the biological treatment of total organic carbon (toc) in water uses activated carbon adsorption .The adsorption accords with 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 the capacity of equilibrium adsorption? (mg.mg/C). note: based on material balance, for every litre of water 100 qe +Ce=200.

Question

GPT-4o mini · Answer

To solve this problem, we need to apply the given information and the Freundlich isotherm equation.

Given:

Freundlich constants: $k = 0.007$ and $n = 1.13$ .
Initial concentration of TOC ( $C_{0}$ ) = 20 mg/L.
Volume of wastewater ( $V$ ) = 10 L.
Amount of activated carbon ( $m$ ) = 1 g.

We are looking to find the equilibrium adsorption capacity ( $q_{e}$ ) in mg/mg of activated carbon.

Step 1: Calculate total TOC in the wastewater

The total amount of TOC in 10 liters of wastewater is calculated as follows: $Total TOC = C_{0} \times V = 20, mg/L \times 10, L = 200, mg$

Step 2: Set up the material balance

From the problem, we know that at equilibrium: $100 \cdot q_{e} + C_{e} = 200$

Where:

$q_{e}$ = amount of TOC adsorbed per unit mass of activated carbon in mg/mg.
$C_{e}$ = equilibrium concentration of TOC in mg/L.

Step 3: Calculate $C_{e}$

Assuming the amount of TOC adsorbed ( $Q$ ) is equal to the total TOC minus what remains in solution: $Q = 200 - C_{e}$

Since we have: $Q = m \cdot q_{e} = 1 \cdot q_{e} = q_{e}, mg$

We have: $C_{e} = 200 - q_{e}$

Step 4: Substitute into the Freundlich equation

According to the Freundlich isotherm: $q_{e} = k C_{e}^{n}$

Substituting $C_{e} = 200 - q_{e}$ into the equation: $q_{e} = 0.007 (200 - q_{e})^{1.13}$

Step 5: Solve the equation iteratively

This is a nonlinear equation, and we can solve it using numerical methods or trial and error. For simplicity, let's conduct an estimate, but realistically, this should be solved numerically or using a mathematical software.

Assuming $q_{e}$ to start with 0:

$C_{e} = 200 - 0 = 200$
$q_{e} = 0.007 (200)^{1.13} \approx 0.007 (285.3) \approx 1.99, mg$

Now using $q_{e} = 1.99$ :

$C_{e} = 200 - 1.99 = 198.01$
Recalculate $q_{e} = 0.007 (198.01)^{1.13} \approx 0.007 (276.6) \approx 1.93, mg$

Continue this process iteratively until convergence. After a few iterations, you would find that:

$q_{e} \approx 1.99, mg$

Step 6: Calculate the final adsorption capacity

Convert $q_{e}$ in mg to mg/mg: Since we started with 1 g of activated carbon (or 1000 mg): $q_{e} = \frac{1.99, mg}{1, g} = 0.00199, mg/mg \approx 0.002, mg/mg$

Conclusion

Thus, the capacity of equilibrium adsorption $q_{e}$ is approximately 1.99 mg/g or 0.00199 mg/mg.

This value would need to be confirmed through more precise iterations or numerical solving to reach a consistent equilibrium value.