The Rates of a first order reaction after 10 and 20 min from initiation of the reaction are 0.04 and `0.03"mol. L"^(-1).s^(-1)` respectively . Find the half - life of the reaction.

Explanation

For a first reaction , rate (r) = k [ reactant].
If the concentration of the reactant after 10 and 20min are `[A]_(1) and [A]_(2)` respectively, then 0.04=k[A] and `0.03=k[A]_(2)`.
These two relation give `([A]_(1))/([A]_(2))=(4)/(3)`
The reaction time for a first order reaction, `t=(2.303)/(k)log.([A]_(0))/([A])`
After 10 min of initiation ,10`t=(2.303)/(k)log.([A]_(0))/([A]_(1))" " ...(1)`
and after 20 min of initiation ,20 `t=(2.303)/(k)log.([A]_(0))/([A]_(2))" " .....(2)`
Subtracting equation [1] from [2] , we get,
`20-10=(2.303)/(k)[log.([A]_(0))/([A]_(2))-log.([A]_(0))/([A]_(1))]`
or , `10=(2.303)/(k)log.([A]_(1))/([A]_(2))=(2.303)/(k)log.(4)/(3)[therefore([A]_(1))/([A]_(2))=(4)/(3)]`
or, `k=0.02877"min"^(-1)`
`therefore"Half-life"=(0.693)/(k)=(0.693)/(0.02877)=24.08"min"`