In a batch adsorption process, 5 g of fresh adsorbent is used to treat 1 liter of an aqueous phenol solution. The initial phenol concentration is 100 mg/liter. The equilibrium relation is given by
$q^* = 1.3C$
where $q^*$ is the amount of phenol adsorbed in mg of phenol per gram of adsorbent; and C is the concentration of phenol in mg/liter in the aqueous solution.
When equilibrium is attained between the adsorbent and the solution, the concentration of phenol in the solution, rounded to 1 decimal place, is ________ mg/liter.
Batch Adsorption Calculation
This problem involves calculating the final equilibrium concentration of phenol in a batch adsorption process. We will use a mass balance approach combined with the given equilibrium isotherm.
Key Parameters
- Mass of adsorbent ($m_{ads}$): 5 g
- Volume of solution ($V_{sol}$): 1 liter
- Initial phenol concentration ($C_0$): 100 mg/liter
- Equilibrium isotherm: $q^* = 1.3C$ (where $q^*$ is in mg phenol/g adsorbent and $C$ is in mg/liter)
Mass Balance Principle
The total amount of phenol initially present in the solution must equal the sum of the phenol adsorbed onto the adsorbent and the phenol remaining in the solution at equilibrium.
Initial phenol mass = Mass of phenol adsorbed + Mass of phenol in solution at equilibrium
Mathematically:
$C_0 \times V_{sol} = (q_{eq} \times m_{ads}) + (C_{eq} \times V_{sol})$
Where:
- $C_{eq}$ is the equilibrium concentration of phenol in the solution (mg/liter).
- $q_{eq}$ is the amount of phenol adsorbed at equilibrium (mg phenol/g adsorbent).
Applying Equilibrium Isotherm
The given equilibrium relationship is $q^* = 1.3C$. At equilibrium, $q_{eq} = 1.3 \times C_{eq}$.
Solving for Equilibrium Concentration ($C_{eq}$)
Substitute the isotherm into the mass balance equation:
$C_0 \times V_{sol} = (1.3 \times C_{eq} \times m_{ads}) + (C_{eq} \times V_{sol})$
Factor out $C_{eq}$:
$C_0 \times V_{sol} = C_{eq} \times (1.3 \times m_{ads} + V_{sol})$
Rearrange to solve for $C_{eq}$:
$C_{eq} = \frac{C_0 \times V_{sol}}{1.3 \times m_{ads} + V_{sol}}$
Calculation
Now, substitute the given values:
$C_{eq} = \frac{100 \text{ mg/liter} \times 1 \text{ liter}}{1.3 \times 5 \text{ g} + 1 \text{ liter}}$
$C_{eq} = \frac{100}{6.5 + 1}$
$C_{eq} = \frac{100}{7.5}$
$C_{eq} = \frac{1000}{75}$
$C_{eq} = \frac{40}{3}$
$C_{eq} \approx 13.333 \text{ mg/liter}$
Final Result
Rounding the equilibrium concentration to 1 decimal place gives 13.3 mg/liter. This value falls within the provided range of 13 to 13.5.
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?
