Question

The extent of adsorption from Langmuir adsorption isotherm is given by
$\frac{x}{m}=\frac{aP}{1+bP}$
The value of a and b are, respectively:

where,
$k_a$ and $k_d$ are rate constant of adsorption and desorption respectively.
$K_1$ is proportionality constant.

• A. $K_1\frac{k_a}{k_d}$ and $\frac{k_a}{k_d}$
  
• B. $\frac{k_a}{k_d}$ and $\frac{k_a}{k_d}$
• C. $K_1\frac{k_d}{k_a}$ and $\frac{k_a}{k_d}$
• D. $K_1\frac{k_a}{k_d}$ and $\frac{k_d}{k_a}$

Answer 1

The correct option is A $K_1\frac{k_a}{k_d}$ and $\frac{k_a}{k_d}$


In Langmuir adsorption isotherm,
$\theta$ : Fraction of surface covered on adsorbent
$1-\theta$ : Free surface are
$P$ : Pressure of gas
By langmuir adsorption isotherm,
$\text{Rate of adsorption}\propto \text{Fraction of surface available for adsorption}$​
$\text{Rate of adsorption}\propto \text{Pressure of gas}$
$\therefore$
$\text{Rate of adsorption}\propto P(1-\Theta )$
$\text{Rate of adsorption}=k_{a}P(1-\Theta )$

$\text{Rate of dessorption}\propto \text{Fraction of surface covered on adsorbent}$​
$\text{Rate of desorption}\propto \Theta$
$\text{Rate of desorption}=k_d\Theta$
$k_a$ and $k_d$ are rate constants.

At equilibrium condition,
$\text{Rate of adsorption}=\text{Rate of desorption}$
$k_{a}P(1-\Theta )=k_d\Theta$
$\frac{k_d}{P{k}_a}=\frac{1-\Theta }{\Theta }$
$1+\frac{k_d}{P{k}_a}=\frac{1}{\Theta }$
$\Theta =\frac{P{k}_a}{P{k}_a+k_d}$
$\Theta =\frac{KP}{KP+1}$

According to Langmuir, extent of adsorption is proportional to $\Theta$
$\frac{x}{m}\propto \Theta$
$\frac{x}{m}\propto \frac{KP}{1+KP}$

$\frac{x}{m}=K^{\prime }\frac{KP}{1+KP}$

$\frac{x}{m}=\frac{K^{\prime }KP}{1+KP}$
$\frac{x}{m}=\frac{aP}{1+bP}$

where,
$a=K^{\prime }K=K^{\prime }\frac{k_a}{k_d}$
$b=K=\frac{k_a}{k_d}$