Question

The number of coliform bacteria is reduced from an initial concentration of 2,000,000 per 100 ml to 400 per 100 ml in a long, narrow chlorination tank under a steady wastewater flow with a hydraulic detention time of 30 min. Assuming second-order kinetics and ideal plug flow, calculate the reaction rate constant. What would be the hydraulic detention time needed to produce the same effluent quality if you assumed perfectly mixed flow in the chlorination tank?

Answer 1

For a second order reaction:   rate = k[A]²

The integrated rate law is 1/[A] = kt + 1/[A_o]

The initial concentration, [A_o].

The final concentration, [A].

2,000,000 per 100 ml means 20000000 per 1000 mL

20000000/6.023 x 10²³ = 3.32 x 10^-17 mol/L

400 per 100 ml means 4000 per 1000 mL

4000/6.023 x 10²³ = 6.64 x 10^-21 mol/L

The rate constant, k, for the reaction

1/6.64 x 10^-21 mol/L = k x 30 min + 1/3.32 x 10^-17 mol/L

1.50 x 10²⁰ L mol^-1 = k x 30 min + 3.01 x 10¹⁶ L mol^-1

1.50 x 10²⁰ L mol^-1 - 3.01 x 10¹⁶ L mol^-1 = k x 30 x 60 sec

1.499 x 10²⁰ L mol^-1 / 1800 sec = k

k = 8.32 x 10¹⁶ L mol^-1sec^-1

0.0025 - 5 x 10^-7 = k x 30 min

0.0024995 = k x 30 min