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Gibbs Free Energy and Feasibility

Use entropy and Gibbs free energy, ΔG = ΔH − TΔS, to decide whether a reaction is feasible.

  • GCE A-Level H2 Chemistry 9476-2027

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An exothermic reaction is not automatically feasible, and an endothermic reaction can be feasible. Gibbs free energy combines the enthalpy and entropy changes so that you can decide the direction favoured under the stated conditions.

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

Connect entropy to feasibility

Start with Predicting entropy changes. Entropy describes how energy can be distributed among the accessible microscopic arrangements of a system. Here, Δ S⦵ is a supplied standard entropy change in J mol⁻¹ K⁻¹; you do not need to calculate it from tabulated molar entropies for H2.

D. Gibbs free energy change, Δ G⦵

At constant temperature and pressure, use: 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.

Key Ideas (What Earns Marks)

  • Unit discipline in Δ G⦵ = Δ H⦵ - TΔ S⦵:
    • Δ H⦵ often in kJ mol⁻¹
    • Δ S⦵ often in J mol⁻¹ K⁻¹
    • convert so units match (common choice: convert Δ S⦵ to kJ mol⁻¹ K⁻¹ by dividing by 1000)

Temperature dependence, treating Δ H and Δ S as approximately constant within the temperature range and phases considered:

Δ HΔ SFeasible when…
−+all positive T in this model
+−no positive T in this model
−−low T
++high T
Quick Recall (Gibbs Unit Check)
  • Use T in K. - If Δ H is in kJ mol⁻¹, convert Δ S to kJ mol⁻¹ K⁻¹ (divide by 1000). - Then TΔ S comes out in kJ mol⁻¹.

ΔG Changes with Temperature (Example Lines)

ΔG Changes with Temperature (Example Lines). ΔH = +35.0, ΔS = +120, ΔH = −50.0, ΔS = −120 plotted as Gibbs free energy change, ΔG against Temperature, T.

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ΔG Changes with Temperature (Example Lines). ΔH = +35.0, ΔS = +120, ΔH = −50.0, ΔS = −120 plotted as Gibbs free energy change, ΔG against Temperature, T.ΔG Changes with Temperature (Example Lines). ΔH = +35.0, ΔS = +120, ΔH = −50.0, ΔS = −120 plotted as Gibbs free energy change, ΔG against Temperature, T.
Example straight lines from ΔG = ΔH − TΔS (with ΔS converted to kJ mol^-1 K^-1): ΔH>0, ΔS>0 becomes feasible at higher T; ΔH<0, ΔS<0 becomes less feasible at higher T.
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Values and uncertainty for ΔG Changes with Temperature (Example Lines)
SeriesTemperature, T (K)Temperature, T uncertaintyGibbs free energy change, ΔG (kJ mol^-1)Gibbs free energy change, ΔG uncertainty
ΔH = +35.0, ΔS = +1202505
ΔH = +35.0, ΔS = +120292-0.04
ΔH = +35.0, ΔS = +120400-13
ΔH = +35.0, ΔS = +120600-37
ΔH = −50.0, ΔS = −120250-20
ΔH = −50.0, ΔS = −1204170.04
ΔH = −50.0, ΔS = −12050010
ΔH = −50.0, ΔS = −12060022

Detailed Explanations

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: the boundary of standard-state feasibility

C. Threshold temperature

At the boundary between negative and positive standard Gibbs energy, Δ G⦵ = 0: T = (Δ H⦵)/(Δ S⦵)

Use this only when Δ H⦵ and Δ S⦵ are approximately constant over the temperature range and no phase change invalidates the model. The threshold must also be positive to represent a physical thermodynamic temperature.

Keep standard and actual conditions separate. At equilibrium the actual reaction Gibbs energy change is Δ G = 0. A standard value Δ G⦵ = 0 does not mean that every mixture is at equilibrium: it identifies the standard-state boundary. A mixture’s composition also affects the direction favoured. Likewise, Δ G⦵ > 0 does not mean that no product can form.

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.

For the ammonia entropy prediction, see Predicting entropy changes.

Common misconception 2

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 3

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 4

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.

Common Mistakes

  • Using °C instead of K in TΔ S.
  • Mixing units: using Δ S in J K⁻¹ mol⁻¹ with Δ H in kJ mol⁻¹ without converting.
  • Incorrect feasibility language (e.g. saying Δ G > 0 is feasible).
  • Saying “not feasible” means “cannot happen” (it means not thermodynamically favourable under the stated conditions).

Use the Energetics Thermodynamics topic check to practise and check your understanding.

Exam Tips

  • Use Δ G = Δ H - TΔ S with T in K.
  • Convert Δ S to kJ K⁻¹ mol⁻¹ if Δ H is in kJ mol⁻¹.
  • State feasibility language: Δ G⦵ < 0 favours the forward reaction under standard conditions; Δ G⦵ > 0 favours the reverse. At equilibrium, the actual Δ G = 0.

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

Set Δ G⦵ = 0, match the energy units, and check which side of the threshold makes Δ G⦵ negative.

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.

Syllabus and review details

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