2
$\begingroup$

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 $(b-x)$?

Improve this question
$\endgroup$
4
  • $\begingroup$ 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

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.