OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2021: Question 20

11 marks · Hard difficulty · Structured Questions

Calculate Gibbs free energy change, maximum feasibility temperature, and equilibrium constant Kp for the reaction of nitrogen monoxide and oxygen to form nitrogen dioxide, and predict the effects of changes on equilibrium and rate.

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Question text

20 This question is about chemical equilibrium.

Nitrogen monoxide, NO, and oxygen, O2, react to form nitrogen dioxide, NO2, in the reversible

reaction shown in Equilibrium 20.1.

Equilibrium 20.1 2NO(g) + O (g) 2NO (g) ΔH = –114 kJ mol–1

ΔS = –147 J mol–1 K–1

(a) A dynamic equilibrium exists in a closed system.

State one other feature of a dynamic equilibrium.

… [1]

(b) (i) Show that the formation of NO2 in Equilibrium 20.1 is feasible at 25 °C.

[2]

(ii) Determine the maximum temperature, in K, for feasibility.

Give your answer to an appropriate number of significant figures.

maximum temperature = … K [1]

The chemist mixes together 1.60 mol of NO(g) and 1.50 mol of O2(g) in a container and the

mixture is allowed to reach equilibrium.

At equilibrium:

• 75% of the NO(g) has been converted to NO2(g)

• the total pressure is 1.21 MPa.

(i) Calculate K , in MPa–1, for Equilibrium 20.1.

Give your answer to 3 significant figures.

K = … MPa–1 [4]

(ii) The chemist then repeats the experiment three times. In each experiment, the chemist

makes one change but uses the same initial amounts of NO and O2.

Complete the table to show the predicted effect of each change compared with the

original experiment.

Only use the words greater, smaller or same.

Equilibrium amount

Change Kp Initial rate

of NO2(g)

Temperature

increase

Pressure

increase

Catalyst

added

[3]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

20 (a) rate of forwards reaction = rate of backwards reaction 1 AO1.1

OR concentrations/pressure/temperature are constant /do not DO NOT ALLOW “are the same”

change

(b) (i) 2 AO2.2

∆G = ∆H – T∆S = –114 – (298 × –0.147) ×2 ALLOW –114000 – (298 × –147)

= –70.194 (kJ mol–1) ALLOW –70 up to calculator value of

AND –70.194 correctly rounded,

statement of ∆G < 0 OR ∆G is –ve OR ∆H < T∆S i.e. –70 OR –70.2 OR –70.19

ALLOW -70000 up to -70194 (J mol–1)

ALLOW ECF for an incorrectly calculated

negative value of ∆G linked to feasibility

statement

IGNORE rounding after 3 SF

ORA for comment about – sign required

for feasibility

(b) (ii) 776 (K) 1 AO2.2

∆H –114

i.e. Maximum temperature = = = 776 (K)

∆S –0.147

3 SF required (appropriate from supplied data)

element

IF answer = 20.7 (MPa–1), award 4 marks ×4 ----------------------------------------------------

-------------------------------------------------------------------- ALLOW ECF throughout

Equilibrium amounts

n(NO) = 0.4 (mol)

AND n(O2) = 0.9 (mol) ALLOW 20.6 from 3 SF partial

AND n(NO2) = 1.2 (mol) pressures, 0.194, 0.436 and 0.581

Total moles at equilibrium

ntot = 2.5(mol) IF there is an alternative answer, check

to see if there is any ECF credit possible

Partial pressures using working below

0.4 -----------------------------------------------------

p(NO) = × 1.21 = 0.1936 (MPa)

2.5

0.9

AND p(O2) = × 1.21 = 0.4356 (MPa)

2.5

1.2 Look for values to 3 SF here:

AND p(NO2) = × 1.21 = 0.5808 (MPa) 0.194, 0.436 and 0.581

2.5

Kp value

0.58082

K = = 20.7 to 3 SF (MPa–1)

p 0.19362 × 0.4356 ALLOW

25.0 as ECF (from omission of partial

pressures for 3 marks)

element

Equilibrium Initial ×3

Change Kp

amount of NO2 rate Mark by COLUMN

Temperature

increased smaller smaller greater

Pressure

increase same greater greater ALLOW obvious alternatives for

greater/smaller/same,

Catalyst e.g.

added same same greater

increases/decreases/

more/less

Total 11

How to answer it

Chemical Equilibrium, Free Energy, and Kp Calculations

📚 What this question tests

This question assesses core physical chemistry concepts including dynamic equilibrium criteria, Gibbs free energy calculations (ΔG = ΔH - TΔS), determining temperature limits for feasibility, multi-step equilibrium mole calculations, partial pressures, equilibrium constants (Kₚ), and applying Le Chatelier's principle to predict shifts in Kₚ, equilibrium yields, and reaction rates.

