AQA A-Level Chemistry Paper 2, 2017: Question 2

Mark scheme

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Question Answers Mark Additional Comments/Guidance

Straight line through (0.00, 0.50) which If ‘tangent’ does not touch 0.5 mol dm-3 then CE=0 for 2.1 and

02.1 cuts time axis at between 5 and 12.5 secs OR 1 2.2.

conc 0.3 at time between 2s and 5s no tangent scores 0 in 2.1 and 2.2.

Mark is for correct calculation of their gradient : If ‘tangent’ does not touch 0.5 mol dm-3 then CE=0 for 2.1 and

e.g. 0.50/11 = 0.045 2.2

02.2 -2 -3 -1 1

or 4.5 × 10 (mol dm s ) Ignore negative sign

(Expect a value between 0.04 and 0.1

new[A]2 = 1.7 × (0.50)2 Award 2 for 0.65

[A] increases by √1.7

= 0.425

Award 1 mark for an AE using a correct method

02.3 new[A] = 1.30 × 0.50

-3 -3 If candidate use their rate then CE=0

= 0.65 (mol dm ) New [A] = 0.65 (mol dm ) 2

2 sfs min 2 sfs min

0.85 scores 1 if √ shown

Total 4

How to answer it

Kinetics: Concentration–Time Curves & Rate Equation Proportionality

📋 WHAT THIS QUESTION TESTS

Core Skills & Knowledge:

Drawing an accurate tangent at t = 0 to determine initial reaction rates from a concentration–time graph.
Calculating gradient ( Δy / Δx ) and quoting answers with appropriate chemical units ( mol dm⁻³ s⁻¹ ).
Applying the rate equation rate = k[A]²[B] when conditions change.
Deducing the effect on concentration when rate changes by a factor involving a non-linear (second order) relationship.

Question 02.1: Drawing the Tangent at t = 0

1 Mark • Graphical Skills

✅ Mark Scheme Requirement

A single, clean, straight ruled line starting at (0.00, 0.50) that represents the initial gradient of the curve.

Acceptable criteria:

Cuts the time axis between 5.0 s and 12.5 s, OR
Passes through [A] = 0.30 mol dm⁻³ at a time between 2.0 s and 5.0 s.

🧠 Exam Technique: Perfect Tangents

Use a clear plastic ruler: Place your ruler edge so it touches the curve exactly at t = 0 , [A] = 0.50 mol dm⁻³ .
Look through the plastic ruler to ensure the angles on either side between the curve and ruler look symmetrical right near the origin point.
Draw a line long enough to cross convenient grid lines to make calculating the gradient easy in the next step.

❌ Common Errors & Examiner Warnings

Missing the intercept: If your tangent line does not originate precisely from (0.00, 0.50) , this is treated as a Contradictory Error (CE = 0), scoring 0 marks for both 02.1 and 02.2!
Drawing a chord instead of a tangent (cutting across the curve into the space below it).
Drawing freehand without a ruler or drawing multiple overlapping lines.

Mark Breakdown [1 Mark]:
• 1 mark: Straight line drawn through (0.00, 0.50) with acceptable gradient (intercepting time axis between 5 and 12.5 s).

Question 02.2: Deducing the Initial Rate

1 Mark • Rate Calculation from Gradient

📐 Calculation Walkthrough

Identify coordinates on your tangent line:
Point 1: (0 s, 0.50 mol dm⁻³)
Point 2: Where the line intercepts the t-axis, e.g., (11.0 s, 0.00 mol dm⁻³) .
Calculate the gradient magnitude:
Gradient = |Δy / Δx| = 0.50 / 11.0 = 0.045 mol dm⁻³ s⁻¹
Alternative: Using scientific notation: 4.5 × 10⁻² mol dm⁻³ s⁻¹ .

✅ Acceptable Answers

Any correctly calculated gradient from a candidate's valid tangent line.
Expected range: between 0.04 and 0.10 mol dm⁻³ s⁻¹.
Examiners ignore the negative sign (rates are expressed as positive numbers).

💡 Key Knowledge

The rate of reaction is the rate of change of concentration:

Rate = - d[A]/dt = - (gradient of [A] vs t)

Initial rate is always the magnitude of the gradient at exactly time t = 0.

❌ Common Errors

Calculating average rate over 25 seconds ( Δ[A]/25 ) instead of using the tangent at t = 0 .
Inverting the gradient formula (dividing time by concentration).
Leaving units off if asked in different questions (note: units are provided on this answer line, but always check).

Mark Breakdown [1 Mark]:
• 1 mark: Correct calculation of the gradient of the student's drawn tangent line (expected value between 0.04 and 0.1).

Question 02.3: Calculating New Initial Concentration of A

2 Marks • Rate Equation & Order of Reaction

📐 Step-by-Step Calculation

Analyze the rate equation:
rate = k[A]²[B]
Temperature is constant, so k is constant. Initial [B] is held constant.
Therefore: rate ∝ [A]²
Determine the scale factor for [A]:
If rate increases by a factor of 1.7 :
(new [A] / old [A])² = 1.7
new [A] / old [A] = √1.7 ≈ 1.30384 (Method Mark 1)
Calculate the new concentration:
From the graph at t = 0 , original [A] = 0.50 mol dm⁻³ .
new [A] = 0.50 × √1.7 = 0.50 × 1.30384 = 0.6519... mol dm⁻³
New [A] = 0.65 mol dm⁻³ (given to 2 significant figures) (Answer Mark 2)

🧠 Exam Technique: Direct Proportionality

Notice that you do NOT need to know k or [B] !

You can set up an algebraic equation directly:

(new [A])² = 1.7 × (0.50)² = 1.7 × 0.25 = 0.425

new [A] = √0.425 = 0.65 mol dm⁻³

Both methods yield full marks quickly without getting bogged down in arbitrary rate calculations.

❌ Severe Traps Highlighted by the Examiner

Using calculated rate from 02.2: If candidates attempt to substitute their initial rate from 02.2 into the rate equation, they cannot proceed because k and [B] are unknown. Doing this leads to a complete breakdown of method (CE = 0).
Treating the reaction as first order: Multiplying 0.50 × 1.7 = 0.85 scores 0 marks (or maximum 1 mark only if the candidate explicitly showed √ in working but forgot to evaluate it).
Squaring instead of square rooting: Multiplying by 1.7² = 2.89 giving 1.45 mol dm⁻³ . Remember: if the rate increases, [A] must increase by √1.7 because rate is proportional to [A]² .

Mark Breakdown [2 Marks]:
• Mark 1: Demonstrating that [A] increases by a factor of √1.7 (or showing new [A]² = 1.7 × (0.50)² = 0.425 ).
• Mark 2: Final answer of 0.65 mol dm⁻³ (minimum 2 significant figures). Award 2 marks directly for 0.65.

Topics

Physical Chemistry · 3.1.9 Rate Equations