Question

Show that for a first-order reaction the time required to complete 75% of reaction is about 2 times more than that required to complete 50% of the reaction.

Answer 1

For a first-order reaction

t = `2.303/k  log  a/(a - x)`

t_75% = `2.303/k  log  100/(100 - 75)`

= `2.303/k  log  100/25`

t_75% = `2.303/k  log 4`    ...(1)

t_50% = `2.303/k log 100/(100 - 50)`

= `2.303/k log  100/50`

t_50% = `2.303/k log 2`    ...(2)

Dividing equation (1) by equation (2), we get

`t_(75%)/t_(50%) = (2.303/k log 4)/(2.303/k log 2)`

= `log 4/log 2`

= `log2^2/log 2`

= `(2 log 2)/(log 2)`

∴ `t_(75%)/t_(50%) = 2/1`

∴ t_75% = 2 × t_50%