The Freundlich adsorption isotherm gives a st | vedclass

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(C) The Freundlich adsorption isotherm equation is given by $\frac{x}{m} = kP^{1/n}$,where $x$ is the mass of the adsorbate,$m$ is the mass of the ads

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$\log(x/m)$ vs $\log P$

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(C) The Freundlich adsorption isotherm equation is given by $\frac{x}{m} = kP^{1/n}$,where $x$ is the mass of the adsorbate,$m$ is the mass of the adsorbent,$P$ is the pressure,and $k$ and $n$ are constants.Taking the logarithm on both sides,we get: $\log(\frac{x}{m}) = \log k + \frac{1}{n} \log P$.This equation is in the form of a straight line $y = mx + c$,where $y = \log(\frac{x}{m})$,$x = \log P$,slope $m = \frac{1}{n}$,and intercept $c = \log k$.Therefore,plotting $\log(\frac{x}{m})$ versus $\log P$ yields a straight line.

The Freundlich adsorption isotherm gives a straight line on plotting:

Similar Questions

Which one of the following gives a straight line for Freundlich adsorption isotherm?

Give an example where physisorption changes to chemisorption with rise in temperature. Explain the reason for change.

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Which of the following relationships is correct regarding the Freundlich adsorption isotherm?
$(1)$ $x/m = \text{constant}$ (at high pressure)
$(2)$ $x/m = k \cdot P^{1/n}$ (at intermediate pressure)
$(3)$ $x/m = k \cdot P$ (at low pressure)

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