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Two substances A (t(1//2)=5 min) and B(t...

Two substances A `(t_(1//2)=5 min)` and B`(t_(t//2)=10min)` are taken in such a way that initially [A]- 4[B]. The time after which both the concentrations will be equal is : (Assume that reaction is first order)

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The correct Answer is:
3
`C_(t)=C_(0) e^(-Kt)`
According to question
`C_(A.t)=C_(B.t)`
`C_(A)e^(-K_(A)t)=C_(B)e^(-K_(B)t)`
`(C_(A))/(C_(B)) =(e^(-K_(B)t))/(e^(-K_(A)t)) rArr (C_(A))/(C_(B))= e^((K_(A)-K_(B))t`
`4=e^([("ln"2)/(5)-(" ln"2)/(15)] xx t)`
`"ln"4= [("ln"2)/(5)-("ln"2)/(15)]t`
`"ln" (2)^(2) = [("ln"2)/(5)-("ln"2)/(15)]t`
`2"ln"2=[("ln"2)/(5)-("ln"2)/(15)]t`
`2=[(1)/(5)-(1)/(15)]t`
`2=(2)/(15) xx t`
`(t=5)`
`t=(15)/(5)=3` minute
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