Edexcel A-Level Chemistry Paper 3, November 2021: Question 9
15 marks · Hard difficulty · Calculations
Calculate the effects of pressure and temperature on equilibrium constants and yields in the Haber process for ammonia synthesis.
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Question text
9 Ammonia is manufactured by the Haber Process.
N2(g) + 3H2(g) ⇌ 2NH3(g)
p(NH )2
K = 3
p 3
p(N2) (pH2)
(a) The pressure used in the Haber Process is 200 atm.
Explain the effect, if any, of increasing the pressure on the equilibrium yield
of ammonia.
(2)
(b) The equilibrium constants for Kp and Kc are related by the equation
Kc
Kp = ∆n
(RT)
where ∆n is the number of moles of reactants minus the number of moles
of products.
Calculate the value of K at 500 K when the value of K = 3.55 × 10−2 atm−2.
c p
Include the units for Kc.
[Use the value of R = 0.0821 dm3 atm K−1 mol−1]
(4)
(c) A mixture of 1.0 mol of nitrogen and 3.0 mol of hydrogen is left to reach
equilibrium at 700K.
Calculate the total pressure, in atmospheres, needed to produce a yield of
0.30mol of ammonia at 700K.
Give your answer to an appropriate number of significant figures.
You must show your working.
[K = 7.76 × 10−5 atm−2 at 700 K]
p
(5)
(d) The value of the equilibrium constant, Kp, varies with temperature.
The equation relating the values of the equilibrium constant at two temperatures is
K2 H 1 1
In
K1 R T1 T2
The equilibrium constant,*P67806A03036*K, for the formation of ammonia is 6.76× 105 atm−2
when the temperature T1 = 298K.
The enthalpy change ∆H = −92 400 J mol−1.
Calculate the value of the equilibrium constant for this reaction at 310K.
[Use the value of R = 8.31 J mol−1 K−1]
(4)
*P67806A03136*
(Total for Question 9 = 15 marks)
Mark scheme
Show the mark scheme
Question Answer Additional Guidance Mark
Number
9(a) An explanation that makes reference to the following Any reference to equilibrium constant changing (2)
points: scores (0) overall
• there are fewer moles / molecules / particles of (gas) (1) Allow 4 moles / molecules of gas on the left and 2
on the right moles / molecules on right
• so (equilibrium) yield of ammonia increases (1) Allow ‘equilibrium shifts to the right’
M2 is conditional on M1 or the idea of fewer
particles on the right / increasing the value of the
quotient / Q
Allow reverse argument
Number
9(b) Example of calculation (4)
• rearrangement of formula (1) K = K x (RT)∆n
c p
(1) K = 3.55 x 10−2 x (0.0821 x 500)2
• substitution of correct values c
• calculation of Kc (1) Kc = 59.821
TE on ∆n
• units (1) Stand alone mark
dm6 mol−2 or mol−2 dm6
Correct value with units and no working scores (4)
Ignore SF except 1 SF
M1 and M2 can be in reverse order
Number
9(c) Example of calculation (5)
N2 H2 NH3
Initial mol 1.0 3.0 -
• calculation of eqm moles (1) Eqm mol 1.0 − 0.15 3.0 – (3 x 0.15) 0.30
=0.85 = 2.55
Total mol at 0.85 + 2.55 + 0.30 = 3.7
eqm
Partial 0.85 x P 2.55 x P 0.30 x P
• expressions for 3 partial pressures (1) pressure 3.7 3.7 3.7
• substitution of values into K (1) K = 7.76 x 10−5 = 0.30 x P 2
p p
expression 3.7
0.85 x P 2.55 x P 3
• rearrangement of Kp expression (1) 3.7 3.7
7.76 x 10−5 = 0.087419
• calculation of total pressure (1) P2
and P2 = 1126.5 (atm2)
answer to 1 / 2 SF
P = 33.564
= 34 / 30 (atm)
Allow any symbol for total pressure
Allow TE throughout
Correct answer to 1 or 2 SF with some working scores (5)
Correct answer to 1 or 2 SF with no working scores (4)
Number
9(d) Example of calculation (4)
ln K2 = −92400 1 − 1
• substitution of numbers into expression (1) 6.76 x 105 8.31 298 310
(1) ln K = −11119.1 x 1.299 x 10−4
• evaluation of ∆H/R and 1/T1 – 1/T2 2
6.76 x 105
= −1.4444
(1) K = 6.76 x 105 x e−1.4444
• rearrangement of expression 2
TE on M2
(1) K = 1.59467 x 105 / 159467(atm−2)
• evaluation of expression 2
TE on M3
Allow answer from earlier correct rounding to 2 or
more SF
Ignore SF except 1 SF
Correct answer with no / some working scores (4)
(Total for Question 9 = 15 marks)
How to answer it
Equilibria & The Haber Process Study Guide
What this question tests
This question assesses advanced physical chemistry concepts involving gaseous equilibria: Le Chatelier's Principle regarding pressure changes, interconverting equilibrium constants ( Kp and Kc ), heterogeneous/homogeneous partial pressure calculations involving mole fractions, and the quantitative temperature dependence of equilibrium constants via the integrated van't Hoff equation.
