Question

For Freundlich adsorption isotherm, a graph of log (x/m) Vs. log (P) gives a straight line. The slope of line and its Y-axis intercept respectively are

• A. \(\log(\frac{1}{n}),K\)
• B. \(\frac{1}{n},\log K\)
• C. \(\log(\frac{1}{n}),\log K\)
• D. \(\frac{1}{n},K\)

Answer 1

The Correct Option is B

The Freundlich adsorption isotherm is given by the equation:

\[ x/m = K \cdot P^{\frac{1}{n}} \]

• Here, \(x\) is the amount of adsorbate adsorbed per unit mass of adsorbent.
• \(m\) is the mass of the adsorbent.
• \(P\) is the equilibrium pressure of the adsorbate.
• \(K\) and \(n\) are constants specific to the adsorbent and adsorbate.

Plotting log (x/m) Vs. log (P):

Taking the logarithm of both sides of the Freundlich equation:

\[ \log(x/m) = \log(K \cdot P^{\frac{1}{n}}) \]

Using the properties of logarithms:

\[ \log(x/m) = \log K + \frac{1}{n} \log P \]

This equation is in the form of a linear equation \(y = mx + b\), where:

• \(y = \log(x/m)\)
• \(x = \log P\)
• \(m = \frac{1}{n}\) (slope of the line)
• \(b = \log K\) (Y-axis intercept)

Conclusion: The slope of the line in the plot of log (x/m) Vs. log (P) is \(\frac{1}{n}\) and the Y-axis intercept is \(\log K\).

Answer 2

Freundlich adsorption isotherm is given by the equation:

\[ \frac{x}{m} = K P^{1/n} \]

Taking the logarithm of both sides:

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

This is the equation of a straight line of the form y = mx + c, where:

• The slope of the line (m) is \( \frac{1}{n} \)
• The Y-intercept (c) is \( \log K \)

Thus, the slope is \( \frac{1}{n} \) and the Y-axis intercept is \( \log K \), which corresponds to option (B).