AQA A-Level Chemistry Paper 3, 2017: Question 1

14 marks · Medium difficulty · Practical Techniques & Data Analysis

Determine the enthalpy change for the hydration of magnesium chloride using Hess's law, describe a calorimetry experiment to find its enthalpy of solution, and calculate entropy change from a Gibbs free-energy graph.

Practise this question

Question

Question 01 consists of four parts based on magnesium chloride. Part 01.1 asks why the enthalpy change for the hydration of MgCl2 to form MgCl2.4H2O cannot be directly measured by calorimetry. Part 01.2 provides a table of enthalpies of solution for MgCl2 (-155 kJ/mol) and MgCl2.4H2O (-39 kJ/mol) to calculate the hydration enthalpy. Part 01.3 is an extended response question asking to describe a calorimetry experiment using 0.8 g of anhydrous MgCl2 to determine the enthalpy of solution and explain how results are calculated. Part 01.4 gives a table of temperature and Gibbs free-energy values for the formation of MgCl2 and provides a grid to plot ΔG against T, determine the gradient, and calculate ΔS in J K^-1 mol^-1.
Question text

01 Anhydrous magnesium chloride, MgCl2, can absorb water to form the hydrated salt

MgCl2.4H2O

MgCl2(s) + 4H2O(l) → MgCl2.4H2O(s)

01.1 Suggest one reason why the enthalpy change for this reaction cannot be determined

directly by calorimetry.

[1 mark]

01.2 Some enthalpies of solution are shown in Table 1.

Table 1

Enthalpy of solution

Salt −1

/ kJ mol

MgCl2(s) −155

MgCl2.4H2O(s) −39

Calculate the enthalpy change for the absorption of water by MgCl2(s) to form

MgCl2.4H2O(s).

[2 marks]

Enthalpy change kJ mol−1

01.3 Describe how you would carry out an experiment to determine the enthalpy of solution

*02* of anhydrous magnesium chloride.

You should use about 0.8 g of anhydrous magnesium chloride.

Explain how your results could be used to calculate the enthalpy of solution.

[6 marks]

01.4 Anhydrous magnesium chloride can be formed by direct reaction between its

elements.

Mg(s) + Cl2(g) → MgCl2(s)

The free-energy change, ∆G, for this reaction varies with temperature as shown in

Table 2.

Table 2

T / K ∆G / kJ mol−1

298 −592.5

288 −594.2

273 −596.7

260 −598.8

240 −602.2

Use these data to plot a graph of free-energy change against temperature on the grid

opposite.

Calculate the gradient of the line on your graph and hence calculate the entropy

change, ΔS, in J K−1 mol−1, for the formation of anhydrous magnesium chloride from its

elements.

Show your working.

[5 marks]

∆S J K–1 mol–1

Mark scheme

Show the mark scheme Mark scheme for Question 01 detailing 14 marks. 01.1 accepts that it is not possible to prevent some dissolving. 01.2 awards 2 marks for using a Hess cycle (-155 - (-39) = -116 kJ/mol). 01.3 is a 6-mark level of response covering Stage 1 (Method: measuring volume of water, insulated cup, known mass), Stage 2 (Measurements: recording initial T, T at timed intervals, extrapolation on T vs time plot), and Stage 3 (Calculations: q = mcΔT, amount = mass/Mr, ΔH = -q/moles). 01.4 awards 2 marks for plotting 5 points and a straight best-fit line, 1 mark for calculating the gradient (0.167 kJ K^-1 mol^-1), and 2 marks for determining ΔS = -gradient x 1000 = -167 J K^-1 mol^-1 (allow -163 to -171).

