Free Energy and the Direction of Spontaneous Change
For a closed system at fixed temperature and pressure, with only pressure-volume work unless another drive is specified, Gibbs free energy determines the direction of spontaneous change. An entropy gain can make a heat-absorbing process favorable.
What a learner can do afterwards
- Writes Gibbs free energy as enthalpy minus temperature times entropy and states the fixed-temperature, fixed-pressure conditions for its spontaneity criterion
- Predicts the direction of a change from the sign of the free energy difference
- Explains how the balance shifts as temperature rises
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The previous partition-function reading gives Helmholtz energy F=-k_B T ln Z at fixed V,N. Also F=U-TS, using internal energy U. Gibbs energy is G=H-TS=F+pV: H=U+pV is enthalpy, p is pressure, S is entropy and T is in kelvin. At fixed T, delta G=delta H-T delta S. Its spontaneity criterion uses fixed T and pressure for a closed system with only pressure-volume work, unless another drive is specified.
Under those conditions, negative delta G favors the forward change. It can proceed without an external drive and can, in principle, supply non-expansion work. Positive delta G does not favor that direction. This does not predict speed. Diamond can persist at room temperature and atmospheric pressure even though conversion to graphite is favored, because that conversion is very slow.
Melting pure ice at atmospheric pressure absorbs heat and increases entropy: delta H>0 and delta S>0. Above the melting temperature, T delta S exceeds delta H, so delta G<0 for melting. Below it, melting has delta G>0. Entropy concerns the number and distribution of accessible microscopic states; visual disorder is only a rough guide.
For melting above its equilibrium temperature, subtracting T delta S gives a negative change.
At equilibrium there is no thermodynamic driving force for an infinitesimal allowed change. For a reaction, its reaction Gibbs energy is zero there. A zero difference between two arbitrarily chosen states alone does not prove equilibrium. Use the actual conditions and composition, not a standard-state value alone, to judge a reaction's direction.
Within a range where delta H and delta S can be treated as constant, their signs give four cases. Negative delta H and positive delta S favor the change at every positive T in that model. Positive delta H and negative delta S never favor it. When both are positive, higher T can favor it; when both are negative, lower T can favor it.
For the constant-parameter model, set delta G=0 to find T=delta H/delta S, provided delta S is nonzero. Use matching energy units: delta H in J/mol and delta S in J/(mol K) give T in K. For delta H=40 J/mol and delta S=0.1 J/(mol K), the boundary is 400 K. Real values may vary with T, so this is an estimate over the stated range.
The equation uses matching energy units and temperature in kelvin.
An endothermic change has delta H>0. It can still be favored if delta S>0 and T delta S>delta H, making delta G negative. Heat absorption alone does not settle the direction. Calculate both contributions at the specified temperature and conditions.
At fixed T and pressure, use delta G=delta H-T delta S under the stated constraints. Its sign gives the favored direction, not the speed. Keep units and temperature assumptions explicit.
Original lesson, questions and flashcards written for this learning dataset.
The rewritten readings were written for this dataset. Videos by Khan Academy.
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