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Consider the following parallel reaction...

Consider the following parallel reactions being given by `A(t_(1//2) = 1.386 xx 10^(2) hours)`, each path being first order.

If the distribution of `B` in the Product mixture is `50%`, the partical half life of `A` for converison into `B` is

A

`A. 346.5 h`

B

`B. 131 h`

C

`C. 115.5 h`

D

`D. 31 h`

Text Solution

Verified by Experts

The correct Answer is:
A
`(2k_(1))/(2k_(1) + 3k_(2)) = 0.5` and
`k_(1) + k_(2) = (0.693)/(1.386 xx 10^(2)) =5 xx 10^(-3)h^(-1)`
Solving `(k_(1))/(k_(2)) = (2)/(3)` and `k_(1) = 2 xx 10^(-3) h^(-1)` and
`k_(2) = 3 xx 10^(-3) h^(-1)`
`t_(1//2(A rarr B)) = (0.693)/(k_(2)) = (0.693)/(3 xx 10^(-3)) = 231 h`
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