Question

The slope of the straight line given in the following diagram for adsorption is:

• A. 1/n ( 0 to 1)
• B. 1/n (0.1 to 0.5)
• C. log n
• D. log (1/n)

Answer 1

The correct option is A 1/n ( 0 to 1)

Answer: (a)

According to Freundlich adsorption isotherm, log (x/m) = log K + 1/n log P.

So, the slope of the graph of log (x/m) vs log P is 1/n.
Here, the factor $\frac{1}{n}$ can have value between 0 to 1 based on the pressure range.

Freundlich isotherm holds good over a limited range of low pressure.
Hence, at low pressure, the isotherm is a straight line
$\therefore \frac{1}{n}=1$

$\Rightarrow \frac{x}{m}=kp$


At intermediate pressure,
$0<\frac{1}{n}<1$
Hence,
$\Rightarrow \frac{x}{m}=k{p}^{\frac{1}{n}}$


At high pressure, the extent of adsorption attains a constant value and becomes independent of pressure.
Thus, $\frac{1}{n}=0$
$\Rightarrow \frac{x}{m}=k{p}^0$
$\Rightarrow \frac{x}{m}=k$


$\therefore$
The value of $\frac{1}{n}$ lies between 0 to 1 under different pressure ranges.