Question Answers Mark Additional Comments/Guidance

ALLOW It is soluble / dissolves / other hydrates may form /

01.1 Not possible to prevent some dissolving 1 suggestions related to difficulty of measuring T (change) of a

solid

(∆hydH =) –155 – (–39) 1 OR labelled cycle

Minimum needed for ‘labelled cycle’

ΔH ΔH

155 39 or 155 (+)39

01.2

–116 (kJ mol–1) 1 – – –

1/2 for (+)116 or for 29 or for seeing –116 that has then be

processed further

01.3 This question is marked using levels of response. Refer to the Mark Indicative Chemistry content

Scheme Instructions for examiners for guidance on how to mark

this question Stage 1 Method

Level 3 All stages are covered and the explanation of each

stage is correct and virtually complete. (1a) Measures water with named appropriate apparatus

5-6 marks Stage 2 must include use of a graphical method for (1b) Suitable volume/mass / volume/mass in range 10 –

200 cm3/g

Level 3 (i.e. ‘highest T reached’ method is max

Level 2) (1c) Into insulated container / polystyrene cup (NOT just

‘lid’)

Answer communicates the whole explanation, (1d) Add known mass of MgCl2(s)

including reference to enthalpy, coherently and (1e) Use of ‘before and after’ weighing method. NOT

shows a logical progression through all three ‘added with washings’

stages.

For the answer to be coherent there must be some Stage 2 Measurements (could mark from diagram)

indication of how the graph is used to find ∆T

Level 2 All stages are covered (NB ‘covered’ means min 2 (2a) Record i10ofnitial temperature (19min 2 measurements)

from each of stage 1 and 3) but the explanation of (2b) Record T at regular timed intervals for 5+ mins / until

3-4 marks each stage may be incomplete or may contain trend seen

6 (2c) Plot T vs time

inaccuracies

OR two stages covered and the explanations are

generally correct and virtually complete Stage 3 Use of Results (3a and 3b could come from

diagram)

Answer is coherent and shows some progression

through all three stages. Some steps in each stage (3a) Extrapolate lines to when solid added (to find initial

may be out of order and incomplete and final T)

Level 1 Two stages are covered but the explanation of (3b) Tfinal – Tinitial = ∆T / idea of finding ∆T from graph at

each stage may be incomplete or may contain point of addition

1-2 marks inaccuracies (3c) q = mc∆T

(3d) amount = mass/M (0.80/95.3 = 8.39 x 10–3 mol)

OR only one stage is covered but the explanation is r

(3e) ∆H = q/8.39 x 10–3 or in words

generally correct and virtually complete soln

Answer shows some progression between two This could all be described in words without showing

stages actual calculations but describing stages

Level 0

Insufficient correct Chemistry to warrant a mark If method based on ‘combustion’ Max Level 1–– –

0 marks

11 of 19

240 250 260 T / K270 280 290 300

-592

-593

-594 M1 = 5 points correctly plotted

-595 M2 = line drawn correctly (NOT if curved, doubled or

-596 kinked)

ΔG / kJ -597

mol-1 -598 2

(Check line of best fit –

-599 if through 250, -600.5 and 280, -595.5 +/- one small

-600

square then award M2, if all crosses on line award M1 as

-601

01.4 -602 well)

-603

Gradient = ∆(∆G)/∆T = 0.167 (kJ K–1 mol–1) 1

(∆G = ∆H – T∆S so gradient = –∆S)

M4 = unit conversion i.e. M3 x1000; M5 = sign (process

–1 –1 marks)

∆S = –167 (J K mol ) 1+1

Correct answer with sign gets M3, M4 and M5

ALLOW 163 to 171

Total 14

How to answer it

Thermodynamics & Calorimetry of Magnesium Chloride

📌 What this question tests

Core A-Level Physical Chemistry skills across Enthalpy, Calorimetry, and Gibbs Free Energy:

  • Hess's Law application: Explaining why some hydration enthalpies cannot be measured directly and calculating indirect enthalpy changes via enthalpy of solution cycles.
  • Required Practical 2 (Calorimetry): Detailed 6-mark experimental method for measuring solution enthalpy, recording cooling curves, extrapolating temperature changes (ΔT), and processing q = mcΔT .
  • Thermodynamics & Graphical Analysis: Plotting ΔG vs T , interpreting the linear form ΔG = -ΔS(T) + ΔH , determining the gradient, and converting units to calculate ΔS in J K⁻¹ mol⁻¹.
Question 01.1 (1 Mark)

Direct Calorimetry Limitations

Suggest why the enthalpy change for MgCl₂(s) + 4H₂O(l) → MgCl₂.4H₂O(s) cannot be determined directly.

