A substance having a half-life period of 30 min decomposes according to firt order rate law.
a) What fraction of this will be decomposed and what will remian behind after 1.5 hr?
b) How long will it take to be `60%` decomposed, If the molar concentration is just doubled?

Explanation

The rate constant for the reaction, k`=(0.693)/(t_(1//2)) = 0.693/(30"min")=0.0231 min^(-1)`
a) After 1.5 h or 90 min, `k=2.303/90 log ([A]_(0))/([A])`
`(0.0231 min^(-1)) = 0.303/(90 "min")log ([A]_(0))/([A])`
`log([A]_(0))/([A]) = (0.0231 min^(-1) xx 90 "min")/(2.303)=0.9027`
`[A]_(0)/[A]="Antilog"(0.9027)=7.993`
The fraction of the substance remaining `=([A])/([A]_(0)) =1/(7.993)=0.125`
The fraction decomposed `=1-0.125=0.875`
b) Time required for `60%` decomposition may be calculated as follow:
`k=(2.303)/t log (a)/(a-x), t=(2.303)/(k) log (a/(a-x))`
`t=(2.303)/(0.0231 min^(-1))log (100/40)=(2.303 xx 0.398)/(0.0231 min^(-1)), t=39.7` min
In a first order reaction, the time taken to complete a certain fration of the reaction is quite independent of the initial concentration of the reactants. Thus, `60%` of the reactant will decomposes in 39.7 min even if the initial concentration is doubled.