Question

What is the value of intercept on y-axis when $\log_{10} \frac{x}{m}$ is plotted against $\log_{10} P$ in Freundlich isotherm?

• A. $\frac{1}{n}$
• B. $\log_{10} k$
• C. $n$
• D. $k$

Hint:

In physical chemistry linear log graphs, multiplication operations turn into additions. Since the constant $k$ was multiplied by the pressure term in the original non-log equation ($\frac{x}{m} = k \cdot P^{1/n}$), taking logs turns it into an additive constant ($\log k$), which sets the starting baseline position or $y$-intercept of the line.

Answer 1

The Correct Option is B

Step 1: Write the Freundlich isotherm.
The empirical equation is $\dfrac{x}{m} = k\,P^{1/n}$, where $\dfrac{x}{m}$ is the amount adsorbed per gram and $P$ is the pressure.
Step 2: Take logarithms of both sides.
$\log_{10}\dfrac{x}{m} = \log_{10}\big(k\,P^{1/n}\big)$.
Step 3: Split the product.
Using $\log(ab) = \log a + \log b$: $\log_{10}\dfrac{x}{m} = \log_{10}k + \log_{10}P^{1/n}$.
Step 4: Bring the power down.
Using $\log a^b = b\log a$: $\log_{10}\dfrac{x}{m} = \dfrac{1}{n}\log_{10}P + \log_{10}k$.
Step 5: Compare with the straight line.
Against $y = mx + c$ with $y = \log_{10}\dfrac{x}{m}$ and $x = \log_{10}P$, the slope is $\dfrac{1}{n}$ and the constant term is $\log_{10}k$.
Step 6: Read off the intercept.
The term independent of $\log_{10}P$ is the y-intercept, which is $\log_{10}k$.
\[ \boxed{\text{Intercept} = \log_{10}k,\ \text{option (2)}} \]