Chemistry - Surface Chemistry Question with Solution | TestHub

$\frac{\text{x}}{\text{m}}\text{=}\frac{\text{ap}}{\text{1}\text{+}\text{bp}}$
at high pressure
$\text{1}\text{+}\text{bp}\simeq \text{bp}$
$\frac{\text{x}}{\text{m}}\text{=}\frac{\text{ap}}{\text{bp}}\Rightarrow \frac{\text{x}}{\text{m}}\text{=}\frac{\text{a}}{\text{b}}\text{=}\text{constant}$
At very high P, rate of adsorption = rate or desorption
$\therefore \frac{\text{X}}{\text{m}}\text{=}\frac{\text{a}}{\text{b}}\text{=}\text{const}\text{.(K)}$
$\therefore {\text{ΔG}}^{\text{o}}\text{=}-\text{2}\text{.303}\text{log}\text{K}$
or${\text{ΔH}}^{\text{o}}-{\text{TΔS}}^{\text{o}}\text{=}-\text{2}\text{.303}\text{log}\text{K}$