Edexcel A-Level Chemistry Paper 3, June 2022: Question 4

13 marks · Hard difficulty · Calculations

Calculate entropy changes and feasibility temperatures for barium carbonate decomposition, compare thermal stabilities of group 2 carbonates, and evaluate a test for carbonate ions using sulfuric acid.

Practise this question

Question

Question text

4 This question is about the white solid barium carbonate.

(a) Barium carbonate decomposes under suitable conditions to form barium oxide

and carbon dioxide.

d –1

BaCO3(s) → BaO(s) + CO2(g) ΔrH = +169.3 kJmol

Standard molar entropy data related to this reaction are shown.

Substance Standard molar entropy, S

/ J K–1 mol–1

BaCO3(s) 112.1

BaO(s) 70.4

CO2(g) 213.6

(i) Show that barium carbonate is thermally stable at 298K, using the data in the

equation and in the table.

(5)

(ii) Calculate the lowest temperature, in °C, at which it is thermodynamically

feasible for barium carbonate to decompose.

Give your answer to three significant figures.

(3)

(b) Explain whether magnesium carbonate is more or less thermally stable than

barium carbonate.

(3)

… 10

… *P67095A01036*

presence of the carbonate ion could be tested for by adding a small amount of

sulfuric acid.

Explain whether or not this suggestion is valid.

(2)

(Total for Question 4 = 13 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

4(a)(i) Marks should be awarded for method 1 or method 2 but not via (5)

mixed methods. If both methods used, then award higher mark.

Example of calculation

• calculation of ΔS (1) (213.6 + 70.4) – 112.1 = 171.9 (J K–1 mol–1)

system

−169.3 / 298 = −0.56812 (kJ K–1 mol–1) or

• calculation of ΔSsurroundings (1)

( −169.3 x 1000) / 298 = −568.12 (J K–1 mol–1)

ΔS converted to J K–1 mol–1

• conversion of ΔSsystem or ΔS surroundings for surroundings

consistent units (1) or

ΔS converted to kJ K–1mol–1

system

M3 could be subsumed as part of either M1 or M2

171.9 + (− 568.12 ) = − 396.22 J K–1 mol–1

• calculation of ΔStotal

and or

0.1719 + (−0.56812) = −0.39622 kJ K–1 mol–1

corresponding units (1)

Allow units to be missing here if correct units given for

ΔSsystem and ΔSsurroundings

Correct answer with units with some or no working scores (4)

Ignore SF except 1 SF

Allow TE throughout calculation

• comment on thermal stability at 298 K (1) Stand alone mark on any negative value for ΔStotal

Negative value / <0 and so reaction is not feasible / it is

thermodynamically stable (at 298 K)/

Ignore just ‘so the reaction is not feasible’

Alternative method on next page No TE for positive values for ΔStotal

4(a)(i) Alternative method using ΔG Example of calculation (5)

continued

• calculation of ΔS (1) (213.6 + 70.4) – 112.1 = 171.9 (J K–1 mol–1)

system

298 x 171.9 = 51226 (J mol–1)

• calculation of TΔSsystem (1)

M2 could be subsumed as part of M3

ΔH converted to J mol–1

• conversion of TΔSsystem or ΔSsystem or ΔH for

consistent units (1) or

ΔS converted to kJ K–1 mol–1

system

TΔS converted to kJ mol–1

system

M3 could be subsumed as part of M4

169300 − 51226 = (+) 118074 J mol–1

• calculation of ΔGtotal –1

and or (+) 118.074 kJ mol

corresponding units (1) Correct answer with units with some or no working scores (4)

Ignore SF except 1 SF

Allow TE from M1 to M4

• comment on thermal stability at 298 K (1) Stand alone mark on any positive value for ΔG

Positive value / >0 and so reaction is not feasible (at 298 K)

Ignore just ‘so reaction is not feasible’

No TE on negative values for ΔG

Question

Answer Additional Guidance Mark

Number

4(a)(ii) Example of calculation (3)

