The biological treatment of Total Organic Carbon (TOC) in water uses activated carbon adsorption.The adsorption Accords with Freundlich Equation, K=0.007 , n=1.13. In the equation,the unit of qe is mg.mg/C, and the unit of Ce is mg/L.To dispose 10L wastewater with 20mg/l TOC, 1g activated carbon is applied. At equilibrium of adsorption, what is capacity of equilibrium adsorption? (mg.mg/C). Note: (based on material balance for every litre of water 100qe + Ce= 200.)

Question

(NOTE: Based on the material balance, for every litre of water 100qe + Ce =200).

Answer 1

To calculate the capacity of equilibrium adsorption (\(qe\)) based on the given relationships, we will use the information provided in the material balance equation and the Freundlich equation.

1. Material Balance Equation: \[ 100qe + Ce = 200 \]

2. Initial Conditions:

   • Volume of wastewater = 10 L
   • Initial concentration of TOC (\(C_i\)) = 20 mg/L
   • Amount of activated carbon used = 1 g

3. Final Water Volume: Since we are disposing of 10 L of wastewater, we focus on the concentration in the final volume of 10 L.

4. Equilibrium Concentration (\(Ce\)): According to the material balance from our equation: \[ Ce = 200 - 100qe \]

   Since we are dealing with 1 g of activated carbon and modifying the original concentration based on \(qe\), we can multiply \(qe\) by the number of grams of carbon to find the equation in terms of total mass adsorbed to the carbon.

5. Instead of focusing on total mg/L in the 10 L volume, we can express everything in terms of 1 L, leading to:
   For 1 g of activated carbon, we need to understand the mass transfer relationship involving the adsorption capacity.

6. Total TOC in 10 L: \[ \text{Total TOC} = 20 \text{ mg/L} \times 10 \text{ L} = 200 \text{ mg} \]

   This is the total mass of TOC that exists before adsorption.

Now, we need to understand that in the presence of \( qe \), where \( qe \) is the amount adsorbed per unit mass of activated carbon.

7. Assuming no other breakdown or reaction, we can consider: At equilibrium and with the given relationships in the material balance for 1 L of water:

   For 1 g of activated carbon: \[ 100qe + Ce = 200 \]

   Substituting \(Ce = 20 - 10qe\) into this balance will yield a straightforward equation to use:

   • For every 1 g of activated carbon (or 1000 mg), the equation results into adsorption related to the mass of carbon and its active sites.

To determine \(qe\):

1. Substitute various \(qe\) back to see if \(Ce\) stays zero; because \[Ce\] simply reduces the TOC.

2. For equilibrium: \[ (100 \times qe) + Ce = 200 \] Equating with the total throughput flowing through.

Rearranging for computation:

1. Solve from: \[ Ce = 200 - 100qe \]

2. \(qe\) can be seen simply as: With empirical observations, let's just push through for: Assume TOC goes down to the limits of:
   Iterate: targeting a \(Ce = 0\) which yields the maximum capacity cleared up.

3. Test for practical conservations. (Final estimation ultimately rounds normally yielding to exact specifics.)

Hence solving iteratively finally leads up, yielding: \[ 100qe + 0 = 200 \implies qe = 2.0 \text{ mg.mg/C} \]

Thus conclude stability at equilibrium: Conclusion: Based on material balance, calculation shows \(qe = 2.0\) mg/mg of carbon.