OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2019: Question 19
23 marks · Hard difficulty · Structured Questions
Calculate enthalpy changes, entropy, free energy, equilibrium constants (Kp), and analyze reaction rates and catalysis in the industrial manufacture of sulfuric acid.
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Question text
19 Sulfuric acid is an important chemical used to make detergents, fertilisers and dyes. It is
manufactured in a multi-step process.
(a) In the first step of the manufacture of sulfuric acid, sulfur dioxide, SO2, can be made from the
combustion of hydrogen sulfide, H2S, shown in Reaction 1.
2H S(g) + 3O (g) 2SO (g) + 2H O(l) ∆ H = –1125 kJ mol−1 Reaction 1
22 2 2 r
(i) Explain why the enthalpy change for Reaction 1 has a negative value.
Use ideas about enthalpy changes associated with bond breaking and bond making.
… [1]
(ii) Some standard entropy values are given below.
Substance H2S(g) O2(g) SO2(g) H2O(l)
Sө / J K−1 mol−1 206 205 248 70
Using calculations, explain whether Reaction 1 is feasible at 20 °C.
Calculations
Explanation for feasible or non feasible …
… [4]
(iii) Calculate the standard enthalpy change of formation, ∆ H ө, of hydrogen sulfide using
f
the enthalpy change for Reaction 1, and the standard enthalpy changes of combustion
below.
Substance ∆ H ө / kJ mol−1
c
S(s) −296.8
H2(g) −285.8
2H S(g) + 3O (g) 2SO (g) + 2H O(l) ∆ H = –1125 kJ mol−1 Reaction 1
22 2 2 r
∆ H ө of hydrogen sulfide = … kJ mol−1 [3]
f
(b) The second step in the manufacture of sulfuric acid is the conversion of SO2 into sulfur
trioxide, SO3, using Equilibrium 1.
2SO (g) + O (g) 2SO (g) ∆H = −197 kJ mol−1 Equilibrium 1
22 3
An industrial chemist carries out some research into Equilibrium 1.
• The chemist fills a 10.2 dm3 container with SO (g) at RTP, and then adds 12.0 g of O (g).
• The chemist adds the vanadium(V) oxide catalyst, and heats the mixture. The mixture is
allowed to reach equilibrium at a pressure of 2.50 atm and a temperature of 1000 K.
• A sample of the equilibrium mixture is analysed, and found to contain 0.350 mol of SO3.
(i) Write an expression for Kp for Equilibrium 1.
Include the units.
units = … [2]
(ii) Determine the value of Kp for Equilibrium 1 at 1000 K.
Show all your working.
Give your answer to 3 significant figures.
Kp = … [5]
(iii) The chemist repeats the experiment in (b) at a different temperature.
The chemist finds that the value of Kp is greater than the answer to (b)(ii).
Explain whether the temperature in the second experiment is higher or lower than
1000 K.
… [2]
(iv) Explain the significance of the expression: Kp & 1.
… [1]
(c) Vanadium(V) oxide, V2O5(s), is used as a catalyst in equilibrium 1.
2SO (g) + O (g) 2SO (g) ∆H = −197 kJ mol−1 Equilibrium 1
22 3
(i) Explain how the presence of V2O5(s) increases the rate of reaction.
Include a labelled sketch of the Boltzmann distribution, on the grid below.
Label the axes.
… [4]
(ii) Explain whether vanadium(V) oxide is acting as a homogeneous or heterogeneous
catalyst.
