= Solution
At <chemical equilibrium>, <adsorption> and <desorption> balance:
$$
(1-\theta_1)Mp k_a=\theta_1Mk_d.
$$
Canceling $M$ and rearranging gives
$$
\frac{\theta_1}{1-\theta_1}=\frac{k_a}{k_d}p=K_{\rm ads}p,
\qquad
\boxed{\theta_1=\frac{K_{\rm ads}p}{1+K_{\rm ads}p}}.
$$
Thus \b[$K_{\rm ads}=b$] when the kinetic and thermodynamic descriptions refer to the same ideal surface and energy convention. In particular,
$$
\frac{k_a}{k_d}=\frac{\Lambda^3}{k_BT}
e^{(\varepsilon_0-\varepsilon_s)/(k_BT)}.
$$
The ratio fixes the <adsorption isotherm>, while the individual <rate constants> also determine relaxation toward equilibrium. Here $k_a$ has units of inverse pressure per time and $k_d$ of inverse time, so $K_{\rm ads}$ is dimensional. This equivalent-site arrival/departure argument is the simple case of the <Langmuir adsorption model>.<The Adsorption of Gases on Plane Surfaces of Glass, Mica and Platinum by Irving Langmuir>[theoretical part, Case I, pp. 1368–1370: equivalent elementary surface spaces and equilibrium adsorption]{ref}
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