OCR A-Level Chemistry AS Breadth in chemistry (01), June 2018: Question 23
10 marks · Medium difficulty · Structured Questions
Calculate the enthalpy change for the reaction between magnesium and silver nitrate, explain how to test if the reaction went to completion using sodium chloride, and use the Boltzmann distribution model to explain the effect of temperature on reaction rate.
Practise this questionQuestion
Question text
23 This question is about energy changes and rate of reaction.
(a) Magnesium reacts with aqueous silver nitrate, AgNO3(aq), as in equation 23.1.
Mg(s) + 2AgNO3(aq) → 2Ag(s) + Mg(NO3)2(aq) Equation 23.1
A student carries out an experiment to determine the enthalpy change of this reaction, ∆rH.
• The student adds 25.0 cm3 of 0.512 mol dm−3 AgNO to a polystyrene cup.
• The student measures the temperature of the solution.
• The student adds a small spatula measure of magnesium powder, stirs the mixture and
records the maximum temperature.
Temperature readings
Initial temperature = 19.5 °C
Maximum temperature = 47.5 °C
(i) Calculate ∆ H, in kJ mol−1, for the reaction shown in equation 23.1.
r
Give your answer to an appropriate number of significant figures.
Assume that the density and specific heat capacity, c, of the solution are the same as for
water and that all the aqueous silver nitrate has reacted.
∆ H = … kJ mol−1 [4]
r
(ii) At the end of the experiment, the student adds a few drops of aqueous sodium chloride
to the reaction mixture in the polystyrene cup to test whether all the aqueous silver nitrate
has reacted.
Explain how the results would show whether all the aqueous silver nitrate has reacted.
Include an equation with state symbols in your answer.
… [2]
(b) Using the Boltzmann distribution model, explain how the rate of a reaction is affected by
temperature.
You are provided with the axes below, which should be labelled.
… [4]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
23 (a) (i) FIRST, CHECK THE ANSWER ON ANSWER LINE 4 FULL ANNOTATIONS MUST BE USED
IF ∆ H = –457 OR –458 (kJ mol–1) award 4 marks
r
IF ∆ H = ±229 OR 457 (kJ mol–1) award 3 marks --------------------------------------
r
----------------------------------- ALLOW ECF throughout
Energy released in J OR kJ ------------------------------------------------------------
= 25.0 × 4.18 × 28.0 = 2926 (J) OR 2.926 (kJ) ALLOW 2930 J OR 2.93 kJ
DO NOT ALLOW < 3 SF
IGNORE any sign and units
Correctly calculates n(AgNO3) i.e. ALLOW correctly calculated number in J OR kJ
25.0 –2
= 0.512 × = 1.28 10 (mol)
1000
-----------------------------------------------------
∆H per mole AgNO3 in kJ AND 3 SF Alternative approach using 1 mol Mg
Answer MUST divide energy by n(AgNO3)
2.926 Energy released = 2926 (J) OR 2.926 (kJ)
± –2 = ±228.59375
1.28 10
n(AgNO ) = 1.28 10–2 (mol)
= ±229 (kJ) 3
3 SF needed Sign NOT needed
1.28 10–2
n(Mg) = = 6.4 10–3 (mol)
∆H for 2 mol AgNO3 AND – sign AND 3 SF 2
∆H = 2 –228.59375 = –457 (kJ mol–1) 2.926
r ∆H = = –457 (kJ mol–1)
r 6.4 10–3
OR 2 –229 = –458 (kJ mol–1) – sign AND 3 SF needed
(a) (ii) 2
Ag+(aq) + Cl–(aq) AgCl(s) ALLOW AgNO (aq) + NaCl(aq) AgCl(s) + NaNO (aq)
State symbols required
White precipitate AND AgNO /Ag+ NOT ALL reacted
OR Observation needs to be linked to conclusion
NO white precipitate AND AgNO /Ag+ ALL reacted
(b) Boltzmann distribution 3 marks 4 FULL ANNOTATIONS MUST BE USED
11 THROUGHOUT
----------------------------------------------------------------
NOTE: Look for marking criteria within annotations on
Boltzmann distribution diagram
Curve
Curve starts within one small square of origin IGNORE slight inflexion on the curve
AND curve does not touch x axis at high energy
AND curve does not increase by more than one For labels,
small square at higher energy ALLOW number of particles
Labels ALLOW amount of molecules/particles
Axes labels correct: IGNORE number of atoms
Number of molecules AND Energy ALLOW kinetic energy
IGNORE enthalpy for energy
Curves for two temperatures
Drawing of two curves with higher and lower IGNORE curves meeting at higher energy BUT
temperature clearly identified in diagram or text DO NOT ALLOW crossing over by more than one small
AND higher T maximum to right AND at least one small square
square lower than lower T max
ALLOW more molecules have the energy to react
Explanation 1 mark IGNORE more successful collisions
More molecules have energy greater than Ea OR collide more frequently
OR
Greater area under curve above Ea DO NOT ALLOW explanation is in terms of two
Could be in diagram activation energies (i.e. ‘catalyst explanation)
Total 10
How to answer it
Energy Changes and Rate of Reaction Study Guide
What this question tests
This question assesses core physical chemistry concepts split into two main areas: Enthalpy Change Calculations from calorimetric data (using q = mcΔT, finding moles, and determining enthalpy change per mole) and Reaction Kinetics using the Boltzmann distribution model to explain the effect of temperature on rate.
