Chemistry 2e · Thermodynamics
Free Energy
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In 30 seconds
The second law gives the complete condition for spontaneity, ΔSuniv > 0, but it is awkward because it requires tracking both the system and its surroundings. Gibbs free energy, defined as G = H - TS, packages the same information into a property of the system alone. At constant temperature and pressure — the conditions of most chemical reactions — the change in free energy is
ΔG = ΔH - TΔS
The sign of ΔG gives the direction: negative means the forward reaction is spontaneous, positive means the reverse is favored, zero means equilibrium. Because ΔH and ΔS both appear, free energy reconciles enthalpy and entropy effects: an endothermic reaction can still be spontaneous if its entropy gain is large enough.
Free energy also connects thermodynamics to equilibrium: under standard conditions, ΔG° determines the equilibrium constant through ΔG°= -RTlnK, and the temperature dependence of ΔG explains why some reactions switch direction as temperature changes.
Why this matters
- One number, one answer: ΔG collapses "system plus surroundings" into a single sign you can compute from tabulated data.
- Engineering and industry: Knowing how ΔG changes with temperature lets engineers choose conditions that push a reaction the wanted way.
- Biochemistry and medicine: Cells run on coupled reactions — ATP hydrolysis drives otherwise nonspontaneous reactions like protein synthesis and muscle contraction.
- Batteries and corrosion: Electrochemistry uses the relationship between free energy and cell potential to predict how much electrical work a battery can do.
- Exams: Expect to compute ΔG° from formation values or from ΔH° and ΔS°, and to relate ΔG° to K.
The college version
Core Concepts
What "free" means
The Gibbs free energy G = H - TS represents the maximum useful (non-expansion) work a process can perform at constant temperature and pressure. "Free" means available for work; the rest is tied up in entropy costs. For a process at constant T,
ΔG = ΔH - TΔS
with all quantities for the system. This is the central equation of the topic.
Spontaneity by sign
- ΔG < 0: the forward process is spontaneous (thermodynamically favored).
- ΔG > 0: the reverse process is spontaneous; the forward direction requires energy input.
- ΔG = 0: the system is at equilibrium.
Because ΔG carries both terms, it resolves apparent paradoxes: ice melting above 0 °C is endothermic (ΔH > 0) yet spontaneous because TΔS outweighs ΔH.
Standard free energy of formation
Standard free energies of formation, ΔG°f, are tabulated for pure substances in standard states (commonly 1 atm, 298 K). By convention, ΔG°f = 0 for elements in their standard states. For a reaction,
ΔG°rxn = ∑nΔG°f(products) - ∑mΔG°f(reactants)
Temperature dependence: the four cases
Because ΔG = ΔH - TΔS, the signs of ΔH and ΔS decide how temperature changes the outcome:
| ΔH | ΔS | Behavior |
|---|---|---|
| negative | positive | Spontaneous at all temperatures |
| positive | negative | Nonspontaneous at all temperatures |
| negative | negative | Spontaneous at low T only |
| positive | positive | Spontaneous at high T only |
The crossover temperature (where ΔG = 0, i.e., equilibrium) is
T = ΔHΔS
assuming ΔH and ΔS are roughly constant over the temperature range.
Free energy and equilibrium
Under standard conditions, the equilibrium constant K relates to ΔG°:
ΔG°= -RTlnK
A very negative ΔG° corresponds to a large K (products favored); a very positive one to a small K. For nonstandard conditions, ΔG = ΔG°+ RTlnQ, where Q is the reaction quotient from the equilibrium chapter.
How It Works / Step-by-Step Process
- Write the balanced equation with physical states.
- Choose a path: ΔG°f values, ΔH°/ ΔS° with T, or K via ΔG°= -RTlnK.
- If using ΔH° and ΔS°, convert ΔH to joules to match TΔS, then subtract.
- Interpret the sign; for a crossover temperature, set ΔG = 0 and solve T = ΔH/ΔS.
- For equilibrium questions, use K = e-ΔG°/(RT) with R = 8.314 J/(mol·K) and T in kelvin.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| ΔG < 0 | "The reaction is fast." | Thermodynamic favorability says nothing about rate; kinetics governs speed. |
| ΔG | ΔG° | ΔG° is for standard-state conditions; ΔG includes nonstandard concentrations via RTlnQ. |
| ΔS°= 0 for elements | ΔG°f = 0 for elements | Elements in standard states have zero formation free energy but nonzero standard entropy. |
| "Endothermic = nonspontaneous" | "Endothermic reactions can be spontaneous" | If ΔS > 0 and TΔS > ΔH, ΔG < 0 even with ΔH > 0. |
| T = ΔH/ΔS always | Only when ΔG = 0 | That expression is the equilibrium/crossover temperature, valid when ΔH and ΔS share a sign. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Free energy is like a "spontaneity score" for a reaction. It combines whether the reaction gives off energy (a ball rolling downhill) and whether it makes more disorder (scattering LEGO bricks). Negative score: the reaction happens on its own. Positive: you have to push it. Zero: balanced like a seesaw.
