Question:
Adsorption of a gas on a metal oxide surface follows Freundlich adsorption isotherm. At $300\text{ K}$, the slope and intercept of this isotherm are 2 and 1.0 respectively. What is the value of $\frac{x}{m}$ when the pressure of the gas is 4 bar?
Adsorption of a gas on a metal oxide surface follows Freundlich adsorption isotherm. At $300\text{ K}$, the slope and intercept of this isotherm are 2 and 1.0 respectively. What is the value of $\frac{x}{m}$ when the pressure of the gas is 4 bar?
Show Hint
Always double check if the intercept given is using base 10 logarithm or natural logarithm.
In standard textbooks, Freundlich log-plots are plotted with $\log_{10}$ by default, meaning Intercept = $1.0 \implies k = 10$.
In standard textbooks, Freundlich log-plots are plotted with $\log_{10}$ by default, meaning Intercept = $1.0 \implies k = 10$.
Updated On: Jul 22, 2026
- 200
- 180
- 160
- 20
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The question is based on the Freundlich adsorption isotherm.
We are given the experimental parameters (slope and intercept) of the logarithmic isotherm plot at 300 K and need to calculate the extent of adsorption ($\frac{x}{m}$) at a specific pressure of 4 bar.
Step 2: Key Formula or Approach:
The Freundlich adsorption isotherm equation is:
\[ \frac{x}{m} = k P^{1/n} \] Taking the logarithm of both sides:
\[ \log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log P \] Comparing this with the equation of a straight line ($y = mx + c$):
Slope = $\frac{1}{n}$
Intercept = $\log k$
Step 3: Detailed Explanation:
• From the given data:
Slope ($\frac{1}{n}$) = 2.
Intercept ($\log k$) = 1.0 $\implies k = 10^{1.0} = 10$.
Pressure ($P$) = 4 bar.
• Let us substitute these values back into the original Freundlich equation:
\[ \frac{x}{m} = k P^{1/n} \] \[ \frac{x}{m} = 10 \times 4^2 \] \[ \frac{x}{m} = 10 \times 16 = 160 \]
• Thus, the value of $\frac{x}{m}$ is 160.
Step 4: Final Answer:
The value of $\frac{x}{m}$ is 160.
The question is based on the Freundlich adsorption isotherm.
We are given the experimental parameters (slope and intercept) of the logarithmic isotherm plot at 300 K and need to calculate the extent of adsorption ($\frac{x}{m}$) at a specific pressure of 4 bar.
Step 2: Key Formula or Approach:
The Freundlich adsorption isotherm equation is:
\[ \frac{x}{m} = k P^{1/n} \] Taking the logarithm of both sides:
\[ \log \left( \frac{x}{m} \right) = \log k + \frac{1}{n} \log P \] Comparing this with the equation of a straight line ($y = mx + c$):
Slope = $\frac{1}{n}$
Intercept = $\log k$
Step 3: Detailed Explanation:
• From the given data:
Slope ($\frac{1}{n}$) = 2.
Intercept ($\log k$) = 1.0 $\implies k = 10^{1.0} = 10$.
Pressure ($P$) = 4 bar.
• Let us substitute these values back into the original Freundlich equation:
\[ \frac{x}{m} = k P^{1/n} \] \[ \frac{x}{m} = 10 \times 4^2 \] \[ \frac{x}{m} = 10 \times 16 = 160 \]
• Thus, the value of $\frac{x}{m}$ is 160.
Step 4: Final Answer:
The value of $\frac{x}{m}$ is 160.
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Adsorption Questions
- The extent of physisorption of an adsorbate increases with
The adsorption of a gas on solid adsorbent follows Freundlich adsorption isotherm. The graph drawn between $ \frac{X}{m} $ and $ p $ at different temperatures X, Y, Z is shown. (X = mass of gas adsorbed, m = mass of adsorbent and p = Pressure of gas). The correct relation between X, Y, Z is:
- Identify the correct statements about chemisorption:
- I) It is irreversible adsorption
- II) Enthalpy of adsorption is in the range of 20-40 kJ mol\(^-1\)
- III) It is unlayered adsorption
- IV) It does not need appreciable activation energy
- From the following, identify the favorable conditions for physical adsorption:
- I) Large surface area
- II) Low temperatures
- III) High temperatures
- IV) High pressures
- The plot of \( \log \frac{x}{m} \) (y-axis) and \( \log p(x) \) (x-axis) is a straight line inclined at an angle of 45°. When the intercept, \( K \) is 10 and pressure is 0.3 atm, the amount of solute in grams adsorbed per gram of adsorbent (log3 = 0.4771) is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions