Chemistry 2e · Thermodynamics
Spontaneity
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A spontaneous process A process that proceeds without continuous outside intervention Full entry → is one that proceeds on its own once it has started, without needing continuous outside intervention. Heat flows from a hot object to a cold one; a gas expands into a vacuum; ice melts above 0 °C. None of these needs a push — each happens by itself. Crucially, spontaneous does not mean fast: a diamond converts to graphite spontaneously at room temperature, yet the process is so slow that diamonds last forever on human timescales. Thermodynamics tells you the direction a process tends to go; kinetics (Chapter 12) tells you how quickly it gets there.
Why does the universe favor some processes and not others? The first law of thermodynamics only conserves energy — it never says which direction a change will take. Direction is decided by a combined accounting of enthalpy (heat at constant pressure, Chapter 5) and entropy (energy dispersal, next topic). That accounting is the Gibbs free energy,
ΔG = ΔH - TΔS
with the simple verdict: ΔG < 0 means the process is spontaneous at constant temperature and pressure; ΔG > 0 means it is not; ΔG = 0 means equilibrium State where ΔG = 0 and forward/reverse rates match Full entry →.
Why this matters
Free-energy reasoning decides whether a reaction can ever be made to work, which is the first question of industrial chemistry. Ammonia synthesis, iron smelting, and cement production are all analyzed as ΔG problems before anyone builds a plant. In the body, thermodynamically unfavorable reactions (building proteins, concentrating ions) are driven by coupling to ATP hydrolysis, which has a large negative ΔG. Batteries work because the cell reaction has ΔG < 0, and the maximum electrical work available is directly proportional to -ΔG. Even everyday questions — will this ice melt, will this salt dissolve, will this metal corrode — are ΔG questions.
The college version
Core Concepts
What "spontaneous" means precisely
A spontaneous process proceeds without outside intervention after it is initiated. The classic clues are qualitative: systems tend toward lower potential energy (exothermic tends to be favorable) and toward more dispersal (more entropy tends to be favorable). But neither tendency alone is decisive — the balance is what matters, which is why some endothermic processes (melting ice, dissolving NH₄NO₃ in a cold pack) are spontaneous.
The first law cannot predict direction
Energy conservation is symmetric: a ball can roll down a hill or up it without violating energy conservation, but only one direction happens by itself. Thermodynamics needs a second criterion beyond energy — the entropy of the universe — to fix the arrow of time. The free-energy criterion ΔG < 0 is the constant-pressure, constant-temperature version of that second-law reasoning.
Gibbs free energy combines enthalpy and entropy
For a process at constant temperature and pressure,
ΔG = ΔH - TΔS
where ΔH is the enthalpy change (J or kJ per mole of reaction), T is the absolute temperature in kelvin, and ΔS is the entropy change (J K⁻¹ per mole of reaction). The sign combinations decide the temperature behavior:
| ΔH | ΔS | ΔG behavior |
|---|---|---|
| negative | positive | always negative — spontaneous at all T |
| positive | negative | always positive — never spontaneous |
| negative | negative | spontaneous only below T = ΔH/ΔS |
| positive | positive | spontaneous only above T = ΔH/ΔS |
Standard free energy changes
Standard molar free energies of formation, ΔG°f, are tabulated (0 for elements in their standard states), and
ΔG°rxn = ∑n ΔG°f(products) - ∑n ΔG°f(reactants)
Equivalently, from enthalpy and entropy data:
ΔG°= ΔH°- TΔS°
And since ΔG°= -RTlnK, the equilibrium constant of Chapter 13 is just free energy in disguise: reactions with very negative ΔG° have very large K.
How It Works / Step-by-Step Process
- Write the balanced reaction and note the physical states.
- Look up ΔH°f and S° (or ΔG°f) for every species; ΔH°f = 0 and ΔG°f = 0 for elements in standard states.
- Compute ΔH°= ∑nΔH°f(products) - ∑nΔH°f(reactants) and similarly for ΔS°.
- Convert units so ΔH and TΔS match (usually kJ), then evaluate ΔG°= ΔH°- TΔS°.
- Apply the verdict: negative → spontaneous at that temperature; positive → not; zero → equilibrium. For mixed signs, solve T = ΔH/ΔS for the crossover temperature.
Common Confusions
| Common Confusion | Correct Understanding |
|---|---|
| "Spontaneous means fast." | It means self-driven. Diamond → graphite is spontaneous but takes eons; activation energy controls speed. |
| "A reaction needs ΔH < 0 to be spontaneous." | Many spontaneous processes are endothermic (melting, dissolving cold packs); ΔS can dominate. |
| "ΔG°< 0 guarantees the reaction runs under any conditions." | ΔG° describes standard conditions only. Under real concentrations, use ΔG = ΔG°+ RTlnQ. |
| "ΔH and ΔS can just be added directly." | Units must match: convert J to kJ (or vice versa) so that TΔS and ΔH are in the same units. |
| "A spontaneous process needs no energy input at all." | It needs none to sustain itself once started, but it may still need activation energy to begin — a spark for methane combustion. |
| "ΔG = 0 means nothing happens." | It means equilibrium: forward and reverse proceed at equal rates; macroscopic concentrations are constant. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A ball on a hill rolls downhill by itself — that's spontaneous. It can be pushed uphill, but it needs help, and the higher the hill, the more help. Gibbs free energy is the height of the hill: if the bottom is lower than the top (ΔG < 0), the ball rolls down on its own. Rolling downhill has nothing to do with how fast the ball rolls — a slow roll is still downhill.
