JEE Main 2021 · 24 February · morning · Single correct · +4 / −1
In Freundlich adsorption isotherm, slope of AB line is :
- (a) 1/n with ( 1/n = 0 to 1 )
- (b) n with (n, 0.1 to 0.5)
- (c) log 1/n with ( n < 1 )
- (d) log n with (n > 1)
Answer and solution
Answer (a) 1/n with ( 1/n = 0 to 1 )
Solution
Chemistry ma’am · 4 steps
Take the log of the Freundlich equation, and it turns into y = mx + c with 1/n as the slope.
- Start from Freundlich. The isotherm is x/m = kP^(1/n), where x/m is the mass of gas adsorbed per gram of adsorbent at pressure P. Here k and n are constants for a given gas, adsorbent and temperature.
log x/m = log k + 1/n log P
- Match with a line. With log x/m on the y-axis and log P on the x-axis, this is y = mx + c. The slope of AB is 1/n, and the intercept on the y-axis (the dashed extension) is log k.
- Range of the slope. For Freundlich adsorption n > 1, so 1/n lies between 0 and 1. If 1/n = 0, adsorption does not depend on pressure. If 1/n = 1, adsorption is directly proportional to pressure.
- So. The slope of AB is 1/n, with 1/n from 0 to 1. That is option (a).
The idea to keep. For Freundlich, plot log x/m against log P: slope is 1/n (between 0 and 1) and intercept is log k. Taking the log never puts a log on the slope itself.
Answer from the NTA final answer key. Solution by PrepWiser.