Edexcel A-Level Chemistry Paper 1, June 2018: Question 9

15 marks · Medium difficulty · Calculations

Calculate entropy changes, Gibbs free energy, and the equilibrium constant for chemical reactions, and explain reaction conditions using kinetic and thermodynamic considerations.

Practise this question

Question

Question 9 on entropy and free energy consisting of four parts totaling 15 marks. Part (a) asks to complete a table giving the sign of ΔS_system for three reactions: CO2(s) to CO2(g), NaCl(s) + aq to NaCl(aq), and N2(g) + 3H2(g) to 2NH3(g). Part (b) asks to calculate ΔS_total for the thermal decomposition of CaCO3 at 298 K given ΔrH = +178 kJ mol^-1 and ΔS_system = +160 J K^-1 mol^-1. Part (c) involves the reaction 2SO2(g) + O2(g) equilibrium to 2SO3(g) with ΔrH = -288.4 kJ mol^-1 and a table of standard molar entropy values (SO2 = +248.1, O2 = +205.0, SO3 = +95.6 J K^-1 mol^-1): (i) calculate ΔS_system, (ii) calculate ΔG at 298 K and deduce feasibility, (iii) calculate K at 700 K where ΔG = -60 kJ mol^-1, and (iv) explain why the reaction is conducted at 700 K instead of 298 K.
Question text

9 This question is about entropy and free energy.

(a) Complete the table by giving the sign of the entropy change of the system,

∆Ssystem, for each reaction.

(2)

Reaction Sign of ∆Ssystem

CO2(s) → CO2(g)

NaCl(s) + aq → NaCl(aq)

N2(g) + 3H2(g) → 2NH3(g)

(b) Calculate the total entropy change, ∆Stotal, for the thermal decomposition of

calcium carbonate at 298K.

CaCO3(s) → CaO(s) + CO2(g)

[Data: ∆ H = +178 kJ mol−1 ∆S = +160 J K−1 mol−1]

r system

(3)

*P52302RA02028*

(iii) In industry, the reaction is carried out at about 700K using a

vanadium(V) oxide catalyst.

Calculate the value of the equilibrium constant, K, at 700K.

∆G at 700 K is –60 kJ mol−1

(3)

(iv) The equilibrium constant has a larger value at 298K than at 700K.

Explain why the reaction is carried out at 700K and not at 298K.

(2)

(Total for Question 9 = 15 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 9 giving full marking criteria. 9(a) gives positive, positive, negative (2 marks). 9(b) awards marks for ΔS_surroundings = -ΔH/T = -597.3 J K^-1 mol^-1 and ΔS_total = -437.3 J K^-1 mol^-1 or -0.437 kJ K^-1 mol^-1 (3 marks). 9(c)(i) shows calculation of ΔS_system as -510 J K^-1 mol^-1 (2 marks). 9(c)(ii) uses ΔG = ΔH - TΔS to give -136 kJ mol^-1 and deduction of feasibility (3 marks). 9(c)(iii) uses ΔG = -RT ln K to find ln K = 10.3 and K = 3.02 x 10^4 (3 marks). 9(c)(iv) awards 2 marks for mentioning that although yield is higher at 298 K, the rate is too slow, so 700 K is an industrial compromise between rate and yield.

Question

Acceptable Answer Additional Guidance Mark

Number

9(a) Example of table (2)

all 3 correct (2)

any 2 correct (1)

Question

Acceptable Answer Additional Guidance Mark

Number

9(b) Example of calculation (3)

use of ∆Ssurroundings = −∆H/T (1) -(178000÷298) / -(178÷298)

calculation of ∆S (1) -597(.315) (J K−1 mol−1)

surroundings

or

-0.597(315) (kJ K−1 mol−1)

TE on equation with minus sign missing

calculation of ∆Stotal

and 160 + (-0.597315) = -0.437(315) kJ K−1 mol−1

sign 1000

and or

units (1) 160 + (-0.597315 x 1000)

