4AB › Chemistry PYQ › CSIR-NET › December 2011 › Q28Q28 · CSIR-NET Chemistry, December 2011 Paper: CSIR-NET December 2011 · Subject: Physical Chemistry · Chapter: Surface Chemistry · Topic: Adsorption Isotherms · 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: D ✓ checked by 4AB · confidence high
Explanation Dissociative adsorption needs two sites: θ = (Kp)^½/(1 + (Kp)^½) (d).
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