Solution
= Solution
$0.1\,\mathrm M=100\,\mathrm{mol\,m^{-3}}=10^5\,\mathrm{mmol\,m^{-3}}$. Therefore
$$
\theta_p=\frac{0.36(10^5)}{1+0.36(10^5)}
=0.999972,
\qquad
v=2.1\theta_p\simeq
\boxed{2.10\,\mathrm{mmol\,m^{-3}s^{-1}}}.
$$
The surface is almost saturated, so the rate approaches the supplied limiting constant. This assumes rapid <adsorption> equilibrium, negligible product coverage, no competing species and unchanged substrate area/photochemical conditions.