Consider a reaction, in which the rate-limiting
elementary reaction step is of the form :A + B → C and occurs at rate k_r when molecules of A and B touch. For a bulk system, the observed reaction rate k is depressed, because molecules of A and B must diffuse towards each other before reacting. At very large values of k_r, the bulk reaction occurs at a rate k_D which is relatively independent of the properties of the reaction itself. The following derivation is adapted from
Foundations of Chemical Kinetics. Consider sphere of radius R_A, centered at a spherical molecule A, with reactant B flowing in and out of it; molecules A and B touch when the distance between the two molecules is R_{AB} apart. Thus [B](R_{AB})k_r = [B]k, where [B](r) is the smoothed "local concentration" of B at position r. If we assume a local steady state, then the average rate at which B reaches R_{AB} corresponds to the observed reaction rate k. This can be written as: {{NumBlk|| [B]k=-4\pi r^2 J_{B}\text{,}|}} where J_{B} is the flux of B into the sphere. By
Fick's law of diffusion, {{NumBlk|| J_{B} = -D_{AB} \left(\frac{d[B](r)}{dr} +\frac{[B](r)}{k_{B}T} \frac{dU}{dr}\right)\text{,}|}} where D_{AB} is the diffusion coefficient, obtained by the
Stokes-Einstein equation. The second term is the positional gradient of the
chemical potential. Inserting into gives {{NumBlk||[B]k = 4\pi r^2 D_{AB}\left(\frac{dB(r)}{dr}+\frac{[B](r)}{k_{B}T} \frac{dU}{dr}\right)\text{.}|}} It is convenient at this point to use the identity \exp\left(-\frac{U(r)}{k_{B}T}\right)\frac{d}{dr}\left([B](r)\exp\left(\frac{U(r)}{k_{B}T}\right)\right) = \frac{d[B](r)}{dr}+\frac{[B](r)}{k_{B}T} \frac{dU}{dr} and rewrite as {{NumBlk||[B]k = 4\pi r^2 D_{AB} \exp\left(-\frac{U(r)}{k_{B}T}\right)\frac{d}{dr}\left([B](r)\exp\left(\frac{U(r)}{k_{B}T}\right)\right)|}} Thus {{NumBlk||k\cdot\frac{[B]}{4\pi r^2 D_{AB}}\exp\left(\frac{U(r)}{k_{B}T}\right) = \frac{d}{dr}\left([B](r)\exp\left(\frac{U(r)}{k_{B}T}\right)\right)|}} which is an
ordinary differential equation in [B](r). Using the boundary conditions that [B](r)\rightarrow [B], ie the local concentration of B approaches that of the solution at large distances, and consequently U(r) \rightarrow 0 as r \rightarrow \infty , we can solve by
separation of variables. Namely: {{NumBlk|| \int_{R_{AB}}^{\infty} \frac{[B]k\,dr}{4\pi r^2 D_{AB}}\exp\left(\frac{U(r)}{k_{B}T}\right) = \int_{R_{AB}}^{\infty} d\left([B](r)\exp\left(\frac{U(r)}{k_{B}T}\right)\right)|}} Defining \beta^{-1} = \int_{R_{AB}}^{\infty} \frac{1}{r^2}\exp\left(\frac{U(r)}{k_B T}\right)\,dr\text{,} simplifies to {{NumBlk|| \frac{[B]k}{4\pi D_{AB}\beta }= [B]-[B](R_{AB})\exp\left(\frac{U(R_{AB})}{k_{B}T}\right)|}} From the definition of k_r, we have . Substituting this into and rearranging yields {{NumBlk|| k = \frac{4\pi D_{AB}\beta k_r }{k_r + 4\pi D_{AB} \beta \exp\left(\frac{U(R_{AB} )}{k_B T}\right) } |}} Taking k_r very large gives the diffusion-limited reaction rate k_D = 4\pi D_{AB} \beta\text{.} can then be re-written as the "diffusion influenced rate constant" {{NumBlk|| k= \frac{k_D k_r}{k_r + k_D \exp\left(\frac{U(R_{AB} )}{k_B T}\right)} |}} If the forces that bind A and B together are weak, i.e. U(r) \approx 0 for all r>R_{AB}, then \beta^{-1} \approx \frac{1}{R_{AB}} In that case, simplifies even further to {{NumBlk|| 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.
Viscosity dependence The
Stokes-Einstein equation describes a frictional force on a sphere of diameter R_A as D_A = \frac{k_BT}{3\pi R_A \eta} where \eta is the viscosity of the solution. Inserting this into gives an estimate for k_D as \frac{8 RT}{3\eta} , where R is the gas constant, and \eta is given in centipoise: == See also ==