Question

In figure, a straight line is given for Freundrich Adsorption\((y=3 x+2505)\) The value of \(\frac{1}{ n }\)and\(\log K\)are respectively

• A. 3 and 2.505
• B. 3 and 0.7033
• C. \(0.3 \  and \  \log 2.505\)
• D. 0.3 and 0.7033

Hint:

Linearizing the Freundlich adsorption isotherm equation by taking the logarithm helps determine the values of \(\frac{1}{n}\) and \(K\) from the slope and intercept of the graph.

Answer 1

The Correct Option is A

Step 1: Recall the Freundlich Adsorption Isotherm}
The Freundlich adsorption isotherm is given by:
\[\frac{x}{m} = KP^{\frac{1}{n}},\]
where \(x\) is the mass of adsorbate, \(m\) is the mass of adsorbent, \(P\) is the pressure, \(K\) is the Freundlich constant, and \(n\) is a constant.
Step 2: Linearize the Equation
Taking the logarithm of both sides, we get:
\[\log \frac{x}{m} = \log K + \frac{1}{n} \log P.\]
This equation represents a straight line with slope \(\frac{1}{n}\) and y-intercept \(\log K\).
Step 3: Compare with the Given Equation
The given equation is:
\[y = 3x + 2.505,\]
where \(y = \log \frac{x}{m}\) and \(x = \log P\). Comparing this with the linearized Freundlich equation, we have:
\[\frac{1}{n} = 3, \quad \log K = 2.505.\]
Conclusion
The value of \(\frac{1}{n}\) is \(3\), and \(\log K\) is \(2.505\)