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

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.