Unit 9 · Topic 9.4 Beta

Thermodynamic and Kinetic Control

A negative ΔG° means a process is thermodynamically favored, not that it is fast.

Practice 4: Model AnalysisPractice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

A diamond is "forever" in the advertisements, but thermodynamics says it should slowly turn into graphite, the gray stuff in a pencil. It never does in any time you could wait. The gap between what is favored and what actually happens is the subject of this short topic.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the activation energy of a reaction?

  1. The minimum energy colliding particles need to react
  2. The total energy released by the reaction
  3. The energy of the products minus the reactants
  4. The energy of the reactants
Show the answer

Only collisions with at least Ea of energy (and the right orientation) lead to reaction.

  • Correct: The minimum energy colliding particles need to react:
  • The total energy released by the reaction:
  • The energy of the products minus the reactants:
  • The energy of the reactants:

2. A process is thermodynamically favored when…

  1. ΔG° < 0
  2. ΔH° > 0
  3. ΔS° < 0
  4. ΔG° > 0
Show the answer

ΔG° = ΔH° − TΔS° < 0 is the test for favorability.

  • Correct: ΔG° < 0:
  • ΔH° > 0:
  • ΔS° < 0:
  • ΔG° > 0:

Part 4 · See it

See it first

An energy profile: reactants on the left sit higher than products on the right, so ΔG° is negative, but a very tall hump labeled large Ea lies between them, so few collisions get over it and the reaction is too slow to observe.
Products lie below reactants (ΔG° < 0), but a tall barrier means almost no collisions get over it at room temperature. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. ΔG° compares only the starting and ending statesit says whether a reaction can go, not how fast
  2. The rate depends on how many collisions have at least the activation energya large Ea makes the rate tiny, however negative ΔG° is
  3. A favored reaction with a very large Ea barely proceeds at room temperatureit is under kinetic control: the reactants are kinetically stable
  4. Heat, a spark or a catalyst gets more particles over the barrierthe favored reaction then runs at a measurable rate; ΔG° is unchanged

Part 6 · Key ideas

Key ideas

  • Thermodynamics (ΔG°) answers can it go? Kinetics (Ea, rate) answers how fast? They are independent.
  • A thermodynamically favored process with a very large activation energy may show no measurable change: it is under kinetic control.
  • Examples: diamond → graphite, paper and gasoline in air, H₂ and O₂ without a spark.
  • A spark, heat or a catalyst can start such a reaction; none of them changes ΔG°.
  • No visible reaction does not prove ΔG° > 0.

Part 7 · Misconception

A common mistake

The wrong idea: If a reaction has a negative ΔG°, it happens quickly, and if nothing happens, ΔG° must be positive.

What actually happens: ΔG° says nothing about rate. A favored reaction with a large activation energy can be unmeasurably slow (diamond to graphite); only thermodynamic data, not watching, tell you the sign of ΔG°.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Three reactions at 298 K

A student compares three reactions at 298 K. Each is mixed under standard conditions and watched for one hour.

Free energy change, activation energy and what is observed
ReactionΔG° (kJ/mol)Ea (kJ/mol)Observed in 1 hour
1: diamond → graphite−2.9very large (over 300)no change
2: H⁺(aq) + OH⁻(aq) → H₂O(l)−79.9very smallcomplete, instantly
3: 2 H₂(g) + O₂(g) → 2 H₂O(l), no spark−474.2largeno detectable change

1. Which reactions in the table are thermodynamically favored at 298 K?

  1. Reaction 2 alone, because it is the one that is observed
  2. All three
  3. Reactions 2 and 3, because their ΔG° values are large
  4. No reaction in the table, because two show no change
Show the answer

Favorability depends on the sign of ΔG° alone. All three have ΔG° < 0.

  • Reaction 2 alone, because it is the one that is observed: Being observed is about rate. Favorability is decided by the sign of ΔG°, which is negative for all three.
  • Correct: All three: Right: every ΔG° is negative, so every reaction is favored, whether or not it is fast.
  • Reactions 2 and 3, because their ΔG° values are large: Any negative ΔG° means favored; −2.9 kJ/mol counts too.
  • No reaction in the table, because two show no change: A reaction can be favored and still too slow to see.

2. Which statement best explains why Reaction 3 shows no change in an hour?

  1. Its ΔG° is positive at 298 K, so the mixture has to be heated before water can form.
  2. The mixture is already at equilibrium, so no further change can occur.
  3. The reaction absorbs too much heat from the surroundings to proceed.
  4. Its activation energy is large, so few H₂–O₂ collisions have enough energy to react.
Show the answer

A favored reaction with a large activation energy is too slow to observe: it is under kinetic control.

  • Its ΔG° is positive at 298 K, so the mixture has to be heated before water can form.: ΔG° is −474.2 kJ/mol, strongly negative. The problem is the rate, not favorability.
  • The mixture is already at equilibrium, so no further change can occur.: With ΔG° this negative, equilibrium lies almost entirely on the product side; the mixture is far from it.
  • The reaction absorbs too much heat from the surroundings to proceed.: Forming water from H₂ and O₂ releases a large amount of heat.
  • Correct: Its activation energy is large, so few H₂–O₂ collisions have enough energy to react.: Right: the reaction is under kinetic control. It is favored, but the rate is negligible.

