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)

पर्याय

• 180

• 900

• 300

• 120

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