) Why can the order with respect to pent-1-ene be assumed to be zero in this investigation?

… [1]

(b)* The student’s experimental procedure shows that the reaction is first order with respect to

iodine.

Show that this statement is true and determine the initial rate of reaction and rate constant.

Assume that the reaction is zero order with respect to pent-1-ene. Show your working on the

graph on page 10 and the lines below as appropriate. [6]

Additional answer space if required.

CH3CH2CH2CH=CH2 AND I2.

(i) Write equations to suggest a two-step mechanism for the reaction.

Slow

Fast

[2]

(ii) Suggest how the investigation could be modified to show that the reaction is first order

with respect to pent-1-ene.

… [2]

Mark scheme

Show the mark scheme

Question Answer Marks element Guidance

4 (a) (Large) excess of pent-1-ene 1 AO3.1 ALLOW

OR pent-1-ene concentration is (much) greater

There is a (large) excess OR

pent-1-ene has a high concentration

(b) Please refer to the marking instructions on page 6 of this 6 AO3.1 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. ×4 Initial rate

Level 3 (5–6 marks) • Evidence of tangent on graph drawn to line

Obtains a comprehensive conclusion to determine AO3.2 at t = 0 s

initial rate AND order AND rate constant k ×2 AND gradient determined in range

4.5 – 6.5 × 10–6

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

and logically structured. The information presented is • initial rate expressed as gradient value with

units of mol dm–3 s–1,

relevant and substantiated. –6 –3 –1

e.g. initial rate = 5.5 × 10 mol dm s

Level 2 (3–4 marks)

Obtains a sound, but not comprehensive conclusion, to Reasoned order of I2

determine Half lives

initial rate AND order • Half life measured on graph OR within text

OR order AND rate constant k OR stated in range 2500 ±10 s

OR initial rate AND rate constant k • Constant half life OR two stated half lives

within ±10 s

There is a line of reasoning presented with some structure. AND conclusion that I2 is 1st order

The information presented is relevant and supported by OR

some evidence. Comparison of rates from gradients

• Rate measured as gradient at a

Level 1 (1–2 marks)

concentration, c

Obtains a simple conclusion to determine

initial rate OR order • Rate measured at c/2

• c halves and rate halves

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

• so order 1

reasoning. The information is in the most part relevant. –3 –1

e.g. initial rate at c = 0.02 = 5.5 × 10–6 mol dm s

rate at c = 0.01 = 2.58 × 10–6 mol dm–3 s–1

0 marks

No response or no response worthy of credit.

14 AO

Determination of k with units

• Rate constant k clearly linked to initial rate

OR half-life:

rate ln 2

k = OR k =

[I2] t1/2

• k determined correctly from measured initial

rate or measured half life with units of s–1,

5.5 × 10–6

e.g. k = = 2.75 × 10–4 s–1

0.02

from initial rate of 5.5 × 10–6 mol dm–3 s–1 OR

from t1/2 of 2500 s

• Typical range 2.25–3.25 × 10–4

× 2

IGNORE state symbols

2 steps that add up to overall equation:

CH2CH2CH=CH2 + I2 → CH3CH2CH2CHICH2I Must be based on slow step, i.e. 2nd mark

e.g. dependent on correct slow step:

CH CH CH CH=CH + I → CH CH CH CHICH + + I– CH3CH2CH2CH=CH2 + I2

32 2 2 2 3 2 2 2

CH CH CH CHICH + + I– → CH CH CH CHICH I IGNORE actual positioning of + charge

32 2 2 3 2 2 2

ALLOW

→ CH3CH2CH2CHICH2 + I (no charge)

CH3CH2CH2CHICH2 + I →

(ii) Repeat experiment with [I2] constant/kept the same 2 AO3.4 ALLOW I2 in (great) excess

OR use (large) excess of I2 ×2

ALLOW initial rates approach of running several

Monitor/measure/plot [CH3CH2CH2CH=CH2] over time experiments with different concentrations of

OR CH3CH2CH2CH=CH2

Monitor/measure how [CH3CH2CH2CH=CH2] affects rate i.e. Measure initial rates for each experiment

AND double concentration → rate doubles

How to answer it

Kinetics: Pent-1-ene and Iodine Reaction

What this question tests

This question assesses your ability to interpret continuous monitoring concentration-time graphs, deduce reaction orders using half-lives or gradients, calculate initial rates and rate constants with correct units, propose multi-step reaction mechanisms, and design experimental modifications to isolate variables.

