Home
Class 12
CHEMISTRY
A plot of log (x/m) against log P for th...

A plot of log `(x/m)` against log P for the adsorption of a gas on a solid gives a straight a straight line with slope equal to :

A

k

B

log k

C

n

D

`(1)/(n)`

Text Solution

Verified by Experts

The correct Answer is:
D
If we plot a graph between log `((x)/(m))` and log p, a straight line will be obtained. The slope of the line is equal to `(1)/(n)` and the intercept is equal to log k.

where, `(x)/(m)` = amount of adsorption
According to Freundlich adsorption isotherm
`(x)/(m)=kp^(1//n)`
Taking log of both sides,
`log.(x)/(m)= log k + (1)/(n) log p `
from y = zx + c
`z = (1)/(n)` (slope)
Promotional Banner

Similar Questions

Explore conceptually related problems

A plot of log (x/m) against log P for the adsorption of a gas on a sloid gives a straight a straight line with slope equal to :

For the adsorption of a gas on a solid, the plot of log(x/m) versus log P is linear with slope equal to:

For the adsorption of a gas on a solid, the plot of log(x/m) versus log p is linear with slope equal to

For a first order reaction the plot of ‘t’ against log c gives a straight line with a slope equal to :

Freundlich adsorption isotherm gives a straight line on plotting :

Adsorption of a gas on a surface follows Freundlich adsorption isotherm. Plot of "log " (x)/(m) versus log P gives a straight line with slope equal to 0.5, then : ( (x)/(m) is the mass of the gas adsorbed per gram of adsorbent)

Rate constant k of a reaction varies with temperature according to the equation logk = constant -E_a/(2.303R)(1/T) where E_a is the energy of activation for the reaction . When a graph is plotted for log k versus 1/T , a straight line with a slope - 6670 K is obtained. Calculate the energy of activation for this reaction

Find slope of a straight line 2x-5y+7=0

A plot of log (a-x) against time 't' is a straight line. This indicates that the reaction is of :

The graph between log k versus 1//T is a straight line.