Effect of Pressure on Equilibrium Yield
✅ Correct Answer
An explanation stating that there are fewer moles/molecules of gas on the right-hand side (2 moles) compared to the left-hand side (4 moles), therefore increasing the pressure shifts the equilibrium to the right, increasing the yield of ammonia.
💡 Key Knowledge
Le Chatelier's principle states that if a system at equilibrium is disturbed, the system tends to shift in a direction that opposes the change. Increasing total pressure causes the equilibrium to shift to the side with fewer gas molecules to reduce pressure.
❌ Common Errors
Students often lose the second mark (M2) if they fail to link the mole ratio (M1) directly to the shift in equilibrium or the yield of ammonia. Vague statements like "pressure increases yield" without referencing particle numbers score zero.
Relating Kp and Kc
📐 Step-by-Step Calculation
- Find Δn: Moles of gaseous products minus moles of gaseous reactants.
Δn = 2 − (1 + 3) = −2. - Rearrange formula:
Kp = Kc / (RT)Δn → Kc = Kp × (RT)Δn - Substitute values:
Kc = (3.55 × 10⁻²) × (0.0821 × 500)⁻² - Evaluate Kc:
Kc = 59.821 (units: dm⁶ mol⁻² or mol⁻² dm⁶)
🧠 Exam Technique & Units
Make sure to calculate Δn carefully. Because Δn is negative (−2), bringing (RT)Δn over via multiplication flips the exponent sign or keeps it in the numerator correctly. Units for Kc must match your working; standalone mark awarded for dm⁶ mol⁻² .
Calculating Total Pressure at Equilibrium
📐 Step-by-Step Calculation
- ICE Table / Equilibrium Moles:
N₂: initial 1.0, eqm = 1.0 − 0.15 = 0.85 mol
H₂: initial 3.0, eqm = 3.0 − (3 × 0.15) = 2.55 mol
NH₃: eqm given as 0.30 mol (so change is +0.30, meaning x = 0.15)
Total moles at eqm = 0.85 + 2.55 + 0.30 = 3.7 mol - Mole Fractions & Partial Pressures:
p(N₂) = (0.85 / 3.7) × P
p(H₂) = (2.55 / 3.7) × P
p(NH₃) = (0.30 / 3.7) × P - Substitute into Kp expression & Solve:
Kp = p(NH₃)² / (p(N₂) × p(H₂)³)
7.76 × 10⁻⁵ = [ (0.30/3.7)P ]² / { [(0.85/3.7)P] × [(2.55/3.7)P]³ }
Simplifies to: P² = 1125.5 atm² → P = 33.564 atm = 34 atm (to 2 SF)
❌ Common Calculation Traps
Significant Figures: The question requests an "appropriate number of significant figures" (match data inputs: 2 SF is accepted based on mark scheme guidelines).
Stoichiometry trap: Remember that 3 moles of H₂ react for every 1 mole of N₂ consumed. Do not subtract 0.30 directly from H₂ without scaling by the 3:1 ratio!
Temperature Dependence of Kp
📐 Step-by-Step Calculation
- Identify parameters:
K1 = 6.76 × 10⁵ atm⁻², T₁ = 298 K, T₂ = 310 K
ΔH = −92400 J mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹ - Substitute into van't Hoff equation:
ln(K₂ / 6.76 × 10⁵) = (−92400 / 8.31) × [ (1 / 298) − (1 / 310) ] - Evaluate right-hand side:
(−11119.1) × (0.0033557 − 0.0032258) = −11119.1 × (1.299 × 10⁻⁴) = −1.4444 - Rearrange and solve for K₂:
K₂ / 6.76 × 10⁵ = e⁻¹·⁴⁴⁴⁴
K₂ = 6.76 × 10⁵ × 0.23588 = 1.59 × 10⁵ atm⁻² (or 159467 atm⁻²)
🧠 Top-Level Exam Strategy
Watch your negative signs carefully! Since ΔH is exothermic (−92400 J mol⁻¹), increasing the temperature from 298 K to 310 K must result in a decrease in the equilibrium constant Kp. Checking this logical trend helps catch arithmetic errors before final submission.
Topics
Physical Chemistry · Topic 10: Equilibrium I · Topic 11: Equilibrium II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.