There is a solution with two chemical species, A and B. Given the equilibrium dissociation constant

$ K_D=\frac{[A][B]}{[AB]} $

how do I relate the mean dissociation time $t_{off}$ to $K_D$, and initial reagent concentrations, assuming that the reaction time it takes for A and B to bind is much lower than the diffusion time from one to the other?

I know that $k_D = t_{off}/t_{on}$, so here I'm looking for $t_{on}$ in the approximation of instantaneous binding reaction time, which I assume should be the time required for a particle of a species to diffuse through the solution from a particle of the other species to the next. How do I do that to solve the question in the previous paragraph?

EDIT: changed notation to fix typo and to incorporate Poutnik's feedback.

  • $\begingroup$ Could you elaborate the question little bit, to avoid confusion, misinterpretation and false assumptions about the question ? As too laconic questions without enough context often lead to too much clarifications effort. $\endgroup$
    – Poutnik
    Commented Oct 13, 2020 at 13:01
  • $\begingroup$ What exactly do you mean by "mean dissociation time" ? Do you mean the reaction half-time, applicable on kinetics of the first order like $\ce{AB -> A + B }$ as $t_{1/2}=\frac{\ln{2}}{k}$ ? $\endgroup$
    – Poutnik
    Commented Oct 13, 2020 at 13:30
  • $\begingroup$ Then you have given K, 1 k from the diffusion control based formula, the other k from these 2, and from it t1/2. $\endgroup$
    – Poutnik
    Commented Oct 14, 2020 at 10:14
  • $\begingroup$ As AB dissociates as $\exp(-k\cdot t)$ the mean time $\langle t\rangle=1/k$. $\endgroup$
    – porphyrin
    Commented Oct 14, 2020 at 14:49

1 Answer 1


Note that convention usually used Capital K for TD equilibrium constants and small k for kinetic rate constants.

Thermodynamic equilibrium constants say nothing about kinetic reaction rate constants, but about their ratio. There can be 2 reactions accidentally with the same equilibrium constant, but one with both forward and backward reaction rate very fast, the other very slow.

If there is reaction $\ce{AB <=> A + B}$ with the equilibrium constant $K_\mathrm{c}=\frac{[A][B]}{[AB]}$

with the forward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = - k_\mathrm{f} \cdot [\ce{AB}]$ and the reaction halftime $t_{1/2}=\frac{\ln{2}}{k_\mathrm{f}}$

and with the backward reaction kinetic $\frac{\mathrm{d}[AB]}{\text{d}t} = k_\mathrm{b} \cdot [\ce{A}][\ce{B}]$

then $K_\mathrm{c}=\frac {k_\mathrm{f}}{k_\mathrm{b}}$

All is assuming the reaction has trivial simple reaction mechanism.

Credit to @Porphyrin:

If $\ce{ A + B -> AB}$ is diffusion controlled, the rate constant would be (for equal sized molecules or approximately so ) $k_\mathrm{d} = k_\mathrm{b} = 8000 \frac {RT}{3η}$
where η is the solvent viscosity,
$k_\mathrm{d}$ evaluates to $≈\pu{5⋅1010dm3/mol/s}$ in low viscosity solvents such as ethanol.

Then :

$$kf=k_\mathrm{b} \cdot K_\mathrm{c}=K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)$$

and the reaction halftime is:

$$t_{1/2,f} = \frac {\ln 2}{k_\mathrm{f}} =\frac{\ln {2}}{K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)} =\frac{\ln {2} \cdot 3η}{ 8000 \cdot K_\mathrm{c} \cdot RT} $$

  • $\begingroup$ In fact it would be easier if A+B were diffusion controlled as the the rate constant would be (for equal sized molecules or approximately so ) $k_d=8000 RT/(3\eta)$ where $\eta$ is the solvent viscosity. $k_d$ evaluates to $\approx 5 \cdot 10^{10}\mathrm{dm^3/mol/s}$ in low viscosity solvents such as ethanol. $\endgroup$
    – porphyrin
    Commented Oct 13, 2020 at 15:30
  • $\begingroup$ @porphyrin thanks, that's exactly what I'm looking for. Can I ask where have you taken the formula from? $\endgroup$ Commented Oct 14, 2020 at 9:54
  • $\begingroup$ Updated answer with info from @porphyrin , giving him credit. $\endgroup$
    – Poutnik
    Commented Oct 14, 2020 at 10:43
  • $\begingroup$ Smoluchowski has calculated the rate constant for diffusion controlled reactions assuming that the solvent was a continuous fluid, i.e. not made of molecules. This is a good approximation if the solvent molecules are small compared to those reacting and still good (surprisingly) even when this is not true. The rate constant is $k_d=4000\pi NDr$ where $D=D_A+D_B,\; r=r_A+r_B$, $r$ being radius. From Stokes-Einstein eqn $D=k_BT/\zeta$ where $\zeta$ is friction $\zeta=6\pi \eta r$ and $\eta$ is viscosity. Substitute for $D$ for both A and B and make $r_A = r_B$. $\endgroup$
    – porphyrin
    Commented Oct 14, 2020 at 14:55

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