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 Oct 13 '20 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 Oct 13 '20 at 13:30
  • $\begingroup$ Yes actually, thankyou $\endgroup$ – Ferdinando Randisi Oct 14 '20 at 9:56
  • $\begingroup$ 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 Oct 14 '20 at 10:14
  • $\begingroup$ As AB dissociates as $\exp(-k\cdot t)$ the mean time $\langle t\rangle=1/k$. $\endgroup$ – porphyrin Oct 14 '20 at 14:49

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 Oct 13 '20 at 15:30
  • $\begingroup$ @porphyrin thanks, that's exactly what I'm looking for. Can I ask where have you taken the formula from? $\endgroup$ – Ferdinando Randisi Oct 14 '20 at 9:54
  • $\begingroup$ Updated answer with info from @porphyrin , giving him credit. $\endgroup$ – Poutnik Oct 14 '20 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 Oct 14 '20 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.