How to calculate the half time of a unimolecular reaction given the Arrhenius coefficient and the activation energy?

Question

Consider the following unimolecular reaction: $$\ce{H2O2 + M -> OH + OH + M}$$

The high pressure limit unimolecular rate coefficient for this reaction is $k_\mathrm{uni} = A\cdot \mathrm{e}^{\frac{-E_\mathrm{a}}{R \cdot T}}$, where $A = \pu{3E14 1/s}$, and $E_\mathrm{a} = \pu{305 kJ/mol}$. Calculate the half-life of this reaction at $\pu{1 atm}$ and $\pu{1000 K}$.

Source: Question 14.1 from John W. Daily: Statistical Thermodynamics: An Engineering Approach. Cambridge University Press, 2018. ISBN: 1108244645, 9781108244640. DOI: 10.1017/9781108233194.

I started by getting that $k_\mathrm{uni}$ through the formula, got $\pu{0.03515 1/s}$ and then, since it's unimolecular: $$t_{\frac{1}{2}} = \frac{\ln(2)}{k_\mathrm{uni}}.$$ But I got it wrong, I don't have the solutions manual to this book to check what is wrong. I also tried using the Troe form, using the $k_\mathrm{uni}$ from the starting formula as the $k_{\inf}$ but still got it wrong.

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Okay, I finally found the solutions manual. And saw that the answer on it is completely wrong, so there was no way that introducing what that math gave would work. The halftime is actually 19.7 s, but in the book it is 9 $\mu$s, which doesn't make sense.

Answer 1

You calculated the high-pressure-limit half life, where the rate of collisions with M is so high that it has no influence on the rate constant. Now all you have to do is to consider that the pressure is 1 atm instead of at the high-pressure limit. That slows down the reaction such that the half-time is in the second instead of the micro second range.

Something similar happens with ozone. It is not very stable at sea level, but has a substantial half-time in the upper atmosphere (unless there are certain man-made substance up there).

To solve numerically, you could assume a Lindemann mechanism (see https://en.wikipedia.org/wiki/Lindemann_mechanism) and that $k_1$ is diffusion-limited (i.e. no activation energy, every collision leads to activated complex).

The textbook cited by the OP has:

Strangely, in an earlier table they show $k_0$ and $k_\infty$ as (along with pre-exponential temperature dependence $T^n$ and ?activation energy? E - units unknown):

Answer 2

For $[A]_{0}$ the initial concentration of $H_{2}O_{2}$, we are looking for the in time in which the half of that concentration is left:

$\frac{[A]_{1}}{[A]_{0}} = \frac{1}{2}$

For unitary reactions like the one on the problem:

$[A]_{1} = [A]_{0}\cdot e^{-k\cdot t}$

To me, there's two ways of proceeding, either you take the $k$ you're given an substitute it on the equation, or you consider it as the $k_{\inf}$ and use the 1 atm pressure to find the $k$. Either way, you'll get that your halftime $\tau_{\frac{1}{2}} = \frac{ln(2)}{k}$ for the $k$ you chose or got.