• recognition that ΔS ≥ –171.9 J K–1 mol–1 for ∆S = ∆S − ∆H = 0

surroundings total system

decomposition to be feasible (1) T

∆Ssystem = ∆H

Allow this equation rearranged

This may be subsumed in M2

• substitution and rearrangement to find T (1) –171.9 = (−169.3 x 1000) / T

T= (−169.3 x 1000) / –171.9

TE on ΔSsystem from 4(a)(i)

• calculation of T

and (= 984.87 K)

conversion to °C (= 711.87°C)

and = 712°C

answer given to 3 SF (1) TE on M1 and M2 but do not award any temperature

below 0°C

Correct answer to 3 SF and in °C scores (3)

Alternative method for M1 and M2

∆G = ∆H − T∆S = 0 or ∆H = T∆S

This may be subsumed in M2 (1)

169300 = T x 171.9

T = 169300/171.9 (1)

TE on ΔSsystem from 4(a)(i)

Question

Answer Additional Guidance Mark

Number

4(b) An explanation that makes reference to the following points: Allow reverse arguments (3)

Ignore reference to ‘covalent character’

(Magnesium carbonate is less thermally stable because) Ignore reference to lattice energies

Size

• the magnesium ion / Mg2+ is smaller / has a greater charge Allow ionic radius of cation increases down the

density (1) group / charge density of cation decreases down

the group

Allow magnesium carbonate has a smaller cation

Allow magnesium ions have fewer shells of

electrons

Ignore ‘magnesium (atom) is smaller’

Ignore atomic radius

Do not award M1 if mention of different /

incorrect charges on magnesium and barium ions

Polarising power

• so more likely to polarise / distort (the carbonate (ion) / Allow ‘magnesium ion has more polarising

anion) (1) power’

Allow polarising power decreases down the group

Allow magnesium ion has more electron pulling

power on (the carbonate (ion) / anion)

Do not award if MgCO3 stated as more stable

Bonds

• and so weaken the C-O bond or the bond(s) within the Allow break (more easily) for weaken

carbonate ion (1) Allow C=O bonds for C−O

Do not award reference to weakening unspecified

bonds

Do not award weakening bond between cation and

anion

Question

Answer Additional Guidance Mark

Number

4(c) An answer that makes reference to the following points: (2)

• (usually carbonates react with acids and ) produce a Allow effervescence / fizzing / bubbles for

(colourless) gas / CO2 (which is an expected observation for observation

the test) (1) Allow little / no gas / CO2 formed when

sulfuric acid is used

Ignore references to limewater / lighted splint

to test for CO2

• (but) the barium sulfate produced is insoluble (so the Allow a (white) precipitate (of barium sulfate)

carbonate may appear to not react / not dissolve in acid) (1) forms

Allow they should have used hydrochloric /

nitric acid as the salts formed are soluble

Accept bubbles of gas would not be expected

because barium sulfate is insoluble for 2 marks

(Total for Question 4 = 13 marks)

How to answer it

Barium Carbonate Thermodynamics & Group 2 Stability

📋 What this question tests

This question assesses your understanding of chemical thermodynamics (entropy changes, enthalpy changes, total entropy, and Gibbs free energy calculations), the thermal stability trends of Group 2 carbonates down the group, and practical analytical chemistry involving ionic precipitation reactions.

Question 4(a)(i)

Thermal Stability at 298 K

Show that barium carbonate is thermally stable at 298 K using entropy and enthalpy data. (5 marks)

📐 Step-by-Step Calculation (Method 1: Total Entropy)

Calculate ΔS(system):
ΔS(system) = ΣS(products) - ΣS(reactants)
ΔS(system) = (70.4 + 213.6) - 112.1 = +171.9 J K⁻¹ mol⁻¹
Calculate ΔS(surroundings):
ΔS(surroundings) = -ΔH / T
ΔS(surroundings) = -(+169.3 × 1000) / 298 = -568.12 J K⁻¹ mol⁻¹
Unit Consistency Check:
Ensure both entropy terms are in J K⁻¹ mol⁻¹ (convert ΔH to Joules by multiplying by 1000).
Calculate ΔS(total):
ΔS(total) = ΔS(system) + ΔS(surroundings)
ΔS(total) = 171.9 + (-568.12) = -396.22 J K⁻¹ mol⁻¹

✅ Final Conclusion & Mark Scheme

Since ΔS(total) is negative (< 0), the reaction is not feasible at 298 K. Therefore, barium carbonate is thermodynamically stable at room temperature.

