The rate of first order reaction is `0.04 mol L^(-1) s^(-1)` at `10 min` and `0.03 mol L^(-1) s^(-1)` at `20 min` after initiation. Find the half life of the reaction.

Explanation

Let the concentration of the reactant after 10 min and 20 min be `C_(1)and C_(2)` respectively .
` :. ` rate after 10 min = `kC_(1)=0.4 xx60`
and rate after 20min =`k.C_(2)=03xx60`
` :. (c_(1))/(c_(2))=4/3`
Supposing the reaction starting after 10 minutes
`k=(2.303)/10 log. (c_(1))/(c_(2)) =(2.303)/(10) log. 4/3 = .02878`
` :. t_(1/2) =(0.6932)/k = (0.6932)/(0.02878)=24.086` min