How do I compute the mean dissociation time from the dissociation constant, assuming instantaneous binding reaction rate?
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 $\ce{AB &lt;=&gt; A + B}$ 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 $\ce{ A + B -&gt; AB}$ 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:
$$t_{1/2,f} = \frac {\ln 2}{k_\mathrm{f}} =\frac{\ln {2}}{K_\mathrm{c} \cdot \left( 8000 \frac {RT}{3η}\right)} =\frac{\ln {2} \cdot 3η}{  8000 \cdot K_\mathrm{c} \cdot RT} $$