OCR A-Level Chemistry AS Breadth in chemistry (01), November 2020: Question 24

8 marks · Medium difficulty · Structured Questions

Calculate the standard enthalpy change of formation of ammonia and explain using Boltzmann distributions why temperature and catalysts increase reaction rate.

Practise this question

Question

Question 24 about making ammonia, containing two parts. Part (a) provides the equation N2(g) + 3H2(g) -> 2NH3(g), a table of standard enthalpy changes of combustion for N2, H2, and NH3, and asks to calculate the standard enthalpy of formation of NH3(g) for 3 marks. Part (b) asks to explain using Boltzmann distributions why increasing temperature and using a catalyst increase reaction rate, with answer lines provided, worth 5 marks.
Question text

24 This question is about making ammonia, NH3.

(a) Ammonia is manufactured by reacting nitrogen with hydrogen:

N2(g) + 3H2(g) → 2NH3(g)

Standard enthalpy changes of combustion, ∆ H o, are given in the table.

c

Substance ∆ H o / kJ mol−1

c

N2(g) +180

H2(g) –286

NH3(g) –293

Calculate the standard enthalpy change of formation, ∆ H o, for NH (g).

f 3

∆ H o for NH (g) = … kJ mol−1 [3]

f 3

(b) The industrial manufacture of NH3 from N2 and H2 is carried out at an increased temperature

and in the presence of a catalyst.

Explain, using Boltzmann distributions, why increasing the temperature and using a catalyst

both increase the reaction rate.

… [5]

Mark scheme

Show the mark scheme Mark scheme for question 24. Part (a) shows the calculation steps for standard enthalpy of formation yielding -46 kJ mol-1 for 3 marks. Part (b) shows expected Boltzmann distribution diagrams including labelled axes, curves for two temperatures, and a catalyst with activation energy, alongside explanation points about molecules having energy above the activation energy, worth 5 marks.

AO

Question Answer Marks Guidance

element

24 (a) FIRST CHECK THE ANSWER ON ANSWER LINE 3 2.6 ×3 FULL ANNOTATIONS MUST BE USED

If answer = –46 (kJ mol–1) award 3 marks

ALLOW ECF throughout

Use of ∆cH values and balancing numbers

± (+180 + (3 × –286)) OR ± 678 COMMON ERRORS

AND –92 omission of ÷2 for ∆fH(NH3) 2 marks

± (2 × –293) OR ± 586 seen anywhere (+)46 Incorrect subtraction 2 marks

(+)92 Incorrect subtraction & no ÷2 1 mark

Correct subtraction using ∆H

(–678) – (–586) –385 no ×2 for –293 and no ÷2 1 mark

= –92 (kJ mol–1) –192.5 no ×2 for –293 2 marks

Calculation of ∆fH(NH3) formation (+)480 no ×3 for –286 and no ÷2 1 mark

–92 –1 (+)240 no ×3 for –286 2 marks

∆fH(NH3) = = –46 (kJ mol )

