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For the following elementary first order...

For the following elementary first order reaction:

If `k_(2)=2k_(1)`, then `%` of B in overall product is:

A

`25%`

B

`50%`

C

`75%`

D

`100%`

Text Solution

Verified by Experts

The correct Answer is:
B
`(1)/(2)(d[B])/(dt)=k_(1)[A] rArr (d[B])/(dt)=2k_(1)[A]`
`(d[C])/(dt)=k_(2)[A]`
`rArr([B])/([C])=(2k_(1))/(k_(2))=1`
`rArr %[B]=(1)/(1+1)xx100=50%`
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