The time required for the decompoistion of `99.9%` fraction of a first order reaction is………….to that of its half-life time.

Options

A. 5
B. 10
C. 15
D. 20

Answer

B

Explanation

For a first order reaction, time required to complete any particular fraction is given as
`t_(f)=(2.303)/(K)log((1)/(1-f))`
Substituting the value of `f=(99.9)/(100)` , we get
`t_(99.9%)=(2.303)/(K)log(1)/(1-0.999)`
`=(2.303)/(K)log((1)/(0.001))`
`=(2.303)/(K)log((10^(3))`
`=(2.303)/(K)3log(10)`
`=(6.903)/(K)(log(10=1))`
Half life of reaction is given as `t_(1//2)=(0.693)/(K)`
Dividing the two result we get
`(t_(99.9%))/(t_(1//2))=(6.909//K)/(0.693//K)`
`=(6.909)/(0.693)`
`=9.97~~10`
Alternatively, we can calculate `yt_(99.9%)` as follows