Two reacants `A` and `B` are present such that `[A_(0)] = 4[B_(0)]` and `t_(1//2)` of `A` and `B` are `5` and `15` mintute respectively. If both decay folliwing `I` order, how much time later will concentrations of both of them would be equal?

Options

A. 15 minute
B. 10 minute
C. 5 minute
D. 12 minute

Explanation

Amount of A left in `n_1` haves `=[A]_0/(2^n1)`
Amount of B left in `n_2` halves `= [B]_0/(2^n2)`
Also if `[A]_0/(2^n1)=[B_0]/2n^2` when A decays to `n_1`halves and B dacays to `n_2` halves.
`:'[A]_0= 4[B]_0`
`:. 4=(2^n1)/(2^n2)=(2)^(n_2-n_2)`
or `(n_1-n_2) = 2`
`:.n _2 = n_1 - 2 ...(i)`
Now, `t=n _1 xx t_(1//2A) and t= n_2xxt_(1//2B)`
`:. (n_1 xx t_(1//2A))/(n_2xx t_(1//2B))=1`
But `t_(1//2)(A)=5 and t_(1//2)(B)= 15`
`or :. (n_1xx 5)/(n_2xx 15)=1`
`or n_1/n_2=3...(ii)`
`:.` By Eqs. (i) and (ii) `n_1 = 3, n_2 = 1`
Thus, `t = 3xx5 = 15` minute