Question

the biological treatment of total organic carbon 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 waste water with 20mg/l toc, 1g activated carbon is applied. at equilibrium of adsorption, what is capacity of equilibrium adsorption?

Answer 1

To determine the capacity of equilibrium adsorption (\(q_e\)), we can use the Freundlich adsorption isotherm equation, which is generally given in the form:

\[
q_e = k \cdot C_e^n
\]

Where:
- \(q_e\) is the amount of adsorbate per unit mass of adsorbent (mg/mg),
- \(C_e\) is the concentration of the adsorbate at equilibrium (mg/L),
- \(k\) is the Freundlich constant,
- \(n\) is the Freundlich exponent.

Given values:
- \(k = 0.007\) (mg/mg)(L/mg)^{1/n}
- \(n = 1.13\)
- The initial concentration of Total Organic Carbon (TOC) in the wastewater is \(C_0 = 20\) mg/L,
- Volume of wastewater = 10 L.

First, we need to find \(C_e\) (the equilibrium concentration of TOC after adsorption). Initially, in 10 L of wastewater at \(20\) mg/L, the total amount of TOC is:

\[
\text{Total TOC} = C_0 \times \text{Volume} = 20 \, \text{mg/L} \times 10 \, \text{L} = 200 \, \text{mg}
\]

Assuming that \(q_e\) is the amount adsorbed by 1 g of activated carbon, we can say that the amount of TOC adsorbed (\(q_e\)) when equilibrium is reached can be written as:

\[
\text{Amount of TOC Adsorbed} = q_e \times \text{mass of activated carbon}
\]

Since we have 1 g of activated carbon, we will consider the mass of the activated carbon as 1 mg for the \(q_e\) calculation.

Let \(m\) be the mass of activated carbon (in mg), and the equation can now represent the equilibrium concentration of TOC as:

\[
C_e = C_0 - q_e
\]

But since we need to express \(C_e\) in terms of \(q_e\), we can rewrite:

Then substituting \(C_e\) into the Freundlich equation gives:

\[
q_e = k \cdot (C_0 - q_e)^n
\]

Rearranging gives us:

\[
q_e = k \cdot (20 - q_e)^{1.13}
\]

However, we can find \(C_e\) using trial values or numerical methods for \(q_e\) knowing \(q_e\) can't exceed the total amount initially present.

Instead, for practical and easier estimation, we can plug in approximate values into the Freundlich equation iteratively. For instance, start by assuming:

1. Assume no TOC adsorbed, so \(C_e \approx 20\) mg/L:
\[
q_e = 0.007 \cdot (20)^{1.13} \approx 0.007 \cdot 22.33 \approx 0.156 \, \text{mg/mg}
\]

Now substituting back to find \(C_e\):
\[
C_e = 20 - q_e \approx 20 - 0.156 = 19.844 \, \text{mg/L}
\]

2. Use this \(C_e\) value back into the Freundlich equation:
\[
q_e = 0.007 \cdot (19.844)^{1.13} \approx 0.164 \, \text{mg/mg}
\]

Repeat until steady.

Through following this calculation up to reach a steady state:

Ultimately, the equilibrium capacity of adsorption after reaching stable iterations will give you the final \(q_e\) in mg/mg.

Continuing doing these iterations until convergence, you'll find a fair estimate for \(q_e\).

Doing it as suggested, when you reach stability, final values for \(q_e\) which could range \( \text{Approx } 0.160 - 0.170 \, \text{mg/mg} \).

So the total adsorption capacity, based on this approximation, might be around \(0.16 \) mg/mg if correct during iterations.