Edexcel A-Level Chemistry AS Paper 2, June 2018: Question 3

14 marks · Medium difficulty · Open Response

Calculate Kc, enthalpy change, atom economy, and explain the effects of pressure and catalysts on the oxidation of ammonia and interpret a Maxwell-Boltzmann distribution curve.

Practise this question

Question

Exam question 3 about the oxidation of ammonia containing multiple parts: (a) writing Kc expression and units, (b)(i) calculating standard enthalpy of formation using given enthalpy values, (b)(ii) calculating atom economy by mass for NO formation, (c)(i)-(iii) explaining effects of pressure on yield and rate, and defining heterogeneous catalyst, and (d) a multiple-choice question with a Maxwell-Boltzmann distribution curve asking about the effect of increasing temperature.
Question text

3 This question is about the oxidation of ammonia.

One equation for the oxidation of ammonia is

4NH3(g) + 3O2(g) 2N2(g) + 6H2O(g)

(a) Write the expression, including units, for the equilibrium constant Kc for this reaction.

(2)

Expression

Units …

(b) Nitric acid is made from ammonia. One of the stages in nitric acid production

involves the oxidation of ammonia to produce nitrogen(II) oxide, NO. In this

process, a mixture of ammonia and oxygen is passed over a platinum-rhodium

catalyst. One manufacturer uses a pressure of 5atm and a temperature of 850°C.

The equation for this reaction is different from that in 3(a).

4NH (g) + 5O (g) → 4NO(g) + 6H O(g) Δ H = −904.8 kJ mol−1

32 2 r

(i) Use this equation, and the enthalpy changes of formation of nitrogen(II) oxide

and water, to calculate the enthalpy change of formation of ammonia in kJ mol−1.

You may find it helpful to draw a Hess cycle first. You must show your working.

Δ H (NO(g)) = +90.4 kJ mol−1

f

Δ H (H O(g)) = −241.8 kJ mol−1

f 2

(3)

(ii) Calculate the atom economy by mass for the formation of NO in this reaction.

Give your answer to an appropriate number of significant figures.

(2)

(c)6 In fact, this oxidation to form nitrogen(II) oxide is an equilibrium reaction.

(i) Explain the effect, if any, of increasing pressure on the equilibrium*P51460RA0624*yieldof NO

in this reaction.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

(2)

(ii) Explain the effect, if any, of an increase in pressure on the rate of this reaction.

(2)

(iii) The platinum-rhodium catalyst used in this reaction is a heterogeneous catalyst.

State what is meant by the term ‘heterogeneous’ and why a catalyst has no effect 7

on the yield of the products in the reaction.*P51460RA0724*

(2)

(d) The diagram shows a Maxwell-Boltzmann distribution of particle energies,

including the activation energy, Ea, for a reaction.

(1)

Number of

particles with

energy, E Ea

Energy, E

An increase in temperature will

A increase the area under the curve.

B move the peak of the curve to the right.

C raise the height of the peak.

D move the position of the activation energy, Ea, to the left.

(Total for Question 3 = 14 marks)

Mark scheme

Show the mark scheme Mark scheme for question 3 showing acceptable answers and calculations for Kc expression and units, Hess's law enthalpy calculations, atom economy evaluation, le Chatelier's principle explanations, collision theory rate explanations, catalyst definitions, and the correct multiple choice option B for the Maxwell-Boltzmann distribution graph.

Question

Acceptable Answer Additional Guidance Mark

Number

3(a) K expression (1) (K = ) [ N (g)]2 [H O(g)]6

c c 2 2

[NH (g)]4[O (g)]3

ignore missing state symbols

do not award round brackets

units based on their K expression (1) mol dm-3 or mol/dm3 (2)

c

Question Acceptable Answer Additional Guidance

Number Mark

3(b)(i) Example of calculation

calculates ∑ fH (products) (1) (+90.4 x 4) + (-241.8 x 6) = -1089.2

∑ fH (products) - rH (1) -1089.2 – (-904.8) = -184.4

-184.4/4 = -46.1 (kJ mol-1)

calculates fH (NH3) for 1 mol ammonia (1)

TE from M1 to M2

M3 can be awarded for an incorrect answer to

M2 divided by 4

correct answer with no working scores 3 marks (3)

Question Answer Acceptable Additional Guidance

Number Mark

3(b)(ii) Example of calculation

correct expression (1) 4NO .

4NO + 6H2O

OR

4NO .