✅ Acceptable Answers (1 Mark)

  • It is not possible to prevent some of the magnesium chloride from dissolving.
  • Magnesium chloride is soluble / it dissolves in water.
  • Other hydrates (e.g. MgCl₂.6H₂O or MgCl₂.2H₂O) may also form.
  • Difficulty of measuring the exact temperature change of a solid reacting with a small amount of liquid.

❌ Common Errors & Pitfalls

  • Vague statements: Saying simply "it is too dangerous" or "activation energy is too high" without addressing the physical reaction mixture.
  • Missing dissolution: Forgetting that adding water to an anhydrous salt easily causes dissolution beyond just crystal hydration.
Mark Scheme Note: 1 mark for any chemically valid physical reason why the solid hydrate cannot be formed cleanly without dissolving or forming a mixture of hydrates.
Question 01.2 (2 Marks)

Hess's Law Cycle: Enthalpy of Hydration

Calculate the enthalpy change for the absorption of water by MgCl₂(s) to form MgCl₂.4H₂O(s).

📐 Step-by-Step Calculation

1 Identify the Target Reaction:
MgCl₂(s) + 4H₂O(l) → MgCl₂.4H₂O(s) [ΔH = ?]

2 Set up the Hess's Law Cycle:
Both the anhydrous salt and the hydrated salt dissolve in excess water to give the exact same aqueous solution: MgCl₂(aq) .

  • Route 1 (Direct dissolution): MgCl₂(s) → MgCl₂(aq), ΔH₁ = -155 kJ mol⁻¹
  • Route 2 (Hydration then dissolution): MgCl₂(s) → MgCl₂.4H₂O(s) [ΔH] followed by MgCl₂.4H₂O(s) → MgCl₂(aq) [ΔH₂ = -39 kJ mol⁻¹]

3 Solve the Expression:
ΔH₁ = ΔH + ΔH₂
ΔH = ΔH₁ - ΔH₂
ΔH = -155 - (-39) = -116 kJ mol⁻¹

✅ Mark Breakdown

  • Mark 1: Correct expression: -155 - (-39) or a correctly constructed & labelled cycle showing correct arrow directions.
  • Mark 2: Final answer of -116 kJ mol⁻¹ .

🧠 Exam Technique: Cycle Diagram

Always draw a triangle! Put both solid salts at the top and MgCl₂(aq) at the bottom. Both arrows point downwards into the aqueous solution. By Hess's law: target = clockwise route = anticlockwise route.

Partial Credit: If you wrote +116 or -29 (from -155 / 4), you score 1/2 marks.
Question 01.3 (6 Marks)

Practical Calorimetry: Enthalpy of Solution

Describe the method to determine the enthalpy of solution of ~0.8 g MgCl₂(s) and explain how results are calculated.

💡 Stage 1: Apparatus & Method

  • Measure Water: Measure a known volume of water (between 10 cm³ and 200 cm³, typically 25 to 50 cm³) using a measuring cylinder or pipette.
  • Insulation: Place water in an insulated container, such as an expanded polystyrene cup with a lid (a beaker alone loses marks).
  • Accurate Weighing: Weigh the weighing boat + ~0.8 g MgCl₂(s), tip into cup, and reweigh the empty boat (weighing by difference). Do not wash out with water, as that alters the measured liquid volume!

💡 Stage 2: Temperature Measurements

  • Initial Readings: Record the temperature of the water every minute for at least 3-4 minutes prior to adding the solid (establishing a stable baseline).
  • Addition: Add the solid at minute 4 (do not measure temperature at minute 4).
  • Subsequent Readings: Stir continuously and record temperature every minute from minute 5 onwards for at least 5-10 minutes until a steady cooling trend is observed.
  • Plot: Plot a graph of Temperature (y-axis) vs Time (x-axis).

