Edexcel A-Level Chemistry Paper 3, June 2024: Question 4

13 marks · Hard difficulty · Calculations

Calculate entropy changes, feasibility temperatures, explain thermal stability trends of nitrates, and complete a redox equation involving nitrate ions.

Practise this question

Question

Question text

4 This question is about the white crystalline solid, barium nitrate.

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

nitrogen dioxide and oxygen.

2Ba(NO ) (s) → 2BaO(s) + 4NO (g) + O (g) ΔH d = +1010 kJ mol–1

32 2 2

Standard molar entropy data related to this reaction are shown.

Substance Standard molar entropy, S d / J K–1 mol–1

Ba(NO3)2(s) 213.8

BaO(s) 70.4

NO2(g) 240.0

O2(g) 205.0

(i) Show that barium nitrate is thermally stable at 298K.

(5)

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

feasible for barium nitrate to decompose.

Give your answer to an appropriate number of significant figures.

(3)

(b) Explain why calcium nitrate is less thermally stable than barium nitrate.

(3)

… 8

… *P74455A0832*

suspected nitrate with aluminium.

(i) Complete the ionic equation for this redox reaction, using oxidation numbers.

State symbols are not required.

(1)

8Al + NO– + OH– + H O → Al(OH)– + NH

… 3 … 2 … 4 … 3

(ii) Describe a test to confirm that ammonia, NH3, has been produced.

(1)

(Total for Question 4 = 13 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

4(a)(i) Example of calculation (5)

• calculation of ΔSo (1) ((2 × 70.4) + (4 × 240.0) + 205.0) – (2 × 213.8 ) = 878.2 (J K–1 mol–1)

system

o (1) (−1010 × 1000) ÷ 298 = −3389.26 / 3389.3 (J K–1 mol–1)

• calculation of ΔS surroundings

(−1010 ÷ 298) = −3.38926 / 3.3893 (kJ K–1 mol–1)

o ΔSo converted to J K–1 mol–1

• consistent units for ΔS system and surroundings

ΔSo (1) or

surroundings

ΔSo converted to kJ K–1mol–1

system

M3 could be subsumed as part of either M1 or M2

• calculation of ΔSo (1) = 878.2+ (−3389.3)

total

= − 2511.1 (J K–1 mol–1)

−2.5111 (kJ K–1 mol–1)

Penalise incorrect units for M4 only

• comment on thermal stability at 298 K (1) Negative value, so reaction is not feasible /compound thermally stable

(at 298 K)

Standalone mark

Do not award on positive values for ΔSo

total

Ignore SF except 1 SF

Allow TE from M1 to M4 provided M4 is negative

A negative value for ΔSo loses M3 to M4 so can only score M2 and M5

system

ΔSo = +874.8 scores (2) for M1 and M2 but the failure to convert to

total

consistent units means a positive ΔSo is obtained that does not match the

total

question

Example of calculation

Alternative method using ΔG

((2 × 70.4) + (4 × 240.0) + 205.0) – (2 × 213.8 ) = (+) 878.2 (J K–1 mol–1)

• calculation of ΔSo (1)

system

298 × 878.2 = 261703.6 (J mol–1)

• calculation of TΔSo (1) M2 could be subsumed as part of either M3 or M4

system

ΔHo converted to J mol–1

• consistent units for TΔSo and ΔHo OR

system (1)

TΔSo converted to kJ mol–1

system

M3 could be subsumed as part M4

( G = H – T Ssystem)

• calculation of ΔG (1) = (+) 1010 000 − 261703.6

= (+) 748296.4 (J mol–1)

= (+) 748.296 (kJ mol–1)

Penalise incorrect units for M4 only

Ignore SF except 1 SF

Allow TE from M1 to M4 provided M4 is positive

• reason why thermally stable at 298 K (1) ΔG positive / > 0 so reaction is not feasible/ compound is stable (at 298 K)

Standalone mark but

Do not award on negative values for ΔGo

A negative value for ΔSo loses M3 to M4 so can only score M2 and M5

system

ΔG = − 260693.6 scores (2) for M1 and M2 but the failure to convert to

consistent units means a negative ΔG is obtained that does not match the

question

Question

Answer Additional Guidance Mark

Number

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

• rearrangement of ΔG expression (1) T = H/ Ssystem

• calculation of T (1) (T = (1010 × 1000) / 878.2)

= 1150.1 (K)

M2 subsumes M1 so the numerical value scores (2)

• conversion to °C to 2/3 SF (1) = 877 (°C) / 880 (°C)

Allow 878 (°C)

Do not award an answer below 25oC for M3

Question

Answer Additional Guidance Mark

Number

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

(calcium nitrate is less thermally stable because)

• (both ions have the same charge but) the calcium ion is Ignore ‘calcium atom is smaller’

smaller / has a greater charge density (1) Do not award references to electronegativity

Do not award reference to electron density

• so polarises (the anion) to a greater extent (1) Allow so more likely to polarise (the nitrate (ion) /

calcium ion has more polarising power

Allow reference to polarisation of the bond

Allow a description of polarisation such as the distortion

of the anion electron cloud to a greater extent

• and so weaken the N−O bond / bond(s) within the

nitrate ion (1)

Do not award reference to breaking of/polarising the

ionic bond between the calcium and the nitrate

Question

Answer Additional Guidance Mark

Number

4(c)(i) An answer that makes reference to the following point: (1)

• 8Al + 3NO – + 5OH– + 18H O → 8Al(OH) – + 3NH

32 4 3

Question

Answer Additional Guidance Mark

Number

4(c)(ii) An answer that makes reference to the following point: (1)

• (ammonia turns damp red) litmus paper blue Allow (ammonia turns damp ) UI paper blue

Do not award if litmus paper bleached after turning blue

Allow (ammonia produces) white smoke with HCl (gas)

Do not award white fumes/misty fumes

If two tests given then both must be correct to score

(Total for Question 4 = 13 marks)

How to answer it

Barium Nitrate Thermodynamics & Ionic Reactions

What this question tests

This question assesses your mastery of A-Level chemical thermodynamics (calculating entropy changes, Gibbs free energy, and feasibility), Group 2 thermal stability trends in terms of ionic radius and charge density (polarisation), and analytical/redox chemistry involving nitrate ions and confirmatory tests for ammonia gas.

