4AB › Chemistry PYQ › CSIR-NET › December 2011 › Q119Q119 · CSIR-NET Chemistry, December 2011 Paper: CSIR-NET December 2011 · Subject: Physical Chemistry · Chapter: Surface Chemistry · Topic: Adsorption Isotherms · Marks: 2 · Difficulty: Medium
The Langmuir adsorption isotherm is given by $\frac{\mathrm{kp}}{1+\mathrm{kp}}$, where P is the pressure of the adsorabate gas. The Langmuir adsorption isotherm for a diatomic gas $\mathrm{A}_{2}$ undergoing dissociative adsorption is:
(c) $\theta=(\mathrm{Kp})^{2} /\left(1+(\mathrm{Kp})^{2}\right)$
(d) $\theta=(\mathrm{Kp})^{1 / 2}(1+(\mathrm{Kp}))^{1 / 2}$
(a) $\theta=\mathrm{Kp}/(1+\mathrm{Kp})$
(b) $\theta=2\mathrm{Kp}/(1+2\mathrm{Kp})$
Answer Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation Dissociative adsorption A₂ → 2A(ads): equilibrium gives θ² ∝ p(1 − θ)², so θ = (Kp)^½/(1 + (Kp)^½).
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