Question

Show that in case of a first order reaction, the time taken for completion of 99% reaction is twice the time required for 90% completion of the reaction. (log 10 = 1)

Answer 1

For a first-order reaction, we have

`t = 2.303/k log  [A]_0/[[A])`

When the reaction is 99% complete,

`[A] = a - (99a)/100 = 0.01a`

∴ `t_99 = 2.303/k log  a/(0.001a)`

= `2.303/k xx 2`   ...(i)

When the reaction is 90% complete,

`[A] = a - (90a)/100 = 0.1a`

`t_90 = 2.303/k log  a/(0.1a)`

`t_90 = 2.303/k xx 1`   ...(ii)

By dividing equations (i) by (ii), we get

 `t_99/t_90 = 2 ⇒ t_99 = 2t_90`