Spontaneous Reactions: When ΔG Goes Negative
Correct one word and a whole topic clicks: in thermodynamics, spontaneity is about direction, not speed.
In thermodynamics, a reaction is spontaneous when ΔG < 0 under the conditions in question: it can proceed without a continuous supply of outside energy. That says nothing about how fast it goes. Since ΔG = ΔH − TΔS, there are four cases depending on the signs of ΔH and ΔS, and in two of them temperature decides the outcome, with the switch near T = ΔH/ΔS.
"Spontaneous" misleads everyone
In everyday speech, spontaneous means sudden. A spontaneous decision. A spontaneous round of applause.
In chemistry it means something narrower. A spontaneous process is one that occurs naturally under certain conditions; a nonspontaneous one won't happen unless it is driven by the continual input of energy from an external source. OpenStax's textbook gives the image: water flows downhill by itself, but uphill flow needs a pump (OpenStax Chemistry 2e, 16.1).
The same text is blunt about the other half: "The spontaneity of a process is not correlated to the speed of the process." Its example is radioactive decay, which is spontaneous by definition. Technetium-99m, used in medical imaging, has a half-life of about six hours. Uranium-238 has a half-life of more than four billion years. Both are spontaneous (OpenStax, 16.1).
The rule: ΔG < 0, under these conditions
The test comes from Gibbs free energy. At constant temperature and pressure:
- ΔG < 0: spontaneous in the forward direction
- ΔG > 0: nonspontaneous forward, which means spontaneous in reverse
- ΔG = 0: equilibrium, with no net change in either direction
The sign of ΔG is the whole rule. It isn't a separate law, either. It is the second law of thermodynamics rewritten for a beaker. The entropy change of the universe is the system's entropy change plus the surroundings', and at constant pressure the surroundings gain −ΔH/T. Multiply through by −T and you get −TΔS_universe = ΔH − TΔS = ΔG (OpenStax, 16.4). A negative ΔG is the universe's entropy going up, expressed in terms you can measure on the system alone.
Notice the phrase "under these conditions". A process that is spontaneous one way is nonspontaneous the other way under the same conditions. At room temperature ice melts spontaneously; at −10 °C it doesn't. Change the conditions and you can change the answer.
The four cases
Because ΔG = ΔH − TΔS, the signs of ΔH and ΔS decide almost everything. OpenStax lays out the four possibilities (16.4):
| ΔH | ΔS | At low T | At high T | Example |
|---|---|---|---|---|
| − | + | Spontaneous | Spontaneous | 2C(s) + O₂(g) → 2CO(g): releases heat and gains a mole of gas |
| + | − | Nonspontaneous | Nonspontaneous | The reverse: 2CO(g) → 2C(s) + O₂(g) |
| − | − | Spontaneous | Nonspontaneous | Iron rusting in a hand warmer: 4Fe + 3O₂ → 2Fe₂O₃; water freezing |
| + | + | Nonspontaneous | Spontaneous | Ice melting; water boiling |
In the first two rows the terms agree, and temperature can't change the verdict. In the last two they disagree, and temperature is the referee. "Low" and "high" here aren't absolute. They mean below or above one specific crossover temperature, which is different for every reaction.
Finding the crossover temperature
At the crossover, ΔG = 0, so ΔH = TΔS and:
T = ΔH / ΔS
OpenStax works this for boiling water using standard data: ΔH° = 44.01 kJ/mol and ΔS° = 118.8 J/(mol·K) (16.4, Example 16.11).
T = 44,010 ÷ 118.8 ≈ 370.5 K, or 97.3 °C
The accepted boiling point is 373.2 K (100.0 °C). Close, and the small miss has a clear cause: the ΔH and ΔS values were measured at 298 K, not at the boiling point, and both drift a little with temperature.
Above 370 K or so, the TΔS term outweighs ΔH and boiling is spontaneous. Below it, ΔH wins and steam condenses. Ice works the same way, with the crossover near 273 K. The Gibbs free energy guide runs those numbers at ±10 °C.
Two cautions when you do this yourself. Convert ΔH to joules first, because ΔS is almost always tabulated in J/(mol·K). And check the signs before dividing: if ΔH and ΔS have opposite signs, T comes out negative, which is the equation telling you there is no crossover.
Spontaneous isn't fast
Here is the confusion worth killing. ΔG tells you which direction is downhill. It says nothing about the height of the hill in between, the activation energy, and that barrier controls the rate.
The textbook case is carbon. Diamond turning into graphite is spontaneous at room temperature and pressure, by about 2.9 kJ/mol, yet its rate is so slow that, as OpenStax says, "for all practical purposes diamonds are indeed 'forever'" (16.1). The diamond essay has the measured barrier, which is hundreds of times bigger than the prize.
