For the Freundlich adsorption isotherm,a grap | vedclass

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Question

(D) The Freundlich adsorption isotherm is given by the equation:$\frac{x}{m} = k p^{\frac{1}{n}}$Taking the logarithm on both sides:$\log \frac{x}{m}

Vedclass · Accepted Answer

$\frac{1}{n}, \log k$

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(D) The Freundlich adsorption isotherm is given by the equation:$\frac{x}{m} = k p^{\frac{1}{n}}$Taking the logarithm on both sides:$\log \frac{x}{m} = \log k + \frac{1}{n} \log p$Comparing this with the equation of a straight line,$y = mx + c$,where $y = \log \frac{x}{m}$,$x = \log p$,$m$ (slope) $= \frac{1}{n}$,and $c$ (intercept) $= \log k$.Therefore,the slope is $\frac{1}{n}$ and the intercept is $\log k$.

For the Freundlich adsorption isotherm,a graph of $\log \frac{x}{m}$ is plotted against $\log P$. The slope of the line and its $y$-axis intercept,respectively,correspond to:

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