A reacts to form P , A plot of the reciprocal of the concentration of A vs time is a straight line . When the initial concentration of A is `1.0 xx 10^(-2) M` , its half-life is found to be 20 min . When initial concentration of A is `3.0 xx 10^(-3) M` , the half -life will be

Options

A. 20 min
B. 40 min
C. 56 min
D. 67 min

Answer

D

Explanation

`A to P`
A plot of 1//[A]vs time is a straight line of second order .
For second order `t_(1//2)=(1)/(k[A]_(0))`
where k=rate constant
`[A]_(0)` = initial concentration of A
When, `[A]_(0)=1xx10^(-2)M`, then `t_(1//2)=20` min
`20=(1)/(k(1xx10^(-2)))impliesk=(1)/(20)(1xx10^(-2))`
when `[A]_(0)=3xx10^(-3)M`, then `t_(1//2)=?`
`t_(1//2)=(1)/(k(3xx10^(-3)))`
Substituting value of k from equation (i)
`t_(1//2)=(1xx20(1xx10^(-2)))/((3xx10^(-3)))=6.66xx10"min~~67"min"`