= -437.(315) J K−1 mol−1

TE on ∆Ssurroundings

Allow correct units shown once in answer for

∆Stotal or ∆Ssurroundings

Ignore SF except 1SF

Correct answer with sign and units without

working scores 3 marks

Question

Acceptable Answer Additional Guidance Mark

Number

9(c)(i) Example of calculation (2)

correct working (1) (2 x 95.6) – ((2 x 248.1) + 205.0) /

(2 x 95.6) – (2 x 248.1) − 205.0

correct answer -510(.0) (J K−1 mol−1 )

and or

sign (1) -0.510 (kJ K−1 mol−1)

TE on working

Ignore SF except 1SF

Correct answer with sign and without

working scores both marks

Question

Acceptable Answer Additional Guidance Mark

Number

9(c)(ii) Example of calculation (3)

use of ∆G = ∆H − T∆Ssystem (1) The equation may be stated or numbers substituted

directly e.g.−288.4 – (298 x −0.510) / −288400 – (298 x

−510)

calculation of ∆G −136(.42) kJ mol−1 / −136420 J mol−1

and

sign TE on ∆Ssystem in (i)

and

units (1) Ignore SF except 1SF

Correct answer with sign and units without working scores

both marks

∆G is negative / less than 0 / <0 Conditional on a stated number

and TE on sign of ∆G:

so the reaction is feasible (1) ∆G is positive / greater than 0 / >0 so the reaction is

not feasible

Question

Acceptable Answer Additional Guidance Mark

Number

9(c)(iii) Example of calculation (3)

use of ∆G = −RTlnK (1) −60000 = − 8.31 x 700 lnK

rearrangement of equation (lnK = −∆G/RT)

and lnK = −(−60000)

substitution of correct values (1) (8.31 x 700)

Allow lnK = 60000

8.31 x 700

Allow lnK = 10.3146 / 10.315 / 10.32 / 10.3 / 10

TE on equation, provided equation involves all of

∆G, K, R and T and no others e.g. S

calculation of K (1) K = e10.315 = 3.016975x 104 / 30169.75

TE on lnK expression / value

Allow answers based on earlier correct rounding

Ignore SF including 1SF

Ignore units

Correct answer without working scores 3 marks

Question

Acceptable Answer Additional Guidance Mark

Number

9(c)(iv) An explanation that makes reference to any two of Allow reverse argument for M1 and M2 (2)

the following points:

Ignore reference to changing the pressure

Yield - even though the (percentage) yield / amount Allow the unused reactants can be recycled to

of SO3 is higher at 298 K / lower temperature (1) increase the yield / products are removed to

increase the yield

Allow the reaction does not reach equilibrium

in industry so there is no effect on the yield

Ignore just a reference to ‘equilibrium

shifting’

Rate - the rate of reaction is slower at 298 K / lower Ignore references to activation energy

temperature (1)

Compromise - so 700 K is a compromise between Allow at 700K the amount of product per unit

a (high) yield and (high) rate (1) time is larger

Ignore just ‘700 K is more economically

viable’

Note

If three points are made related to yield, rate

and compromise and one of these is

incorrect, maximum mark is (1) for 1 correct

point

(Total for Question 9 = 15 marks)

How to answer it

Thermodynamics: Entropy Changes, Gibbs Free Energy & Equilibrium

What this question tests
  • Predicting entropy changes (ΔSsystem): Relating changes in state and number of gaseous moles to disorder.
  • Entropy calculations: Determining ΔSsurroundings from -ΔH/T and combining to find ΔStotal.
  • Standard molar entropies: Calculating ΔSsystem using ΣS°(products) - ΣS°(reactants).
  • Gibbs Free Energy & Feasibility: Using ΔG = ΔH - TΔSsystem and understanding that ΔG < 0 defines feasibility.
  • Quantitative Equilibrium: Linking ΔG to the equilibrium constant using ΔG = -RT ln K.
  • Industrial compromises: Balancing thermodynamic yield against reaction kinetics/rate.

