Question

Freundlich adsorption isotherm fails at which of the following condition?

• A. High temperature
• B. Low temperature
• C. High Pressure
• D. Low pressure

Hint:

Freundlich adsorption isotherm: $x/m = k P^{1/n}$.
It is an empirical model valid for a limited range of pressures.
It fails at high pressures because it does not account for the saturation of the adsorbent surface (monolayer formation, as described by Langmuir isotherm).
At high pressure, $x/m$ should become constant (saturation), but Freundlich predicts $x/m$ keeps increasing.

Answer 1

The Correct Option is C

The Freundlich adsorption isotherm is an empirical equation that describes the adsorption of a gas onto a solid surface. It is given by: $\frac{x}{m} = k P^{1/n}$ where: $x$ is the mass of the gas adsorbed $m$ is the mass of the adsorbent $P$ is the pressure of the gas $k$ and $n$ are constants that depend on the nature of the adsorbent and the gas, and on the temperature ($n>1$). The Freundlich isotherm has limitations: \begin{itemize} \item It is an empirical equation and does not have a strong theoretical basis. \item It is valid over a limited range of pressures. \item At high pressures: The amount of gas adsorbed ($x/m$) approaches a saturation value, meaning it becomes independent of pressure (i.e., $x/m$ becomes constant). However, the Freundlich equation ($x/m = kP^{1/n}$) predicts that $x/m$ will continue to increase with pressure (though at a decreasing rate since $1/n<1$). Thus, the Freundlich isotherm fails to accurately describe adsorption at high pressures where saturation occurs. The Langmuir isotherm provides a better description at high pressures and predicts monolayer adsorption leading to saturation. \item At very low pressures: The Freundlich isotherm often fits well. Here $x/m \approx kP$, which means adsorption is nearly linear with pressure (if $1/n \approx 1$). \item Temperature dependence: The constants $k$ and $n$ change with temperature. The isotherm itself is usually applied at a constant temperature. High temperature generally leads to less adsorption (desorption is favored), and low temperature favors adsorption. The failure is primarily related to pressure ranges rather than temperature absolutes, although the constants change with temperature. \end{itemize} The primary condition under which the Freundlich isotherm fails is at high pressures, where it does not predict the observed saturation of adsorption. Option (c) High Pressure is the condition where it fails. \[ \boxed{\text{High Pressure}} \]