Asked by Maria

the concentration of a chemical degrades in water according to order kinetics. the degradation constant is 0.2 day^-1. if the initial concentration is 100.0 mg/L, how may days are required for the concentration to reach 0.14 mg/L?
Answered by DrBob222
You didn't post the order.
Answered by Maria
The concentration of a chemical degrades in water according to order kinetics. the degradation constant is 0.2 day^-1. if the initial concentration is 100.0 mg/L, how many days are required for the concentration to reach 0.14 mg/L?
Answered by DrBob222
You still didn't post the order. If we assume first order (a guess on my part) then
k = 0.693/t<sub>1/2</sub>
k = 0.693/0.2 = ??

ln(No/N) = kt
No = 100.0 mg/L
N =0.14 mg/L
k = from above.
t = unknown
Solve for t in days.
Answered by freed
the answer should be 32.9 days...but im not getting tht
Answered by freed
problem was it should be ln(N/No)=kt
thx tho DrBob222
Answered by freed
dC/dt = - k Cln(C/Co) = - ktt = - ln(0.14/100) / 0.2 = 32.9 days
Answered by freed
dC/dt = - k Cln(C/Co) = - kt
t = - ln(0.14/100) / 0.2 = 32.9 days
sorry this one written better
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