Edexcel A-Level Chemistry Paper 2, June 2025: Question 5
14 marks · Medium difficulty · Synoptic Questions
Calculate reacting volumes and gas amounts, explain reaction rates using collision theory and Maxwell–Boltzmann distributions, and determine an initial rate from experimental data.
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How to answer it
Kinetics & Ideal Gas Analysis of Metal Carbonate Reactions
WHAT THIS QUESTION TESTS
This question examines core practical and theoretical physical chemistry concepts across 14 marks:
- Stoichiometry & Titration Math: Finding reactant volumes using reacting ratios ( 1 : 2 ) and significant figures.
- Collision Theory: Effect of surface area on reaction rate.
- The Ideal Gas Equation ( pV = nRT ): Unit conversions ( kPa → Pa , °C → K , m³ → cm³ ) and evaluating experimental limitations.
- Maxwell–Boltzmann Distribution: Drawing curves at elevated temperatures and explaining kinetic energy shifts with activation energy ( Eₐ ).
- Graphical Rates Analysis: Drawing tangents at t = 0 to calculate initial rates with correct units.
PART (a) • 4 MARKS
Reacting Volume of Hydrochloric Acid
Calculate the minimum volume (cm³) of 0.100 mol dm⁻³ HCl needed to react with 0.355 g of CaCO₃
📐 Step-by-Step Calculation
Step 1: Calculate moles of CaCO₃
Moles = mass / Mᵣ = 0.355 / 100.1
Moles of CaCO₃ = 3.5465 × 10⁻³ mol
Moles = mass / Mᵣ = 0.355 / 100.1
Moles of CaCO₃ = 3.5465 × 10⁻³ mol
Step 2: Use stoichiometry (1 : 2 ratio)
From equation: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Moles of HCl = 3.5465 × 10⁻³ × 2
Moles of HCl = 7.0929 × 10⁻³ mol
From equation: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Moles of HCl = 3.5465 × 10⁻³ × 2
Moles of HCl = 7.0929 × 10⁻³ mol
Step 3: Calculate volume in dm³
Volume = moles / concentration = 7.0929 × 10⁻³ / 0.100
Volume in dm³ = 0.070929 dm³
Volume = moles / concentration = 7.0929 × 10⁻³ / 0.100
Volume in dm³ = 0.070929 dm³
Step 4: Convert to cm³ & apply significant figures
Volume in cm³ = 0.070929 × 1000 = 70.929 cm³
To 3 SF: 70.9 cm³ (or 71 cm³ to 2 SF)
Volume in cm³ = 0.070929 × 1000 = 70.929 cm³
To 3 SF: 70.9 cm³ (or 71 cm³ to 2 SF)
✅ Mark Scheme Breakdown
- Mark 1: Moles of CaCO₃ = 3.55 × 10⁻³ mol (allow use of Mᵣ = 100).
- Mark 2: Moles of HCl = 7.09 × 10⁻³ mol (multiplying by 2).
- Mark 3: Volume in dm³ = 0.0709 dm³ .
- Mark 4: Final conversion to cm³ = 70.9 cm³ (3 SF) or 71 cm³ (2 SF).
⚠️ Significant Figures Trap: 0.355 g is 3 SF, 0.100 mol dm⁻³ is 3 SF. Do not write 71.0 if using 100.1 as Mᵣ.
❌ Common Student Errors
- Missing the 1:2 ratio: Forgetting that 1 mole of carbonate requires 2 moles of acid, leading directly to half the required volume (35.5 cm³).
- Leaving volume in dm³: The question specifically asks for the answer in cm³. Always re-read the units requested in bold.
- Premature rounding: Rounding intermediate values to 1 or 2 SF too early, introducing rounding errors in the final answer.
🧠 Exam Technique: Checking Plausibility
Always double-check your answer against the apparatus shown. The syringe holds 100 cm³. An acid volume of ~71 cm³ fits comfortably within typical conical flasks used in these kinetic setups.
PART (b) • 1 MARK
Effect of Particle Size on Reaction Rate
State why powdered metal carbonate reacts faster than large lumps.
✅ Correct Answer
Powdered metal carbonate has a greater surface area (or larger surface area to volume ratio).
💡 Collision Theory Context
Increasing surface area exposes more reactant particles to the acid solution, resulting in a higher frequency of successful collisions per second. (Note: Only stating "greater surface area" is required for this 1 mark).
PART (c)(i) • 4 MARKS
Ideal Gas Calculation & Experimental Justification
Calculate the volume of CO₂ gas at 74.0 °C and 100 kPa from 4.19 × 10⁻³ mol of MgCO₃, and explain why the apparatus would not give a measurable result.
📐 Step-by-Step Calculation
Step 1: Unit conversions
Pressure ( p ) = 100 kPa = 1 × 10⁵ Pa
Temperature ( T ) = 74.0 + 273 = 347.0 K
Gas constant ( R ) = 8.31 J K⁻¹ mol⁻¹
Moles of CO₂ ( n ) = 4.19 × 10⁻³ mol (1:1 ratio with MgCO₃)
Pressure ( p ) = 100 kPa = 1 × 10⁵ Pa
Temperature ( T ) = 74.0 + 273 = 347.0 K
Gas constant ( R ) = 8.31 J K⁻¹ mol⁻¹
Moles of CO₂ ( n ) = 4.19 × 10⁻³ mol (1:1 ratio with MgCO₃)
Step 2: Rearrange ideal gas equation
pV = nRT ➔ V = (nRT) / p
pV = nRT ➔ V = (nRT) / p
Step 3: Calculate volume in m³
V = (4.19 × 10⁻³ × 8.31 × 347.0) / (1 × 10⁵)
V = 1.2082 × 10⁻⁴ m³
V = (4.19 × 10⁻³ × 8.31 × 347.0) / (1 × 10⁵)
V = 1.2082 × 10⁻⁴ m³
Step 4: Convert volume to cm³ and state conclusion
V = 1.2082 × 10⁻⁴ × 10⁶ = 120.8 cm³ ≈ 121 cm³
Conclusion: The gas syringe has a maximum capacity of 100 cm³. Since 121 cm³ > 100 cm³, the volume produced exceeds the syringe capacity (plunger would blow out/not measure total volume).
