Favorable doesn’t mean it happens
ΔG° answers “how much product is there at equilibrium?”, not “how fast is equilibrium reached?” The rate depends on the activation energy Ea (U5.6): when Ea is very high, very few collisions at ordinary temperatures have enough energy, and the reaction is too slow to observe.
When a thermodynamically favorable reaction effectively doesn’t proceed because its activation energy is too high, the reaction is under kinetic control动力学控制. Such a substance is thermodynamically unstable but can be stored for a long time; it is kinetically stable动力学稳定.
To change this, you increase the rate rather than change ΔG°:
- Raise the temperature or ignite it: more molecules have enough energy, so the rate increases; once an exothermic reaction starts, the heat it releases keeps it going.
- Add a catalyst: it provides a pathway with a lower activation energy, speeding up both forward and reverse reactions; ΔG°, ΔH° and K don’t change (U5.11).
Drag ΔG°, the activation energy and the temperature, then look at K and the half-life separately to decide whether the reaction is observed:
Model: a first-order reaction with A = 1.0 × 10¹³ s⁻¹, for comparing orders of magnitude only.
Example 1: why doesn’t alcohol catch fire by itself?
For C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔG° = −1325 kJ (calculated from ΔfG° values in OpenStax Appendix G). Yet an unlit alcohol burner can sit in contact with air for months without burning. Explain this, and explain what lighting it does.
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Thermodynamics: how favorable? Find K from ΔG° (ΔG° in J, T = 298 K):
K is too large to write as an ordinary number; at equilibrium it would be almost entirely products: very favorable thermodynamically.
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Kinetics: how fast? Combustion first requires breaking bonds such as O=O and C–H, so the activation energy is high; at room temperature almost no molecules get over the barrier, and the rate is extremely slow.
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Connect to what you see. Thermodynamically favorable means “can happen”, not “happens quickly”. The reaction is under kinetic control, so no burning is seen for months.
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What lighting does. A flame makes the local temperature very high, so enough molecules get over the barrier; the heat released by burning then heats the nearby alcohol vapor, and the reaction keeps going. Lighting changes neither ΔG° nor K; it just supplies the energy to get over the barrier.
Example 2: NO in car exhaust
For 2NO(g) → N₂(g) + O₂(g), ΔG° = −175.2 kJ and K ≈ 5 × 10³⁰ (25 °C). Yet NO released into the air in exhaust doesn’t decompose quickly. A car’s three-way catalytic converter contains catalysts such as platinum and rhodium. Explain why NO can exist, and what the catalyst does.
- Thermodynamics: how favorable is decomposition? ΔG° = −175.2 kJ < 0 and K ≈ 5 × 10³⁰ ≫ 1: NO decomposing to N₂ and O₂ is very favorable thermodynamically.
- Kinetics: why NO survives. The decomposition has a high activation energy, so without a catalyst it is very slow, and NO released into the air lasts a long time: it is under kinetic control.
- What the catalyst does. Platinum and rhodium surfaces provide a pathway with a much lower activation energy, so NO decomposes during its brief time in the converter.
- What doesn’t change. A catalyst speeds up the forward and reverse reactions equally and doesn’t change ΔG°, ΔH°, or K; it only changes how fast equilibrium is reached.
- The other way around. If a reaction had ΔG° > 0 and K ≪ 1, a catalyst couldn’t make it produce much product: a catalyst only speeds up a reaction that is already favorable.
- ΔG° = −120 kJ/mol, half-life about 10⁵ years at room temperature
Favorable but far too slow.
- ΔG° = +35 kJ/mol
K < 1.
- ΔG° = −20 kJ/mol, half-life about 2 s
Favorable and fast.
- Freshly cut sodium metal tarnishes (oxidizes) within seconds
Both favorable and fast.
- N₂ and O₂ in air form NO at room temperature (ΔG° = +175 kJ)
ΔG° > 0.
- NO decomposes into N₂ and O₂ at room temperature without a catalyst
K is huge, but the activation energy is high.
- NO decomposes rapidly on a platinum–rhodium catalyst
The catalyst lowers the activation energy.
- Gasoline vapor mixed with air doesn’t burn until ignited
Combustion is favorable, but the activation energy is high.
