Worked example 1
A reaction has ΔH = -92.2 kJ/mol and ΔS = -198.1 J/mol K. Is it favoured at 298 K?
Try it first: Convert the entropy value before substituting anything.
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Thermochemistry & ThermodynamicsGibbs Free EnergyContent level: Challenge 26 min
Gibbs Free Energy and Thermodynamic Favorability
What you'll be able to do: Use ΔG = ΔH - TDS to decide whether a reaction is thermodynamically favoured and find the temperature where the answer changes.
Best after: Standard Enthalpies of Formation, Entropy and the Second Law
Introduction
Enthalpy and entropy each push a reaction in their own direction, and they do not always agree. Gibbs free energy is the single quantity that settles the argument, with temperature deciding how much weight the entropy term carries.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Apply ΔG = ΔH - TDS with consistent units
- Classify reactions using the four sign combinations of ΔH and ΔS
- Calculate the crossover temperature where ΔG changes sign
- Distinguish thermodynamic favourability from reaction rate
Lesson
The combining equation
Gibbs free energy weighs the enthalpy drive against the entropy drive, with temperature setting the weight. A negative value means the forward reaction is thermodynamically favoured under the stated conditions.
ΔG = ΔH - T ΔS
Units before anything else
Enthalpies come in kJ/mol and entropies in J/mol K, so one has to be converted. Temperature must be in kelvin. Almost every wrong answer in this topic is a factor of 1000 or a forgotten 273.
Divide ΔS by 1000 to convert J/mol K into kJ/mol K, then work entirely in kilojoules.
The four cases
Negative ΔH with positive ΔS is favoured at all temperatures. Positive ΔH with negative ΔS is favoured at none. Negative ΔH with negative ΔS is favoured only at low temperature, where the TDS penalty is small. Positive ΔH with positive ΔS is favoured only at high temperature, where the TDS term becomes large enough to win.
Ask which term is helping. If enthalpy helps and entropy hurts, low temperature wins. If entropy helps and enthalpy hurts, high temperature wins.
Crossover temperature
Setting ΔG to zero identifies the temperature where the two drives exactly balance. Above or below it, whichever term dominates flips the sign of ΔG. At that temperature the system is at equilibrium under standard conditions, which is exactly what happens at a normal boiling point.
T(crossover) = ΔH / ΔS
Favourability is not speed
The conversion of diamond to graphite has a negative ΔG at room temperature, yet diamonds do not visibly change, because the activation energy is enormous. Thermodynamics tells you the destination, kinetics tells you the travel time.
Collision theory and activation energyKey ideas
Definition
Gibbs free energy change
ΔG = ΔH - TDS, the combined measure of thermodynamic favourability at constant temperature and pressure.
Rule
Sign of ΔG
Negative means favoured forward, positive means favoured in reverse, zero means equilibrium under those conditions.
Rule
Crossover temperature
Setting ΔG to zero gives T = ΔH/ΔS, the temperature where favourability switches.
Key concept
Temperature weighting
High temperature amplifies the entropy term, low temperature lets enthalpy dominate.
Key concept
Not a rate
A negative ΔG says nothing about how fast the reaction occurs.
Equation
Gibbs free energy
ΔG = ΔH - T ΔS
- ΔH = enthalpy change in kJ/mol
- T = absolute temperature in K
- ΔS = entropy change in kJ/mol K after conversion from J/mol K
Equation
Crossover temperature
T = ΔH / ΔS
- T = temperature in K at which ΔG = 0
Worked examples
Worked example 2
For the same reaction, above what temperature does it stop being favoured?
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Worked example 3
CaCO₃(s) → CaO(s) + CO₂(g) has ΔH = +178 kJ/mol and ΔS = +161 J/mol K. Find the minimum temperature for favourability.
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Common mistakes
Using ΔS in J/mol K alongside ΔH in kJ/mol.
Why it's wrong: The entropy term comes out 1000 times too large and swamps the answer.
Check instead: Convert ΔS to kJ/mol K before substituting.
Adding TDS instead of subtracting it.
Why it's wrong: The equation is ΔH minus TDS; subtracting a negative TDS is what makes the term positive.
Check instead: Write the minus sign explicitly, then substitute the signed value.
Using Celsius in the TDS term.
Why it's wrong: Thermodynamic temperature must be absolute.
Check instead: Add 273.15 before multiplying.
Concluding that a negative ΔG means a fast reaction.
Why it's wrong: Rate depends on activation energy, not on ΔG.
Check instead: Treat favourability and rate as separate questions.
Assuming a reaction with positive ΔH can never be favoured.
Why it's wrong: A large positive ΔS wins at high enough temperature.
Check instead: Check both signs, then find the crossover temperature.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
ΔG = ΔH - TDS combines both drives into one number. A negative ΔG means the reaction is thermodynamically favoured in the forward direction. When ΔH and ΔS have opposite effects, the reaction is favoured at every temperature or none. When they compete, temperature decides, and the crossover point is T = ΔH/ΔS. Favourability says nothing about rate: a reaction with a very negative ΔG can still be immeasurably slow because of a large activation energy.
- ΔG = ΔH - TDS decides thermodynamic favourability
- Convert ΔS to kJ/mol K and use kelvin
- Negative ΔH with positive ΔS is favoured at every temperature
- Competing signs give a crossover temperature T = ΔH/ΔS
- A negative ΔG does not imply a fast reaction
Sources and further reading
- Chemistry 2e, Section 16.4: Free EnergyPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 16.4View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License
- Chemistry 2e, Section 16.2: EntropyPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 16.2View source
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License