Mark Breakdown (5 marks):
1 mark for ΔS(system)
1 mark for ΔS(surroundings)
1 mark for unit conversion consistency
1 mark for calculating ΔS(total) with correct units
1 mark for correct conclusion linking negative ΔS(total) to stability.

❌ Common Errors & Exam Technique

Unit Trap: Forgetting to multiply ΔH by 1000 to convert kJ mol⁻¹ into J mol⁻¹ before dividing by temperature.
Sign Errors: Forgetting the negative sign in the formula -ΔH / T for surroundings entropy.

Question 4(a)(ii)

Minimum Feasible Temperature

Calculate the lowest temperature, in °C, at which barium carbonate decomposes. (3 marks)

📐 Step-by-Step Calculation

Understand the condition for feasibility:
For a reaction to just become feasible, ΔS(total) = 0, which means ΔS(system) = ΔH / T .
Rearrange to solve for T (Kelvin):
T = ΔH / ΔS(system)
T = (169.3 × 1000) / 171.9 = 984.87 K
Convert Kelvin to Celsius and apply Significant Figures:
T(°C) = 984.87 - 273.15 = 711.72 °C = 712 °C (to 3 sig figs).

🧠 Exam Technique & Guidance

Always double-check your rounding requirements. The question explicitly asks for three significant figures. Leaving the answer in Kelvin or failing to convert to Celsius will cost you the final accuracy mark.

Mark Breakdown (3 marks):
1 mark for recognizing that ΔS(total) = 0 or setting up ΔH = TΔS.
1 mark for correct substitution and rearrangement.
1 mark for final temperature converted to °C and formatted to 3 SF.

Question 4(b)

Thermal Stability Trend in Group 2

Explain whether magnesium carbonate is more or less thermally stable than barium carbonate. (3 marks)

💡 Key Knowledge (The "Why")

Mg²⁺ size & charge density: The magnesium ion is smaller than the barium ion and has a greater charge density.
Polarising power: Mg²⁺ has a higher polarising power, distorting the electron cloud of the carbonate ion (CO₃²⁻).
Bond weakening: This polarization weakens the C-O bonds within the carbonate ion, making it decompose much more easily at lower temperatures.

✅ Core Answer Summary

Magnesium carbonate is LESS thermally stable than barium carbonate.

Mark Breakdown (3 marks):
1 mark: Stating Mg²⁺ is smaller / has higher charge density.
1 mark: Explaining it polarises the carbonate anion more strongly.
1 mark: Concluding that this weakens the internal C-O bond.

Question 4(c)

Analytical Testing & Practical Pitfalls

A student tests for carbonate by adding sulfuric acid. Explain whether this suggestion is valid. (2 marks)

❌ Why the test fails

Adding sulfuric acid (H₂SO₄) to barium carbonate produces an insoluble layer of barium sulfate (BaSO₄) on the solid surface:

BaCO₃(s) + H₂SO₄(aq) → BaSO₄(s) + H₂O(l) + CO₂(__g__)

This insoluble coating forms a barrier that prevents further acid contact, causing the reaction to stop prematurely. It may appear as though no carbonate is present.

✅ Examiner Guidance & Valid Alternatives

The suggestion is not valid because the expected effervescence (bubbling) will quickly stop.

Mark Breakdown (2 marks):
1 mark: Acknowledging that carbonates normally react with acids to produce CO₂ gas.
1 mark: Explaining that barium sulfate is insoluble, creating a barrier that stops the reaction / prevents proper observation. (Note: hydrochloric or nitric acid should be used instead since their barium salts are soluble).

Topics

Physical Chemistry · Inorganic Chemistry · Topic 4: Inorganic Chemistry and the Periodic Table · Topic 13: Energetics II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.