The following derivation is adapted from Foundations of Chemical Kinetics.2 This derivation assumes the reaction A + B → C {\displaystyle A+B\rightarrow C} . Consider a sphere of radius R A {\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 R A B {\displaystyle R_{AB}} apart.
If we assume a local steady state, then the rate at which B reaches R A B {\displaystyle R_{AB}} is the limiting factor and balances the reaction.
Therefore, the steady state condition becomes
1. k [ B ] = − 4 π r 2 J B {\displaystyle k[B]=-4\pi r^{2}J_{B}}
where
J B {\displaystyle J_{B}} is the flux of B, as given by Fick's law of diffusion,
2. J B = − D A B ( d B ( r ) d r + [ B ] k B T d U d r ) {\displaystyle J_{B}=-D_{AB}({\frac {dB(r)}{dr}}+{\frac {[B]}{k_{B}T}}{\frac {dU}{dr}})} ,
where D A B {\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. k [ B ] = 4 π r 2 D A B ( d B ( r ) d r + [ B ] ( r ) k B T d U d r ) {\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 exp ( − U ( r ) / k B T ) ⋅ d d r ( [ B ] ( r ) exp ( U ( r ) / k B T ) = ( d B ( r ) d r + [ B ] ( r ) k B T d U d r ) {\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. k [ B ] = 4 π r 2 D A B exp ( − U ( r ) / k B T ) ⋅ d d r ( [ B ] ( r ) exp ( U ( r ) / k B T ) {\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. k [ B ] exp ( U ( r ) / k B T ) 4 π r 2 D A B = d d r ( [ B ] ( r ) exp ( U ( r ) / k B T ) {\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 [ B ] ( r ) → [ B ] {\displaystyle [B](r)\rightarrow [B]} , ie the local concentration of B approaches that of the solution at large distances, and consequently U ( r ) → 0 {\displaystyle U(r)\rightarrow 0} , as r → ∞ {\displaystyle r\rightarrow \infty } , we can solve 5 by separation of variables, we get
6. ∫ R A B ∞ d r k [ B ] exp ( U ( r ) / k B T ) 4 π r 2 D A B = ∫ R A B ∞ d ( [ B ] ( r ) exp ( U ( r ) / k B T ) {\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. k [ B ] 4 π D A B β = [ B ] − [ B ] ( R A B ) exp ( U ( R A B ) / k B T ) {\displaystyle {\frac {k[B]}{4\pi D_{AB}\beta }}=[B]-[B](R_{AB})\exp(U(R_{AB})/k_{B}T)} (where : β − 1 = ∫ R A B ∞ 1 r 2 exp ( U ( r ) k B T d r ) {\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 k r {\displaystyle k_{r}} , so [ B ] ( R A B ) = k [ B ] / k r {\displaystyle [B](R_{AB})=k[B]/k_{r}} . Substituting this into 7 and rearranging yields
8. k = 4 π D A B β k r k r + 4 π D A B β exp ( U ( R A B ) k B T ) {\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 k r {\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 k D = 4 π D A B β {\displaystyle k_{D}=4\pi D_{AB}\beta } . 8 can then be re-written as the "diffusion influenced rate constant" as
9. k = k D k r k r + k D exp ( U ( R A B ) k B T ) {\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 U ( r ) ≈ 0 {\displaystyle U(r)\approx 0} for all r except very small r, β − 1 ≈ 1 R A B {\displaystyle \beta ^{-1}\approx {\frac {1}{R_{AB}}}} . The reaction rate 9 simplifies even further to
10. k = k D k r k r + k D {\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 R A {\displaystyle R_{A}} as D A = k B T 3 π R A η {\displaystyle D_{A}={\frac {k_{B}T}{3\pi R_{A}\eta }}} where η {\displaystyle \eta } is the viscosity of the solution. Inserting this into 9 gives an estimate for k D {\displaystyle k_{D}} as 8 R T 3 η {\displaystyle {\frac {8RT}{3\eta }}} , where R is the gas constant, and η {\displaystyle \eta } is given in centipoise. For the following molecules, an estimate for k D {\displaystyle k_{D}} is given:
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Atkins, Peter (1998). Physical Chemistry (6th ed.). New York: Freeman. pp. 825–8. ↩
Roussel, Marc R. "Lecture 28:Diffusion-influenced reactions, Part I" (PDF). Foundations of Chemical Kinetics. University of Lethbridge (Canada). Retrieved 19 February 2021. http://people.uleth.ca/~roussel/C4000foundations/slides/28diffusion_influencedI.pdf ↩
Berg, Howard, C. Random Walks in Biology. pp. 145–148.{{cite book}}: CS1 maint: multiple names: authors list (link) /wiki/Template:Cite_book ↩