Question

Which among the following statement are correct with respect to adsorption of gases on a solid?

• A. According to Freundiich isotherm,
  Extent of adsorption $=k{p}^n$
• B. According to Freundiich isotherm,
  Extent of adsorption $=k{p}^{1/n}$
• C. According to Langmuir isotherm,
  $\text{Extent of adsorption}=(1+bp)/ap$
• D. According to Langmuir isotherm,
  $\text{Extent of adsorption}=ap/(1+ap)$

Answer 1

Answer: (B), (D)

According to Langmuir isotherm,
$\text{Extent of adsorption}=ap/(1+ap)$

According to Freundlich isotherm,

$\frac{x}{m}\propto p^{\frac{1}{n}}$

$\frac{x}{m}=k{p}^{{}^{\frac{1}{n}}}n=1\dots ..\infty$

$\frac{x}{m}$ = Extent of adsorption

Langmuir isotherm:

$A\left(g\right)+M\left(s\right)\rightleftharpoons AM\left(s\right)$

$R_a\propto \text{Pressure}\times \text{Number of sites available}{R}_a\propto p\times \left[n\right(1-\theta \left)\right]R_a=K_a\times p\times n(1-\theta )$

$R_d\propto \text{Number of sites occupied}{R}_d\propto n\times \theta R_d=K_d\times n\times \theta$

At equilibrium,
$\text{Rate of adsorption}=\text{Rate of desorption}$

$K_a\times p\times n(1-\theta )=K_d\times n\times \theta$

$\frac{K_a}{K_d}\times p(1-\theta )=\theta$

$a\times p(1-\theta )=\theta$

$\frac{1-\theta }{\theta }=\frac{1}{ap}$

$\frac{1}{\theta }-1=\frac{1}{ap}$

$\theta =\frac{ap}{1+ap}$

$\theta$ = Extent of adsorption