Common mistakes
Mistake 1: “The more negative ΔG°, the faster the reaction.” ΔG° and rate aren’t linked. Combustion reactions with extremely negative ΔG° don’t proceed until ignited.
Mistake 2: “A catalyst can make a thermodynamically unfavorable reaction happen.” A catalyst speeds up the forward and reverse reactions equally and doesn’t change ΔG° or K; it only reaches the same equilibrium faster.
Mistake 3: “If no reaction is observed, K must be small.” K may be large while the rate is extremely slow (kinetic control). Consider thermodynamics and kinetics separately.
Quick check
A reaction has ΔG° = −250 kJ, yet a week after the reactants are mixed, no product can be detected. The best explanation is:
A very negative ΔG° means a large K; the missing product is a kinetic effect.
A catalyst is added to a reaction that is thermodynamically favorable but slow. What does it change?
A catalyst lowers the activation energy and speeds up the reaction; ΔG°, K and ΔH° don’t change.
At 25 °C a reaction has ΔG° = +40 kJ. Which approach could make it produce a large amount of product?
A, B and C only affect the rate, not the equilibrium; coupling can make the overall ΔG° negative (U9.6).
Whether a favorable reaction is observed within an hour at room temperature depends mainly on:
The rate depends on the activation energy.
Raising the temperature of a kinetically controlled reaction mainly:
Heating gives more molecules energy above the activation energy; the activation energy itself stays about the same.
Which statement about kinetic stability is correct?
Alcohol in air is an example: it would burn thermodynamically, but its high activation energy lets it be stored for a long time.
AP-style free response
Ozone in the stratosphere can decompose: 2O₃(g) → 3O₂(g). Data: for O₃(g), ΔfH° = 142.7 kJ/mol, ΔfG° = 163.2 kJ/mol and S° = 238.9 J/(mol·K); for O₂(g), S° = 205.2 J/(mol·K) (OpenStax Appendix G).
(a) Calculate ΔG° for this reaction at 25 °C.
(b) Estimate the order of magnitude of K at 25 °C, and state which side the equilibrium favors.
(c) In that case, why can ozone exist in the stratosphere for a long time?
(d) Cl atoms in the stratosphere catalyze the decomposition of ozone. What effect do Cl atoms have on ΔG°, K, the activation energy and the rate?
(e) Calculate ΔH° and ΔS°, and state whether the reaction is thermodynamically favorable at all temperatures.
Model answer and scoring
(a) ΔG° = 3 × 0 − 2 × 163.2 = −326.4 kJ.
(b) K = e326400/(8.314 × 298) ≈ e132 ≈ 1057, so the equilibrium strongly favors O₂.
(c) The decomposition has a high activation energy, so without a catalyst it is very slow; meanwhile ozone is continually formed from O₂ by ultraviolet light. Ozone’s existence is a result of kinetic control.
(d) ΔG° and K don’t change; Cl atoms provide a pathway with a lower activation energy, so the rate increases.
(e) ΔH° = 0 − 2 × 142.7 = −285.4 kJ; ΔS° = 3 × 205.2 − 2 × 238.9 = +137.8 J/K. With ΔH° < 0 and ΔS° > 0, ΔG° is negative at every temperature, so the reaction is favorable at all temperatures.
Scoring: (a) ΔG°; (b) order of magnitude and judgment; (c) high activation energy, slow rate; (d) all four quantities; (e) calculations and sign reasoning.
Summary
- ΔG° decides how much product there is at equilibrium; the activation energy decides how fast the reaction goes. Judge the two separately.
- A reaction that is favorable but too slow to observe because of a high activation energy is under kinetic control; such a substance is kinetically stable.
- Heating, igniting and catalysts change only the rate; a catalyst doesn’t change ΔG° or K.
AP tip: When a question asks why a reaction with ΔG° < 0 isn’t observed, write “the activation energy is high, so the rate is slow (kinetic control),” and make clear it isn’t because K is small. Don’t write “the reaction isn’t spontaneous.”
Enrichment (thermodynamic and kinetic products): If the same reactants can form two products, the one with the lower activation energy forms faster (the kinetic product), and the one with the more negative ΔG° is more stable (the thermodynamic product). At low temperatures and short times the kinetic product dominates; at high temperatures and long times the system can reach equilibrium, and the thermodynamic product dominates.