Question

The slope of the straight line given in the following diagram for adsorption is:

• A. $\frac{1}{\mathrm{n}}(0\mathrm{to}1)$
• B. $\frac{1}{\mathrm{n}}(0.1\mathrm{to}0.5)$
• C. $\log n$
• D. $\log \left(\frac{1}{n}\right)$

Answer 1

The correct option is A

$\frac{1}{\mathrm{n}}(0\mathrm{to}1)$

Explanation for the correct answer:-

Option (A) $\frac{1}{\mathrm{n}}(0\mathrm{to}1)$

Step 1: Finding the equation

From Freundlich adsorption isotherm we know that,

$\log \left(\frac{x}{m}\right)=\log k+\frac{1}{n}\log P$

At an intermediate pressure, the variation in $\frac{\mathrm{x}}{\mathrm{m}}vsP$ can be expressed as"

$$\begin{gathered} \frac{x}{m}\propto p^{\frac{1}{n}} \\ \Rightarrow \frac{x}{m}=K{p}^{\frac{1}{n}} \\ \Rightarrow \log \left(\frac{x}{m}\right)=\log k+\frac{1}{n}\log P\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{[}taking\mathit{}\mathit{log}\mathit{}on\mathit{}both\mathit{}sides\mathit{]}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{.}\mathit{\left(}i\mathit{\right)} \end{gathered}$$

Step 2: Comparing the equation with straight line equation

Equation of a straight line as we know is:

$\mathrm{y}=\mathrm{mx}+\mathrm{c}.........\left(\mathrm{ii}\right)$

$\therefore$Comparing both equations (i) and (ii) we get

$$\begin{gathered} \mathrm{y}=\log \left(\frac{\mathrm{x}}{\mathrm{m}}\right) \\ \mathrm{m}=\frac{1}{\mathrm{n}} \\ \mathrm{x}=\log \mathrm{P} \\ \mathrm{c}=\log \mathrm{K} \end{gathered}$$

Thus we can see that the value of the slope is $\mathrm{m}=\frac{1}{\mathrm{n}}(0\mathrm{to}1)$

Therefore, option (A) $\frac{1}{\mathrm{n}}(0\mathrm{to}1)$ is the correct answer.