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I have the following reversible bimolecular reactions

$$\begin{align} \ce{A_0 +B &<=>[$3k_+$][$k_-$] A_1} \\[5pt] \ce{A_1 +B &<=>[$2k_+$][$2k_-$] A_2} \\ \ce{A_2 +B &<=>[$k_+$][$3k_-$] A_3} \\ \end{align}$$

that can be written as follows

$$\begin{align} \ce{A_0 +3B &<=>[$k_1$][$k_2$] A_3} \\[5pt] \end{align}$$

The initial conditions are $[\ce{A_1}]_0 = [\ce{A_2}]_0 = [\ce{A_3}]_0 = 0$. I know rate constants $k_+$ and $k_-$ and would like to write rate constants $k_1$ and $k_2$ in terms of $k_+$ and $k_-$.

It would be great if you have any ideas. I am quite new to the topic. So far, I can only think of using some fitting algorithms to find $k_1$ and $k_2$. Thanks!

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    $\begingroup$ I'm not getting your point. What data you have? $\endgroup$
    – Mithoron
    Nov 30, 2017 at 18:52
  • $\begingroup$ The slowest reaction in the 'chain' is the so-called rate limiting reaction, a bottleneck, if you like. In this case the third one. So $k_1=k_+$. $\endgroup$
    – Gert
    Nov 30, 2017 at 20:18
  • $\begingroup$ Your terminology is sloppy here. $3k_+$ is the reaction rate coefficient. If we assume first order kinetics then the reaction rate, $r$, for the first reaction is $$ r = 3k_+\ce{[A_0][B]}$$ $\endgroup$
    – MaxW
    Nov 30, 2017 at 20:49
  • $\begingroup$ Is it appropriate to assume that all the reactions orders follow the stoichiometry? For example for the first reaction the forward reaction is second order and the backward reaction is first order? If so then $K_{eq} = 3k_+ / k_-$ for the first reaction. Overall $K_{eq, total} = k_1 / k_2$. $\endgroup$
    – MaxW
    Nov 30, 2017 at 21:51
  • $\begingroup$ and $K_eq =6k_{+}^{3} / 6k^{3}_{-} $ for the chain? so $k_1=6k_{+}^{3} $ ? $\endgroup$
    – Sp_J
    Dec 1, 2017 at 16:04

1 Answer 1

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Finding the equilibrium constant of your overall reaction is straightforward, as you've done in the comments, and we have $$K_\text{tot} = K_1K_2K_3 = \left(\frac{3k_+}{k_-}\right)\left(\frac{2k_+}{2k_-}\right)\left(\frac{k_+}{3k_-}\right) = \frac{k_+^3}{k_-^3}.$$ But this is insufficient to conclude that $k_1 = k_+^3$ or that $k_2 = k_-^3$. Indeed, let us consider what is stated when we have, as in your last reaction, rate constants for the overall reaction. Such a scenario presumes an overall rate law of the form $$\frac{\text{d}\ce{[A_3]}}{\text{d}t} = k_1\ce{[A_0][B]^3} - k_2\ce{[A_3]},$$ and there is no a priori reason to suggest that this is likely. (Often this is possible in certain limiting regimes, but not in general.) To investigate $k_1$ and $k_2$, we will have to solve our system fully. We can write out the rate equations for our system as \begin{align*} \frac{\text{d}\ce{[A_1]}}{\text{d}t} &= 3k_+\ce{[A_0][B]}-k_-\ce{[A_1]}-2k_+\ce{[A_1][B]}+2k_-\ce{[A_2]}\\ \frac{\text{d}\ce{[A_2]}}{\text{d}t} &= 2k_+\ce{[A_1][B]}-2k_-\ce{[A_2]}-k_+\ce{[A_2][B]}+3k_-\ce{[A_3]}\\ \frac{\text{d}\ce{[A_3]}}{\text{d}t} &= k_+\ce{[A_2][B]}-3k_-\ce{[A_3]}, \end{align*} a coupled system of differential equations, which can be solved using matrix methods---diagonalize the coupling matrix, find the eigenvectors and eigenvalues, and solve---but these calculations are often quite messy, so I will leave them to you if you're interested.

Once $\ce{[A_2]}$ and $\ce{[A_3]}$ are known, we can plug them into the last differential equation and compare it with the presumed form of the overall rate law to determine $k_1$ and $k_2$.

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  • $\begingroup$ @RodrigodeAzevedo, sorry, I'm not seeing it. Can you write out what you think should be the corrected form of the differential equations? $\endgroup$
    – a-cyclohexane-molecule
    Dec 16, 2017 at 17:05
  • $\begingroup$ @RodrigodeAzevedo, oh, yes, that's correct. Thanks! $\endgroup$
    – a-cyclohexane-molecule
    Dec 16, 2017 at 17:19
  • $\begingroup$ @ a-cyclohexane-molecule thank you for your reply! Do mean to solve the system in the steady state? $\endgroup$
    – Sp_J
    Dec 19, 2017 at 11:16
  • $\begingroup$ @Sp_J, no, a steady-state assumption is not required, though it would make the calculations simpler. $\endgroup$
    – a-cyclohexane-molecule
    Dec 19, 2017 at 15:39

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