OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2024: Question 18

12 marks · Hard difficulty · Structured Questions

Calculate Kp for the decomposition of sulfur trioxide, explain the effect of temperature and pressure on the equilibrium, and explain the shapes and polarity of sulfur dioxide and sulfur trioxide molecules.

Practise this question

Question

Question 18 about sulfur dioxide and sulfur trioxide. Part (a)(i) gives the equilibrium equation SO3(g) <=> SO2(g) + 1/2 O2(g) with delta H = +99 kJ mol-1 and asks to calculate Kp given initial moles and total pressure. Part (a)(ii) provides a table of Kp values at temperatures T1 and T2 and asks to explain why T2 is higher. Part (a)(iii) has a table asking to tick boxes for the effect of no catalyst and increased pressure on Kp. Part (b) asks to explain the bond angle in SO3 and why both have polar bonds but only SO2 is a polar molecule.
Question text

18 This question is about two oxides of sulfur: sulfur dioxide, SO2, and sulfur trioxide, SO3.

(a) SO3 decomposes to form SO2 and O2, as shown in Equilibrium 18.1.

SO (g) SO (g) + 1O (g) ∆H = +99 kJ mol–1 Equilibrium 18.1

32 2 2

(i) 2.25 moles of SO3 is heated to 550°C in the presence of a catalyst and the resulting mixture

allowed to reach equilibrium.

The equilibrium mixture contains 0.900 mol of SO2 and the total pressure is 2.80 atm.

Calculate the numerical value for Kp for Equilibrium 18.1 under these conditions and state the

units of Kp.

Give your answer to 3 significant figures.

Kp = …

17 units … [5]

(ii) The numerical values of Kp for Equilibrium 18.1 at temperatures T1 and T2 are shown below.

Temperature Kp

T 3.3 × 10–5

T 7.7 × 10–2

Explain why T2 is a higher temperature than T1.

… [2]

(iii) Suggest how the value of Kp would change if the reaction was repeated with no catalyst added

and the pressure of the system increased.

Tick (✓) one box in each row.

Change Decrease No change Increase

No catalyst

Increased pressure

[2]

(b) SO2 and SO3 both have molecules with sulfur in the centre and bond angles of approximately

120°.

(i) Explain why the bond angles in SO3 are 120°.

… [1]

(ii) Explain why both SO2 and SO3 have polar bonds, but only SO2 has polar molecules.

… [2]

Mark scheme

Show the mark scheme Mark scheme for Question 18. Details step-by-step calculation for Kp including equilibrium moles, partial pressures, expression, units (atm^1/2), and acceptable alternative formats. Gives marking points for temperature and pressure effects, and explains the 3 electron regions in SO3 and the shapes and dipole cancellation for SO2 and SO3 polarity.

Question Answer Marks Guidance

1 (a) (i) FIRST CHECK THE ANSWER ON ANSWER LINE 5 IF there is an alternative answer, check for any

8 IF answer = 0.455 award 4 marks ECF credit possible using working below.

AND IF units = atm1/2 award 5 marks -------------------------------------------------------------

Equilibrium moles ✓

N SO3 = 1.35 , n O2 = 0.45(0) AND n total = 2.7(0) ALLOW 3SF or more unless there is a trailing zero

e.g. ALLOW p(SO3) = 1.4, n total =2.7

Partial pressures ✓

p(SO3) 1.35 ALLOW all marks to be awarded if atmospheres are converted

2.7(0) 2.80 OR 1.4(0) into other pressure units e.g. to kPa.

p(SO2) 0.900

2.7(0) 2.80 OR 0.933

ALLOW use of fractions for intermediate working

p(O2) 0.450

2.7(0) 2.80 OR 0.467

p(SO₂) p¹ᐟ² (O₂)

p(SO₂) p(O₂)¹ᐟ² ALLOW (Kp) =

(Kp) = p(SO₃)

p(SO₃)

ALLOW

(0.933) x (0.467)¹ᐟ² p(SO₂)² x p(O₂)

OR (Kp = ) (1.40) Kp2 =

p(SO₃)²

Answer to 3 SF IGNORE [ ] (we are just looking for the calculation)

Kp= 0.455 ✓

ALLOW ECF for units of an incorrect Kp expression

Units

Substitution of units into correct Kp expression ALLOW atm0.5

atm¹ x atm¹ᐟ² ½

atm¹ = atm

DO NOT ALLOW atm

Common errors

4 marks

(3 marks for calculation + unit mark)

p(SO₂)² x p(O₂)

0.207 (from expression ) Unit: atm

p(SO₃)²

2.20 (from inverted expression) Unit: atm-1/2

(ii) H is +ve / endothermic (in forward direction). 2 ORA throughout

AND

(At higher temperature,) equilibrium shifts to right

hand side ALLOW towards the products for right hand side

ALLOW increases yield of products

(T2 ) has greater Kp value

OR 7.7 10–2 > 3.3 10–5 DO NOT ALLOW T1 has greater Kp value

(iii) One mark per correct row 2

Change Decrease No Increase

change

No

catalyst

Increased

pressure

(b) (i) There are 3 bonding regions 1 ALLOW electron regions / areas of electron density

OR 3 double bonds (round the S atom).

ALLOW - It has a resonance structure with all 3 bonds being

the same/inbetween a single and double bond OR has 3

bonds.

