Surface Chemistry - Result Question 4

####4. A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation $\frac{x}{m}=K p^{0.5}$

Adsorption of the gas increases with

(2019 Main, 10 April I)

(a) increase in $p$ and increase in $T$

(b) increase in $p$ and decrease in $T$

(c) decrease in $p$ and decrease in $T$

(d) decrease in $p$ and increase in $T$

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Answer:

Correct Answer: 4. (b)

Solution:

  1. For physisorption or physical adsorption,

Adsorption isotherm (Temperature, $T=$ constant) is shown below:

where, $x=$ amount of adsorbate, $m=$ amount of adsorbent,

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

$\frac{1}{n}=$ order of the reaction, where, $0<\frac{1}{n}<1$ and so,

$1<n<\infty$

$$ \begin{array}{ll} \text { Here, } & \frac{x}{m}=K p^{\frac{1}{2}}, \ \text { i.e. } & \frac{x}{m} \propto p^{\frac{1}{2}} \end{array} $$

Adsorption isobar (Pressure, $p=$ constant)

So, the rate of physical adsorption of the gas, increases with $p$ (when, $T$ is constant) and decreases with $T$ (when $p$ is constant).



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