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 k+[A]=k−[B]k_+[\mathrm{A}] = k_-[\mathrm{B}] 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 NOX2\ce{NO2} molecules keep pairing up into colourless NX2OX4\ce{N2O4}, and NX2OX4\ce{N2O4} keeps splitting back into NOX2\ce{NO2}. 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) NX2OX4\ce{N2O4}, the other pure NOX2\ce{NO2} (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 A⇌B\ce{A} \mathrel{\ce{<=>}} \ce{B}, 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.

k+[A]eq=k−[B]eq⟹K=[B]eq[A]eq=k+k−k_+[\mathrm{A}]_\mathrm{eq} = k_-[\mathrm{B}]_\mathrm{eq}\quad\Longrightarrow\quad K = \frac{[\mathrm{B}]_\mathrm{eq}}{[\mathrm{A}]_\mathrm{eq}} = \frac{k_+}{k_-}
The equilibrium condition: forward rate equals reverse rate. The equilibrium constant is the ratio of the two rate constants (for an elementary step).
k+k_+
forward rate constant
k−k_-
reverse rate constant
[A]eq[\mathrm{A}]_\mathrm{eq}
concentration of A at equilibrium
KK
equilibrium constant: set by temperature alone, not by the starting amounts
Molecule-by-molecule stochastic simulation (Gillespie algorithm)

Reversible reaction: reaching equilibrium from both sides

A⇌B\ce{A <=> B}
A 39 B 81 t = 16.0 s
Q right now 2.08
K predicted from k₊/k₋ 2

Last second: 25 forward, 23 reverse

Product molecules over time

0246810121416Model time t / s020406080100120Number of B
  • Simulation: from pure reactant
  • Simulation: from pure product
  • Mass action (average)

Forward and reverse rates (counted per 0.5 s)

0246810121416Model time t / s0204060Events per second
  • Forward
  • Reverse
  • Mass action (average)
What is simulated is a genuine random process: the time and kind of every reaction event is drawn by the Gillespie algorithm from the rate constants, using seeded random numbers (the same run can be replayed). Dashed curves are the average predicted by mass action. k₊, k₋ and the time unit are model values, not those of a particular molecule; in dimer mode K is counted in molecules. Particle positions and motion are illustrative; the window tint is computed from the NO₂ absorption band.
With k₊ = 0.6 and k₋ = 0.3 both runs settle with about two-thirds as B, and the forward and reverse rate curves below merge. Now cut the number of molecules to 20: the fluctuations become obvious. At 200 the curves are much steadier.

Worked example

Equilibrium composition of A ⇌ B and how fast it is reached

For the elementary step A ⇌ B, k+=0.60 s−1k_+ = 0.60\ \mathrm{s^{-1}} and k−=0.30 s−1k_- = 0.30\ \mathrm{s^{-1}}; 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, k+=0.020 s−1k_+ = 0.020\ \mathrm{s^{-1}} and k−=0.0050 s−1k_- = 0.0050\ \mathrm{s^{-1}}. 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 1/N1/\sqrt{N}: about 10 % for 100 molecules, but only about 1.3×10−121.3\times10^{-12} for a mole (6.0×10236.0\times10^{23} 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, K=k+/k−K = k_+/k_-; the approach to equilibrium is governed by k++k−k_+ + k_-.
  • Random fluctuations in molecule numbers scale as 1/N1/\sqrt{N} and are undetectable on the macroscopic scale.
Progress is stored only in this browser and never uploaded.

Data and references

  1. P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford University Press (2018), Topics 6A and 17B (approach to equilibrium).
  2. D. T. Gillespie, “Exact stochastic simulation of coupled chemical reactions”, J. Phys. Chem. 81, 2340–2361 (1977).
  3. D. D. Wagman et al., “The NBS tables of chemical thermodynamic properties”, J. Phys. Chem. Ref. Data 11, Suppl. 2 (1982) (N₂O₄ and NO₂).