Question

Adsorption of a gas on a solid surface follows the Langmuir isotherm. If \(\frac{ k_a}{k_d}\)=1.0 bar^-1 , the fraction of adsorption sites occupied by the gas at equilibrium under 2.0 bar pressure of the gas at 25 °C is 
( k_a and k_d are the rate constants for adsorption and desorption processes, respectively, at 25 °C)

• A. \(\frac{1}{4}\)
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{2}\)
• D. \(\frac{2}{3}\)

Answer 1

The Correct Option is D

To find the fraction of adsorption sites occupied by the gas at equilibrium given that the adsorption follows the Langmuir isotherm, we use the Langmuir adsorption equation. This equation states:

\[\theta = \frac{k_a p}{k_d + k_a p}\]

where:

• \(\theta\) is the fraction of occupied sites.
• \(k_a\) and \(k_d\) are the adsorption and desorption rate constants, respectively.
• \(p\) is the gas pressure.

In this problem, we are given:

• \(\frac{ k_a}{k_d} = 1.0  \text{bar}^{-1}\)
• Pressure: 2.0 bar

Let's reformulate the equation using the provided values:

Given \(\frac{k_a}{k_d} = 1.0 \, \text{bar}^{-1}\), we can write:

\[\theta = \frac{\left(\frac{k_a}{k_d}\right) p}{1 + \left(\frac{k_a}{k_d}\right) p}\]

Substituting the given pressure and ratio of rate constants:

\[\theta = \frac{1.0 \times 2.0}{1 + 1.0 \times 2.0}\]\[\theta = \frac{2.0}{1 + 2.0} = \frac{2.0}{3.0} = \frac{2}{3}\]

Thus, the fraction of adsorption sites occupied by the gas at equilibrium is \(\frac{2}{3}\), which matches the provided correct answer.