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Entropy and Gibbs Free Energy

Learn and apply Entropy and Gibbs Free Energy in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Entropy and Gibbs Free Energy: Orientation

Gibbs questions are “sign + units + decision” questions. The safest method is: predict signs first (from gas moles/state changes), then calculate carefully (K, J↔kJ), then state the feasibility conclusion.

Keep Enthalpy Changes and Energy Profiles and the Energetics and Thermodynamics hub in view, because most questions mix definitions with cycle reasoning.

Definitions (Must Know)

A. Entropy, S

Entropy measures the dispersal of energy / disorder (number of ways energy can be arranged).

B. Standard molar entropy, S⦵

S⦵ is the entropy of 1 mol of a substance under standard conditions (units: J mol⁻¹ K⁻¹).

C. Standard entropy change, Δ S⦵

For a reaction: Δ S⦵ = ∑ ν S⦵(products) - ∑ ν S⦵(reactants)

D. Gibbs free energy change, Δ G⦵

Δ G⦵ = Δ H⦵ - TΔ S⦵

Feasible (under standard conditions) when Δ G⦵ < 0.

Feasible does not mean fast: a reaction can be feasible but have a high activation energy (slow) unless conditions/catalysts are provided.

E. Temperature, T

Temperature must be in kelvin (K): T(K) = T(°C) + 273.15.

Detailed Explanations

A. Calculating Δ S⦵ from standard entropies (workflow)

  1. Write the balanced equation.
  2. Use coefficients in the sum: Δ S⦵ = ∑ ν S⦵(products) - ∑ ν S⦵(reactants)
  3. Keep units as J mol⁻¹ K⁻¹ unless told otherwise.

Mini example (gas moles decrease): N₂(g) + 3H₂(g) → 2NH₃(g) Because 4 mol gas becomes 2 mol gas, Δ S⦵ is expected to be negative.

B. Using Δ G⦵ (workflow)

  1. Convert T to K.
  2. Make units consistent (kJ with kJ, or J with J).
  3. Calculate Δ G⦵ = Δ H⦵ - TΔ S⦵.
  4. Interpret:
    • Δ G⦵ < 0: feasible
    • Δ G⦵ > 0: not feasible
    • Δ G⦵ = 0: equilibrium

C. Threshold temperature

At the boundary between feasible/not feasible, Δ G⦵ = 0: T = (Δ H⦵)/(Δ S⦵)

Use this only when Δ H⦵ and Δ S⦵ are (approximately) constant over the temperature range considered.

Worked Examples

Modelled example 1

Test feasibility at a stated temperature

Core

Problem

A reaction has Δ H⦵ = +20.0 kJ mol⁻¹ and Δ S⦵ = +80.0 J mol⁻¹K⁻¹. Is it feasible under standard conditions at 298 K?
Study the worked solution

  1. Make units consistent

    Method

    Convert the entropy change to kilojoules per mole per kelvin.

    Reason

    TΔ S must use the same energy unit as Δ H.

    Working

    Δ S⦵ = 80.0/1000 = 0.0800 kJ mol⁻¹K⁻¹.

  2. Calculate Gibbs free energy

    Method

    Substitute into Δ G⦵ = Δ H⦵-TΔ S⦵.

    Reason

    Both energy terms are now expressed in kilojoules per mole.

    Working

    Δ G⦵ = 20.0-(298)(0.0800) = -3.84 kJ mol⁻¹.

  3. Interpret the sign

    Method

    State feasibility for the specified conditions.

    Reason

    A negative standard Gibbs free-energy change is thermodynamically feasible.

    Working

    Δ G⦵ < 0, so the reaction is feasible at 298 K.

Guided practice 2

Predict an entropy-change sign qualitatively

About 5 min

Problem

Predict the sign of Δ S⦵ for N₂(g) + 3H₂(g) → 2NH₃(g). Explain using gaseous-particle evidence; do not calculate from tabulated molar entropies.

Try this before viewing the solution

Hints

Hint 1: count gaseous particles
Compare the total gaseous coefficients on the two sides.
Hint 2: link to dispersal
Decide whether changing from four gaseous particles to two increases or decreases dispersal.
View solution step by step

  1. Compare gaseous amounts

    Method

    Add the gaseous stoichiometric coefficients on each side.

    Reason

    All species are gases, so the change in gaseous-particle amount is useful qualitative evidence.

    Working

    Reactants: 1 + 3 = 4 mol gas; products: 2 mol gas.

  2. Predict the sign

    Method

    State a negative entropy change.

    Reason

    Fewer gaseous particles means less dispersal and fewer accessible arrangements.

    Working

    Δ S⦵ < 0.