… [1]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
19 (a) (i) More energy is released by forming bonds 1 1.2 Response needs link between energy, breaking and
than energy required when breaking bonds making bonds
OR Eg ‘bond breaking is endothermic’ AND ‘bond
bond enthalpy of bonds being made is higher than making is exothermic’ AND ‘exothermic change
bond enthalpy of bonds being broken outweighs endothermic change’
IGNORE more bonds made than broken
(ii) FIRST CHECK ΔG 4 2.2 ×3
If ΔG = –1010 (kJ mol–1) award first 3 marks ALLOW ecf
∆S = (2 × 248 + 2 × 70) – (2 × 206 + 3 × 205)
= –391 (J K–1 mol–1 ) OR –0.391 (kJ K–1 mol–1)
ΔG = ΔH – TΔS = –1125 – (293 × –0.391) –1
ALLOW –1010000 (J mol )
= –1010 (kJ mol–1) ALLOW 3 SF up to calculator value –1010.437
Common errors
ALLOW:
Two calculation marks for:
–1117 to 3 SF up to calculator value of
–1117.179865
(use of 20 instead of 293)
(+)113438 (kJ mol–1) or 113000, 113400, 113440
(mix of J and kJ)
–1008 up to calculator value of –1008.482
(use of T = 298)
–1018 up to calculator value of –1018.257
(use of T = 273)
3.2 ×1
Feasible AND ∆G < 0 OR ∆G is negative ALLOW ECF for from incorrect ∆G,
eg Non feasible AND ∆G > 0 OR ∆G is +ve
19 a (iii) FIRST CHECK THE ANSWER ON ANSWER LINE 3 2.2 ×3
If answer = –20 (kJ mol–1) award 3 marks
H432/01 Mark Schemes June 2019
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15 element
First mark may be awarded from data on a cycle
Using Both ∆cHo values multiplied by 2
2 × (–296.8) or –593.6
AND
2 × (–285.8) or –571.6 (= –1165.2)
Use of –1125 and correctly processed: ALLOW – 20.1(0)
2∆fH(H2S) = [2 × (–296.8) + 2 × (–285.8)] – (–1125)
= –40.2 (kJ mol–1) ALLOW ECF: third mark is for dividing by 2
and use of all three values
Division by 2
∆ H(H S) = –20 (kJ mol–1) Common errors
f 2
Two marks for (+)20(.1)
ALLOW ecf if no multiplication by two occurred
[(–296.8)+(–285.8)]–(–)1125 = (+)542.4 for 2nd mark
Leading to ∆ H(H S) = (+) 271(.2) for 3rd mark
f 2
ALLOW –296.8 –285.8 = – 582.6 for 1st mark if –
OR – 562.5 is seen in 2nd mark
(b) (i) p(SO )2(g) 2 ALLOW species without state symbols and without
31.2 ×2
(Kp) = 2 brackets.
p(SO2(g)) × p(O2(g))
e.g., pSO 2, ppSO 2, PSO 2, p(SO )2 (pSO )2etc.
33 3 3 3
DO NOT ALLOW square brackets
atm–1 ALLOW atm as ECF if K is upside down
p
ALLOW use of any pressure unit
eg Pa–1 or kPa–1
H432/01 Mark Schemes June 2019
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element
19 b (ii) FIRST CHECK THE ANSWER ON ANSWER LINE 5 2.6 ×5 IF there is an alternative answer, check to see if
if answer = 27.2 award 5 marks there is any ECF credit possible using working
-------------------------------------------------------------------------16 below.
Initial amounts
10.2 Common errors
n(SO2) = ( =) 0.425 (mol) AND Allow 4 marks for 1.45/1.46 (depending upon
24.0
12 rounding)
n(O2) = ( =) 0.375 (mol) Initial amounts
32.0
Equilibrium amounts in moles n(SO2) = 2 x n(O2)
n(SO2) = (0.425 – 0.350 =) 0.075 (mol) AND n(O2) = 0.375 and n(SO2) = 0.75(0)
n(O2) = (0.375 – 0.350/2 =) 0.200 (mol) Equilibrium moles
n(SO2) 0.75 – 0.350 = 0.4(0)
Total moles n(O2) = 0.2(0)
ntot = 0.625 (mol) total moles
ntot = 0.95
Partial pressures partial pressures
0.075 pSO2 = 1.05
pSO2 = ( × 2.50 =) 0.3 (atm) AND pO2 = 0.526
0.625
0.2 pSO3 = 0.921
pO2 = ( × 2.50 =) 0.8 (atm) AND
0.625
0.350 Allow 4 marks for 15.1/15.0
pSO3 = ( × 2.50 =) 1.4 (atm) Initial amounts
0.625
n(O2) = 12/16 = 0.75
Equilibrium moles
n(O2) = 0.575
total moles
ntot = 1.00
partial pressures
pSO2 = 0.188
pO2 = 1.438
K to 3 SF pSO3 = 0.88
p
1.42
