Chemistry: Atoms First 2e · Thermodynamics
Free Energy
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In 30 seconds
Free energy is the thermodynamic quantity that decides the direction of change at constant temperature and pressure — the conditions of most laboratory, industrial, and biological chemistry. For any process, the Gibbs free energy change is
ΔG = ΔH - TΔS
where ΔH is the enthalpy change, T is the absolute temperature in kelvin, and ΔS is the entropy change. The verdict is simple: ΔG < 0 means the process is spontaneous (product-favored) as written, ΔG > 0 means it is not, and ΔG = 0 marks equilibrium. Free energy also links thermodynamics to equilibrium chemistry: ΔG°= -RTlnK connects the standard free-energy change to the equilibrium constant of Chapter 13.
Why this matters
Before an engineer builds a plant or a chemist designs a synthesis, the first question is thermodynamic: can this reaction ever go? A negative ΔG answers yes. Batteries and fuel cells convert a spontaneous reaction's free energy into electrical work — the maximum useful work equals -ΔG. In biochemistry, thermodynamically uphill reactions such as protein synthesis are driven by coupling Running an unfavorable reaction alongside a favorable one Full entry → to ATP hydrolysis, whose ΔG is large and negative. On exams, free energy unifies entropy, enthalpy, and equilibrium constants: compute ΔG and you can predict reaction direction at any temperature.
The college version
Core Concepts
The Gibbs free energy as a direction meter
At constant T, P, a process is spontaneous only if the total entropy of the universe increases (second law). ΔG is the bookkeeping that applies that law to the system alone: a negative ΔG means the universe's entropy increases if the process runs. It packages enthalpy and entropy effects into one number, with temperature weighting the entropy term.
The four sign combinations of ΔH and ΔS
Because ΔG = ΔH - TΔS, the signs of ΔH and ΔS control how temperature changes the verdict:
| ΔH | ΔS | Behavior of ΔG |
|---|---|---|
| negative | positive | negative at every T — spontaneous at all temperatures |
| positive | negative | positive at every T — never spontaneous |
| negative | negative | spontaneous only below T = ΔH/ΔS |
| positive | positive | spontaneous only above T = ΔH/ΔS |
The two "mixed" rows give a crossover temperature T = ΔH/ΔS where ΔG = 0 — for a phase change, that is the melting or boiling point.
Standard free energies of formation
Tabulated ΔG°f values — the free energy of forming a compound from its elements in standard states, with ΔG°f = 0 for elements — give standard reaction free energies without experiments:
ΔG°rxn = ∑n ΔG°f(products) - ∑n ΔG°f(reactants)
Equivalently, ΔG°= ΔH°- TΔS° uses enthalpy and entropy tables. Standard state: 1 bar pressure (often approximated as 1 atm), 1 M solutes, pure phases, usually 298.15 K.
Free energy, equilibrium, and coupling
ΔG°= -RTlnK means a strongly negative ΔG° corresponds to a large equilibrium constant, and ΔG°= 0 corresponds to K = 1. Under non-standard conditions,
ΔG = ΔG°+ RTlnQ
where Q is the reaction quotient. Cells exploit Q by keeping concentrations far from equilibrium values. Coupling an unfavorable reaction (positive ΔG) to a strongly favorable one (negative ΔG) makes the pair net-spontaneous — how ATP hydrolysis drives biosynthesis.
How It Works / Step-by-Step Process
- Write the balanced reaction with physical states.
- Look up ΔH°f and S° (or ΔG°f) for every species; elements are 0.
- Compute ΔH°= ∑nΔH°f(products) - ∑nΔH°f(reactants), and ΔS° the same way.
- Convert ΔS° from J to kJ, evaluate ΔG°= ΔH°- TΔS°, and apply the verdict; for mixed signs, solve T = ΔH/ΔS.
Common Confusions
| Common Confusion | Correct Understanding |
|---|---|
| "Spontaneous means fast." | It means self-driven once started; diamond → graphite is spontaneous yet glacially slow. |
| "A reaction needs ΔH < 0 to be spontaneous." | Endothermic processes (melting, dissolving cold packs) are spontaneous when TΔS dominates. |
| "ΔH and ΔS can just be added directly." | Units must match: convert J to kJ (or vice versa) before computing ΔH - TΔS. |
| "ΔG°< 0 guarantees the reaction runs as written." | ΔG° is standard-state only; real concentrations require ΔG = ΔG°+ RTlnQ. |
| "ΔG = 0 means nothing happens." | It means equilibrium: forward and reverse rates are equal and concentrations hold steady. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A ball rolls downhill by itself and uphill only with a push. Gibbs free energy is the reaction's "hill height": if the bottom is lower than the top (ΔG < 0), the reaction rolls forward by itself. Temperature changes the hill's height — why water freezes when cold and melts when warm.