Heat and catalysts speed reactions up by getting more molecules over the barrier, or by lowering it. Neither changes ΔG. A catalyst can make a spontaneous reaction fast; it can't make a nonspontaneous reaction go.
ΔG versus ΔG°: why positive ΔG° doesn't mean "never"
The rule uses ΔG, the free-energy change under the actual conditions. Tables give ΔG°, the value under standard conditions: 1 bar for gases, 1 M for solutes. They are linked by:
ΔG = ΔG° + RT ln Q
where Q is the reaction quotient, the ratio of products to reactants right now (OpenStax, 16.4). When Q is small (few products), ln Q is negative and pulls ΔG down. When Q is large, it pushes ΔG up. At equilibrium Q equals K and ΔG is zero, which is the subject of the essay on chemical equilibrium.
Take ammonia decomposing: 2NH₃(g) → 3H₂(g) + N₂(g). Its ΔG° is +33.0 kJ/mol, so under standard conditions it doesn't go (OpenStax, Example 16.12). OpenStax's worked example, with plenty of ammonia and little hydrogen or nitrogen around, still gives ΔG = +9.68 kJ/mol at 25 °C: nonspontaneous.
Now make the products scarcer. With ammonia at 1 atm and hydrogen and nitrogen each at 0.01 atm, Q = (0.01³ × 0.01) ÷ 1² = 10⁻⁸. Then:
ΔG = 33.0 + (8.314 × 298 × ln 10⁻⁸) ÷ 1,000 ≈ 33.0 − 45.6 ≈ −12.6 kJ/mol
Negative. With products swept away, the "nonspontaneous" reaction runs. That is the same lever Le Chatelier's principle describes: remove a product and the system shifts to make more. A positive ΔG° tells you where the standard state sits, not what will happen in your flask.
Spontaneity in cells
Living cells constantly run reactions with positive ΔG. They do it by coupling. Making glutamine from glutamic acid and ammonia costs +14.2 kJ/mol on its own; hydrolysing ATP releases about −30.5 kJ/mol under standard conditions. Glutamine synthetase links the two through a phosphorylated intermediate, so the combined reaction is downhill: 14.2 − 30.5 = −16.3 kJ/mol (Alberts et al., Essential Cell Biology, ch. 3). The free energies add because G is a state function, the same reason OpenStax can couple the uphill step of getting zinc from zinc sulfide to a downhill one (16.4).
The checklist
A reaction is spontaneous when ΔG is negative under the actual conditions. Not when it is fast, not when ΔG° is negative, and not forever: it runs until ΔG reaches zero. Check the signs of ΔH and ΔS, find the crossover if they disagree, and then ask the separate question of whether kinetics will let it happen.
The pebble: spontaneity is a compass, not a clock.
FAQ
Does spontaneous mean fast?
No. Spontaneous means ΔG < 0, so the process can run without continuous outside energy. Speed depends on the activation energy. Diamond turning into graphite and uranium-238 decaying are both spontaneous and both extraordinarily slow.
Can a reaction with positive ΔH be spontaneous?
Yes, if ΔS is positive and the temperature is high enough that TΔS outweighs ΔH. Ice melting and water boiling both absorb heat and are spontaneous above their crossover temperatures.
How do you find the temperature at which a reaction becomes spontaneous?
Set ΔG = 0 and solve T = ΔH/ΔS, with ΔH in joules and ΔS in J/(mol·K). This only works when ΔH and ΔS have the same sign. For boiling water, standard data give about 370.5 K, close to the real 373.2 K.
What is the difference between ΔG and ΔG°?
ΔG° is the free-energy change at standard conditions (1 bar, 1 M). ΔG is the change under the actual conditions. They are linked by ΔG = ΔG° + RT ln Q, so a reaction with positive ΔG° can still run when products are scarce.
Is a spontaneous reaction irreversible?
Not necessarily. A spontaneous reaction runs until it reaches equilibrium, where ΔG = 0. Change the conditions, by adding products or changing the temperature, and the reverse direction can become the spontaneous one.
Further Reading
- Flowers, P. et al. Chemistry 2e, OpenStax. Section 16.1, "Spontaneity." https://openstax.org/books/chemistry-2e/pages/16-1-spontaneity
- Flowers, P. et al. Chemistry 2e, OpenStax. Section 16.4, "Free Energy." https://openstax.org/books/chemistry-2e/pages/16-4-free-energy
- Alberts, B. et al. Essential Cell Biology, 6th ed. W. W. Norton. Chapter 3, section "Activated Carriers and Biosynthesis." https://nerd.wwnorton.com/ebooks/epub/ecb6/EPUB/content/3.3-chapter03.xhtml
This is Part 3 of the Gibbs Free Energy series. Next: Temperature's Hidden Power: The Gibbs-Helmholtz Equation
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