Answer:
[tex]k =0.101 \ min^{-1}[/tex]
Explanation:
From the given information:
The initial concentration of a chemical [tex]C_{AO} = 15 \ mg/L[/tex]
The final concentration is [tex]C_A = 5 \ g/m^3[/tex] = 5 mg/L
Volume flow rate [tex]V_o = 0.42 \ m^3/sec[/tex]
Volume of the tank V = 500 000 L = 500 m³
The time t is determined by using the formula:
[tex]t= \dfrac{V}{V_o}[/tex]
[tex]t= \dfrac{500 \ m^3}{0.42 \ m^3/sec}[/tex]
t = 1190.47 sec
t ≅ 19.8 min
∴
The rate of the decay constant is:
[tex]kt= \dfrac{C_{AO}-C_{A}}{C_A} \\ \\ k = \dfrac{1}{19.8}( \dfrac{15-5}{5})[/tex]
[tex]k =0.101 \ min^{-1}[/tex]