Common misconception 3

Correct a Gibbs unit mismatch

Find and correct the mistake

Learner substitution

A learner substitutes Δ H = 20.0 kJ mol⁻¹, T = 298 K and Δ S = 80.0 J mol⁻¹K⁻¹ directly as 20.0-(298)(80.0). Diagnose the setup.

Choose the valid correction

Before subtraction

View solution step by step

  1. Audit dimensions

    Method

    Identify that TΔ S is in joules per mole while Δ H is in kilojoules per mole.

    Reason

    Quantities in different energy units cannot be subtracted.

    Working

    (298)(80.0) = 23840 J mol⁻¹, not 23840 kJ mol⁻¹.

  2. Choose one energy unit

    Method

    Convert Δ S to 0.0800 kJ mol⁻¹K⁻¹.

    Reason

    Then both terms in the Gibbs equation are in kilojoules per mole.

    Working

    Δ G = 20.0-(298)(0.0800) = -3.84 kJ mol⁻¹.

Examiner practice 4

Explain low-temperature feasibility

3 marks

Problem

A reaction has Δ H⦵ = -50.0 kJ mol⁻¹ and Δ S⦵ = -120 J mol⁻¹K⁻¹. Explain its feasibility at low and high temperature. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Identify competing terms

    1 mark

    Method

    Recognise that negative Δ H favours negative Δ G, while negative Δ S makes -TΔ S positive.

    Reason

    The enthalpy and entropy contributions oppose one another.

    Working

    Δ G = negative + positive temperature term.

  2. Consider low temperature

    1 mark

    Method

    Make the positive entropy term small.

    Reason

    At low T, negative Δ H can dominate.

    Working

    Δ G < 0 is favoured, so the reaction is feasible at sufficiently low temperature.

  3. Consider high temperature

    1 mark

    Method

    Increase the positive -TΔ S contribution.

    Reason

    It can outweigh the negative enthalpy term.

    Working

    At sufficiently high temperature, Δ G > 0 and the reaction is not feasible under the stated conditions.

Challenge 5

Find the boundary temperature

Minimal support

Problem

For the reaction with Δ H⦵ = -50.0 kJ mol⁻¹ and Δ S⦵ = -120 J mol⁻¹K⁻¹, calculate the temperature at which Δ G⦵ = 0 and state the feasible temperature range.

Try this before viewing the solution

Hints

Hint 1: boundary equation
At the boundary, set 0 = Δ H⦵-TΔ S⦵.
Hint 2: keep signs and units
Use -0.120 kJ mol⁻¹K⁻¹ for Δ S⦵, then check one temperature below the result.
View solution step by step

  1. Convert entropy units

    Method

    Divide the entropy value by 1000 while retaining its sign.

    Reason

    The enthalpy is expressed in kilojoules per mole.

    Working

    Δ S⦵ = -0.120 kJ mol⁻¹K⁻¹.

  2. Solve the boundary equation

    Method

    Use T = Δ H⦵/Δ S⦵ at Δ G⦵ = 0.

    Reason

    Both negative signs cancel to give a physically meaningful positive temperature.

    Working

    T = (-50.0)/(-0.120) = 417 K to three significant figures.

  3. State the feasible range

    Method

    Select temperatures below the threshold.

    Reason

    For negative Δ S, increasing temperature raises Δ G.

    Working

    The reaction is feasible when T < 417 K under the stated approximation.

Mind Stretchers

Mind stretcher 1Extension

A reaction has Δ H⦵ = +35.0 kJ mol⁻¹ and Δ S⦵ = +120 J mol⁻¹K⁻¹. Calculate the temperature at which Δ G⦵ = 0, and state the temperature range where the reaction is feasible.

Show Hint

Predict the sign of entropy from phase and gaseous-particle evidence before combining units in the Gibbs equation.

Show Answer

Mark scheme:

  • Convert Δ S⦵ to kJ: 120 J mol⁻¹K⁻¹ = 0.120 kJ mol⁻¹K⁻¹
  • At Δ G⦵ = 0: T = (Δ H⦵)/(Δ S⦵) = 35.0/0.120 = 292 K (3 s.f.)
  • Since Δ H > 0 and Δ S > 0, it is feasible at high T, so feasible when T > 292 K.

Mind stretcher 2: Finding a feasibility thresholdExtension

Question. For a process, Δ H° = +42.0 kJ mol⁻¹ and Δ S° = +120 J mol⁻¹K⁻¹. Estimate the temperature above which Δ G° < 0 and state one limitation of the prediction.

Show Hint

At the threshold, set ΔG° to zero and convert ΔS° to kJ mol⁻¹ K⁻¹.

Show Answer

T = Δ H°/Δ S° = 42.0/0.120 = 350 K. The process is thermodynamically feasible above about 350 K under the stated standard-data approximation, but this does not imply a measurable rate and assumes ΔH° and ΔS° do not vary significantly with temperature.