Question

Adsorption of a gas on solids follows the Freundlich adsorption isotherm. The graph drawn between log $\frac{x}{m}$ (on the y-axis) and log \( p \) (on the x-axis) is a straight line with a slope equal to 3 and an intercept equal to 0.30. What is the value of $\frac{x}{m}$ at a pressure of 2 atm?
(Given: log 2 = 0.3)

• A. 48
• B. 32
• C. 16
• D. 8

Hint:

The Freundlich adsorption isotherm is given by $\frac{x}{m} = K p^{1/n}$. Its linear form is $\log \left(\frac{x}{m}\right) = \log K + \frac{1}{n} \log p$. This linear equation is critical for solving problems where slope and intercept are provided. Remember that the slope directly gives the value of $1/n$, and the intercept gives the value of $\log K$.

Answer 1

The Correct Option is C

Step 1: Understand the Freundlich adsorption isotherm equation.
The Freundlich adsorption isotherm is empirically expressed as:
$$ \frac{x}{m} = K p^{1/n} $$ where:
$x$ = mass of adsorbate (gas)
$m$ = mass of adsorbent (solid)
$p$ = pressure of the gas
$K$ and $1/n$ are constants for a given adsorbent and adsorbate at a particular temperature.
Taking the logarithm on both sides, we get:
$$ \log \left(\frac{x}{m}\right) = \log K + \frac{1}{n} \log p $$ Step 2: Relate the given information to the linear equation of the Freundlich isotherm.
The problem states that the graph drawn between $\log \left(\frac{x}{m}\right)$ (on the y-axis) and $\log p$ (on the x-axis) is a straight line. This matches the logarithmic form of the Freundlich isotherm, which is in the form of $y = c + mx'$ (where $m'$ is the slope).
Comparing:
$y = \log \left(\frac{x}{m}\right)$
$x' = \log p$
Slope ($m'$) = $\frac{1}{n}$
Intercept ($c$) = $\log K$
Given:
Slope = 3
Intercept = 0.30
Pressure, $p = 2$ atm
$\log 2 = 0.3$
Step 3: Use the given slope and intercept to find $K$ and $1/n$.
From the given information:
$\frac{1}{n}$ (slope) = 3
$\log K$ (intercept) = 0.30
We know that $\log 2 = 0.3$. Therefore, from $\log K = 0.30$, we can deduce:
$K = \operatorname{antilog}(0.30)$
$K = 2$
Step 4: Calculate the value of $\frac{x}{m}$ at a pressure of 2 atm.
Now, substitute the values of $K$, $1/n$, and $p$ into the original Freundlich adsorption isotherm equation:
$$ \frac{x}{m} = K p^{1/n} $$ $$ \frac{x}{m} = 2 \cdot (2)^3 $$ $$ \frac{x}{m} = 2 \cdot 8 $$ $$ \frac{x}{m} = 16 $$ The final answer is $\boxed{16}$.