For a first order for the reaction `A rarr` products `:` the rate of reaction at `[A]=0.2 M ` is `1.0 xx 10^(-3) mol L^(-1)s^(-1)`. The reaction will occur to `75%` completion in .

Options

A. `138.65`
B. `277.2s`
C. `44.25`
D. `822.5`

Answer

2

Explanation

`k=(rate)/([A])=(1.0xx10^(-3))/(0.2)=5xx10^(-3)s^(-1),t_(1//2)=(0.693)/(k)=(0.693)/(5xx10^(-3))=138.6s`
Fraction remaining `,((1)/(2))^(n)=((100-75)/(100))=(25)/(100)=(1)/(4)=((1)/(2))^(2)`
Hence, `n=2` half lives,` t=2xx138.6=277.2s`.