If the constant of a first reaction at a certain temperature is `1.5xx10^(-1)s^(-1)andt_(1)andt_(2)` are the respective times for 50% and 75% completion of the reaction , determine the ratio of `t_(2) andt_(1)`

Explanation

The rate constant of a first order reaction is given by ,
`k=(2.303)/(t)log.(a)/(a-x)`
As given in the problem , time `t_(1)` is required for 50% completion of the reaction.
Therefore, `x=axx(50)/(100)=(a)/(2)and(a-x)=a-(a)/(2)=(a)/(2)`
`therefore t_(1)=(2.303)/(k)log.(a)/(a-x)=(2.303)/(k)log.(a)/(a//2)=(2.303)/(k)log2`
`=(0.693)/(k)=(0.693)/(0.15)s^(-1)[because=1.5xx10^(-1)s^(-1)]`
Now , time `t_(2)` is needed for 75% completion of reaction.
Therefore, `x=axx(75)/(100)=(3a)/(4)and (a-x)=a-(3a)/(4)=(a)/(4)`
`thereforet_(2)=(2.303)/(k)log.(a)/(a-x)=(2.303)/(k)log.(a)/(a//4)=(2.303)/(k)log4`
`= (2xx0.693)/(k)=(1.386)/(0.15)s^(-1)`
`therefore(t_(2))/(t_(1))=(1.386//0.15)/(0.693//0.15)=(1.386)/(0.693)=2`
Alternative method:

Half-life of a first order reaction does not depend on the initial concentration of the reactant and at a given temperature , its value is constant.
`therefore" "t_(2)-t_(1) =t_(1)" or, "t_(2)=2t_(1) " "therefore(t_(2))/(t_(1))=2`