Part (a) Predicting Signs of Entropy Change

Qualitative assessment of ΔSsystem (2 Marks)

✅ Correct Table Entries

Reaction Sign of ΔSsystem
CO₂(s) → CO₂(g) Positive (+)
NaCl(s) + aq → NaCl(aq) Positive (+)
N₂(g) + 3H₂(g) → 2NH₃(g) Negative (-)
Mark Scheme: All 3 correct = 2 marks; any 2 correct = 1 mark. Words (positive/negative) or symbols (+/-) are fully acceptable.

💡 Key Knowledge

  • Solid to Gas: Gas particles possess far higher translational energy and positional disorder than a rigid solid lattice → ΔS > 0.
  • Dissolving an ionic solid: Dissolving frees ions from a fixed lattice into solution, increasing dispersal of matter/energy → ΔS > 0.
  • Gas mole change: 4 moles of gas react to form 2 moles of gas. A decrease in gaseous particles means fewer microstates and less disorder → ΔS < 0.

Part (b) Calculating Total Entropy Change (ΔStotal)

Decomposition of CaCO₃(s) at 298 K (3 Marks)

📐 Step-by-Step Calculation

  1. Calculate ΔSsurroundings:
    Formula: ΔSsurroundings = -ΔH / T
    Convert ΔH to J mol⁻¹: +178 kJ mol⁻¹ = +178 000 J mol⁻¹
    ΔSsurroundings = -(+178 000) / 298 = -597.315 J K⁻¹ mol⁻¹ (or -0.5973 kJ K⁻¹ mol⁻¹)
    Award 1 mark for formula usage, 1 mark for correct value.
  2. Calculate ΔStotal:
    Formula: ΔStotal = ΔSsystem + ΔSsurroundings
    Ensure identical units! Given ΔSsystem = +160 J K⁻¹ mol⁻¹
    ΔStotal = +160 + (-597.315) = -437 J K⁻¹ mol⁻¹ (or -0.437 kJ K⁻¹ mol⁻¹)
    Award 1 mark for final answer with sign AND matching units. (Allow 3-4 SF: -437 or -437.3)

❌ Common Errors

  • Unit mismatch: Adding +160 directly to -0.597 without converting both to J K⁻¹ mol⁻¹ or both to kJ K⁻¹ mol⁻¹.
  • Sign dropped: Forgetting the negative sign in the formula -ΔH / T .
  • Omitting units: Mark 3 strictly requires correct sign AND correct units (e.g. J K⁻¹ mol⁻¹).

🧠 Exam Technique

Always write down units next to each intermediate number. If ΔStotal is negative at 298 K, this correctly reflects that calcium carbonate does not spontaneously decompose at room temperature!

Part (c)(i) Calculating ΔSsystem from Molar Entropies

Reaction: 2SO₂(g) + O₂(g) → 2SO₃(g) (2 Marks)

📐 Step-by-Step Calculation

  1. Apply Stoichiometry:
    ΔSsystem = ΣS°(products) - ΣS°(reactants)
    Reactants = (2 × 248.1) + 205.0 = 496.2 + 205.0 = 701.2 J K⁻¹ mol⁻¹
    Products = 2 × 95.6 = 191.2 J K⁻¹ mol⁻¹
    Award 1 mark for correct working expression.
  2. Subtract Reactants from Products:
    ΔSsystem = 191.2 - 701.2 = -510 J K⁻¹ mol⁻¹ (or -0.510 kJ K⁻¹ mol⁻¹)
    Award 1 mark for answer with sign and units.

❌ Common Errors

  • Swapping reactants and products: Calculating Reactants - Products gives +510 J K⁻¹ mol⁻¹ (incorrect sign).
  • Ignoring stoichiometric coefficients: Forgetting to multiply SO₂ and SO₃ entropy values by 2.

🧠 Sanity Check

3 moles of gas → 2 moles of gas. Since gaseous moles decrease, ΔSsystem must be negative. If your answer is positive, re-check your subtraction order.

