In the figure,a straight line is given for th | vedclass

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(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K \cdot P^{1/n}$.Taking logarithm on both sides,we get: $\log(\frac{x}

Vedclass · Accepted Answer

$0.3$ and $0.7033$

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(B) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K \cdot P^{1/n}$.Taking logarithm on both sides,we get: $\log(\frac{x}{m}) = \log K + \frac{1}{n} \log P$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(\frac{x}{m})$,$x = \log P$,the slope $m = \frac{1}{n}$,and the intercept $c = \log K$.Given the equation $y = 0.3x + 0.7033$,we have:Slope $(m)$ = $\frac{1}{n} = 0.3$.Intercept $(c)$ = $\log K = 0.7033$.Thus,the values of $\frac{1}{n}$ and $\log K$ are $0.3$ and $0.7033$ respectively.

In the figure,a straight line is given for the Freundlich Adsorption isotherm equation $(y = mx + c)$,where the equation is $y = 0.3x + 0.7033$. The values of $\frac{1}{n}$ and $\log K$ are respectively:

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