In a second order reaction, `20%` of a substance is dissociated in `40 min`. The time taken by `80%` of its dissociation is

Options

A. `160 min`
B. `640 min`
C. `200 min`
D. `320 min`

Answer

B

Explanation

For second order reaction,
When `x = 20%, (a-x) = 80%, a = 100%`
`k_(2) = (1)/(t).(x)/(a(a-x))`
`= (1)/(40) xx (20)/(100 xx (100 - 20))`
`= (1)/(20 xx 100 xx 80)` …(i)
When `x = 80%, (a-x) = 20%`
`k_(2) = (1)/(t) xx (80)/(100xx20) = 1//25t` ...(ii)
form Eqs. (i) and (ii),
`t = (2 xx 100 xx 80)/(25) = 640 min`