The half life period of a first order reactions is 10 mins. What percentage of the reactant will remain after one hour ?

Explanation

Given data : Half life period `(t_(1//2))=10` mins
Formula : `K=(2.303)/(t)"log"(a)/(a-x)`
Solution :
`k=(0.693)/(t_(1//2))=(0.693)/(10)`
`=0.0693 "min"^(-1) = 6.93xx10^(-2)"min"^(-1)`
Time taken = 1 hour = 60 minutes
a = 100 %
x = ?
`k=(2.303)/(t)"log"(a)/(a-x)`
`=(2.303)/(60)[log 100-log(a-x)]`
`log 100 - log (a-x)=(6.93xx10^(-2)xx60)/(2.303)`
`log 100 - log (a-x)=1.8060`
`log(a-x)=log 100 - 1.8060`
`= 2-1.8060`
`log (a-x)=0.1940`
`a-x=` Antilog of 0.1940
`a-x=1.563%`