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Is it possible to describe a reaction of the $$\ce{A + B <=> AB}$$ exactly using the solution of an isomerization reaction? $$\ce{A <=> B}$$

I just can not figure it out the equations seem so similar but the solution in my script is so convoluted that I can not prove they are equal.

Equations for top reaction: $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ and for the bottom: \begin{align} {\partial c_\ce{A} \over \partial t} &= -k_{\ce{A -> B}} c_\ce{A} + k_{\ce{B -> A}} c_\ce{B}\\ {\partial c_\ce{B} \over \partial t} &= k_{\ce{A -> B}} c_\ce{A} - k_{\ce{B -> A}} c_\ce{B}\\ \end{align}

Could we not just represent $\ce{A + B <=> AB}$ as a new reaction with $\ce{C $=$ A + B}$ and $\ce{D $=$ AB}$ and thus use the isomerization results on $\ce{C <=> D}$?

I could write a set of equations for the $ A+B \rightleftharpoons AB $ reaction : $${\partial c_{AB} \over \partial t} = -k_{off} c_{AB} + k_{on} c_A c_B$$ $${\partial c_{A}c_{B} \over \partial t} = +k_{off} c_{AB} - k_{on} c_A c_B$$

which looks suspiciously like the $$A \rightleftharpoons B$$ reaction.

A starting point might to prove they can be used to express each other would be to use the equilibrium relation $$k_d = { c_\ce{A} c_\ce{B} \over c_\ce{AB}}$$ with the relation for the total number of species A $c_\mathrm{total} = c_\ce{A} + c_\ce{AB}$ to get : $$c_\ce{A} c_\ce{B} = k_d (c_\mathrm{total} -c_\ce{A})$$

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    $\begingroup$ This is a bit like "can't we describe a kitten as a dolphin"? Sure, both are cute, but other than that, they are hardly similar. The same applies to your reactions. $\endgroup$ – Ivan Neretin Mar 29 '16 at 9:47
  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. A bimolecular reaction will never be a unimolecular reaction. They occur with a completely different mechanism, e.g. UC Davis. However, you can tweak the experiment to run a bimolecular reaction in a pseudo-first order fashion. $\endgroup$ – Martin - マーチン Mar 29 '16 at 9:55
  • $\begingroup$ If you are only concerned about the fact that they are mammals that this might be a good example. It might depend.. But is there a mathematical reason why these can't be equivalent? $\endgroup$ – pindakaas Mar 29 '16 at 9:59
  • $\begingroup$ not shure what happened to my edit just now sorry $\endgroup$ – pindakaas Mar 29 '16 at 10:10
  • $\begingroup$ Thanks for the link martin but in the article the isomerisation is not talked about. maybe that is the difference? $\endgroup$ – pindakaas Mar 29 '16 at 10:15
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TL;DR; Addition in a chemical reaction does not mean mathematical addition, and translating this operation to a rate expression makes no physical sense.

Elaboration: Your idea $\ce{C = A + B}$ changes the order of reaction, and has no physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation: $${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$ has the physical interpretation that $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

To help clarify, consider the following reaction: $$\ce{Br^{\cdot} + Br^{\cdot} <=> Br2}$$

In this case A would be $\ce{Br^{\cdot}}$ and B would be $\ce{Br^{\cdot}}$, and thus your C would be $\ce{[Br^{\cdot}] + [Br^{\cdot}] = 2 [Br^{\cdot}]}$. The rate expressions would therefore be first order in $\ce{Br^{\cdot}}$ as opposed to second order.

Other comments:

  • You also forgot to include the other reactions for the bimolecular case: $${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$Omitting these may be leading to some confusion.

  • Little k is a rate constant. Equilibria are described with big K. It is unclear what $\ce{k_{d}}$ means in your question.

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