Question

In Freundlich adsorption isotherm, slope of AB line is :

• A. $\log n$ with $( n >1)$
• B. $n$ with $( n , 0.1$ to 0.5$)$
• C. $\log \frac{1}{n}$ with $(n<1)$
• D. $\frac{1}{ n }$ with $\left(\frac{1}{ n }=0\right.$ to $\left.1\right)$

Answer 1

The Correct Option is D

 The Freundlich adsorption isotherm is an empirical relationship describing the adsorption of solutes on a solid surface. It is mathematically expressed as:

\(x/m = k \cdot C^{1/n}\)

where:

• \(x\) is the mass of the adsorbate.
• \(m\) is the mass of the adsorbent.
• \(C\) is the equilibrium concentration of the solute.
• \(k\) and \(n\) are constants, with \(n > 1\).

By taking the logarithm on both sides, the equation becomes:

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

This is a linear equation with the form \(y = mx + c\), where:

• \(\log \left(\frac{x}{m}\right)\) is the dependent variable (y-axis).
• \(\log C\) is the independent variable (x-axis).
• (m).
• (c).

The question concerns the slope of the line AB, which represents the relationship between \(\log \left(\frac{x}{m}\right)\) and \(\log C\). According to the equation, the slope is \log n with n with )

• (n<1)
• \left(\frac{1}{n}=0\right. to \frac{1}{n} with \left.1\right). This matches the determined slope from the Freundlich equation.