Diffusion-controlled reaction

The following derivation is adapted from Foundations of Chemical Kinetics.^[2] This derivation assumes the reaction ${\displaystyle A+B\rightarrow C}$. Consider a sphere of radius ${\displaystyle R_{A}}$, centered at a spherical molecule A, with reactant B flowing in and out of it. A reaction is considered to occur if molecules A and B touch, that is, when the distance between the two molecules is ${\displaystyle R_{AB}}$ apart.

If we assume a local steady state, then the rate at which B reaches ${\displaystyle R_{AB}}$ is the limiting factor and balances the reaction.

Therefore, the steady state condition becomes

1. ${\displaystyle k[B]=-4\pi r^{2}J_{B}}$

where

${\displaystyle J_{B}}$ is the flux of B, as given by Fick's law of diffusion,

2. ${\displaystyle J_{B}=-D_{AB}({\frac {dB(r)}{dr}}+{\frac {[B]}{k_{B}T}}{\frac {dU}{dr}})}$,

where ${\displaystyle D_{AB}}$ is the diffusion coefficient and can be obtained by the Stokes-Einstein equation, and the second term is the gradient of the chemical potential with respect to position. Note that [B] refers to the average concentration of B in the solution, while [B](r) is the "local concentration" of B at position r.

Inserting 2 into 1 results in

3. ${\displaystyle k[B]=4\pi r^{2}D_{AB}({\frac {dB(r)}{dr}}+{\frac {[B](r)}{k_{B}T}}{\frac {dU}{dr}})}$.

It is convenient at this point to use the identity ${\displaystyle \exp(-U(r)/k_{B}T)\cdot {\frac {d}{dr}}([B](r)\exp(U(r)/k_{B}T)=({\frac {dB(r)}{dr}}+{\frac {[B](r)}{k_{B}T}}{\frac {dU}{dr}})}$ allowing us to rewrite 3 as

4. ${\displaystyle k[B]=4\pi r^{2}D_{AB}\exp(-U(r)/k_{B}T)\cdot {\frac {d}{dr}}([B](r)\exp(U(r)/k_{B}T)}$.

Rearranging 4 allows us to write

5. ${\displaystyle {\frac {k[B]\exp(U(r)/k_{B}T)}{4\pi r^{2}D_{AB}}}={\frac {d}{dr}}([B](r)\exp(U(r)/k_{B}T)}$

Using the boundary conditions that ${\displaystyle [B](r)\rightarrow [B]}$, ie the local concentration of B approaches that of the solution at large distances, and consequently ${\displaystyle U(r)\rightarrow 0}$, as ${\displaystyle r\rightarrow \infty }$, we can solve 5 by separation of variables, we get

6. ${\displaystyle \int _{R_{AB}}^{\infty }dr{\frac {k[B]\exp(U(r)/k_{B}T)}{4\pi r^{2}D_{AB}}}=\int _{R_{AB}}^{\infty }d([B](r)\exp(U(r)/k_{B}T)}$ or

7. ${\displaystyle {\frac {k[B]}{4\pi D_{AB}\beta }}=[B]-[B](R_{AB})\exp(U(R_{AB})/k_{B}T)}$ (where : ${\displaystyle \beta ^{-1}=\int _{R_{AB}}^{\infty }{\frac {1}{r^{2}}}\exp({\frac {U(r)}{k_{B}T}}dr)}$)

For the reaction between A and B, there is an inherent reaction constant ${\displaystyle k_{r}}$, so ${\displaystyle [B](R_{AB})=k[B]/k_{r}}$. Substituting this into 7 and rearranging yields

8. ${\displaystyle k={\frac {4\pi D_{AB}\beta k_{r}}{k_{r}+4\pi D_{AB}\beta \exp({\frac {U(R_{AB})}{k_{B}T}})}}}$

Suppose ${\displaystyle k_{r}}$ is very large compared to the diffusion process, so A and B react immediately. This is the classic diffusion limited reaction, and the corresponding diffusion limited rate constant, can be obtained from 8 as ${\displaystyle k_{D}=4\pi D_{AB}\beta }$. 8 can then be re-written as the "diffusion influenced rate constant" as

9. ${\displaystyle k={\frac {k_{D}k_{r}}{k_{r}+k_{D}\exp({\frac {U(R_{AB})}{k_{B}T}})}}}$

If the forces that bind A and B together are weak, ie ${\displaystyle U(r)\approx 0}$ for all r except very small r, ${\displaystyle \beta ^{-1}\approx {\frac {1}{R_{AB}}}}$. The reaction rate 9 simplifies even further to

10. ${\displaystyle k={\frac {k_{D}k_{r}}{k_{r}+k_{D}}}}$ This equation is true for a very large proportion of industrially relevant reactions in solution.

The Stokes-Einstein equation describes a frictional force on a sphere of diameter ${\displaystyle R_{A}}$ as ${\displaystyle D_{A}={\frac {k_{B}T}{3\pi R_{A}\eta }}}$ where ${\displaystyle \eta }$ is the viscosity of the solution. Inserting this into 9 gives an estimate for ${\displaystyle k_{D}}$ as ${\displaystyle {\frac {8RT}{3\eta }}}$, where R is the gas constant, and ${\displaystyle \eta }$ is given in centipoise. For the following molecules, an estimate for ${\displaystyle k_{D}}$ is given:

Solvents and ${\displaystyle k_{D}}$

Solvent	Viscosity (centipoise)	${\displaystyle k_{D}({\frac {\times 1e9}{M\cdot s}})}$
n-Pentane	0.24	27
Hexadecane	3.34	1.9
Methanol	0.55	11.8
Water	0.89	7.42
Toluene	0.59	11

^[3]

• Diffusion limited enzyme