For a first order reaction, the time required for `99.9%` of the reaction to take place is nearly

Options

A. `10` times that required for half of the reaction
B. `100` times that required for two-third of the reaction
C. `10` times that required for one-fourth of the reaction
D. `20` times that required for half of the reaction

Answer

A

Explanation

Time required to complete `99.9%` of reaction
`t_(99.9)=(2.303)/(k)log.(100)/(100-99.9)`
= `(2.303)/(k)log1000.=(6.909)/(k)`
and `t_(1//2)=(2.303)/(k)log.(100)/(100-50)`
= `(2.303)/(k)log2.=(0.693)/(k)`
So, `t_(99.9)` is 10 times that of required for half of the reaction .