Worked example
Example 1: ΔG° from formation values
For the oxidation of nitrogen monoxide, 2NO(g) + O2(g) → 2NO2(g), use ΔG°f(NO) = +86.6 kJ/mol and ΔG°f(NO2) = +51.3 kJ/mol; oxygen (an element in its standard state) has ΔG°f = 0.
Formula:
ΔG°rxn = ∑nΔG°f(products) - ∑mΔG°f(reactants)
Substitution:
ΔG°rxn = 2(+51.3) - [2(+86.6) + 1(0)] = 102.6 - 173.2 = -70.6 kJ
ΔG°< 0: the reaction is spontaneous under standard conditions — which is why NO is thermodynamically unstable toward oxidation in air.
Example 2: Crossover temperature
A reaction has ΔH°= +177.8 kJ and ΔS°= +160.5 J/K (both positive — spontaneous only at high temperature). Find the temperature above which it becomes spontaneous.
Formula:
T = ΔH°ΔS°
Substitution (converting kJ to J so units cancel):
T = 177.8 × 103 J160.5 J/K = 1108 K
Above ~1108 K, ΔG < 0 and the reaction proceeds; below it, the reverse is favored.
Example 3: ΔG° from K
A reaction has K = 1.5 × 1010 at 298 K. Find ΔG°.
Formula:
ΔG°= -RTlnK
Substitution:
ΔG°= -(8.314 J/(mol·K))(298 K)ln(1.5 × 1010)
ΔG°= -(2478 J/mol)(23.4) = -5.8 × 104 J/mol = -58 kJ/mol
A large K means a strongly negative ΔG° — products dominate.
Key takeaways
- ΔG = ΔH - TΔS; ΔG < 0 spontaneous, > 0 nonspontaneous, = 0 equilibrium.
- ΔG°rxn = ΣnΔG°f(products) - ΣmΔG°f(reactants); elements have ΔG°f = 0.
- ΔG°= -RTlnK: negative ΔG° means large K; positive means small K.
- Crossover temperature T = ΔH/ΔS marks where a reaction switches spontaneity (when signs of ΔH and ΔS match).
- When ΔH and ΔS have opposite signs, temperature does not change the direction of spontaneity.
- ΔG says nothing about rate: a spontaneous reaction can be imperceptibly slow (diamond → graphite).
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Write the equation for ΔG at constant temperature and state the spontaneity rule.
Show answer
ΔG = ΔH - TΔS; ΔG < 0 spontaneous, ΔG > 0 nonspontaneous, ΔG = 0 equilibrium.
Why is ΔG°f = 0 for an element like O2(g) in its standard state?
Show answer
By convention, the free energy of formation of an element in its standard state is defined as zero — the reference point for all ΔG°f values.
A reaction has ΔH°= -100 kJ and ΔS°= +50 J/K. Is it spontaneous at all temperatures? Why?
Show answer
Yes. ΔH < 0 and ΔS > 0 both favor spontaneity at every temperature (opposite signs of ΔH and ΔS make the direction temperature-independent).
How does a very large K (e.g., 1010) manifest in ΔG°?
Show answer
A very large K gives a very negative ΔG°= -RTlnK, meaning products strongly dominate at equilibrium.
What is the crossover temperature for a reaction with ΔH°= +200 kJ and ΔS°= +100 J/K?
Show answer
T = 200 × 103 J / 100 J/K = 2000 K.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Gibbs free energy, G
- H - TS; energy available to do useful work at constant T, P.
- standard free energy of formation, Δ G°f
- Free energy change when one mole of a compound forms from its elements in standard states.
- standard free energy change, Δ G°
- ΔG for a reaction with all reactants and products in standard states.
- equilibrium constant, K
- Ratio of product to reactant activities at equilibrium.
- coupled reaction
- A spontaneous reaction (e.g., ATP hydrolysis) driving a nonspontaneous one.
- equilibrium constant (K)
- A temperature-dependent ratio of product-to-reactant concentrations (or pressures) at equilibrium
Sources & references
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