OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 14

1 mark · Medium difficulty · Multiple Choice

Calculate the rate of reaction at 200 s from a concentration-time graph of hydroxide ions.

Practise this question

Question

Multiple choice question 14 featuring a concentration-time graph for hydroxide ions against time in seconds. A curve slopes downward from an initial concentration of approximately 0.18 mol dm-3 at 0 s down towards near zero at 600 s. Four multiple choice options are provided: A, 2.2 x 10^-4; B, 2.8 x 10^-4; C, 1.8 x 10^-3; D, 4.4 x 10^-2.
Question text

14 A student measures how the OH– concentration changes over time for a reaction.

The student plots the graph below.

0.20

0.15

[OH–(aq)]

/moldm–3 0.10

0.05

0.00

0 100 200 300 400 500 600

time/s

What is the rate of reaction, in mol dm–3 s–1, at 200 s?

A 2.2 × 10–4

B 2.8 × 10–4

C 1.8 × 10–3

D 4.4 × 10–2

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer for question 14 is B, worth 1 mark.

14 B 1 AO2.8

How to answer it

Determining Rate from a Concentration-Time Graph

What this question tests

This question assesses your ability to find the instantaneous rate of reaction at a specific time from a concentration-time graph. You must demonstrate skills in constructing tangents to curves, calculating gradients (change in y divided by change in x), handling standard form numbers, and applying correct units ( mol dm⁻³ s⁻¹ ).

Question 14 (Multiple Choice)

Exam Breakdown

✅ Correct Answer

B: 2.8 × 10⁻⁴

Marks: 1 / 1 (AO2.8)

💡 Key Knowledge

  • The rate at any specific time on a concentration-time graph is represented by the gradient of the tangent at that point.
  • Units for rate are derived from concentration / time , giving mol dm⁻³ s⁻¹ .

🧠 Exam Technique

  • Use a sharp pencil and a clear plastic ruler to draw a precise tangent touching the curve strictly at t = 200 s .
  • Choose large coordinates far apart on your tangent line to minimise percentage error when calculating the gradient.

❌ Common Errors

  • Reading a single point directly off the curve at 200 s and dividing by time (calculating average rate instead of instantaneous rate).
  • Inverting the gradient fraction ( Δx / Δy instead of Δy / Δx ).
  • Misreading graph gridline intervals (each small square on the y-axis represents 0.0025 mol dm⁻³ ).

📐 Step-by-Step Calculation

  1. Locate the target time: Find t = 200 s on the x-axis and move up vertically to intersect the curve.
  2. Draw the tangent: Construct a straight line that skims the curve precisely at t = 200 s , matching the slope of the curve.
  3. Select coordinate points on the tangent:
    Choose two easy-to-read points along your drawn tangent line. For example:
    Point 1: x₁ = 0 s, y₁ = 0.125 mol dm⁻³
    Point 2: x₂ = 450 s, y₂ = 0.000 mol dm⁻³ (or similar large span based on your drawn line).
    Using standard mark scheme accepted values for the tangent at 200 s:
    Change in concentration ( Δy ) = approx 0.125 - 0.005 = 0.120
    Change in time ( Δx ) = approx 430 - 0 = 430
  4. Calculate Gradient:
    Rate = Δ[OH⁻] / Δt = 0.12 / 430 = 2.79 × 10⁻⁴
  5. Round to appropriate significant figures: Matches option B ( 2.8 × 10⁻⁴ mol dm⁻³ s⁻¹ ).

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Practical Activity Groups · 3.2 Physical chemistry · 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, AS Breadth in chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.