Edexcel A-Level Chemistry Paper 1, June 2024: Question 9

5 marks · Medium difficulty · Calculations

Calculate ΔG at 298 K to show non-feasibility for the thermal decomposition of sodium hydrogencarbonate, and determine the minimum temperature in °C for feasibility.

Practise this question

Question

Question 9 starts with context on sodium hydrogencarbonate used as a raising agent in baking. Part (a) asks to show that the reaction 2NaHCO3(s) -> Na2CO3(s) + CO2(g) + H2O(l) (with ΔrH = +91.6 kJ mol^-1) is not feasible at 298 K by calculating ΔG, using a table providing standard molar entropy values in J K^-1 mol^-1: NaHCO3(s) = 101.7, Na2CO3(s) = 135.0, CO2(g) = 213.6, and H2O(l) = 69.9. Part (b) asks to calculate the minimum temperature, in degrees Celsius, at which an oven should be set for this reaction to be thermodynamically feasible.
Question text

9 Sodium hydrogencarbonate is used as a raising agent in baking as carbon dioxide gas

is released when it undergoes thermal decomposition.

(a) Show that this reaction is not feasible at 298K by calculating ΔG.

2NaHCO (s) → Na CO (s) + CO (g) + H O(l) Δ H = +91.6 kJ mol−1

32 3 2 2 r

(3)

Compound NaHCO3(s) Na2CO3(s) CO2(g) H2O(l)

Standard molar entropy

−1 −1 101.7 135.0 213.6 69.9

/JK mol

(b) Calculate the minimum temperature, in degrees Celsius (°C), at which an oven

should be set for this reaction to be thermodynamically feasible.

(2)

Minimum temperature = … °C

(Total for Question 9 = 5 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 9. Part (a) awards 1 mark for calculating ΔSsystem = (135 + 213.6 + 69.9) - (2 x 101.7) = +215.1 J K^-1 mol^-1, 1 mark for calculating ΔG = (+91.6 x 1000) - (298 x 215.1) = +27500 J mol^-1 (+27.5 kJ mol^-1) with correct units, and 1 mark for stating that ΔG is positive / > 0 so the reaction is not feasible. Part (b) awards 1 mark for rearranging to T = ΔH / ΔSsystem = 91.6 / 0.2151 = 425.8 K, and 1 mark for converting to Celsius: 425.8 - 273 = 153 °C.

Question

Acceptable Answer Additional Guidance Mark

Number

9(a) Example of calculation (3)

• calculation of Ssystem (1) = (135 + 213.6 + 69.9) – (2 × 101.7)

= (+) 215.1 (J K−1 mol−1) /

(+) 0.2151 (kJ K−1 mol−1)

• calculation of Go with appropriate units (1)

( G = H – T Ssystem)

= (+ 91.6 × 1000) – (298 × 215.1)

= (+) 27500 J mol−1 / 27500.2 J mol−1 /

(+) 27.500 kJ mol−1 / 27.5002 kJ mol−1

TE on answer to M1

Ignore SF except 1SF

• reason why not feasible (1) G positive / ≥ / >0 so not feasible (standalone

mark)

Question

Acceptable Answer Additional Guidance Mark

Number

9(b) Example of calculation (2)

• rearrangement of G expression and calculation T = H/ Ssystem

of feasible temperature (1) T = 91.6 / 0. 2151 = 425.8 K

(1) 425.8 K – 273 = 152.8/ 153 (oC)

• conversion to degrees Celsius

Allow 160 (oC)

Do not award 150 (oC) if rounded down

Do not award a negative answer for the

temperature.

Credit can be given for working that uses Stot = 0

Stot = 0 = 215.1 + (−91600 / T)

T = −91600 / −215.1 = 425.8 (1)

Conversion to oC = 425.8 −273 = 152.8 oC (1)

TE from answer to 9(a)

Ignore SF except 1SF

(Total for Question 9 = 5 marks)

How to answer it

Thermodynamics & Feasibility: Decomposition of NaHCO₃

📋 What this question tests
  • Entropy change of system (ΔSsystem): Calculating entropy changes from standard molar entropies and stoichiometric ratios.
  • Gibbs free energy equation (ΔG = ΔH − TΔSsystem): Handling unit conversion between kJ and J, calculating ΔG, and linking its sign to reaction feasibility.
  • Threshold temperature for feasibility: Setting ΔG = 0 (or ΔStotal = 0) to solve for minimum temperature, followed by Kelvin-to-Celsius conversion.
Part (a) 3 Marks