Worked examples
Consider H2(g) + 12O2(g) → H2O(l) at 298.15 K.
Step 1 — entropy change. Standard molar entropies: S°(H2) = 130.7, S°(O2) = 205.2, S°(H2O, l) = 69.9 J mol⁻¹ K⁻¹.
ΔS°= S°(H2O,l) - [S°(H2) + 12S°(O2)]
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. Convert ΔS° to kJ:
ΔG°= ΔH°- TΔS°= -285.8 kJ mol-1 - (298.15 K)(-0.1634 kJ mol-1K-1)
ΔG°= -285.8 + 48.7 = -237.1 kJ mol-1
The units cancel correctly (K × kJ mol⁻¹ K⁻¹ = kJ mol⁻¹), and the result matches the tabulated ΔG°f(H2O,l) = -237.1 kJ/mol — a nice self-check. Water formation is strongly spontaneous, which is why hydrogen and oxygen mixtures react so violently once ignited.
The Haber reaction N2(g) + 3H2(g) → 2NH3(g) has ΔH°= -91.8 kJ per mole of reaction and ΔS°= -198.1 J mol⁻¹ K⁻¹. At 298.15 K:
ΔG°= -91.8 kJ - (298.15 K)(-0.1981 kJ K-1) = -91.8 + 59.1 = -32.7 kJ
Spontaneous at room temperature. But both signs are negative, so there is a crossover temperature:
T = ΔHΔS = -91.8 kJ-0.1981 kJ K-1 = 463 K
Above about 463 K (at standard pressures) the reaction becomes non-spontaneous — yet industrial ammonia plants run near 700 K! The resolution is kinetic: below ~600 K the reaction is too slow without a catalyst, so plants accept a thermodynamically less favorable temperature and use high pressure plus an iron catalyst to push the equilibrium and speed the reaction. This is the classic lesson that thermodynamics sets the limit while kinetics sets the practice.
Freezing is the reverse of fusion: ΔH = -6.01 kJ/mol and ΔS = -22.0 J mol⁻¹ K⁻¹ for H2O(l) → H2O(s).
At 263 K (below the freezing point):
ΔG = -6.01 kJ - (263 K)(-0.0220 kJ K-1) = -6.01 + 5.79 = -0.22 kJ
Freezing is spontaneous — water freezes. At 298 K:
ΔG = -6.01 kJ - (298 K)(-0.0220 kJ K-1) = -6.01 + 6.56 = +0.55 kJ
Now melting is spontaneous — ice melts. The same reaction flips direction purely because T changed, exactly as the sign table predicts for ΔH < 0, ΔS < 0.
Key takeaways
- Spontaneous = proceeds without outside intervention once started; not a statement about speed.
- ΔG < 0: spontaneous (product-favored) at constant T, P. ΔG > 0: non-spontaneous. ΔG = 0: equilibrium.
- ΔG = ΔH - TΔS — memorize the sign table for the four ΔH/ΔS combinations.
- Exothermic (ΔH < 0) and entropy-increasing (ΔS > 0) processes are spontaneous at every temperature.
- Endothermic processes can still be spontaneous when ΔS > 0 is large (melting, dissolving).
- Standard state is 1 bar (often approximated as 1 atm) and usually 298.15 K; ΔG° refers to standard conditions only.
- Temperature crossover: T = ΔH/ΔS (when both signs match) is where ΔG changes sign.
- ΔG°= -RTlnK: big negative ΔG° → large K; ΔG°= 0 → K = 1.
- Units trap: ΔH in kJ and ΔS in J/K — convert ΔS to kJ/K before combining.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
Define spontaneous process and give one example that is spontaneous but slow.
Show answer
A spontaneous process proceeds without outside intervention once started. Diamond converting to graphite is spontaneous but extremely slow at room temperature.
For each (ΔH, ΔS) pair — (+, +), (−, −), (+, −) — state when ΔG < 0.
Show answer
(+, +): spontaneous only above T = ΔH/ΔS. (−, −): spontaneous only below T = ΔH/ΔS. (+, −): never spontaneous at any temperature.
Why must you convert ΔS from J/K to kJ/K before using ΔG = ΔH - TΔS?
Show answer
Because ΔH is in kJ and TΔS would otherwise be in J; matching units is required before subtracting. (100 J = 0.1 kJ.)
A reaction has ΔH°= +50.0 kJ and ΔS°= +100.0 J K⁻¹. Above what temperature does it become spontaneous?
Show answer
T > ΔH/ΔS = 50.0 kJ / 0.100 kJ K-1 = 500 K.
How is the equilibrium constant K related to ΔG°?
Show answer
ΔG°= -RTlnK. Large negative ΔG° → large K; ΔG°= 0 → K = 1.
Ice melts spontaneously at 10 °C but not at −10 °C. Which quantity in ΔG = ΔH - TΔS changes, and what direction does it flip the sign?
Show answer
T changes; the -TΔS term grows with temperature. With ΔH < 0 and ΔS < 0, raising T eventually makes -TΔS positive enough to flip ΔG from negative to positive.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- spontaneous process
- A process that proceeds without continuous outside intervention
- Gibbs free energy (G)
- Energy available to do useful work; its change decides spontaneity
- Δ G = Δ H - TΔ S
- The free-energy equation linking enthalpy, temperature, and entropy
- standard free energy (Δ G°)
- Free-energy change with all species in standard states
- equilibrium
- State where ΔG = 0 and forward/reverse rates match
- activation energy
- Kinetic barrier a reaction must climb before proceeding
- Gibbs free energy, G
- H - TS; energy available to do useful work at constant T, P.
Sources & references
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