Question

What is the order of the reaction `A -> B` if changing the concentration of `A` from `"0.05 M"` to `"0.1 M"` changes the rate from `2 xx 10^(-3) "M/s"` to `1.6 xx 10^(-2) "M/s"`?

Answer 1

The reaction is:

`"A" -> "B"`

The trick with this is that the order of the reactant `"A"`, given that it is the only reactant, is the order of the reaction.

The order of a reactant is basically the contribution of its concentration to the change of reaction rate.

To put this in a larger context, the reaction rate `r(t)` (as a function of time) is related to the concentration `["A"]` for reactant `"A"` by a rate constant `k`, as follows:

`\mathbf(r(t) = k["A"]^(m) = -(d[A])/(dt))`

where `m` is the order of reactant `A` and `-(d[A])/(dt)` is the rate of disappearance of `A` (you don't need the rate of disappearance for this problem, but I thought I'd provide that for a brief review).

The above equation is known as a rate law.

Note that you have two rates and two initial concentrations (you do not need equilibrium concentrations---in fact, those are not practical for kinetics studies, because the reaction is most active when it first starts).

These are for each trial run for a particular kinetics study, and can be used to determine the order of the reaction as follows.

Suppose we compared rate `2` with rate `1` relative to the change in initial concentration `["A"]_i` of `"A"`. Then what we really do is divide the rate laws for trials `2` and `1`.

`(r_2(t))/(r_1(t)) = (([A]_(i,2))/([A]_(i,1)))^m`

`(1.6xx10^(-2))/(2xx10^(-3)) = (("0.1 M")/("0.05 M"))^m`

`2^m = 8`

`color(blue)(m = 3)`

For this process, we thus describe it by saying it is a third-order unimolecular reaction.

What this order says is that if you double the concentration of reactant `"A"`, you make the rate of reaction multiply by `\mathbf(8)`.