Part (c)(ii) Gibbs Free Energy & Feasibility

Calculate ΔG at 298 K (3 Marks)

📐 Step-by-Step Calculation

  1. Formula & Unit Alignment:
    ΔG = ΔH - TΔSsystem
    Convert ΔSsystem to kJ K⁻¹ mol⁻¹: -510 J K⁻¹ mol⁻¹ = -0.510 kJ K⁻¹ mol⁻¹
    Given: ΔH = -288.4 kJ mol⁻¹, T = 298 K
    Award 1 mark for formula or correct substitution of values.
  2. Calculate ΔG:
    ΔG = -288.4 - [298 × (-0.510)]
    ΔG = -288.4 - (-151.98) = -288.4 + 151.98 = -136.42 kJ mol⁻¹ (or -136 kJ mol⁻¹)
    Award 1 mark for correct value with negative sign and units (kJ mol⁻¹ or J mol⁻¹).
  3. Feasibility Deduction:
    Because ΔG < 0 (negative), the reaction is feasible at 298 K.
    Award 1 mark for linking negative ΔG to feasibility. (Transfer of error applies if ΔG was positive).

❌ Common Errors

  • Double negative blunder: -(-151.98) becomes +151.98. Students often accidentally subtract 151.98 instead of adding it.
  • Unit mixing: Combining -288.4 (kJ) directly with 298 × (-510) (J) without dividing by 1000.

🧠 Exam Technique

Always state the condition explicitly: "Since ΔG is negative (< 0), the reaction is feasible." Never simply write "yes".

Part (c)(iii) Calculating Equilibrium Constant (K)

Using ΔG = -RT ln K at 700 K (3 Marks)

📐 Step-by-Step Calculation

  1. State Formula & Substitute:
    ΔG = -RT ln K
    Convert ΔG into J mol⁻¹ because R is in J K⁻¹ mol⁻¹: ΔG = -60 kJ mol⁻¹ = -60 000 J mol⁻¹
    -60 000 = -(8.31) × 700 × ln K
    Award 1 mark for formula and correct substitution.
  2. Rearrange for ln K:
    ln K = -ΔG / (R × T)
    ln K = -(-60 000) / (8.31 × 700) = 60 000 / 5817 = +10.315
    Award 1 mark for correct rearrangement and calculating ln K (accept 10.3 to 10.32).
  3. Solve for K:
    K = e10.315 = 3.02 × 10⁴ (or 30 200)
    Award 1 mark for final calculated value of K. (K has no required units here; any reasonable SF accepted).

❌ Common Errors

  • Neglecting J vs kJ: Using ΔG = -60 instead of -60 000. This results in ln K = 0.0103 and K ≈ 1.01, which is completely wrong.
  • Inverse log error: Using 10x instead of ex ( ln is base e, not base 10).
  • Losing the double negative: Notice that -(-60 000) produces a positive ln K value.

💡 Key Rule

Whenever gas constant R = 8.31 J K⁻¹ mol⁻¹ appears in a formula, free energy ΔG MUST be converted into Joules (multiply kJ by 1000).

Part (c)(iv) Industrial Compromise: Kinetics vs Equilibrium

Why run at 700 K instead of 298 K? (2 Marks)

✅ Acceptable Points (Any Two)

  • Yield / Equilibrium: The yield of SO₃ is lower at 700 K (or higher at 298 K) because the forward reaction is exothermic. [1 mark]
  • Rate of reaction: At 298 K the rate of reaction is far too slow / at 700 K the rate of reaction is much faster. [1 mark]
  • Compromise: 700 K is an economic compromise between acceptable rate and acceptable yield. [1 mark]

🧠 Examiner Insight

  • Avoid vague statements: Writing "700 K is more economically viable" earns zero marks unless you explain that it provides a faster rate of production.
  • Contradiction trap: If you write three points and one is scientifically incorrect, the mark scheme caps you at a maximum of 1 mark! Be concise and accurate.
  • Mention both axes: High-scoring students explicitly state that 298 K gives better yield (thermodynamics) but 700 K gives acceptable rate (kinetics).

Topics

Physical Chemistry · Topic 13: Energetics II · Topic 10: Equilibrium I · Topic 9: Kinetics I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.