3. A spark is passed through the mixture in Reaction 3 and it reacts explosively. What does the spark do?

  1. It gets some molecules over the barrier; the heat they release lets others react.
  2. It changes ΔG° from positive to negative, which makes the reaction favored.
  3. It lowers the activation energy permanently, the way a catalyst does.
  4. It shifts the equilibrium constant to favor products for as long as the spark lasts.
Show the answer

The barrier, not ΔG°, kept the mixture unreacted. A spark supplies enough energy to start the reaction, and the energy it releases keeps it going.

  • Correct: It gets some molecules over the barrier; the heat they release lets others react.: Right: the spark supplies the activation energy to start; the exothermic reaction then supplies its own.
  • It changes ΔG° from positive to negative, which makes the reaction favored.: ΔG° was already negative; the spark changes the rate, not the thermodynamics.
  • It lowers the activation energy permanently, the way a catalyst does.: A spark adds energy; it does not provide a new pathway, so Ea is unchanged.
  • It shifts the equilibrium constant to favor products for as long as the spark lasts.: K depends on temperature, not on a spark, and was already very large.

4. Diamonds kept for centuries do not turn into graphite. Which statement is correct?

  1. Diamond is thermodynamically more stable than graphite at 298 K.
  2. The conversion is not favored, because its ΔG° is small.
  3. Diamond is kinetically stable: the change is favored but has a huge Ea.
  4. The conversion is at equilibrium, so equal amounts of diamond and graphite are present.
Show the answer

Kinetic stability (a slow rate) and thermodynamic stability (the sign of ΔG°) are different. Diamond is the first, not the second.

  • Diamond is thermodynamically more stable than graphite at 298 K.: ΔG° for diamond → graphite is negative, so graphite is the more stable form.
  • The conversion is not favored, because its ΔG° is small.: A small negative ΔG° is still negative: the conversion is favored, just slowly.
  • Correct: Diamond is kinetically stable: the change is favored but has a huge Ea.: Right: ΔG° < 0, yet breaking and rearranging the strong covalent network is extremely slow.
  • The conversion is at equilibrium, so equal amounts of diamond and graphite are present.: There is no graphite forming at a measurable rate, so this is not an equilibrium mixture.

5. What does a negative ΔG° tell you about a reaction?

  1. It is favored, but it may be fast or very slow.
  2. It will happen quickly once the reactants are mixed.
  3. It has a small activation energy barrier.
  4. It needs a continuous input of energy to keep going.
Show the answer

Thermodynamics (ΔG°) answers "can it go?"; kinetics (Ea) answers "how fast?".

  • Correct: It is favored, but it may be fast or very slow.: Right: ΔG° says whether a reaction can go on its own, not how fast.
  • It will happen quickly once the reactants are mixed.: ΔG° says nothing about rate; rate depends on the activation energy.
  • It has a small activation energy barrier.: ΔG° and Ea are independent; a very favored reaction can have a huge barrier.
  • It needs a continuous input of energy to keep going.: That describes a process with ΔG° > 0.

6. Paper burns in air with ΔG° far below zero, yet a book can sit on a shelf for a hundred years. Which explanation is best?

  1. ΔG° for burning paper becomes positive at room temperature.
  2. Paper is at equilibrium with the oxygen in the air.
  3. Burning paper has a large Ea, so at 25 °C its rate is negligible.
  4. There is not enough oxygen in the air for the reaction to be favored.
Show the answer

Favored processes can be kinetically stable when the activation energy is large.

  • ΔG° for burning paper becomes positive at room temperature.: Combustion has ΔH° < 0 and ΔS° > 0, so ΔG° is negative at every temperature.
  • Paper is at equilibrium with the oxygen in the air.: Equilibrium would lie almost entirely on the side of CO₂ and H₂O; the book is far from it.
  • Correct: Burning paper has a large Ea, so at 25 °C its rate is negligible.: Right: it is favored, but under kinetic control at 298 K.
  • There is not enough oxygen in the air for the reaction to be favored.: Air holds plenty of oxygen; the reaction is favored but slow.

7. A student mixes two solutions, sees no reaction after an hour, and concludes "ΔG° for this reaction must be positive." Which reply is best?

  1. That is correct, because any reaction with ΔG° < 0 happens within minutes.
  2. That is correct, because a reaction that does not happen has ΔG° = 0.
  3. That does not follow, because ΔG° is negative for any reaction that releases heat.
  4. It does not follow: a favored reaction can be too slow to see; ΔG° comes from data.
Show the answer

An observation of no change cannot separate "not favored" from "favored but slow". You need ΔG° data to decide.

  • That is correct, because any reaction with ΔG° < 0 happens within minutes.: Many favored reactions are extremely slow (diamond to graphite, paper burning at room temperature).
  • That is correct, because a reaction that does not happen has ΔG° = 0.: ΔG° = 0 means neither direction is favored under standard conditions; it is not what "no visible change" shows.
  • That does not follow, because ΔG° is negative for any reaction that releases heat.: Not so: an exothermic reaction with ΔS° < 0 can have ΔG° > 0 at high temperature.
  • Correct: It does not follow: a favored reaction can be too slow to see; ΔG° comes from data.: Right: no visible change could mean ΔG° > 0 or could mean a large Ea.

Part 9 · Summary

Summary

A negative ΔG° means a process is thermodynamically favored, not that it is fast. When the activation energy is very large, a favored reaction can be too slow to observe; it is under kinetic control. Rate is set by Ea; favorability by ΔG°.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections