Question

At T(K), adsorption of a gas on 1 g of solid adsorbent follows Freundlich adsorption isotherm and is represented by the given graph. What is the value of \(k\)?

• A. 0.1
• B. 1.259
• C. 1.023
• D. 1.2

Hint:

For Freundlich adsorption isotherm, \[ \log\left(\frac{x}{m}\right) = \log k+\frac{1}{n}\log P \] Always remember: \[ \text{Intercept}=\log k \] \[ \text{Slope}=\frac{1}{n} \] Hence, if the intercept is known, \(k\) can be obtained directly by taking antilog.

Answer 1

The Correct Option is B

Concept: Freundlich adsorption isotherm is given by \[ \frac{x}{m}=kP^{1/n} \] Taking logarithm on both sides, \[ \log\left(\frac{x}{m}\right) = \log k+\frac{1}{n}\log P \] This equation represents a straight line. \[ y=c+mx \] where \[ \text{Intercept}=\log k \] and \[ \text{Slope}=\frac{1}{n} \]

Step 1: Identify the intercept from the graph.
From the graph, the line cuts the y-axis at \[ 0.1 \] Therefore, \[ \log k = 0.1 \]

Step 2: Calculate the value of \(k\).
Taking antilogarithm, \[ k=\text{antilog}(0.1) \] Using the given value, \[ k=1.259 \]

Step 3: Verify with the options.
The obtained value matches option (B). \[ \boxed{k=1.259} \]