1
$\begingroup$

I have some questions about radical chain reactions. Let's take an example:

enter image description here And here are my questions:

  1. How can we define in the above case at which level the reaction 2) is located (initiation or propagation)? I understand initiation as: creation of new radicals and propagation as: left and right with equal number of radicals. My difficulty comes from the fact that for reaction 2), there is both the creation of a new radical, and an equality of the number of radicals. What confuses me is that I have the following example in my course:

initiation propagation

We can see that there is radicals used as reactant in one of the initiation step, as well as in the propagation step. How to decide which is the right choice in this case ?

  1. In a reaction like this, can we both use the pre-equilibrium hypothesis (pseudo-equilibrium due to quasi-stationarity of intermediate reactant) and the quasi-stationary state -for radicals- hypothesis?

  2. Regarding the pre-equilibrium hypothesis, at what stage can we consider a pseudo-equilibrium and calculate an equilibrium constant? Only at the step of creating the intermediate reaction? Or also at the stage of its disappearance?

  3. Concerning the hypothesis of quasi-stationarity of the radicals, can we equalize the speed of appearance and disappearance of the radicals only between the initiation and the termination, or is it also possible between initiation and propagation and/or between propagation and termination?

Many thanks in advance.

PS: my apologies if I'm not using to best words for the hypotheses, English is not my native language. I tried to find the corresponding words on internet.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Take new as more. Propagation, aside of keeping radicals, is regenerating cycle, producing one of its reactants, like here the ethyl radical. $\endgroup$
    – Poutnik
    Commented Dec 31, 2022 at 19:14

1 Answer 1

Reset to default
2
$\begingroup$

Your step with $k_{init2}$ is technically a propagation step ( one radical in one radical out) but the reaction is given as 'fast' so that $\ce{CH3^*}$ is consumed faster than it is produced and we may assume $\ce{[CH3^*]}\to 0$ and takes no further part in the reaction. The chain carrier is thus $\ce{C2H5^*}$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.