KCET · Chemistry · General Organic Chemistry
For Freundlich isotherm a graph of \( \log \frac{x}{m} \) is plotted against log \( P \). The slope of the line and its
\( y \)-axis intercept, respectively corresponds to
- A \( \log \frac{1}{n}, k \)
- B \( \log \frac{1}{n}, \log k \)
- C \( \frac{1}{n}, k \)
- D \( \frac{1}{n}, \log k \)
Answer & Solution
Correct Answer
(D) \( \frac{1}{n}, \log k \)
Step-by-step Solution
Detailed explanation
Freundlich adsorption isotherm,
\[
\frac{x}{m}=k p^{\frac{1}{n}} \rightarrow(1)
\]
Taking log of Eq. (1), we get
\( \log \frac{x}{m}=\log k+\frac{1}{n} \log p \)
Now, compare with equation for straight line, \( y=m x+c \)
By plotting \( \log \frac{x}{m} \) against \( \log p \), a straight line with slope \( =1 / n \) and intercept logk.
\[
\frac{x}{m}=k p^{\frac{1}{n}} \rightarrow(1)
\]
Taking log of Eq. (1), we get
\( \log \frac{x}{m}=\log k+\frac{1}{n} \log p \)
Now, compare with equation for straight line, \( y=m x+c \)
By plotting \( \log \frac{x}{m} \) against \( \log p \), a straight line with slope \( =1 / n \) and intercept logk.
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