Feasibility of Decomposition at 298 K

Calculate ΔG to demonstrate why the reaction is non-feasible at room temperature

📐 Step-by-Step Calculation

  1. Calculate ΔSsystem:
    Reaction: 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(l)
    ΔSsystem = ΣS°(products) − ΣS°(reactants)
    ΔSsystem = (135.0 + 213.6 + 69.9) − (2 × 101.7)
    ΔSsystem = 418.5 − 203.4 = +215.1 J K⁻¹ mol⁻¹
  2. Convert units & calculate ΔG at 298 K:
    ΔH = +91.6 kJ mol⁻¹ = +91 600 J mol⁻¹
    ΔG = ΔH − TΔSsystem
    ΔG = 91 600 − (298 × 215.1)
    ΔG = 91 600 − 64 099.8 = +27 500 J mol⁻¹ (or +27.5 kJ mol⁻¹ )
  3. State feasibility conclusion:
    Since ΔG > 0 (ΔG is positive), the reaction is not feasible at 298 K.

✅ Mark Scheme Breakdown

  • Mark 1: ΔSsystem = +215.1 J K⁻¹ mol⁻¹ (or +0.2151 kJ K⁻¹ mol⁻¹)
  • Mark 2: Value for ΔG with correct units: +27.5 kJ mol⁻¹ or +27 500 J mol⁻¹ (allow 27.500 kJ mol⁻¹ or 27 500.2 J mol⁻¹). Transfer error (TE) allowed from M1.
  • Mark 3 (Standalone): Explicitly stating that ΔG is positive (or ΔG > 0) therefore it is not feasible.

❌ Common Errors & Traps

  • Stoichiometry trap: Forgetting to multiply S°(NaHCO₃) by 2.
  • Unit mismatch: Directly doing 91.6 − (298 × 215.1) without converting ΔH to J or ΔS to kJ. This results in a nonsensical value.
  • Missing units: Writing +27.5 without kJ mol⁻¹ or writing +27500 without J mol⁻¹ forfeits Mark 2.
  • Vague conclusion: Writing "not feasible because it is too low" without mentioning that ΔG > 0.

🧠 Exam Technique & Examiner Tips

  • Even if you make an arithmetic error in calculating ΔS, you can still gain Marks 2 and 3 through transferred error (TE), provided your calculation of ΔG is algebraically correct and you draw the correct conclusion based on your ΔG sign.
  • Always check: thermal decomposition is endothermic (ΔH > 0) and produces gas from a solid (ΔS > 0), so it must only be feasible at high temperatures.
Part (b) 2 Marks

Minimum Temperature for Feasibility

Determine the threshold temperature in degrees Celsius (°C)

📐 Step-by-Step Calculation

  1. Set condition for feasibility:
    At the point of feasibility, ΔG = 0
    0 = ΔH − TΔSsystem ⟹ T = ΔH / ΔSsystem
  2. Calculate T in Kelvin:
    T = 91.6 / 0.2151 (or 91 600 / 215.1)
    T = 425.85 K
  3. Convert Kelvin to Celsius:
    T(°C) = T(K) − 273
    T(°C) = 425.85 − 273 = 152.8 °C or 153 °C

✅ Mark Scheme Breakdown

  • Mark 1: Rearranging ΔG = 0 to get T = ΔH / ΔS and calculating feasible temperature in Kelvin = 425.8 K (or via ΔStotal = 0).
  • Mark 2: Converting temperature to Celsius = 152.8 °C or 153 °C (allow 160 °C if correctly rounded from 2 SF).

💡 Alternative Method: Using ΔStotal

A reaction is feasible when ΔStotal ≥ 0.

  • ΔStotal = ΔSsystem + ΔSsurroundings = ΔSsystem − (ΔH / T)
  • Set ΔStotal = 0: 0 = 215.1 − (91 600 / T)
  • T = 91 600 / 215.1 = 425.8 K ⟹ T = 425.8 − 273 = 152.8 °C

❌ Common Errors & Penalties

  • Leaving answer in Kelvin: Calculating 425.8 K and stopping there loses the final mark. Always double-check the requested units!
  • Incorrect conversion: Adding 273 instead of subtracting ( 425.8 + 273 gives 698.8 °C, which is wrong).
  • Severe rounding down: The mark scheme specifically states: "Do not award 150 °C if rounded down" and "Do not award a negative temperature".

Topics

Physical Chemistry · Topic 13: Energetics II

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.