The half-life period of a first order reaction is 60 minutes. What percentage of the reactant will be left behind after 120 minutes ?

Answer

0.25

Explanation

`k=( 0.693 )/( 60 min ) = 0.0155 min""^(-1) , t_f = ( 2.303)/(k) log"" (1)/(1-f)`
` or 120 min = ( 2.303 )/(0.01155 min ) log"" (1)/(1-f)`
` or ( 120 xx 0.01155 )/( 2.303 ) = log "" (1)/(1-f)`
` or 0.6018 = log 1- log (1-f)^(-1)`
` = log 1-(-1) log (1-f)`
` or 0.6018 =- log f `
` or f- ` antilog `- 0.6018 =0.25`
Percentage of reactant left behind = 25%