Freundlich adsorption isotherm equation is

Question

Freundlich adsorption isotherm equation is (A)  log (m/x)= log k+(1/n) log p  (B)  log (x/m)= log k +n log p  (C)  log (m/x)= log k+n log p  (D)  log (x/m)= log k+(1/n) log p

Tardigrade Admin · Accepted Answer

Freundlich adsorption isotherm equation is as log (x/m)= log k+(1/n) log p where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p. k and n are constants.

Q. Freundlich adsorption isotherm equation is

$lo g \frac{m}{x} = lo g k + \frac{1}{n} lo g p$

$lo g \frac{x}{m} = lo g k + n lo g p$

$lo g \frac{m}{x} = lo g k + n lo g p$

$lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g p$

Freundlich adsorption isotherm equation is as
$lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g p$
where $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent at pressure $p . k$ and $n$ are constants.

Q. Freundlich adsorption isotherm equation is

logxm​=logk+n1​logp

logmx​=logk+nlogp

logxm​=logk+nlogp

logmx​=logk+n1​logp

Freundlich adsorption isotherm equation is as logmx​=logk+n1​logp where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p.k and n are constants.

$lo g \frac{m}{x} = lo g k + \frac{1}{n} lo g p$

$lo g \frac{x}{m} = lo g k + n lo g p$

$lo g \frac{m}{x} = lo g k + n lo g p$

$lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g p$

Freundlich adsorption isotherm equation is as
$lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g p$
where $x$ is the mass of the gas adsorbed on mass $m$ of the adsorbent at pressure $p . k$ and $n$ are constants.