Discussion
According to Langmuir model of adsorption, there is an equilibrium between adsorption and desorption as given below :
\[ \mathrm{A_{(g)} + S_{(s)} \rightleftharpoons A{-}S_{(s)}} \]
where A & S represent the adsorbate and the solid surface sites and A-S is the adsorbed species.
The adsorption coefficient K is defined as ka/kd where ka and kd are the rate constants for adsorption and desorption respectively.
Which of the following is correct ?
(I) \(\mathrm{K = \frac{[A{-}S]}{[A][S]}}\)
(II) \(\mathrm{K = \frac{[A][S]}{[A{-}S]}}\)
(III) A large value of K indicates strong adsorption
(IV) A large value of K indicates strong desorption
Answer: A
The correct statements are (I) and (III).
At equilibrium the rates of adsorption and desorption are equal:
\[ \begin{aligned} k_{a}[\mathrm{A}][\mathrm{S}] &= k_{d}[\mathrm{A{-}S}] \\ K = \frac{k_{a}}{k_{d}} &= \frac{[\mathrm{A{-}S}]}{[\mathrm{A}][\mathrm{S}]} \end{aligned} \]
So (I) is right and (II), its reciprocal, is wrong. A large K means adsorption is fast compared with desorption and most of the adsorbate stays bound, which is strong adsorption (III), not strong desorption (IV).
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