A substance having a half-life period of 30 minutes decomposes according to the first order rate law. The fraction decomposed, the balance remaining after 1.5 hours and time for 60% decomposition on its doubling the initial concentration will be.

We know that K=0.693t1/2=0.69330=0.023/min⁡−1Let the initial concentration of the substance = 100 and the substance decomposed in 1.5 hours (90 min) = x then, a−x=100−xSubstituting these values in K=2.303tlog⁡(aa−x)⇒ 0.0231=2.30390log⁡100100−xlog⁡100100−x=0.0231×902.303⇒ x=87.4So the fraction decomposed in 1.5 hours =87.4100=0.874Fraction remaining behind after 1.5 hours =1−0.874=0.126The time required for 60% of decomposition K=2.303tlog⁡(aa−x)⇒ 0.023=2.303tlog⁡(10040)t=2.3030.0231log⁡(104)=39.7 minutesSince the reaction is of first order the time required to complete specific fraction is independent of initial concentration (or pressure). Hence 60% of the reaction will decompose in 39.7 mins in this case also.