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  4. The rate of chemisorption of a gas

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Q. The rate of chemisorption of a gas

AMUAMU 2017

Solution:

Rate of chemisorption of a gas increases with gas increases in pressure because more molecules are adsorbed on the surface at higher pressure. This can also understood by the following relation:
$\log \frac{x}{m}=\log\,K+\frac{1}{n} \log\,p$
where, $\frac{x}{m}$= mass adsorbed over per unit surface of adsorbent
$n$ and $k$ are constants $p$= pressure,
i.e., $\frac{x}{m}\propto p$