3.2 More than one species involved
In the previous section we assumed that only one reactant specie A is involved in thereaction. This is of course too restrictive; in reality we much more often encounter
reactions involving more than one species. This implies that the data analysis will be
more complicated. For example, how can we determine the orders n and m when therate law is given by
Rate = k[A]n [B]m(35)which means that the differential equation now contains two unknown time-dependent
variables [A]t and [B]t ? Two representative strategies to work around this problem are
the isolation method and the initial rate method.
3.2.1 Isolation Method
Suppose that we have prepared a large excess of one of the reactants, say A, comparedto B. That is, the initial concentrations are such that [A]0  [B]0 . In this case, we may
consider that [A] will stay large in the course of the reaction so that its concentration
change may be neglected approximately.
[A]t ' [A]0(= constant)

Then the rate law will be approximated asRate ' k[A]n0 [B]m ≡ k 0 [B]mwhere k 0 is defined by k 0 = k[A]n0 . In this way, the unknown variable [A] has been(approximately) eliminated from the rate law. Now we can apply the methods in theprevious section in order to determine the order m and the new rate constant k 0 .
In the next set of experiments, we prepare a large excess of B compared to A, andrepeat similar procedure to determine the order n. Finally, from n and k 0 we determine
the original rate constant k.
In this way, the contributions from each of the reactant species can be isolated one
by one.
3.2.2 Initial Rate Method
Another method to avoid the complication coming from the involvement of multiple
species is to consider only the initial rates (rates at t = 0).

17The essential point is that the concentrations of all the reactant species are precisely
known at t = 0 as we have prepared them.
(Initial Rate) = k[A]n0 [B]m0
Therefore, we do not have to be concerned with the complicated changes of the concentrations of the multiple species that will occur in t > 0.
For example, to determine the order n, we measure the initial rate of the decay of[A], that is, −d[A]/dt at t = 0. We repeat this for different initial concentrations [A]0 as
shown in the diagram below. It is important to keep [B]0 to the same value in this seriesof experiments. (Otherwise we will mess up the analysis!)18

Example:

The following data have been obtained for a reaction A + B → P.
[A]0/ mol dm−3
0.2
0.4
0.2

[B]0/ mol dm−3
0.2
0.2
0.6

Initial Rate/ mol dm−3 s−1
5 × 10−5
1 × 10−41.5 × 10−4

Find the orders with respect to A and B and the rate constant k.
As illustrated by this Example, the analysis is easy when the orders are immediately found to be simple integers. For otherwise and more general cases, the following
procedure is useful.
By taking the logarithms of both sides of(Initial Rate) = k[A]n0 [B]m0
we obtain∗∗log(Initial Rate) = log k + n log[A]0 + m log[B]0Therefore, the rate constant k and order n can be determined from a plot of log(Initial Rate)against log[A]0 . Note that [B]0 should be kept to the same value in this analysis. By exchanging A and B, we can determine the order m from another set of experiments.

∗∗

Remember that log XY = log X + log Y and log X n = n log X. Here the base can be anything, so
we can use either log10 or ln depending on our particular conveniences.

19