💡 Stage 3: Graphical Analysis & Calculation

Finding Accurate ΔT:

  • Extrapolate the baseline temperature line forward to the minute of addition (e.g., minute 4).
  • Extrapolate the cooling curve backward to the minute of addition.
  • Measure vertical difference between both lines at the point of addition to find true ΔT (compensating for heat loss).

Mathematical Processing:

  1. q = m × c × ΔT (where m = mass of water, c = 4.18 J g⁻¹ K⁻¹ )
  2. Amount of MgCl₂ (n) = 0.80 / 95.3 = 8.39 × 10⁻³ mol
  3. ΔH_soln = -(q / n) (expressed in kJ mol⁻¹; note the negative sign because dissolving MgCl₂ is exothermic).

🧠 Level 3 Requirements (5-6 Marks)

  • All 3 stages must be covered logically and virtually completely.
  • Crucial: To access Level 3, you must detail a graphical extrapolation method. Simply stating "record the highest temperature reached" caps your score at Level 2 (max 4 marks)!

❌ Common Errors That Cost Marks

  • Using mass of solid + water for m in mcΔT without specifying that water provides the bulk heat capacity.
  • Omitting the negative sign in the final enthalpy change.
  • Failing to mention weighing by difference.
Question 01.4 (5 Marks)

Gibbs Free Energy Plot: Determining Entropy Change (ΔS)

Mg(s) + Cl₂(g) → MgCl₂(s)

💡 Thermodynamic Relationship

The Gibbs Free Energy equation is: ΔG = ΔH - TΔS

Rearranging in the form of a straight line equation ( y = mx + c ):

ΔG = (-ΔS) × T + ΔH

When ΔG is plotted on the y-axis and T on the x-axis:

  • Gradient (m) = -ΔS
  • y-intercept (c) = ΔH

📐 Step-by-Step Gradient & Entropy Calculation

1 Plot Points & Best Fit Line:
Points: (240, -602.2), (260, -598.8), (273, -596.7), (288, -594.2), (298, -592.5).
Draw a straight line of best fit through all five points using a ruler.

2 Calculate Gradient of the Line:
Using values from the line (e.g. across the extremes):
Gradient = Δy / Δx = [-592.5 - (-602.2)] / [298 - 240]
Gradient = +9.7 / 58 = +0.167 kJ K⁻¹ mol⁻¹

3 Relate Gradient to ΔS:
Because gradient = -ΔS :
-ΔS = +0.167 kJ K⁻¹ mol⁻¹ ⇒ ΔS = -0.167 kJ K⁻¹ mol⁻¹

4 Unit Conversion (kJ to J):
Multiply by 1000 to get J K⁻¹ mol⁻¹ :
ΔS = -0.167 × 1000 = -167 J K⁻¹ mol⁻¹
(Acceptable range: -163 to -171 J K⁻¹ mol⁻¹)

✅ Mark Breakdown (5 Marks)

  • Mark 1: All 5 points plotted accurately to within half a small square.
  • Mark 2: Straight line of best fit drawn cleanly with a ruler (not kinked, doubled, or curved).
  • Mark 3: Gradient correctly evaluated from the plotted line (~0.167 kJ K⁻¹ mol⁻¹).
  • Mark 4: Correct unit conversion: multiplying gradient by 1000.
  • Mark 5: Correct negative sign applied ( ΔS = -gradient ).

❌ Common Pitfalls

  • Sign error: The gradient is positive (+0.167), which means ΔS MUST be negative (-167). A gas reacts to form a solid, so entropy must decrease!
  • Missing ×1000: Giving the final answer as -0.167 without converting to J K⁻¹ mol⁻¹.
  • Small triangle: Using points too close together to measure the gradient. Use at least half the length of your line.

Topics

Physical Chemistry · Required Practicals · 3.1.4 Energetics · 3.1.8 Thermodynamics · Required Practical 2: Measurement of an enthalpy change

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.