ABC26PH0212 · Surface Chemistry

Subject: Physical Chemistry · Chapter: Surface Chemistry · Topic: Adsorption Isotherms · Exam: CSIR-NET DEC 2011 · Marks: 2 · Difficulty: Hard

The Langmuir adsorption isotherm is given by $\theta=\dfrac{kp}{1+kp}$, where P is the pressure of the adsorabate gas. The Langmuir adsorption isotherm for a diatomic gas $A_{2}$ undergoing dissociative adsorption is:

(a)$\theta=Kp/(1+Kp)$

(b)$\theta=2Kp/(1+2Kp)$

(c)$\theta=(\mathrm{Kp})^{2} /\left(1+(\mathrm{Kp})^{2}\right)$

(d)$\theta=(K p)^{1 / 2}(1+(K p))^{1 / 2}$

Answer

Answer (as printed): D

Explanation

For dissociative adsorption, $\theta=(Kp)^{1/2}(1+(Kp))^{1/2}$.

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