(K = =) 27.2 (atm–1) IGNORE units
p 0.32 ×0.8
H432/01 Mark Schemes June 2019
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element
19 b (iii) (greater K value means) equilibrium position shifted to17 2 ALLOW greater/higher amount of SO /product
p 3.2 ×2 3
right/RHS ALLOW greater Kp means larger numerator
Lower temperature because (forward) reaction is
exothermic
(iv) equilibrium position (far) to the right 1 3.2 ALLOW (very) high yield of products or of SO3
ALLOW reaction is nearly complete / irreversible
ALLOW Forward reaction is (greatly) favored
ALLOW (far) more product(s) than reactant(s) or
ALLOW equilibrium (greatly) favours product
H432/01 Mark Schemes June 2019
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element
19 (c) (i) 4 1.1
×4
Correct drawing of Boltzmann distribution DO NOT ALLOW two curves
Curve starts within one small square of origin Confusion with effect of temperature
AND not touching the x axis at high energy
DO NOT ALLOW ‘enthalpy’ for x-axis label
Axes labels: DO NOT ALLOW ‘atoms’ as y-axis label
y: (number of) molecules/particles
AND x: (kinetic) energy
Catalyst and activation energy ALLOW ECF for atoms (instead of
Catalyst provides a lower activation energy OR Ec molecules/particles) if y axis labelled as ‘atoms’
shown to the left of Ea on Boltzmann distribution
Particles with E > Ea
more or a greater proportion of molecules / particles / IGNORE (more) successful collisions
collisions have (energy above) activation energy (with IGNORE response implying ‘more collisions’
catalyst) OR more molecules have enough energy to (confusion with effect of greater temperature)
react OR greater area under curve above activation
energy
(ii) heterogeneous (catalyst) AND 1 1.2 ALLOW catalyst is a solid AND not a gas /
catalyst in a different phase/state (from other everything else is a gas
substances)
Total 23
How to answer it
Manufacture of Sulfuric Acid: Energetics, Equilibria & Catalysis
What this question tests
This comprehensive multi-part question assesses core physical chemistry modules: Enthalpy changes (bond breaking/making), Entropy and Gibbs free energy feasibility calculations, Enthalpy of formation cycles, Equilibrium constants (Kp) expressions and mole fraction/partial pressure calculations, temperature effects on equilibria, and Boltzmann distribution curves with heterogeneous catalysis.
Enthalpy Change and Bond Energetics
✅ Correct Answer
More energy is released by forming bonds than energy required when breaking bonds.
Alternative phrasing: Bond enthalpy of bonds being made is higher than bond enthalpy of bonds being made.
❌ Common Errors
Saying "bond breaking is exothermic" or "bond making is endothermic". Remember: Breaking = endothermic (+), Making = exothermic (-).
Feasibility and Gibbs Free Energy (ΔG)
📐 Step-by-Step Calculation
- Find total entropy change (ΔS):
ΔS = ΣS(products) - ΣS(reactants)
ΔS = [ (2 × 248) + (2 × 70) ] - [ (2 × 206) + (3 × 205) ]
ΔS = [ 496 + 140 ] - [ 412 + 615 ] = 636 - 1027 = -391 J K⁻¹ mol⁻¹ (or -0.391 kJ K⁻¹ mol⁻¹) - Convert temperature to Kelvin:
T = 20 + 273 = 293 K - Calculate Gibbs Free Energy change (ΔG):
ΔG = ΔH - TΔS
ΔG = -1125 - [ 293 × (-0.391) ]
ΔG = -1125 - (-114.563) = -1010 kJ mol⁻¹ (3 SF) - State Feasibility:
Feasible because ΔG < 0 (it is negative).
🧠 Exam Technique & Traps
- Unit mismatch trap: ΔH is typically given in kJ mol⁻¹ while ΔS is given in J K⁻¹ mol⁻¹ . Always divide ΔS by 1000 before putting it into the ΔG equation!
- Temperature trap: Do not forget to convert Celsius to Kelvin by adding 273.
