How to graph a second order reaction with two different reactants

In a laser flash photolysis experiment, $$\ce{C6H5NH2}$$ $$(C_0 = \pu{60 \mu M})$$ is oxidized to the corresponding radical cation. The disappearance of radical cation is monitored spectrophotometrically and the results are shown in the table below. There are three possible reaction pathways for the radical cation formed:

3. Reaction between radical cation and solvent. Using experimental data, determine the most probable path of reaction and the rate constant of the reaction. $$\begin{array}{c|c} t/\pu{\mu s} & [\text{Radical}]/\pu{\mu M} \\ \hline 0\ & 20\\ 5 & 11.12\\ 10\ & 6.86\\ 15\ & 4.47\\ 20\ & 3.00\\ 25 & 2.06 \\ 30 & 1.43 \\ 35 & 1\\ \hline \end{array}$$
I know how to use the given data to graph a first order and a second order with the same reactants, but I don't know how to do it for a second order with reactants A + B (which is the second pathway: reaction between radical cation and starting material). I know that the graph should be ln$$\frac{a-x}{b-x}$$ against $$t$$ but I don't understand how to get $$(a-x)$$ and $$(b-x)$$. From what I understand $$a = \pu{20 \mu M})$$ and $$b = \pu{60 \mu M})$$ but what is $$x$$?
• $x$ is the conversion variable. If the reaction is A+B $\to$ P+Q at time $t$ the concentration of A is $A_0-x$ and of B, $B_0-x$. Then product is $P_0+x$ and $Q_0+x$ where the zero subscript is the initial amount. You have 60 micromol to begin with and 20 is converted instantly by the flash so 40 is left unreacted at time zero. The integrated rate equation is $\displaystyle \ln\left(\frac{A_0-x}{B_0-x}\right)=kt(A_0-B_0)+\ln(A_0/B_0)$. In (3) the solvent conc'n is huge so pseudo first order. May 30 at 16:31
• @porphyrin I don't quite understand how to calculate ($A_0 -x$) and ($B_0 -x$). I would have assumed that ($A_0 -x$) is calculated as ($60 - 20$), ($60-11.12$) and so on. But that isn't the concentration of the radical. So how exactly is ($A_0 -x$) and ($B_0 -x$) calculated? May 31 at 6:56