A reactant has a half-life of 10 minutes.
What fraction of the reactant will be left after an hour of the reaction has occurred ?

Explanation

`t_(1//2) = 10 min`
Let the initial conc. be 100
t = 1 hour = 60 min, (a-x) = ?
For a first order reaction,
` k=( 2.303)/(t) log "" (a)/((a-x))`
` or 0.0693 min""^(-1) = ( 2.303 )/( 60 min ) log ""(100)/( (a-x))`
` or log ""(100)/( (a-x)) = ( 0.0693 min""^(-1) xx 60 min ) /( 2.303 ) = 1.805`
` or log 100 - log (a-x) = 1.805 or 2- log (a-x) = 1.805`
` or log (a-x) =2-1.805 = 0.195`
` (a-x) =` Antilog `0.195 = 1.56`
thus fraction of reactant of reactant left undecomposed `= 1.56 %`