Standard Enthalpy of Formation (ΔfH) Calculation
📐 Step-by-Step Calculation
- Recall the equation for Reaction 1: 2H₂S(g) + 3O₂ → 2SO₂(g) + 2H₂O(l) (ΔH = -1125 kJ mol⁻¹)
- Set up the enthalpy of formation expression:
ΔH = Σ ΔfH(products) - Σ ΔfH(reactants)
-1125 = [ 2(ΔfH SO₂) + 2(ΔfH H₂O) ] - [ 2(ΔfH H₂S) + 3(ΔfH O₂) ] - Substitute known values (remember O₂ is an element, so its ΔfH = 0):
-1125 = [ 2(-296.8) + 2(-285.8) ] - [ 2(ΔfH H₂S) + 0 ]
-1125 = [ -593.6 - 571.6 ] - 2(ΔfH H₂S)
-1125 = -1165.2 - 2(ΔfH H₂S) - Rearrange to solve for 2ΔfH(H₂S):
2ΔfH(H₂S) = -1165.2 + 1125 = -40.2 - Divide by 2 to find ΔfH for 1 mole of H₂S:
ΔfH(H₂S) = -40.2 / 2 = -20 kJ mol⁻¹
💡 Key Knowledge
Elements in their standard states (like O₂(g) and S(s) ) have an enthalpy of formation of exactly 0 kJ mol⁻¹ .
Equilibrium Constant Expression (Kp) & Units
✅ Correct Answer
Expression:
Kp = p(SO₃)² / ( p(SO₂)² × p(O₂) )
Units: atm⁻¹ (or equivalent pressure units like kPa⁻¹ or Pa⁻¹ )
❌ Common Errors
Do NOT use square brackets [ ] in Kp expressions; square brackets denote concentration (Kc). Always use round parentheses or partial pressure notation p(...) .
Calculating Kp from Initial Masses and Equilibrium Moles
📐 Step-by-Step Calculation
- Calculate initial moles:
n(SO₂) initial = 10.2 g / 64.1 g mol⁻¹ = 0.425 mol
n(O₂) initial = 12.0 g / 32.0 g mol⁻¹ = 0.375 mol - Determine equilibrium moles using stoichiometry:
Equilibrium n(SO₃) = 0.350 mol. From balanced equation ( 2:1:2 ), 0.350 mol of SO₃ formed requires 0.350 mol of SO₂ and 0.175 mol of O₂.
Equilibrium n(SO₂) = 0.425 - 0.350 = 0.075 mol
Equilibrium n(O₂) = 0.375 - (0.350 / 2) = 0.200 mol - Calculate total moles at equilibrium (ntot):
ntot = 0.075 + 0.200 + 0.350 = 0.625 mol - Calculate mole fractions and partial pressures (Total Pressure = 2.50 atm):
p(SO₂) = (0.075 / 0.625) × 2.50 = 0.3 atm
p(O₂) = (0.200 / 0.625) × 2.50 = 0.8 atm
p(SO₃) = (0.350 / 0.625) × 2.50 = 1.4 atm - Calculate Kp to 3 significant figures:
Kp = (1.4)² / [ (0.3)² × (0.8) ] = 1.96 / [ 0.09 × 0.8 ] = 1.96 / 0.072 = 27.2 atm⁻¹
🧠 Exam Technique
Always layout your calculation in a clear ICE table (Initial, Change, Equilibrium) format to prevent arithmetic errors and pick up error-carried-forward (ecf) marks if you slip up early on.
Effect of Temperature on Kp and Equilibrium Significance
✅ Correct Answers
(iii) Temperature: Higher than 1000 K.
Reason: A higher Kp value means the equilibrium position has shifted to the right. Since the forward reaction is exothermic, a lower temperature increases Kp, therefore a larger Kp at a *different* temperature implies a lower temperature was used (or if Kp is greater, explain via Le Chatelier's principle).
(iv) Significance of Kp >> 1: The equilibrium position lies far to the right (favours products heavily / near completion / high yield of SO₃).
Boltzmann Distribution and Catalysis
💡 Sketching the Boltzmann Distribution
- Axes: y-axis = Number of molecules / particles ; x-axis = Energy .
- Curve shape: Starts at the origin (0,0), rises to a peak, and asymptotically approaches the x-axis at high energy without ever touching it.
- Catalyst effect: Draw a vertical activation energy line labelled E_c (or lower Ea) to the left of the original activation energy ( E_a ). Shade or reference the area under the curve past E_c to show a greater proportion of molecules have E > E_c .
✅ Catalyst Type ((c)(ii))
Heterogeneous catalyst (because V₂O₅ is a solid while the reactants are gases - it exists in a different phase).
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · 5.2 Energy · 5.3 Transition elements
Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.