Adsorption Calculation for Phenol Reduction
The problem requires calculating the minimum amount of activated carbon (solid) needed to reduce phenol concentration in wastewater from an initial value ($C_1$) to a final value ($C_2$) using a given adsorption isotherm.
Key Data
- Initial Phenol Concentration ($C_1$): $0.04$ mol/l
- Final Phenol Concentration ($C_2$): $0.008$ mol/l
- Adsorption Isotherm: $q = 0.025C^{1/3}$ (where $q$ is mol phenol/g solid, $C$ is mol phenol/l)
- Volume of wastewater considered: $1$ L
Calculating Phenol Removed
The amount of phenol that needs to be removed from $1$ liter of wastewater is the difference between the initial and final concentrations.
- Phenol removed per liter = $V \times (C_1 - C_2)$
- Phenol removed per liter = $1 \text{ L} \times (0.04 \text{ mol/L} - 0.008 \text{ mol/L})$
- Phenol removed per liter = $1 \text{ L} \times 0.032 \text{ mol/L} = 0.032$ mol
Calculating Adsorption Capacity ($q$)
The adsorption capacity of the solid ($q$) is determined by the phenol concentration in water at equilibrium ($C_2$), using the given isotherm equation.
- At equilibrium, $C = C_2 = 0.008$ mol/l.
- Calculate $q$ at $C_2$: $q_2 = 0.025 \times (0.008)^{1/3}$
- First, find the cube root of $0.008$: $(0.008)^{1/3} = 0.2$
- Now, substitute this back into the equation for $q_2$: $q_2 = 0.025 \times 0.2 = 0.005$ mol/g solid
This means $0.005$ moles of phenol can be adsorbed by $1$ gram of activated carbon when the water concentration is $0.008$ mol/l.
Determining Minimum Solid Amount
The minimum amount of solid required is the total amount of phenol to be removed divided by the amount of phenol adsorbed per gram of solid.
- Minimum solid amount (grams) = $\frac{\text{Total phenol removed (mol)}}{\text{Adsorbed phenol per gram solid (mol/g)}}$
- Minimum solid amount = $\frac{0.032 \text{ mol}}{0.005 \text{ mol/g}}$
- Minimum solid amount = $6.4$ g
Therefore, $6.4$ grams of activated carbon are required per liter of wastewater.
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
- 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?
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)