Question Part (a)

Pseudo-Zero Order Assumptions

✅ Correct Answer

Pent-1-ene is present in a large excess (or has a much greater/high concentration).

💡 Key Knowledge

When a reactant is in massive excess, its concentration remains practically constant throughout the reaction. Therefore, any change in its concentration does not affect the rate, allowing us to treat it as zero order (pseudo-zero order).

Mark: 1 mark (AO3.1)

Question Part (b)

Proving Order, Initial Rate, and Rate Constant (Extended Response)

🧠 Exam Technique (Level of Response)

Step 1: Initial Rate. Draw a tangent to the curve at t = 0 s where [I₂] = 0.0200 mol dm⁻³ . Calculate the gradient ( Δy / Δx ). Expected range: 4.5 – 6.5 × 10⁻⁶ mol dm⁻³ s⁻¹ .
Step 2: Order of Reaction. Show that half-lives are constant (e.g., t₁/₂ ≈ 2500 s as concentration drops from 0.0200 to 0.0100 , and from 0.0100 to 0.0050 ), proving first order with respect to iodine. Alternatively, compare gradients at halved concentrations.
Step 3: Rate Constant ( k ). Use k = rate / [I₂] or k = ln(2) / t₁/₂ .

📐 Step-by-Step Calculation

Initial Rate Example: 5.5 × 10⁻⁶ mol dm⁻³ s⁻¹
Rearranging Rate Equation: Rate = k[I₂] so k = Rate / [I₂]
Substitution: k = (5.5 × 10⁻⁶) / 0.0200
Final Answer & Units: 2.75 × 10⁻⁴ s⁻¹ (Acceptable range: 2.25 – 3.25 × 10⁻⁴ s⁻¹ )

❌ Common Errors & Traps

Poor Tangents: Drawing tangents with a ruler placed carelessly instead of ensuring equal space of the curve on either side of the pivot point.
Incorrect Units for k : For a first-order reaction, the unit is simply s⁻¹ (or min⁻¹ ). Students frequently lose the final mark by writing mol⁻¹ dm³ s⁻¹ .

Marks: Up to 6 marks (Level 3 response evaluated on comprehensive conclusions, structured reasoning, and accurate working).

Question Part (c)(i)

Proposing a Two-Step Mechanism

✅ Correct Answer & Equations

Slow Step (Rate-Determining Step):

CH₃CH₂CH₂CH=CH₂ + I₂ → CH₃CH₂CH₂CHICH₂⁺ + I⁻ (or formation of carbocation/iodine intermediate matching stoichiometry).

Fast Step:

CH₃CH₂CH₂CHICH₂⁺ + I⁻ → CH₃CH₂CH₂CHICH₂I

💡 Key Knowledge

The rate equation ( Rate = k[pent-1-ene][I₂] ) tells us exactly what species are involved in the slow (rate-determining) step. Both reactants must feature in the first step's equation.

Marks: 2 marks

Question Part (c)(ii)

Experimental Modification

✅ Correct Answer

1. Keep the concentration of I₂ constant (or use a large excess of I₂ ).

2. Monitor/measure the concentration of pent-1-ene over time (or run multiple experiments with different initial pent-1-ene concentrations to show rate is proportional to [pent-1-ene] ).

🧠 Exam Technique

To isolate and prove the order of one reactant, you must effectively "flush out" its effect by keeping it in large excess, while continuously tracking the other reactant. Mentioning initial rates methods with varying concentrations also gains full credit.

Marks: 2 marks (AO3.4)

Topics

Module 5: Physical chemistry and transition elements · Practical Activity Groups · 5.1 Rates, equilibrium and pH · PAG 9: Rates of reaction – continuous monitoring method

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.