2. (3 points) The following laboratory data were collected to establish an equil

Question

2. (3 points) The following laboratory data were collected to establish an equilibrium isotherm for an SOC (synthetic organic compound) and a sample of PAC (powered activated carbon). The isotherm bottles contained 200 ml of water and the initial SOC concentration was 9.2 mg/L Mass of PAC (mg) 0 5 1025 50 100150 200 Equilibrium SOC 9.2 7.36 6.86 3.86 1.13 0.22 0.18 0.11 concentration (ma/L) 1) Find the Ereundlich coefficients for the isotherm. Plot the Ereuodlich model and the 2) Find the PAC dosage (g/L) that would give anequilibium SOC concertatioo of 0.25 3) If this PAC costs $0.7/kg, estimate the annual cost for treating water with a lowrate data on the same set of axes to assess the fit of the calculated model. mg/L. of 1000 L/min (or 0.38 x 106 gallon/day) at this dosage. (Please check Example 11.19 on Page 474 and class examples. To stabilize the results, please use this form of Freundlich equation: q = KrCen. Please note that Ce is the equilibrium concentration of adsorbate not the initial concentration).

Explanation / Answer

1. For freudlich isotherm

q =x/m = KF (Ce)1/n                 where q = x mass of adsorbate adsorbed on m mass of adsorbent

                                                               Ce = equllibrium concentration of adsorbate

                                                               KF and n are freudlich coefficients

Volume of total solution is 200 ml. Hence by multiplying this volume with each concentration value (in mg/ml)of SOC, we will get mass of SOC(adsorbate) correspond to each concentration.

x/m

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0.2944

0.1372

0.031

0.00452

0.00044

0.00024

0.00011

Ce (mg/L)

9.2

7.36

6.86

3.86

1.13

0.22

0.18

0.11

Now     log(x/m) = logKF + 1/n logCe

So putting two sets of value for (x/m,Ce) [e.g. (0.2944,7.36) and (0.1372,6.86)] in the above equation, we will get the values for freudlich coefficients

n = 0.0934

KF = 0.1212

2.   log(x/m) = logKF + 1/n logCe

This equation will give a straight line when graph is drawn between x/m vs Ce. So a linear fitted graph is drawn using the given values. Afterthat extrapolating the graph, we can obtain the x/m valuecorrespond to 0.25 mg/L of SOC

log(x/m)

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-0.53

-0.86

-1.51

-2.34

-3.36

-3.62

-3.96

Ce(mg/L)

9.2

7.36

6.86

3.86

1.13

0.22

0.18

0.11

Now from graph, log(x/m) correspond to Ce = 0.25 mg/L is -3.37. So x/m = 0.00043

Hence m = 0.05/0.00043 = 116.3 mg

So concentration of PAC (0.1163 × 1000)/200 = 0.5815 g/L

3. Total amount of water treated annually is 1387 × 105 gallon.

1 gallon = 3.785 L

So PAC needed annually is 305275579.3 g = 305275.5793 kg

Cost = $ 213692.9055

x/m

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0.2944

0.1372

0.031

0.00452

0.00044

0.00024

0.00011

Ce (mg/L)

9.2

7.36

6.86

3.86

1.13

0.22

0.18

0.11

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