Question

For a reversible physisorption process,
At pressure of $1atm$, fraction of surface covered is 0.3 i.e. $(\Theta =0.3)$.
If $K=\frac{k_a}{k_d}$, then calculate the value of $K$

• A. $1.22$
• B. $0.43$
• C. $0.02$
• D. $0.3$

Answer 1

The correct option is B $0.43$

$\text{Rate of adsorption}=k_{a}P(1-\Theta )$
$\text{Rate of desorption}=k_d\Theta$
$\text{Rate of adsorption}=\text{Rate of desorption}$
$k_{a}P(1-\Theta )=k_d\Theta$
$\Theta =\frac{k_{a}P}{k_d+k_{a}P}$

$\Theta =\frac{\frac{k_a}{k_d}\times P}{1+(\frac{k_a}{k_d}\times P)}$

$\Theta =\frac{KP}{1+KP}$
From given conditions:
$0.3=\frac{K\times 1}{1+K\times 1}$
$K=\frac{k_a}{k_d}=0.428\approx 0.43$