Worked examples
Consider H2(g) + 12O2(g) → H2O(l) at 298.15 K.
Step 1 — entropy change. With S°(H2) = 130.7, S°(O2) = 205.2, S°(H2O, l) = 69.9 J mol⁻¹ K⁻¹:
ΔS°= 69.9 - (130.7 + 12(205.2)) J mol-1K-1 = 69.9 - 233.3 = -163.4 J mol-1K-1
Step 2 — combine with enthalpy. ΔH°f(H2O, l) = -285.8 kJ mol⁻¹. Convert the entropy term to kJ: -163.4 J mol-1K-1 = -0.1634 kJ mol-1K-1, then substitute:
ΔG°= -285.8 kJ mol-1 - (298.15 K)(-0.1634 kJ mol-1K-1) = -285.8 + 48.7 = -237.1 kJ mol-1
Dimensional analysis: K × kJ mol⁻¹ K⁻¹ = kJ mol⁻¹, so the subtraction is valid, and the result matches the tabulated ΔG°f(H2O, l) = -237.1 kJ mol⁻¹ — a built-in self-check. Water formation is strongly spontaneous, though H₂–O₂ mixtures wait for a spark (activation energy).
Freezing, H2O(l) → H2O(s), has ΔH = -6.01 kJ mol⁻¹ and ΔS = -22.0 J mol⁻¹ K⁻¹ — both signs negative, so spontaneous only below T = ΔH/ΔS.
At 263 K (below the freezing point):
ΔG = -6.01 kJ mol-1 - (263 K)(-0.0220 kJ mol-1K-1) = -6.01 + 5.79 = -0.22 kJ mol-1
At 263 K freezing is spontaneous. At 298 K:
ΔG = -6.01 kJ mol-1 - (298 K)(-0.0220 kJ mol-1K-1) = -6.01 + 6.56 = +0.55 kJ mol-1
Now melting is spontaneous — the same substances flip direction purely because T changed, exactly as the sign table predicts for ΔH < 0, ΔS < 0.
For CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), with ΔG°f = −50.8 (CH₄), −394.4 (CO₂), −228.6 (H₂O, g) kJ mol⁻¹:
ΔG°rxn = [(-394.4) + 2(-228.6)] - [(-50.8) + 2(0)] = -851.6 + 50.8 = -800.8 kJ mol-1
One mole of methane releases about 801 kJ of free energy.
Key takeaways
- ΔG = ΔH - TΔS; ΔG < 0 spontaneous, ΔG > 0 non-spontaneous, ΔG = 0 equilibrium (constant T, P).
- Spontaneous ≠ fast. Activation energy controls speed; ΔG controls direction.
- Memorize the four sign cases and the crossover temperature T = ΔH/ΔS.
- ΔG°rxn = ∑nΔG°f(products) - ∑nΔG°f(reactants); elements have ΔG°f = 0.
- ΔG°= -RTlnK: negative ΔG° → K > 1; ΔG°= 0 → K = 1.
Check yourself
4 review questions from the chapter. Try each one, then open the answer.
State the spontaneity criterion in terms of ΔG, and name the conditions under which it applies.
Show answer
ΔG < 0: spontaneous; ΔG > 0: non-spontaneous; ΔG = 0: equilibrium — all at constant temperature and pressure.
A reaction has ΔH°= +40.0 kJ and ΔS°= +0.200 kJ K⁻¹. Above what temperature does it become spontaneous?
Show answer
T > ΔH/ΔS = 40.0 kJ / 0.200 kJ K-1 = 200 K.
Why must you convert ΔS from J/K to kJ/K before using ΔG = ΔH - TΔS?
Show answer
Because ΔH is in kJ while TΔS would otherwise be in J; both terms must share units before subtracting (163.4 J = 0.1634 kJ).
Ice melts spontaneously at 10 °C but not at −10 °C. Which quantity in ΔG = ΔH - TΔS changes with temperature, and how does it flip the sign?
Show answer
The temperature in the -TΔS term changes. With ΔH < 0 and ΔS < 0, raising T makes -TΔS more positive until ΔG flips from negative (spontaneous freezing) to positive (spontaneous melting).
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Gibbs free energy (G)
- Energy available to do useful work at constant T, P
- spontaneous process
- A process that proceeds without outside intervention once started
- Δ G°f
- Free energy of forming a compound from its elements in standard states
- coupling
- Running an unfavorable reaction alongside a favorable one
- Gibbs free energy (ΔG°)
- The maximum useful work available from a reaction under standard conditions.
Sources & references
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