In the Freundlich adsorption isotherm,the slo | vedclass

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

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$1/n$ where $(1/n = 0 	ext{ to } 1)$

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(D) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = K(P)^{1/n}$Taking the logarithm on both sides:$\log \left(\frac{x}{m}\right) = \log K + \frac{1}{n} \log P$This equation is in the form of a straight line $y = mx + c$,where:$y = \log \left(\frac{x}{m}\right)$$x = \log P$$c = \log K$$m = \frac{1}{n}$ (slope)In the Freundlich adsorption isotherm,the value of $\frac{1}{n}$ ranges between $0$ and $1$ (i.e.,$0 \leq \frac{1}{n} \leq 1$).Thus,the slope of the line is $\frac{1}{n}$ where $\frac{1}{n}$ lies between $0$ and $1$.

In the Freundlich adsorption isotherm,the slope of the $AB$ line is:

Similar Questions

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