AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-4)
Unit 9 · Topic 9.4
9.4 Thermodynamic and Kinetic Control
A reaction can be thermodynamically favored and still be far too slow to notice, usually because its activation energy is very high. Chemists call this kinetic control. A system where nothing seems to be happening may simply be stuck, not at equilibrium.
Key terms
- kinetic control
- activation energy
- reaction rate
- thermodynamically favored
Favored doesn't mean fast
Thermodynamics (ΔG°) tells you which direction a reaction would go and where it would end up. Kinetics (rate, activation energy) tells you how fast it gets there. These are separate questions.
Many reactions with very negative ΔG° happen too slowly to observe at room temperature. Paper burning in air is favored, but a book on a shelf doesn't catch fire. Diamond turning into graphite is favored at room temperature (ΔG° ≈ −2.9 kJ/mol), but diamonds last for billions of years.
Kinetic control
Chemists say a favored-but-stuck reaction is under kinetic control: its speed, not its thermodynamics, decides what you actually observe. The usual reason is a high activation energy: very few collisions have enough energy to get over the barrier (topic 5.5).
So on the exam, if you're told ΔG° is negative but no reaction is seen, blame the rate: the process is under kinetic control.
Kinetic control is also why fuels like gasoline and natural gas can be stored safely. They are thermodynamically unstable in air, but they react only when a spark or flame gets them over the activation barrier.
Not reacting is not the same as equilibrium
A balloon of H₂ and O₂ at room temperature doesn't change, but it isn't at equilibrium: the equilibrium mixture would be almost all water. The system is simply stuck, waiting for enough energy to get over the barrier.
At true equilibrium, forward and reverse reactions are both happening at equal rates. In a kinetically controlled system, essentially no reaction is happening at all.
Seeing kinetic control on an energy diagram
On an energy diagram, a kinetically controlled reaction has products well below the reactants (favored) but a very tall hump between them (high activation energy). Thermodynamics compares the starting and ending levels; kinetics depends on the height of the hump. A reaction can be strongly favored and still extremely slow if the hump is tall enough.
Speeding up a stuck reaction
You can't change ΔG° with a catalyst, but you can change the rate. Raising the temperature gives more particles enough energy to react. A spark or flame provides a local burst of energy that starts a self-sustaining reaction. A catalyst provides a lower-energy pathway (topic 5.11).
Living things depend on kinetic control. Many molecules in your body are thermodynamically unstable in the presence of oxygen but react only when enzymes (catalysts) let them.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Explaining an observation
For the combustion of glucose, C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), ΔG° is strongly negative. Yet a spoonful of sugar exposed to air does not visibly react. Explain, and say whether the sugar–air system is at equilibrium.
Show the solutionHide the solution
- Step 1: A negative ΔG° means the reaction is thermodynamically favored: the products are strongly favored at equilibrium.
- Step 2: Not reacting at a measurable rate means the reaction is very slow at room temperature. The most likely reason is a high activation energy, so almost no collisions between sugar and O₂ molecules have enough energy to react.
- Step 3: The reaction is under kinetic control.
- Step 4: The system is not at equilibrium. At equilibrium it would be almost entirely CO₂ and H₂O. It just hasn't gotten there.
Answer: The reaction is favored but under kinetic control because of a high activation energy; the system is not at equilibrium.
- Example 2
Trap: concluding equilibrium from 'no change'
A sealed flask of N₂ and H₂ at room temperature shows no change in pressure for a week. ΔG° for N₂ + 3H₂ → 2NH₃ is −33 kJ/mol at 298 K. A student concludes that the mixture must be at equilibrium. Evaluate this conclusion.
Show the solutionHide the solution
- Step 1: ΔG° is negative, so K > 1 and an equilibrium mixture at 298 K would contain a lot of NH₃.
- Step 2: The flask has no NH₃ forming at a measurable rate, so it is far from that equilibrium composition.
- Step 3: The very strong N≡N triple bond gives the reaction a high activation energy, so it's under kinetic control. (Industrially, the reaction needs a catalyst and high temperature to run at a useful rate.)
- Step 4: Constant pressure here means 'not reacting', not 'at equilibrium'.
Answer: The conclusion is wrong: the system is under kinetic control (too slow to react), not at equilibrium.
Common mistakes
- Assuming a negative ΔG° means a reaction will happen quickly.
- Concluding that a system is at equilibrium because nothing is changing.
- Thinking a catalyst can make an unfavored reaction favored. It only changes the rate.
- Explaining kinetic control without mentioning activation energy.
On the exam
- Questions in this topic are explanations. A strong answer says two things: ΔG° < 0, so products are favored; and the activation energy is so high that almost no collisions succeed, so the reaction is too slow to notice.
- Be ready to connect to kinetics: draw or read an energy profile with a large hump between reactants and lower-energy products.
Connected topics
Videos
Check yourself
4 questions on 9.4 Thermodynamic and Kinetic Control. Pick an answer to see if you got it, and why.
For C(diamond) → C(graphite), ΔG° ≈ −2.9 kJ/mol at 25 °C, yet diamonds show no detectable change into graphite over centuries. Which of the following best explains this?
A sealed flask of H₂ and O₂ gas can sit at room temperature for years with no measurable formation of water, but a single spark makes it explode. ΔG° for forming water from these gases is large and negative. Which statement best describes the mixture before the spark?
A reaction has ΔG° = −120 kJ/mol but is so slow at room temperature that no product can be detected after a day. A catalyst is added, and the product forms quickly. Which of the following is true about the effect of the catalyst?
Paper sits in air for years without burning, even though its combustion is thermodynamically favored. Once a corner is lit with a match, the whole sheet burns. Which of the following best explains why the paper keeps burning after the match is removed?
0 of 4 answered