DO NOT ALLOW bonding pairs

(ii) S/Sulfur and O/Oxygen have different 2 ALLOW if partial charges are seen on diagram.

electronegativities (and S–O bonds are polar)

DO NOT ALLOW sulfur is more electronegative than oxygen

(SO2 lone pair gives) non-linear shape / For non-linear, ALLOW bent OR v-shaped

asymmetrical AND dipoles don’t cancel / dipoles do

not act in opposite directions IGNORE shapes seen in diagrams, treat as rough working

OR IGNORE polar bonds cancel

uneven electron charge density AND dipoles don’t IGNORE polarity cancels

cancel DO NOT ALLOW charges cancel

ORA

e.g.

SO3 trigonal planar shape/symmetrical AND dipoles cancel /

dipoles act in opposite directions

OR

even electron charge density AND dipoles cancel

How to answer it

Equilibria, Kp Calculations, and Molecular Bonding Study Guide

What this question tests

This multi-step physical and inorganic chemistry question evaluates your ability to construct and calculate equilibrium constant expressions (Kp), derive units, interpret the effect of temperature on Kp values in relation to enthalpy changes, predict Le Chatelier's principle outcomes regarding catalysts and pressure, and apply VSEPR theory to explain molecular shapes, bond angles, bond polarity, and overall molecular polarity.

Question 1 (a) (i)

Calculating Kp and Deriving Units

Total: 5 Marks

✅ Correct Answer

Kp = 0.455 (to 3 sig figs)

Units: atm⁰·⁵ or atm½

💡 Key Knowledge

  • Mole fraction = (moles of component) / (total moles).
  • Partial pressure = mole fraction × total pressure.
  • Kp expression must match the stoichiometric coefficients: p(SO₂) · p(O₂)½ / p(SO₃) .
  • Catalysts alter the rate of reaction but have zero effect on equilibrium position or Kp values.

📐 Step-by-Step Calculation

  1. Find equilibrium moles:
    Start SO₃ = 2.25 mol. Equilibrium SO₂ = 0.900 mol.
    SO₃ reacted = 0.900 mol. Therefore, equilibrium SO₃ = 2.25 - 0.900 = 1.35 mol.
    O₂ produced = 0.900 × (1/2) = 0.450 mol.
    Total moles = 1.35 + 0.900 + 0.450 = 2.70 mol.
  2. Calculate partial pressures (Total P = 2.80 atm):
    p(SO₃) = (1.35 / 2.70) × 2.80 = 1.40 atm
    p(SO₂) = (0.900 / 2.70) × 2.80 = 0.933 atm
    p(O₂) = (0.450 / 2.70) × 2.80 = 0.467 atm
  3. Substitute into Kp expression:
    Kp = (0.933 × (0.467)⁰·⁵) / 1.40 = 0.455

❌ Common Errors & Traps

  • Inverted Kp expression: Putting reactants on top instead of products loses calculation marks.
  • Stoichiometry omission: Forgetting to raise p(O₂) to the power of 0.5 (or ½).
  • Unit mistakes: Writing √atm instead of fractional indices like atm½ .
  • Significant figures: Giving answers to 2 sig figs instead of the requested 3 sig figs.
Question 1 (a) (ii)

Temperature Dependence of Kp

Total: 2 Marks

✅ Correct Answer

ΔH is positive / endothermic in the forward direction. At higher temperatures, equilibrium shifts to the right hand side, therefore T₂ has a greater Kp value (7.7 × 10⁻² is greater than 3.3 × 10⁻⁵).

🧠 Exam Technique

When linking temperature to Kp, always state two things clearly:

  1. The sign of ΔH and how the forward reaction responds (endothermic shifts right on heating).
  2. Explicitly compare the numeric values of Kp provided in the table to prove which temperature is higher.
Question 1 (a) (iii)

Effects of Catalysts and Pressure on Kp

Total: 2 Marks

✅ Correct Answer Table

Change Decrease No change Increase
No catalyst ✔
Increased pressure ✔

💡 Key Knowledge

Crucial Rule: The equilibrium constant Kp is only affected by changes in temperature. Changing pressure, concentration, or adding a catalyst changes the position of equilibrium or the rate of reaction, but Kp remains completely constant.

Question 1 (b) (i)

Bond Angles and VSEPR Theory in SO₃

Total: 1 Mark

✅ Correct Answer

There are 3 bonding regions (or 3 double bonds) around the sulfur atom.

❌ Common Errors

Do NOT write "bonding pairs" — examiners strictly penalize this terminology at A-Level. Use electron regions, areas of electron density, or bonding regions.

Question 1 (b) (ii)

Bond Polarity vs. Molecular Polarity

Total: 2 Marks

✅ Correct Answer

1st mark: Sulfur and oxygen have different electronegativities, making S-O bonds polar.
2nd mark: SO₂ has a non-linear (bent) shape so dipoles do not cancel out (asymmetrical), whereas SO₃ is symmetrical so individual bond dipoles cancel out.

🧠 Top-Level Examiner Guidance

To secure full marks in molecular polarity questions, your explanation must address two distinct features:

  • The presence of polar bonds (due to electronegativity differences).
  • The 3D shape and symmetry of the molecule determining whether dipoles cancel or reinforce each other.

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · 2.2 Electrons, bonding and structure

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.