V = 1.2082 × 10⁻⁴ × 10⁶ = 120.8 cm³ ≈ 121 cm³
Conclusion: The gas syringe has a maximum capacity of 100 cm³. Since 121 cm³ > 100 cm³, the volume produced exceeds the syringe capacity (plunger would blow out/not measure total volume).
✅ Mark Breakdown
- Mark 1: Both conversions correct ( p = 100 000 Pa AND T = 347 K ).
- Mark 2: Correct rearrangement: V = nRT / p .
- Mark 3: Correct calculated value of volume ( 1.21 × 10⁻⁴ m³ or 121 cm³ ).
- Mark 4: Correct units for volume and explicit deduction that the volume exceeds the 100 cm³ gas syringe capacity.
❌ Common Traps in Ideal Gas Calculations
- Forgetting that ideal gas V is in m³: Many students multiply by 1000 thinking it converts m³ to cm³ (it converts m³ to dm³!). To go from m³ → cm³ , multiply by 10⁶ .
- Missing the comparison: Calculating 121 cm³ without explaining why this causes the apparatus to fail. You must explicitly link back to the diagram which labels a 100 cm³ gas syringe.
- Using Celsius directly: Forgetting to add 273 to get Kelvin.
PART (c)(ii) • 3 MARKS
Maxwell–Boltzmann Distribution at Higher Temperature
Explain, by adding another line and annotating the graph, how heating results in a faster reaction rate.
💡 Visual Representation: What to Draw
On the axis of Number of particles with energy E against Energy, E:
- Draw second curve: Start at the origin (0,0) . Its peak must be shifted to the right and be lower than the original curve's peak.
- Right-hand tail: Crosses the original curve once, then stays above the original curve at higher energies, approaching but never touching the x-axis.
- Vertical Line (Eₐ): Draw a vertical dashed or solid line labeled Eₐ to the right of both peaks.
- Shaded Area: Shade the region under the higher-temperature curve to the right of Eₐ.
✅ Mark Scheme Requirements
- Mark 1 (Curve): Curve peak shifted to the right and lower than the original peak. Must start at origin, cross original curve only once, and remain above original at high energy without cutting the x-axis.
- Mark 2 (Activation Energy): Vertical activation energy line ( Eₐ ) positioned to the right of both curve peaks.
- Mark 3 (Annotation / Explanation): Clear label or statement indicating that a greater proportion of particles / molecules have energy greater than or equal to the activation energy (E ≥ Eₐ).
❌ Severe Examiner Red Flags
- Saying activation energy changes: Eₐ is a fixed barrier for a reaction. Heating does NOT change or lower the activation energy! Any mention of Eₐ decreasing loses Mark 3 automatically.
- Sloppy curves: Crossing the original curve twice, touching the x-axis at high energy, or curving upwards at the end forfeits Mark 1.
- Vague collision claims: Stating only "more collisions happen" gets 0 marks. You must specify that a greater proportion/fraction of collisions have E ≥ Eₐ.
🧠 Key Phrase to Memorise
"At higher temperatures, a significantly greater proportion of molecules have kinetic energy equal to or exceeding the activation energy (E ≥ Eₐ), leading to more frequent successful collisions per unit time."
PART (d) • 2 MARKS
Determining Initial Rate from a Kinetic Curve
Determine the initial rate of this reaction from the graph of (Vfinal − Vt) against Time.
📐 Step-by-Step Gradient Calculation
Step 1: Draw a straight tangent at t = 0
Place a ruler against the curve at t = 0, y = 91 cm³ . Draw a long tangent line extending across the grid.
Place a ruler against the curve at t = 0, y = 91 cm³ . Draw a long tangent line extending across the grid.
Step 2: Read intercept coordinates
Tangent line starts at (0, 91) and hits the time axis around t = 75 s at y = 0 .
Tangent line starts at (0, 91) and hits the time axis around t = 75 s at y = 0 .
Step 3: Calculate gradient (rate)
Rate = |Δy / Δx| = 91 cm³ / 75 s = 1.21 cm³ s⁻¹
Allowed range: 1.1 to 1.3 cm³ s⁻¹
Rate = |Δy / Δx| = 91 cm³ / 75 s = 1.21 cm³ s⁻¹
Allowed range: 1.1 to 1.3 cm³ s⁻¹
✅ Mark Scheme Breakdown
- Mark 1: Tangent drawn accurately at t = 0 with working visible on the graph.
- Mark 2: Value in the range 1.1 – 1.3 with correct units: cm³ s⁻¹ (ignore sign; negative or positive accepted).
Note: Examiners allow transfer of error (TE) if you drew a tangent at a point other than t = 0, provided your calculation and units are correct.
❌ Common Errors on Tangent Questions
- Not drawing a tangent: Simply taking the first data point (30 s, 55 cm³) and calculating (91 - 55) / 30 gives a chord, not a tangent. You will lose the construction mark.
- Wrong or missing units: Omitting cm³ s⁻¹ or writing mol dm⁻³ s⁻¹ out of habit. Always check the axis labels: y-axis is cm³ , x-axis is s .
- Drawing too short a tangent: Short tangent lines lead to massive reading errors. Draw your tangent line so it intersects both axes.
Topics
Physical Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 9: Kinetics I · Topic 16: Kinetics II
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.