(+)187 no ×3 for –286 AND no ×2 for –293

AND no ÷2 1 mark

(+)93.5 no ×3 for –286 AND no ×2 for –293

2 marks

(b) Boltzmann distribution (seen anywhere) 2 marks 5 FULL ANNOTATIONS THROUGHOUT

NOTE: Look for marking criteria within

annotations on Boltzmann distribution diagram

IGNORE slight inflexion on the curve

Curve IGNORE small increase at end of curve

Curve starts close to origin (ALLOW flexibility) For labels,

AND curve does not touch x axis at high energy 1.1×2 ALLOW kinetic energy

Labels IGNORE number of atoms

(Number of) molecules/particles AND Energy IGNORE enthalpy for energy

AO

Question Answer 11 Marks Guidance

element

Curves for two temperatures 1 mark Temperature

Drawing of two labelled curves

AND higher temperature peak

at higher energy

AND lower on molecules

1.2×3 IGNORE curves meeting at higher energy

Higher temperature curve must cross over

ASSUME that T2 is higher temperature than T1

Catalyst and activation energy 1 mark Catalyst

Ec shown at lower energy than Ea on

Boltzmann distribution

IGNORE catalyst provides a lower

activation energy

Boltzmann distribution not used

Molecules and activation energy, Ea 1 mark

Explanation ALLOW more molecules have energy to react

At higher temperature OR in presence of catalyst

ALLOW Ea for activation energy

More molecules/particles/collisions

ALLOW Ec for activation energy with catalyst

• have energy above activation energy

OR have enough energy to overcome Ea

IGNORE more successful collisions

OR collide more frequently

Could be shown on diagram(s) using shaded area

with annotations

How to answer it

Enthalpy Changes of Combustion & Boltzmann Distributions

OCR AS Level Chemistry • Enthalpy & Reaction Rates

What this question tests

This question assesses your ability to apply Hess's Law cycles using standard enthalpy changes of combustion (ΔcH°) to calculate standard enthalpy changes of formation (ΔfH°). It also tests your qualitative understanding of reaction kinetics using Boltzmann distribution curves to explain the effects of changing temperature and adding a catalyst.

Part (a): Enthalpy of Formation Calculation

Calculate the standard enthalpy change of formation, ΔfH°, for NH₃(g). [3 marks]

✅ Correct Answer

ΔfH° for NH₃(g) = -46 kJ mol⁻¹

Full marks awarded directly if the final correct value with correct sign and units is stated on the answer line.

💡 Key Knowledge

  • Combustion cycle rule: Enthalpy of formation is calculated using: ΣΔcH(reactants) - ΣΔcH(products) .
  • Stoichiometry matters: You must multiply individual combustion values by the balancing numbers in the full equation: N₂(g) + 3H₂(g) → 2NH₃(g) .
  • Definition check: Enthalpy of formation is for one mole of product, hence the final division by 2.

📐 Step-by-Step Calculation

  1. Calculate Reactants:
    N₂ + 3H₂ = (+180) + (3 × -286)
    = +180 - 858 = -678 kJ mol⁻¹
  2. Calculate Products:
    2NH₃ = 2 × -293 = -586 kJ mol⁻¹
  3. Subtract Products from Reactants:
    ΔrH° = (-678) - (-586) = -92 kJ mol⁻¹ (this is for 2 moles of NH₃)
  4. Divide by 2 for 1 mole of NH₃:
    ΔfH° = -92 / 2 = -46 kJ mol⁻¹

❌ Common Errors & Traps

  • Forgetting to divide by 2: Leaving the answer as -92 loses the final accuracy mark (common trap giving 2/3 marks).
  • Sign inversion errors: Miscalculating (+678) - (-586) or mixing up reactant/product subtraction order.
  • Stochiometric slips: Forgetting to multiply the hydrogen combustion value (-286) by 3.

Part (b): Boltzmann Distributions & Rates

Explain, using Boltzmann distributions, why increasing the temperature and using a catalyst both increase the reaction rate. [5 marks]

✅ Marking Points Breakdown (5 Marks)

  • Mark 1 & 2 (Axes & Curve): Correct Boltzmann distribution axes (Number of molecules / particles vs Energy) with a curve starting at origin and not touching the x-axis at high energy.
  • Mark 3 (Temperature): Second curve drawn with a lower peak shifted to the right (higher temperature).
  • Mark 4 (Catalyst): Activation energy line shifted to the left ( E_catalyst or E_c ) compared to uncatalyzed E_a .
  • Mark 5 (Explanation): Statement that more molecules/particles have energy greater than or equal to the activation energy ( E ≥ E_a ), leading to a higher frequency of successful collisions.

🧠 Exam Technique & Diagram Guidelines

  • Drawing Temperature Curves: The higher temperature curve must have a lower peak and be shifted to the right. Crucially, the area under both curves must remain equal (conservation of total particles).
  • Catalyst Representation: Draw a vertical line labeled E_c or catalyst to the left of the original activation energy ( E_a ).
  • Key Phrasing: Always use precise terminology: "greater proportion of molecules have energy equal to or greater than the activation energy", rather than vague statements like "molecules move faster".

❌ Common Misconceptions

  • Drawing the higher temperature curve with a higher peak (violates conservation of particle number and fails mark scheme criteria).
  • Failing to explicitly label the axes ("Number of molecules" and "Energy").
  • Stating that a catalyst "lowers activation energy" without showing it graphically or explaining the consequence on particle energy proportions.

💡 Examiner Commentary

Top-level responses clearly distinguished between the two effects on the diagram. They annotated the shaded area under the curve beyond the activation energy to visually prove why rate increases. Students frequently lost marks by omitting axis labels or drawing the temperature curves crossing incorrectly.

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.