JEE Main202124 Feb 2021Morning ShiftChemistrySurface ChemistryActual
In Freundlich adsorption isotherm, slope of AB line is:
Options
- A1 n with 1 n = 0   to   1
- Bn with n = 0 . 1   to   0 . 5
- Clog 1 n with n < 1
- Dlogn with n > 1
Correct answer
A. 1 n with 1 n = 0   to   1
Step-by-step solution
Freundlich adsorption isotherm is : x m = kp 1 / n x = mass of adsorbate m = mass of adsorbent P = eq. pressure k 1 n = 1 n logp + logk y = mx + c compairing m = 1 n = slope 1 n = 0   to   1 n > 1 log x m = logK + 1 n logP y = c + mx m = 1 / n so slope will be equal to 1 / n Hence 0 ≤ 1 n ≤ 1