Enthalpy Change Calculation
Calculate ΔrH, in kJ mol⁻¹, for the reaction shown in equation 23.1
📐 Step-by-Step Calculation
- Calculate temperature change (ΔT): 47.5 - 19.5 = 28.0 °C
- Calculate energy released (q): q = m × c × ΔT = 25.0 × 4.18 × 28.0 = 2926 J (or 2.926 kJ)
- Calculate moles of limiting reagent (AgNO₃): n(AgNO₃) = (25.0 / 1000) × 0.512 = 1.28 × 10⁻² mol
- Calculate enthalpy change per mole (ΔH): ΔH = ±q / n = 2.926 / (1.28 × 10⁻² mol) = ±228.59 kJ mol⁻¹.
Alternatively, per 2 moles of AgNO₃ (matching equation stoichiometry): 2 × (-228.59) = -457 kJ mol⁻¹.
✅ Correct Answer & Mark Scheme
Final Answer: -457 kJ mol⁻¹ (or -458 kJ mol⁻¹ depending on rounding steps).
- Mark 1: Correct energy calculation (2926 J or 2.926 kJ)
- Mark 2: Correct moles of AgNO₃ (1.28 × 10⁻² mol)
- Mark 3: Division of energy by moles (gives ±229 or ±228.6 kJ mol⁻¹)
- Mark 4: Correct value for 2 moles of AgNO₃ with negative sign and 3 significant figures ( -457 kJ mol⁻¹ )
❌ Common Errors & Calculation Traps
- Sign omission: Enthalpy changes for exothermic reactions must include a negative sign. Losing the minus sign in final answers drops marks at top levels.
- Significant figures: The question asks for an "appropriate number of significant figures". Less than 3 SF is penalised. Stick to 3 SF as given by the data values (19.5, 25.0, 0.512).
- Stoichiometry confusion: Students often forget to scale the enthalpy value to match the stoichiometric coefficient of 2 for AgNO₃ in the equation (-457 kJ mol⁻¹ vs -229 kJ mol⁻¹). Both approaches are credited if correctly executed.
🧠 Exam Technique
Always write down individual steps clearly (q first, then moles, then division). This allows examiners to award error carried forward (ECF) marks even if an arithmetic error occurs early on.
Testing for Remaining Reactants
Explain how results show whether all aqueous silver nitrate has reacted
✅ Correct Answer & Equation
Ionic Equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Observation: A white precipitate forms if AgNO₃ is in excess (not all reacted). No white precipitate forms if all AgNO₃ has reacted.
- Mark 1: Correct equation with valid state symbols. (Allow molecular equation with NaCl and AgNO₃).
- Mark 2: Valid observation linked directly to the conclusion (e.g., white precipitate means Ag⁺ is still present).
💡 Key Knowledge
Silver ions ( Ag⁺ ) react with halide ions like chloride ( Cl⁻ ) to produce insoluble silver chloride, which visibly crashes out of solution as a characteristic white precipitate.
❌ Common Errors
Students frequently describe a colour change instead of a precipitate formation, or they state "a precipitate forms" without specifying the colour (white). State symbols must be included in the equation to secure full credit.
Boltzmann Distribution and Temperature
Explain how the rate of a reaction is affected by temperature using the Boltzmann distribution model
✅ Diagram Requirements & Explanation
- Axes: Y-axis labeled Number of molecules (or amount), X-axis labeled Energy (or kinetic energy).
- Curve T₁ (Lower Temp): Starts at origin, does not touch x-axis at high energy.
- Curve T₂ (Higher Temp): Peak shifts lower and to the right. The new peak must be lower in height and shifted right compared to T₁. Curves can cross once.
- Explanation point: At higher temperature, a greater proportion of molecules have energy greater than or equal to the activation energy ( E ≥ Ea ), leading to a higher frequency of successful collisions.
- Mark 1: Properly shaped Boltzmann curve starting at origin.
- Mark 2: Correct axis labels (Number of molecules & Energy).
- Mark 3: Second temperature curve correctly drawn (lower peak, shifted to the right).
- Mark 4: Explanation citing more molecules with energy ≥ Ea / greater area under curve past Ea.
🧠 Exam Technique & Top Tips
When sketching curves for two temperatures, make sure the high-temperature peak is clearly displaced to the right and lower down. Never let the tail of the curve touch or cross the x-axis asymptotically—examiners are strict on this detail!
❌ Common Errors
Do not explain temperature effects by talking about activation energy changing—activation energy remains constant with temperature! The change is entirely in the kinetic energy distribution of the molecules.
Topics
Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · Practical Activity Groups · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · PAG 3: Enthalpy determination
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.