#### Question

Describe Freundlich adsorption isotherm.

#### Solution 1

**Freundlich adsorption isotherm:**

Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature.

From the given plot it is clear that at pressure *P*_{S}, x/m reaches the maximum valve. *Ps* is called the saturation pressure. Three cases arise from the graph now.

**Case I- At low pressure:**

The plot is straight and sloping, indicating that the pressure in directly proportional to x/m i.e., `x/m alpha P`

`x/m = kP` (k is constant)

**Case II- At high pressure:**

When pressure exceeds the saturated pressure, x/m becomes independent of P values

`x/m alpha P^@`

`x/m = kP^@`

**Case III- At intermediate pressure:**

At intermediate pressure, x/m depends on *P* raised to the powers between 0 and 1. This relationship is known as the Freundlich adsorption isotherm.

`x/m alpha P^(1/n)`

`x/m = kP^(1/n)` n > 1

Now taking log: `log x/m =log k + 1/n log P`

On plotting the graph between log (x/m) and log *P*, a straight line is obtained with the slope equal to 1/n and the intercept equal to log k

#### Solution 2

Freundlich Adsorption isotherm:

Freundlich, in 1909, gave an empirical relationship between the quantity, of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. The relationship can be expressed by the following equation:

where x is the mass of the gas adsorbed by mass ‘m’ of the adsorbent at pressure P, k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature. The relationship is generally represented in the form of a curve where mass of the gas adsorbed per gram by the adsorbent is plotted against pressure. These curves indicate that at a fixed pressure, there is a decrease in physical adsorption with increase in temperature. These curves always seem to approach saturation at high pressure.

Taking log of equation (i), we get

`log x/m =log k + 1/n log P`