AQA A-Level Chemistry Paper 2, 2017: Question 3

7 marks · Medium difficulty · State/Explain/Numerical

Calculate the rate constant and its units from rate equation data, then determine the activation energy using the Arrhenius equation.

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Question

Question 03 is divided into two parts. The intro states: A series of experiments is carried out with compounds C and D. Using the data obtained, the rate equation is deduced to be rate = k[C][D]. In one experiment at 25 °C, the initial rate of reaction is 3.1 × 10⁻³ mol dm⁻³ s⁻¹ when [C] = 0.48 mol dm⁻³ and [D] = 0.23 mol dm⁻³. Part 03.1 asks to calculate a value for the rate constant at this temperature and give its units (3 marks). Part 03.2 presents the Arrhenius equation ln k = (-Ea / RT) + ln A and asks to calculate Ea in kJ mol⁻¹ at 25 °C, given ln A = 16.9 and R = 8.31 J K⁻¹ mol⁻¹, using either the calculated k or an alternative value of 3.2 × 10⁻³ (4 marks).
Question text

03 A series of experiments is carried out with compounds C and D. Using the data

obtained, the rate equation for the reaction between the two compounds is

deduced to be

rate = k[C][D]

In one experiment at 25 °C, the initial rate of reaction is 3.1 × 10−3 mol dm−3 s−1

when the initial concentration of C is 0.48 mol dm−3 and the initial concentration

of D is 0.23 mol dm−3

03.1 Calculate a value for the rate constant at this temperature and give its units.

[3 marks]

Do

Rate constant 7 Units ou

03.2 An equation that relates the rate constant, k, to the activation energy, Ea, and

the temperature, T, is

− Ea

lnk = + lnA

RT

Use this equation and your answer from Question 3.1 to calculate a value, in

kJ mol−1, for the activation energy of this reaction at 25 °C.

For this reaction lnA = 16.9

The gas constant R = 8.31 J K−1 mol−1

(If you were unable to complete Question 3.1 you should use the value of

3.2 × 10−3 for the rate constant. This is not the correct value.)

[4 marks]

Activation energy kJ mol−1

Mark scheme

Show the mark scheme Mark scheme for Question 03. Part 03.1 gives 1 mark for substitution k = (3.1 × 10⁻³) / (0.48 × 0.23), 1 mark for 2.8 × 10⁻² (minimum 2 significant figures), and 1 mark for the units mol⁻¹ dm³ s⁻¹. Part 03.2 awards M1 for evaluating ln k = ln(2.8 × 10⁻²) = -3.58; M2 for rearranging to Ea = RT(ln A - ln k); M3 for substitution Ea = 8.31 × 298 × (16.9 + 3.58) = 50716 J mol⁻¹; and M4 for final value 51 kJ mol⁻¹. An alternative calculation using k = 3.2 × 10⁻³ gives Ea = 56 kJ mol⁻¹.

Question Answers Mark Additional Comments/Guidance

3.1 10-3

Mark is for insertion of numbers into correctly rearranged rate

k = (rate/[C][D]) =) 1

(0.48) (0.23) equation

03.1

= 2.8 × 10-2 min 2sfs 1

mol–1 dm3 s–1 1 Mark units separately in any order.

M1 = ln (their k)

-2 Alternative value

ln k = ln 2.8 × 10 (= - 3.58) M1 If incorrect then award -3

ln k = ln 3.2 × 10 = - 5.74

M2 and M3 only

if ln 16.9 used max 3

Ea = RT(ln A - ln k)

If temp used 25 max 2

OR M2

Incorrect

03.2 Ea = RT(ln k – ln A) rearrangement then

M1 only

-1 Ea = 8.31 × 298 (16.9 + 5.74)

Ea = 8.31 × 298 (16.9 + 3.58) ( = 50716 J mol ) M3 -1

( = 56076 J mol )

E = 56 kJ mol-1

-1 – 50.7 or -51 scores a

Ea = 51 kJ mol M4

max 2

Total 7

How to answer it

Kinetics: Rate Constants & The Arrhenius Equation

📌 What This Question Tests

This question assesses quantitative mastery of reaction kinetics, mathematical rearrangement of rate expressions, and applying the Arrhenius equation to calculate activation energy.

  • Rate Constant Calculation: Rearranging a rate expression and correctly substituting values.
  • Unit Derivation: Deducing the correct units for a second-order rate constant using dimensional analysis.
  • Arrhenius Equation: Rearranging the logarithmic form ln k = -Ea/RT + ln A to make activation energy ( Ea ) the subject.
  • Unit Conversions: Converting Celsius into Kelvin ( + 273 ) and Joules into kilojoules ( ÷ 1000 ).
  • Precision & Attention to Detail: Handling natural logs of decimal numbers and noting whether A or ln A was provided.
Question 03.1 • 3 Marks

Calculating the Rate Constant (k) and its Units

Rate equation: rate = k[C][D]

