Question

In Freundlich adsorption isotherm, slope of AB line is : 

• A. n with (n, 0.1 to 0.5)
• B. log n with (n>1)
• C. log 1/n with (n<1)
• D. 1/n with (1/n = 0 to 1)

Hint:

When $1/n = 0$, adsorption is independent of pressure. When $1/n = 1$, adsorption varies linearly with pressure.

Answer 1

The Correct Option is D

The Freundlich adsorption isotherm is given by the equation:

\frac{x}{m} = k P^{1/n}

Taking the logarithm on both sides, we get:

\log \left(\frac{x}{m}\right) = \log k + \frac{1}{n} \log P

This equation can be rearranged in the form of a straight line:

y = mx + c

where:

• y = \log \left(\frac{x}{m}\right)
• x = \log P
• The slope (m) is \frac{1}{n}
• c = \log k (intercept)

In the given graph of \log \left(\frac{x}{m}\right) versus \log P, the slope of the line AB is \frac{1}{n}.

Since the slope is \frac{1}{n}, the correct answer is:

1/n with (1/n = 0 to 1)