# How do I compute the mean dissociation time from the dissociation constant, assuming instantaneous binding reaction rate?

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.

• 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. – Poutnik Oct 13 '20 at 13:01
• 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 $t_{1/2}=\frac{\ln{2}}{k}$ ? – Poutnik Oct 13 '20 at 13:30
• Yes actually, thankyou – Ferdinando Randisi Oct 14 '20 at 9:56
• Then you have given K, 1 k from the diffusion control based formula, the other k from these 2, and from it t1/2. – Poutnik Oct 14 '20 at 10:14
• As AB dissociates as $\exp(-k\cdot t)$ the mean time $\langle t\rangle=1/k$. – 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:

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

• 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. – porphyrin Oct 13 '20 at 15:30
• @porphyrin thanks, that's exactly what I'm looking for. Can I ask where have you taken the formula from? – Ferdinando Randisi Oct 14 '20 at 9:54
• Updated answer with info from @porphyrin , giving him credit. – Poutnik Oct 14 '20 at 10:43
• 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 $r_A = r_B$. – porphyrin Oct 14 '20 at 14:55