📐 Step-by-Step Calculation

  1. Rearrange the rate expression:
    k = rate / ([C][D])
  2. Substitute numerical values:
    k = (3.1 × 10⁻³) / (0.48 × 0.23)
    k = (3.1 × 10⁻³) / (0.1104) = 0.02808...
  3. Round appropriately (min 2 sig figs):
    k = 2.8 × 10⁻² (or 0.028 )
  4. Deduce the units:
    units = (mol dm⁻³ s⁻¹) / ((mol dm⁻³)(mol dm⁻³))
    Cancelling one mol dm⁻³ gives:
    units = s⁻¹ / (mol dm⁻³) = mol⁻¹ dm³ s⁻¹

✅ Mark Scheme Allocation

  • Mark 1: Correct substitution into rearranged equation:
    (3.1 × 10⁻³) / (0.48 × 0.23)
  • Mark 2: Correct calculated value for k :
    2.8 × 10⁻² (or 0.028 / 0.0281 ) [minimum 2 sig figs].
  • Mark 3: Correct units:
    mol⁻¹ dm³ s⁻¹ (indices in any order, e.g. dm³ mol⁻¹ s⁻¹ ).

❌ Common Errors to Avoid

  • Inverting indices in units: Writing mol dm⁻³ s⁻¹ (the units of rate) or mol dm⁻⁶ s⁻¹ instead of mol⁻¹ dm³ s⁻¹ .
  • Incorrect rounding: Writing 0.03 (1 significant figure), which loses the answer mark.
  • Rearrangement slip: Multiplying rate by concentrations instead of dividing.

🧠 Exam Technique Tip

Always treat numbers and units completely separately! First, determine the numerical value on your calculator. Then, write out the unit fraction explicitly and cancel identical terms step by step to ensure your negative indices are correct.

Question 03.2 • 4 Marks

Arrhenius Equation: Determining Activation Energy (Ea)

Equation: ln k = (-Ea / RT) + ln A

📐 Step-by-Step Calculation

  1. Find ln k:
    ln k = ln(2.8 × 10⁻²) = -3.576 (or -3.58 )
  2. Convert temperature to Kelvin:
    T = 25 °C + 273 = 298 K
  3. Rearrange to make Ea the subject:
    ln k - ln A = -Ea / RT
    Ea = RT(ln A - ln k)
  4. Substitute and calculate Ea in J mol⁻¹:
    Ea = 8.31 × 298 × [16.9 - (-3.576)]
    Ea = 2476.38 × (16.9 + 3.576)
    Ea = 2476.38 × 20.476 = +50 706 J mol⁻¹
  5. Convert to kJ mol⁻¹ (÷ 1000):
    Ea = 50.7 kJ mol⁻¹ or +51 kJ mol⁻¹

✅ Mark Scheme Allocation

  • Mark 1 (M1): Correctly evaluating ln(their k) :
    ln(2.8 × 10⁻²) = -3.58
    (If backup value used: ln(3.2 × 10⁻³) = -5.74 ).
  • Mark 2 (M2): Correct algebraic rearrangement:
    Ea = RT(ln A - ln k) or -Ea = RT(ln k - ln A) .
  • Mark 3 (M3): Correct substitution yielding value in J mol⁻¹:
    8.31 × 298 × (16.9 + 3.58) = 50 716 J mol⁻¹
    (Using backup: 8.31 × 298 × (16.9 + 5.74) = 56 076 J mol⁻¹ ).
  • Mark 4 (M4): Final answer in kJ mol⁻¹:
    +51 kJ mol⁻¹ (or 50.7 kJ mol⁻¹ ; backup yields +56 kJ mol⁻¹ ).

❌ Major Calculation Traps

  • Taking the natural log of ln A: The question gives ln A = 16.9 , NOT A = 16.9 . Calculating ln(16.9) is a catastrophic mistake that caps your score at max 3 marks.
  • Negative Activation Energy: Ea is an energy barrier and must be positive! Forgetting that subtracting a negative is an addition ( 16.9 - (-3.58) = 16.9 + 3.58 ) gives -51 kJ mol⁻¹ , capping marks at max 2.
  • Leaving T in °C: Using T = 25 instead of 298 K caps your score at max 2 marks.
  • Forgetting to divide by 1000: The question asks for kJ mol⁻¹ . The gas constant R = 8.31 J K⁻¹ mol⁻¹ outputs Joules, so you must divide by 1000.

💡 Key Knowledge: The Arrhenius Equation

  • k = A e^(-Ea / RT)
  • In logarithmic form: ln k = (-Ea / R)(1 / T) + ln A
  • This has the straight-line form y = mx + c , where:
    • y = ln k
    • x = 1 / T
    • gradient (m) = -Ea / R
    • y-intercept (c) = ln A
Examiner Insight: Notice the alternative value provided ( 3.2 × 10⁻³ ). If you get completely stuck on Part 1, you can still gain 100% of the 4 marks on Part 2 using this backup figure! Never leave a multi-step kinetics question blank.

Topics

Physical Chemistry · 3.1.9 Rate Equations

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.