In a laser flash photolysis experiment, $\ce{C6H5NH2}$ $(C_0 = \pu{60 \mu M})$ is oxidize to its corresponding radical cations. There are three possible reaction pathways for the formation of the cations:

  1. Dimerization of the radical cations;
  2. Reaction between the cations and the initial substance; and
  3. Reaction between the cations and the solvent.

Determine with the help of the experimental data which is the most likely pathway and the rate constant of the reaction. $$ \begin{array}{l|cccccccc} \hline t \text{ after the pulse (in µs)} & 0 & 5 & 10 & 15 & 20 & 25 & 30 & 35 \\ \hline \text{[Radical cation] (in µM)} & 20 & 11.12 & 6.86 & 4.47 & 3 & 2.06 &1.43 &1 \\ \hline \end{array} $$

I understand how to plot for (1) and for (3), however how do I plot for the second pathway? I know that I will have the reaction type:

$$\ce{ A + B -> P}$$

and then I should plot $\ln\left(\frac{a-x}{b-x}\right)$ against $t$ however, how do I actually calculate $(a-x)$ and $(b-x)$?

  • $\begingroup$ Does it make sense to consider that two radical cations may dimerize ? As they are charged, they repel one another, which prevents a dimerization ... $\endgroup$
    – Maurice
    Aug 25, 2021 at 19:12
  • $\begingroup$ @Maurice my question is more, how do I get (a-x) and (b-x) I don't understand how they are calculated. I tried to say that $a = 20$ µM and $b = 60$µM then for $b$ I thought to get (b-x) I should take (b- the concentrations of the radical ion) while (a-x) is just the concentration of the radical ion (which I know is not correct but it is the closest I get to the correct answer). But what is the correct way of calculating (a-x) and (b-x)? $\endgroup$
    – katara
    Aug 26, 2021 at 8:30
  • $\begingroup$ @Katara form the stoichiometry of the reaction I guess that $$x = [\ce{C6H5NH2+}]_0 - [\ce{C6H5NH2+}]_t$$ $\endgroup$
    – PAEP
    Aug 28, 2021 at 11:26
  • $\begingroup$ @Katara, are you sure that you got the right units for the concentration of the radical cation? I would have expected that the concentration of the radical cation generated in the excitation step was small compared to the concentration of aniline. In that case, $\ce{[C6H5NH2]} \gg \ce{[C6H5NH2+]}$ and you would observe pseudo-first order kinetics A semi-log plot of the data will show a linear behaviour. $\endgroup$
    – PAEP
    Aug 28, 2021 at 12:03


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