Question

A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is ______. (Use ln 2 = 0.693)

Options

• 120

• 300

• 180

• 900

Answer 1

A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is 900.

Explanation:

C_t = `C_0 e^(- kt)`; k = `(ln 2)/(t_(1//2))`

`(C_t)_A = (C_0)_(A^(e^(- k_(A^t)))`; k_A = `(ln 2)/300`

`(C_t)_B = (C_0)_(B^(e^(- k_(B^t)))`; k_B = `(ln 2)/180`

`((C_t)_B)/((C_t)_A) = ((C_0)_B)/((C_0)_A) xx e^((k_B - k_A) t)`

⇒ 4 = `e^((k_B - k_A)t)`

⇒ 2 ln 2 = `[(ln 2)/180 - (ln 2)/300] t`

⇒ 2 ln 2 = `ln 2 [1/180 - 1/300] t`

⇒ 2 = `(120/(180 xx 300)) t`

⇒ t = `(2 xx 180 xx 300)/120`

⇒ 900 sec