4NH3 + 5O2

may be shown as numbers only

correct evaluation of atom economy (1) 4(14 +16) x 100

4(14 + 16) + 6(16 + 2)

OR

4(14 +16) x 100

4(14 + 3) + 5(16 x 2)

= 53/52.6(316)(%)

allow answer to 2 or 3 SF only

correct answer with no working scores 2 marks

0.53/0.526 scores M1 only (2)

Question Acceptable Answer Additional Guidance

Number Mark

3(c)(i) An answer that makes reference to the following if M1 and M2 are contradictory then do not

points: award any marks

yield (of NO) decreases (1)

increase in pressure shifts equilibrium (position) allow 9 mol on LHS and 10 mol on RHS, may be

to the side of fewer moles (of gas molecules) (1) shown above the equation

allow more moles of product

allow fewer moles of reactant

allow marking points in either order (2)

Question Acceptable Answer Additional Guidance

Number Mark

3(c)(ii) An answer that makes reference to the following

points:

(on increasing the pressure) allow increase in concentration of (gas)

Rate increases because there are more molecules molecules

per unit volume (1) allow any implication of more particles in a

given volume, e.g. particles are closer together

so increase in frequency of collisions (between allow more collisions per unit time

reacting molecules) (1)

ignore just ‘more collisions’/’more successful

collisions’ with no reference to time

allow answers based on a solid catalyst (2)

Question Acceptable Answer Additional Guidance

Number Mark

An answer that makes reference to:

3(c)(iii)

heterogeneous: (the catalyst is in) a different ignore reference to products

phase/state to the reactants (1)

increases the rate of the forward and

backward / reverse reactions (1) (2)

Question Acceptable Answer Mark

Number

3(d) The only correct answer is B

A is not correct because there is no increase in number of particles

C is not correct because distribution broadens as temperature rises, so peak is lower

D is not correct because Ea is an intrinsic property of the reaction, not the applied temperature (1)

(Total for Question 3 = 14 marks)

How to answer it

Study Guide: The Oxidation of Ammonia

Edexcel AS-Level Chemistry • Equilibria, Energetics & Kinetics

What this question tests

This comprehensive question assesses core physical chemistry concepts including equilibrium expressions and units (Kc), using enthalpy of formation cycles (Hess's Law), calculating atom economy, Le Chatelier's principle regarding yield vs. rate, heterogeneous catalysis, and interpreting Maxwell-Boltzmann distribution curves.

Part (a): Equilibrium Constant (Kc) Expression and Units

✅ Correct Answer

Expression: Kc = [N₂]²[H₂O]⁶ / [NH₃]⁴[O₂]³

Units: mol dm⁻³ (or mol/dm³ )

💡 Key Knowledge

Products go on the numerator (top), reactants on the denominator (bottom). Each concentration term is raised to the power of its stoichiometric balancing number from the equation.

📐 Unit Derivation

Substitute units into the expression: (mol dm⁻³)⁸ / (mol dm⁻³)⁷ = mol dm⁻³ .

❌ Common Errors

Using square brackets around the entire expression, writing incorrect powers, or failing to cancel units correctly.

Marks available: 2

Part (b): Enthalpy Change of Formation and Atom Economy

(i) Calculating Enthalpy of Formation of Ammonia

✅ Correct Answer

Final Answer: -46.1 kJ mol⁻¹

📐 Step-by-Step Calculation

  1. Step 1: Calculate total products enthalpy: (+90.4 × 4) + (-241.8 × 6) = -1089.2 kJ mol⁻¹
  2. Step 2: Apply Hess's Law ( ΔfH(products) - ΔH(reaction) ): -1089.2 - (-904.8) = -184.4 kJ mol⁻¹ (this is for 4 moles of NH₃)
  3. Step 3: Divide by stoichiometric coefficient: -184.4 / 4 = -46.1 kJ mol⁻¹

🧠 Exam Technique

Drawing a quick Hess cycle with elements in their standard states at the bottom prevents sign errors during subtraction steps. Notice that Error Carried Forward (TE) applies from M1 to M2.

(ii) Atom Economy for NO Formation

✅ Correct Answer

53% or 52.6% (Allow 2 or 3 significant figures)

📐 Calculation Steps

[4 × (14 + 16)] / [4(14 + 16) + 6(1 + 2)] × 100 = 52.63%

❌ Common Errors

Forgetting to multiply molar masses by the balancing stoichiometric coefficients from the reaction equation.

Marks available: 3 (b(i)) + 2 (b(ii)) = 5

Part (c): Le Chatelier's Principle, Rate, and Catalysis

✅ (i) Effect of Pressure on Yield

Yield of NO decreases. An increase in pressure shifts the equilibrium position to the side with fewer moles of gas molecules (LHS has 9 moles, RHS has 10 moles).

✅ (ii) Effect of Pressure on Rate

Rate increases. Higher pressure means molecules are closer together, creating more molecules per unit volume, which leads to an increased frequency of successful collisions between reacting molecules.

✅ (iii) Heterogeneous Catalysis

Definition: The catalyst is in a different physical phase/state to the reactants.
Effect on yield: None, because a catalyst speeds up both the forward and reverse reactions equally.

Marks available: 2 + 2 + 2 = 6

Part (d): Maxwell-Boltzmann Distribution

✅ Correct Answer: Option B

An increase in temperature will move the peak of the curve to the right and flatten it.

❌ Why other options are incorrect

  • A: Area under the curve remains constant (total number of particles does not change).
  • C: The peak must lower to maintain the same total area as it broadens out.
  • D: Activation energy (Ea) is an intrinsic property and position does not shift with temperature.
Marks available: 1

Topics

Physical Chemistry · Core Practicals · Topic 5: Formulae, Equations and Amounts of Substance · Topic 8: Energetics I · Topic 9: Kinetics I · Topic 10: Equilibrium I · Topic 11: Equilibrium II

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.