I am an astrophysicist working in exoplanetary atmosphere chemistry right now, in particular modelling the chemical kinetics taking place within the atmosphere. Based on this IOP article, we have that given a certain chemical reaction,

$\ce{\alpha A + \beta B <=> \gamma C + \eta D}$

In which its forward rate coefficient is $k_{\mathrm{f}}$ and the reverse counterpart is noted as $k_{\mathrm{r}}$ the equilibrium constant let us to compute the reverse reaction rate coefficient as

$$\frac{k_{\text{f}}}{k_{\mathrm{r}}} = K_{\text{eq}}\left(\frac{k_{\mathrm{b}}T}{P_{0}}\right)^{\Delta \mu}$$

$$K_\text{eq} = \exp\left(-\frac{\Delta H^{0}-T\Delta s^{0}}{RT}\right)$$

The previously cited paper is pretty clear at this statement and it is relatively easy to find this information to me in other sources. Nevertheless I came up with a certain kind of reactions, so called "Combination and Dissociation Reactions" in which a third body is involved, in my case representing the atmospheric bulk, for example molecular hydrogen for a Jupiter-like planet and it is represented by a letter "M". For example we could have

$\ce{A + M -> C + D + M}$

For these kind of reactions the rate coefficient is parametrized based upon on to limiting to zero and infinite pressure, $k_{0}$, $k_{\inf}$ and an "F" coefficient leading to the equivalente rate coefficient

$$k = \frac{k_{0}M}{1+\frac{k_{0}M}{k_{\inf}}}F$$

My question is, is this kind of reactions reversible?. Can we reverse the reaction using the previous formalism just exchanging $k_\mathrm{f}$ by $k$?

  • $\begingroup$ While not a direct answer you should look at the 'Collision theory of Unimolecular reactions' also called 'Lindemann Theory' (in most under-graduate phys. chem. textbooks). These reactions have the form $\ce{A + M <=> A^* + M; \quad A^* \to products }$. You should be able to generalise from this:) $\endgroup$ – porphyrin May 15 '18 at 16:36
  • $\begingroup$ Thank you a lot for the reference!. I will have a look at it :) $\endgroup$ – Juan Luis Gómez González May 15 '18 at 16:58

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.