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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^([("In 2")/(5)-("In 2")/(15)]xxt)`
In `4=[("In 2")/(5)-("In 2")/(15)]t`
In `(2)^(2)=[("In 2")/(5)-("In 2")/(15)]t`
2 In `2[("In 2")/(5)-("In 2")/(15)]t`
`2=[(1)/(5)-(1)/(15)]t`
`2=(2)/(15)xxt`
(t=5)
`t=(15)/(5) =3` minute
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