Rate laws, half-lives and diffusion-controlled reactions

A rate law says how fast something is used up, in one of two forms

Intuition

Think of a bath draining. You can say how fast the water level is falling right now (the rate). Or you can say what the level will be after ten minutes (the level against time). Kinetics uses both. The first is the differential rate law. The second is the integrated rate law.

In chemical kinetics we want to know how the concentration of a species changes over time, from the starting state to equilibrium. A differential rate law gives the rate of change of concentration in terms of one or more concentrations, for example $d[A]/dt = -k[A]$. An integrated rate law gives the concentration as a function of time, for example $[A] = [A]_0\exp(-kt)$. If you know the differential law you can find the integrated one by hand for simple cases, or numerically otherwise. The lecture also shows a differential law that is not a simple order, the hydrogen-bromine type:

$\frac{d[\mathrm{HBr}]}{dt} = \frac{k_a[\mathrm{H}_2][\mathrm{Br}_2]^{3/2}}{[\mathrm{Br}_2] + k_b[\mathrm{HBr}]}$

Integrating the first-order law, one plain step at a time

Take A → B with a first-order law, $d[A]/dt = -k[A]$. In words: the more A there is, the faster it goes. Here is every step.

Try it: worked example

Integrate d[A]/dt = -k[A] to get [A] as a function of time, given [A] = [A]0 at t = 0.

Show answer and steps

Answer[A] = [A]0 exp(-kt), and a plot of ln[A] against t is a straight line with gradient -k.

1. Start with $\dfrac{d[A]}{dt} = -k[A]$. In words, the rate of loss is proportional to what is left.
2. Put everything with $[A]$ on one side and $t$ on the other: $\dfrac{d[A]}{[A]} = -k\,dt$.
3. Integrate both sides: $\displaystyle\int \frac{d[A]}{[A]} = \int -k\,dt$. The left side gives $\ln[A]$ and the right gives $-kt$ plus a constant.
4. At $t = 0$, $[A] = [A]_0$. That fixes the constant, so $\ln[A] = -kt + \ln[A]_0$.
5. Move $\ln[A]_0$ across: $\ln\left(\dfrac{[A]}{[A]_0}\right) = -kt$.
6. Take the exponential of both sides: $[A] = [A]_0\exp(-kt)$.
7. A plot of $\ln[A]$ against $t$ is a straight line. Its gradient is $-k$, so you can read $k$ off the graph.

The lecture’s table gives all three simple orders side by side.

	0th order	1st order	2nd order
Rate equation	$\frac{d[A]}{dt} = -k$	$\frac{d[A]}{dt} = -k[A]$	$\frac{d[A]}{dt} = -k[A]^2$
Integrated	$[A] = [A]_0 - kt$	$[A] = [A]_0\exp(-kt)$	$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$
Plot to test and get $k$	$[A]$ vs $t$	$\ln[A]$ vs $t$	$1/[A]$ vs $t$
Half-life	$t_{1/2} = \frac{[A]_0}{2k}$	$t_{1/2} = \frac{\ln 2}{k}$	$t_{1/2} = \frac{1}{k[A]_0}$

The half-life is the time for the concentration to halve. Only the first-order half-life is independent of the starting concentration.

Exam detail

Learn the table. In a question, spot the order from which plot is straight, and pick the half-life that goes with it.

The fastest reactions are limited only by how quickly molecules find each other

Intuition

Imagine a crowded dance floor where two people, one in orange and one in green, instantly pair up the moment they touch. How often does that happen? It depends only on how fast people shuffle around and how big they are. That is a diffusion-controlled reaction. Each meeting works. The only slow part is the meeting.

Suppose a reaction in solution happens every time an orange molecule touches a green one. Solvent molecules fill the rest of the space. The lecture builds the rate in steps. Each step is below.

First, how long does an orange molecule take to swap places with a neighbour? The average time to diffuse a distance $x$ is the Einstein–Smoluchowski result:

$\text{time} = \frac{x^2}{2D}$

If every molecule has radius $r$, the neighbour is $x = 2r$ away, so the exchange time is

$\tau = \frac{(2r)^2}{2D} = \frac{2r^2}{D}$

In a time $\Delta t$, each orange molecule makes about $\Delta t/\tau$ exchanges. Next, how likely is each swap to bring a green neighbour? There are about 12 neighbours in 3-d, and the chance is set by the fraction of sites that are green:

$P \sim 12\,\frac{n_g V_g}{V} = 12\,\frac{n_g \frac{4\pi}{3} r^3}{V}$

The successful number of collisions in $\Delta t$ is exchanges times chance times the number of oranges:

$Z = n_o\frac{\Delta t}{\tau}P = \frac{n_o n_g}{V}\,4\pi\,2r\,D\,\Delta t$

Letting green molecules diffuse too, and allowing different radii, gives

$Z = \frac{n_o n_g}{V}\,4\pi\,(r_o + r_g)(D_o + D_g)\,\Delta t$

The instantaneous rate is then a second-order differential rate law,

$\frac{dc_o}{dt} = -k\,c_o c_g \qquad k = 4\pi\,(r_o + r_g)(D_o + D_g)\,N_A$

So the rate of a diffusion-controlled reaction depends on the diffusion coefficients of the partners ($D_o + D_g$) and the distance at which they touch and can react ($r_o + r_g$).

Pitfall

Do not mix up the symbols. $\tau$ is a time (the exchange time), $P$ is a chance, $Z$ is a count of collisions, and $k$ is the rate coefficient. Also note that the rate coefficient uses $r_o + r_g$ and $D_o + D_g$ as sums, not products.

Optional aside

Slide 6 writes the encounter probability with a tilde, meaning it is an approximation, not an exact result. The lecture does not give numbers for $D$ or $r$ here, so no numeric example is given.

Recall

What are the two things that set the speed of a diffusion-controlled reaction?

How fast the partners diffuse ($D_o + D_g$) and how close they need to be to react ($r_o + r_g$). Bigger and faster-moving partners meet more often.