Lesson 1 · 20 min
Dynamic equilibrium: both directions at once
Reversible reactions never “stop”: the forward reaction slows and the reverse speeds up until the two are equal. Watch a molecule-by-molecule simulation reach the same equilibrium from both ends.
In this lesson
- Tell “the reaction has stopped” apart from dynamic equilibrium: forward and reverse rates equal and both non-zero
- Explain why starting from reactants or from products leads to the same equilibrium
- Use to find the equilibrium composition of an elementary step A ⇌ B
- Explain why a macroscopic equilibrium looks still while molecule counts fluctuate
Put a little nitrogen dioxide in a sealed flask and it is red-brown at room temperature. Leave it all day and the colour does not change — nothing seems to happen. Yet molecules keep pairing up into colourless , and keeps splitting back into . The colour holds only because the two directions run at exactly the same rate.
Isotope tracers give the same evidence. Drop a few crystals containing radioactive iodide into a saturated solution of the same salt: the concentration does not change, yet the radioactivity soon turns up in the solution — crystals keep dissolving while an equal amount of ions crystallises.
Predict first
Two identical sealed flasks sit at 25 °C: one holds pure (colourless) , the other pure (dark brown) containing the same number of nitrogen atoms. What do you see after a long wait?
Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.
The forward reaction slows, the reverse speeds up
Take the simplest case, an isomerisation , and assume both directions are single elementary steps. The forward rate is proportional to the concentration of A, the reverse rate to that of B. At the start there is only A: the forward reaction is fast and the reverse is zero. As A is used up the forward reaction slows; as B builds up the reverse reaction speeds up. The two rates must meet — and from then on, as much B forms each second as is used up, so the concentrations stop changing.
That is dynamic equilibrium: the macroscopic properties (concentrations, colour, pressure) are constant, while at the molecular level both reactions carry on at the same rate.
- forward rate constant
- reverse rate constant
- concentration of A at equilibrium
- equilibrium constant: set by temperature alone, not by the starting amounts
Worked example
Equilibrium composition of A ⇌ B and how fast it is reached
For the elementary step A ⇌ B, and ; the start is 1.00 M A only. Find the equilibrium concentrations and the time to get halfway there.
Concept check
Which statement about a closed system at equilibrium is true?
Try the calculation
For the elementary step A ⇌ B, and . What percentage of the total is B at equilibrium?
Why a macroscopic equilibrium looks still
The simulation has only a hundred or so molecules, so the number of B keeps hopping a few above and below its average. Random fluctuations scale roughly as : about 10 % for 100 molecules, but only about for a mole ( molecules) — far below anything an instrument can detect. That is why a macroscopic equilibrium looks perfectly still, and the equilibrium constant describes the average of countless random events.
Key takeaways
- A reversible reaction in a closed system reaches dynamic equilibrium: equal forward and reverse rates, constant macroscopic properties.
- From reactants or from products, the same atoms at the same temperature reach the same equilibrium composition.
- For an elementary step A ⇌ B, ; the approach to equilibrium is governed by .
- Random fluctuations in molecule numbers scale as and are undetectable on the macroscopic scale.
Data and references
- P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford University Press (2018), Topics 6A and 17B (approach to equilibrium).
- D. T. Gillespie, “Exact stochastic simulation of coupled chemical reactions”, J. Phys. Chem. 81, 2340–2361 (1977).
- D. D. Wagman et al., “The NBS tables of chemical thermodynamic properties”, J. Phys. Chem. Ref. Data 11, Suppl. 2 (1982) (N₂O₄ and NO₂).