NTA Abhyas JEE Main2020ChemistrySurface ChemistryPractice
Which of the following option is correct for given curve?
Options
- Ax m ∝ P
- Bx m ∝ P 2
- Cx m ∝ P 1 / 2
- Dx m ∝ P 0
Correct answer
C. x m ∝ P 1 / 2
Step-by-step solution
According to the Freundlich adsorption isotherm, x m = K p 1 n Taking log on bot sides. l o g x m = l o g K + 1 n log P Slope = 1 n Slope from graph = y 2 - y 1 x 2 - x 1 = 2 4 = 1 2 or, 1 n = 1 2 ∴ x m = K p 1 2 or x m ∝ p 1 2