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JEE Main 20231 FebruaryEvening Shift

Exam: JEE Main
Year: 2023
Date: 1 February
Shift: Evening Shift
Subject: Chemistry
Chapter: Surface Chemistry

JEE Main 2023 · 1 February · Evening Shift · Chemistry · Surface Chemistry

In figure, a straight line is given for Freundrich Adsorption ( y =...

In figure, a straight line is given for Freundrich Adsorption $( y = 3x + 2 . 505 )$ . The value of $\dfrac{1}{n}$ and $logK$ are respectively.

In figure, a straight line is given for Freundrich Adsorption ( y = 3x + 2 . 505 ) . The value of (1)/(n) and logK are

Options

A $0 . 3$ and $\log 2 . 505$
B $0 . 3$ and $0 . 7033$
C $3$ and $2 . 505$
D $3$ and $0 . 7033$

Correct answer

C. $3$ and $2 . 505$

Step-by-step solution

The expression for the Freundlich isotherm can be represented by the following equation

$\dfrac{x}{m} = {K}_{p}{P}^{1 / n}$

Taking log on both sides

$log\dfrac{x}{m} = logk + \dfrac{1}{n}logP$

On comparing with given equation of straight line, we get

$Y = 3x + 2 . 505, \dfrac{1}{n} = 3, logK = 2 . 505$

So, the answer is 3 and 2.505

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