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(D) The Freundlich adsorption isotherm is given by the relation: $\frac{x}{m} = kP^{1/n}$,where $0 < \frac{1}{n} < 1$.Adsorption is generally an exoth

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(D) The Freundlich adsorption isotherm is given by the relation: $\frac{x}{m} = kP^{1/n}$,where $0 Adsorption is generally an exothermic process. According to Le Chatelier's principle,an increase in temperature will shift the equilibrium in the backward direction,leading to a decrease in the extent of adsorption $(\frac{x}{m})$.Therefore,at a constant pressure $P$,the extent of adsorption $\frac{x}{m}$ will be higher at a lower temperature $(T_{2})$ compared to a higher temperature $(T_{1})$.Thus,the curve for $T_{2}$ will lie above the curve for $T_{1}$ (given $T_{1} > T_{2}$),which is correctly represented in option $D$.

Select the graph that correctly describes the adsorption isotherms at two temperatures $T_{1}$ and $T_{2}$ $\left(T_{1}>T_{2}\right)$ for a gas :
($x-$ mass of the gas adsorbed ; $m-$ mass of adsorbent ; $P -$ pressure)

Similar Questions

In the Freundlich adsorption isotherm equation $(x/m) = kp^{1/n}$,the value of $n$ is:

Physical adsorption is essentially quite appreciable:

According to the Freundlich adsorption isotherm,which of the following is correct?

The correct option representing a Freundlich adsorption isotherm is

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Select the graph that correctly describes the adsorption isotherms at two temperatures $T_{1}$ and $T_{2}$ $\left(T_{1}>T_{2}\right)$ for a gas :($x-$ mass of the gas adsorbed ; $m-$ mass of adsorbent ; $P -$ pressure)