At 273 K and 10 bar, the Langmuir adsorption of a gas on a solid surface gave the fraction of surface coverage as 0.01. The Langmuir adsorption isotherm constant is ________ bar$^{-1}$. (Give the answer to the third decimal place)
This solution explains how to calculate the Langmuir adsorption isotherm constant (K) using the provided data on fractional surface coverage (θ) and pressure (P).
Calculating Langmuir Adsorption Constant
The Langmuir adsorption isotherm describes the relationship between the pressure of a gas and the extent of adsorption onto a solid surface at a constant temperature. The simplified form relating fractional surface coverage (θ), pressure (P), and the adsorption constant (K) is:
$ \frac{\theta}{1-\theta} = KP $
We are given:
- Pressure, P = 10 bar
- Fractional surface coverage, θ = 0.01
We need to find the Langmuir adsorption isotherm constant, K.
Step-by-Step Calculation
- Rearrange the formula: To solve for K, we rearrange the Langmuir isotherm equation:
$ K = \frac{\theta}{(1-\theta)P} $
- Substitute the values: Plug the given values of θ and P into the rearranged formula:
$ K = \frac{0.01}{(1-0.01) \times 10 \text{ bar}} $
- Calculate the denominator: First, calculate the term $(1-\theta)$:
$ 1 - 0.01 = 0.99 $
Now, multiply by the pressure:
$ 0.99 \times 10 \text{ bar} = 9.9 \text{ bar} $
- Compute the constant K: Divide the coverage by the result from the previous step:
$ K = \frac{0.01}{9.9 \text{ bar}} $
$ K \approx 0.0010101... \text{ bar}^{-1} $
- Round to the required precision: The question asks for the answer to the third decimal place. Rounding $0.0010101...$ to three decimal places gives $0.001$.
$ K = 0.001 \text{ bar}^{-1} $
The calculated value of the Langmuir adsorption isotherm constant is approximately $0.001 \text{ bar}^{-1}$. This value falls within the range specified by the correct answer (0.0009 to 0.0011).
Important Questions from Adsorption
- A volatile organic compound (VOC) is to be adsorbed from air onto a bed of activated carbon. The equilibrium capacity of activated carbon at the feed conditions is $0.4 \text{ grams}$ VOC per gram of activated carbon. The column contains $4 \text{ grams}$ of activated carbon per $\text{cm}^2$ of cross-section. The feed rate into the adsorber column is $0.2 \text{ grams}$ VOC $\text{cm}^{-2}\text{ h}^{-1}$. Breakthrough time is defined as the time at which the concentration at the exit of the bed ($c$) reaches a value of $0.05c_0$, where $c_0$ is the concentration of VOC in the feed. The breakthrough time for the bed is $2.1 \text{ h}$. The area under $c / c_0$ curve between the initial and breakthrough times is $0.1 \text{ h}$. Which one of the following is the fraction of unused bed at breakthrough?
- The correct statement(s) for a surface catalyzed unimolecular gaseous reaction is(are)
- 500 mg of a dry adsorbent is added to a beaker containing 100 mL solution of concentration 100 mg phenol/(L solution). The adsorbent is separated out after 5 h of rigorous mixing. If the residual concentration in the solution after separating the adsorbent is 30 mg phenol/(L solution), the amount of phenol adsorbed (in mg per gram of dry adsorbent) is
- Adsorption on activated carbon is to be used for reducing phenol concentration in wastewater from 0.04 mol/l to 0.008 mol/l. The adsorption isotherm at the operating temperature can be expressed as $q = 0.025C^{1/3}$; where $q$ is the phenol concentration in solid (mol/g solid) and $C$ is the phenol concentration in water (mol/l). The minimum amount of solid (in grams) required per liter of wastewater (up to one decimal place) is ________
- At 30°C, the amounts of acetone adsorbed at partial pressures of 10 and 100 mmHg are 0.1 and 0.4 kg acetone/kg activated carbon, respectively. Assume Langmuir isotherm describes the adsorption of acetone on activated carbon. What is the amount of acetone adsorbed (in kg per kg of activated carbon) at a partial pressure of 50 mmHg and 30°C?
