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Little Laboratory/ III

Reaction Kinetics

How quickly does a reactant disappear when its rate is its own concentration?

An idealised first-order reaction, A → products. Set the starting concentration and the rate constant, then follow the exponential decay, its half-life, and the moment-by-moment rate.

0.00.51.01.52.0036912t / s[A] / mol L⁻¹1t½2t½3t½
  • [A](t)
  • half-lives
100 of 100 remain — a tally of the fraction, not a simulation.
[A](t)
1.000 mol/L
Fraction remaining
100.0 %
Half-life t½
1.98 s
Rate k[A]
0.350 mol L⁻¹ s⁻¹

Every 1.98 s the remaining concentration halves, whatever its value — so after 0 half-lives 100.0 % is left.

A → products, first order. The curve is [A]₀·e^(−kt); dashed lines mark successive half-lives, equally spaced because each halving takes the same time. Units are illustrative.

The model

An elementary reaction A→products\mathrm{A} \to \text{products} is first order when its rate is proportional to how much A remains:

d[A]dt=−k [A],k≥0,\frac{d[\mathrm{A}]}{dt} = -k\,[\mathrm{A}], \qquad k \ge 0,

which integrates to an exponential decay from the initial concentration,

[A](t)=[A]0 e−kt.[\mathrm{A}](t) = [\mathrm{A}]_0\, e^{-kt}.

The half-life

Setting [A]=12[A]0[\mathrm{A}] = \tfrac{1}{2}[\mathrm{A}]_0 gives

t1/2=ln⁡2k,t_{1/2} = \frac{\ln 2}{k},

which does not involve [A]0[\mathrm{A}]_0 at all. Whatever the concentration, the next halving takes the same time as the last — the signature of first-order kinetics, and the reason the dashed markers on the plot are equally spaced. When k=0k = 0 nothing reacts, the concentration stays where it started, and there is no half-life to speak of.

What is assumed

Units are illustrative — mol L⁻¹ for concentration, seconds for time, s⁻¹ for kk — and no particular real reaction is meant. The particles in the lower panel are a tally, not a simulation: at each instant the number shown is the fraction of A remaining, rounded to one in a hundred.