Question 4(a)(i)

Thermal Stability at 298 K

Show that barium nitrate is thermally stable at 298 K.

📐 Step-by-Step Calculation (ΔS Method)

Calculate ΔS°(system):
ΣS°(products) - ΣS°(reactants)
= (2 × 70.4 + 4 × 240.0 + 205.0) - (2 × 213.8)
= 878.2 J K⁻¹ mol⁻¹
Calculate ΔS°(surroundings):
-ΔH° / T = (-1010 × 1000) / 298
= -3389.26 J K⁻¹ mol⁻¹
Unit Consistency Check:
Ensure both ΔS° terms are in J K⁻¹ mol⁻¹ before combining.
Calculate ΔS°(total):
ΔS°(system) + ΔS°(surroundings)
= 878.2 + (-3389.26) = -2511.1 J K⁻¹ mol⁻¹

✅ Correct Conclusion

Because ΔS°(total) is negative (-2511.1 J K⁻¹ mol⁻¹), the reaction is not feasible at 298 K. Therefore, barium nitrate is thermally stable under standard conditions.

Awarded 5 marks total (1 for system entropy, 1 for surroundings entropy, 1 for unit matching, 1 for total entropy, 1 for correct thermodynamic conclusion).

🧠 Exam Technique

You can alternatively use the Gibbs Free Energy equation: ΔG = ΔH - TΔS .
If using ΔG, calculate TΔS°(system) = 298 × 878.2 = 261703.6 J mol⁻¹ .
Then find ΔG = 1010000 - 261703.6 = +748296.4 J mol⁻¹ (+748 kJ mol⁻¹). A positive ΔG proves stability.

❌ Common Errors

Unit Mismatch: Forgetting to multiply ΔH° by 1000 to convert kJ to J when combining with J K⁻¹ entropy values.
Stoichiometry: Forgetting to multiply standard molar entropies by the balancing coefficients from the equation.

Question 4(ii)

Minimum Decomposition Temperature

Calculate the minimum temperature, in °C, at which it is thermodynamically feasible to decompose barium nitrate.

📐 Step-by-Step Calculation

Condition for Feasibility:
Reaction becomes feasible when ΔG ≤ 0 , meaning ΔH = TΔS .
Rearrange for Temperature (T in Kelvin):
T = ΔH° / ΔS°(system)
T = (1010 × 1000) / 878.2
Calculate Kelvin Value:
T = 1150.1 K
Convert to Celsius:
1150.1 - 273.15 = 876.95 °C = 877 °C (to 3 sig fig)

✅ Correct Answer

877 °C (or 878 °C depending on intermediate rounding). Valid to 2 or 3 significant figures.

Awarded 3 marks: 1 for rearranging expression, 1 for calculating Kelvin temperature, 1 for correct conversion to °C with appropriate sig fig.

Question 4(b)

Group 2 Thermal Stability Trend

Explain why calcium nitrate is less thermally stable than barium nitrate.

💡 Key Knowledge

Thermal stability of Group 2 nitrates increases down the group as cations get larger and their polarising power decreases.

✅ Mark Scheme Breakdown

Point 1: The calcium ion (Ca²⁺) is smaller than the barium ion (Ba²⁺) and has a higher charge density.
Point 2: Therefore, the Ca²⁺ ion polarises the nitrate anion (NO₃⁻) to a greater extent.
Point 3: This distorts the electron cloud, weakening the N–O bonds within the nitrate ion, requiring less thermal energy to decompose.

Awarded 3 marks total for linking cation size/charge density to increased polarisation and bond weakening.

❌ Common Errors & Penalties

Do not refer to electronegativity or electron density of the cation.
Do not state that the ionic bond between the metal and nitrate breaks; it is the covalent bonds within the nitrate ion that weaken.

Question 4(c)(i)

Redox Equation for Nitrate Testing

Complete the ionic equation for this redox reaction. State symbols are not required.

✅ Balanced Ionic Equation

8Al + 3NO₃⁻ + 5OH⁻ + 18H₂O → 8Al(OH)₄⁻ + 3NH₃

Awarded 1 mark for fully correct balancing coefficients.

🧠 Examiner Insight

This reaction is used in the qualitative detection of nitrates (often via Devarda's alloy or direct reduction with aluminium in NaOH). Aluminium acts as the reducing agent, being oxidised from oxidation state 0 to +3 in the aluminate ion, while nitrogen is reduced from +5 in nitrate to -3 in ammonia.

Question 4(c)(ii)

Confirmatory Test for Ammonia

Describe a test to confirm that ammonia, NH₃, has been produced.

✅ Correct Answer

Hold damp red litmus paper in the gas.

Result: The litmus paper turns blue because ammonia is alkaline.

Awarded 1 mark for reagent (damp red litmus) and correct observation (turns blue).

❌ Common Errors

The litmus paper must be damp so the ammonia gas can dissolve to form hydroxide ions (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻). Dry litmus will show no change.
Do not confuse with hydrogen chloride testing (which produces white fumes with concentrated HCl).

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 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.