Worked example 1
A reaction has ΔH° = −92.2 kJ/mol and ΔS° = −198.1 J/(mol·K). Is the forward reaction favored in the standard state 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 combines enthalpy and entropy, with temperature setting the weight of the entropy term. A negative reaction ΔG means the forward direction is thermodynamically favored for the mixture under the stated conditions. For a standard-state calculation, write ΔG°; the actual value also depends on composition.
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
For the standard-state reaction Gibbs energy, assuming ΔH° and ΔS° are approximately constant, the signs predict these temperature trends. A negative ΔH° with positive ΔS° gives ΔG° < 0 at all temperatures; positive ΔH° with negative ΔS° gives ΔG° > 0 at all temperatures. Negative values for both favor the standard-state forward reaction at low temperature, while positive values for both favor it at high temperature. For a mixture away from standard conditions, the reaction direction also depends on Q.
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° = 0 identifies the standard-state crossover temperature where the standard reaction Gibbs energy changes sign. This estimate assumes ΔH° and ΔS° stay roughly constant over the temperature range. At that point K = 1; a real mixture is at equilibrium only when its reaction quotient Q equals K. At other compositions or pressures, determine direction from ΔG = ΔG° + RT ln Q.
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 energyAdd free-energy changes for coupled steps
Gibbs energy is a state function, so ΔG values add when reactions are added with matching conditions and stoichiometry. An unfavorable step can be driven by coupling it to a sufficiently favorable step, provided the two are actually linked in the overall process. A negative total ΔG says the coupled process is thermodynamically favorable; it does not say how quickly it occurs.
Sum ΔG for the complete coupled reaction, not for one step in isolation.
Key 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
Using the same ΔH° and ΔS° values and assuming they are constant, estimate the temperature at which the standard reaction Gibbs energy crosses zero.
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Worked example 3
For CaCO₃(s) → CaO(s) + CO₂(g), ΔH° = +178 kJ/mol and ΔS° = +161 J/(mol·K). Estimate the standard-state crossover temperature, assuming ΔH° and ΔS° are constant.
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Worked example 4
A step has ΔG = +20 kJ/mol and is coupled one-to-one to a step with ΔG = -35 kJ/mol. Find the net ΔG.
Try it first: Add the two steps as written.
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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.
After the lesson · Question bank
Independent practice
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What you should now know
ΔG = ΔH − TΔS combines enthalpy and entropy. A negative reaction ΔG favors the forward direction for the mixture under the stated conditions; favorability does not determine rate. In standard-state calculations, ΔG° crosses zero at an estimated T = ΔH°/ΔS° when those values are treated as constant. At that point K = 1, but a real mixture is at equilibrium only when Q = K. Away from standard conditions use ΔG = ΔG° + RT ln Q.
- 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