Question

The half-life for the first order decomposition of the pesticide methyl bromide (CH3Br) in human blood is about 12 days. How long after exposure will the blood concentration be the following percentages of its initial concentration?

a. 50% ___________________

b. 25% ___________________

c. 12.5% __________________

d. 33% ___________________

e. 1% ___________________

Answer 1

12 days 12 days 12 days   

100 ----------- 50 -------------- 25 ---------------12.5

Therefore,

a. 50% = 12 days

b. 25% = 12 + 12 = 24 days

c. 12.5% = 12 + 12 + 12 = 36 days

d) 33 % Ans = 19.2 days

[Ao] = 100 %

[At] = 33 %

For first order recation,

half life t_1/2 = 0.693 /k where k is rate constant

k = 0.693/ t_{1/2 --- Eq (1)}

k = 1/t ln { [A]_o/[A]_t} -----Eq (2)

From Eqs (1) and (2),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t}

0.693 / 12 = (1/t) In (100/33)

t = 19.2 days

e) 1% Ans: 79.7 days

[Ao] = 100 %

[At] = 1 %

For first order recation,

half life t_1/2 = 0.693 /k where k is rate constant

k = 0.693/ t_{1/2 --- Eq (1)}

k = 1/t ln { [A]_o/[A]_t} -----Eq (2)

From Eqs (1) and (2),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t}

0.693 / 12 = (1/t) In (100/1)

t = 79.7 days