Answer to Question #330405 in General Chemistry for lisa

Question

The first order rate of constant for the decomposition of a certain insecticide in water at 12degrees is 1.45/yr. A quantity of this is washed into the pond leading to a concentration of 5.0*10^-7M.calculate the halflife.

Answer 1

The easiest way to do this since it is first order decay is to use the following expression:

fraction remaining = 0.5ⁿ where n = the number of half lives that have elapsed.

First we must find the half life, which for a first order reaction is 0.693/k. Thus...

t_1/2 = 0.693 / 1.45 yr^-1 = 0.4779 yrs

Now, applying the above formula for fraction remaining:

fraction remaining = 0.5ⁿ and since the time interval is 1 year (June 1 to June 1), the number of half lives will be n = 1 yr/0.4779 yr/half life = 2.092 half lives have elapsed

fraction remaining = 0.5^2.092 = 0.2346

To find the resulting concentration, we have...

5.0x10^-7 g/cm³ x 0.2346 = 1.17x10^-7 g/cm³ remaining after 1 year

How long will it take for the concentration to reach 3.0x10^-7 g/cm³? So, here we want to find time needed to reach a certain fraction remaining.

Fraction remaining = 3.0x10^-7 / 5.0x10^-7 = 0.60 remaining

0.6 = 0.5ⁿ and we can find the number of half lives...

log 0.6 = n log 0.5

-0.222 = -0.301 n

n = 0.738 half lives

0.738 half lives x 0.4779 yrs/ half life = 0.35 years for the insecticide to decrease to 3.0x10^-7 g/cm³