The rate of a first - order is `0.04 "mol l"^(-1) s^(-1)` at 10 seconds and `0.03 "mol l"^(-1) s^(-1)`. At 20 seconds after initiation of the reaction. The half - life period of the reaction is

Answer

`t_(1//2) = 24.27 s`

Explanation

For a first order reaction,
Rate `= k[A]`
`r_(1) = k[A_(1)]`
or `0.04 = k[A_(1)]`
`r_(2) = k[A_(2)]`
or `0.03 = k[A_(2)]`
`:. ([A_(1)])/([A_(2)]) = (4)/(3)`
Let `[A_(1)]` be the initial and `[A_(2)]` be the final concentrations after a given gap `10 min`. Therefore,
`k = (2.303)/(t) log.([A_(1)])/([A_(2)])`
`= (2.303)/(10)log.(4)/(3)`
`= 2.855 xx 10^(-2) s^(-1)`
For a first order reaction.
Half life `= (0.693)/(k)`
`= (0.693)/(2.855 xx 10^(-2))`
`= 24.27 s`