Part (a) — Features of Dynamic Equilibrium

State one other feature of a dynamic equilibrium [1 mark]

✅ Correct Answer

Rate of forwards reaction = rate of backwards reaction
OR concentrations/pressure/temperature of reactants and products remain constant (do not change)

❌ Common Errors

Students frequently write "concentrations of reactants and products are the same". This will lose the mark! They must be constant, not necessarily equal.

Part (b) — Gibbs Free Energy & Feasibility

(i) Show that the formation of NO₂ is feasible at 25 °C [2 marks]

📐 Step-by-Step Calculation

Convert units: Ensure ΔS is in kJ mol⁻¹ K⁻¹ by dividing by 1000: -147 ÷ 1000 = -0.147 kJ mol⁻¹ K⁻¹ .
Convert temperature: 25 °C + 273 = 298 K.
Substitute into Gibbs equation: ΔG = ΔH - TΔS = -114 - (298 × -0.147) .
Calculate value: ΔG = -70.194 kJ mol⁻¹ .
Conclusion: Since ΔG < 0 (negative), the reaction is feasible.

🧠 Exam Technique & Mark Scheme

1 mark is awarded for the correct calculation of ΔG (allowing ECF for unit conversion errors). The 2nd mark requires both stating the numerical value and explicitly linking a negative ΔG to feasibility.

(ii) Determine the maximum temperature, in K, for feasibility [1 mark]

✅ Correct Answer & Method

Set ΔG = 0, meaning T = ΔH / ΔS .

Calculation: T = -114 / -0.147 = 775.51... K

Final Answer: 776 K (to 3 significant figures, matching the precision of the supplied data).

Part (c) — Equilibrium Calculations & Kₚ

(i) Calculate Kₚ, in MPa⁻¹, for Equilibrium 20.1 [4 marks]

📐 Step-by-Step Calculation

Find equilibrium moles:
- Initial NO = 1.60 mol. 75% reacted = 1.60 × 0.75 = 1.20 mol used.
- Equilibrium NO = 1.60 - 1.20 = 0.40 mol
- Equilibrium O₂ = 1.50 - (1.20 / 2) = 0.90 mol (using stoichiometry ratio 2:1)
- Equilibrium NO₂ = 1.20 mol (formed in 2:2 ratio)
Total moles (n_tot): 0.40 + 0.90 + 1.20 = 2.50 mol
Calculate partial pressures (Total P = 1.21 MPa):
- p(NO) = (0.40 / 2.50) × 1.21 = 0.1936 MPa
- p(O₂) = (0.90 / 2.50) × 1.21 = 0.4356 MPa
- p(NO₂) = (1.20 / 2.50) × 1.21 = 0.5808 MPa
Construct Kₚ expression & calculate:
Kₚ = (p(NO₂))² / ((p(NO))² × p(O₂))

Kₚ = (0.5808)² / ((0.1936)² × 0.4356) = 20.697...
Final Answer: 20.7 MPa⁻¹ (to 3 significant figures).

❌ Common Calculation Traps

Forgetting to square the partial pressure of NO and NO₂ in the Kₚ expression.
Incorrect stoichiometry deduction for O₂ (forgetting to halve the moles reacted).
Omitting or miscalculating units (MPa⁻¹).

(ii) Predict the effect of changes on equilibrium [3 marks]

Marked strictly by column. Only the words greater, smaller, or same are permitted.

Change	Kₚ	Equilibrium amount of NO₂ (g)	Initial rate
Temperature increase	smaller	smaller	greater
Pressure increase	same	greater	greater
Catalyst added	same	same	greater

💡 Examiner Commentary for Table

Temperature: Forward reaction is exothermic (ΔH is negative). Increasing temperature shifts equilibrium in the endothermic direction (backwards), decreasing yield of NO₂ and reducing Kₚ. Rate increases due to more frequent successful collisions.
Pressure: Kₚ only changes with temperature. Increasing pressure shifts position towards fewer moles of gas (products side, 2 moles vs 3 moles on reactant side), increasing NO₂ amount. Rate increases due to higher concentration/collision frequency.
Catalyst: Has zero effect on position of equilibrium, equilibrium yield, or Kₚ. It increases initial rate by providing an alternative pathway with a lower activation energy.

Topics

Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH · 5.2 Energy

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.