The half-life of a first order reaction is 30 min.
What fraction of the reactant remains after 70 min?

Answer

`19.85 %`

Explanation

`t=(2.303 )/(k) log "" (a)/(a-f) = ( 2.303 )/(k) log (1)/(f) = ( 2.303 )/(k ) log (-f)`
` or 70 min =( 2.303 )/( 0.0231 min""^(-1)) log ""(-f)`
` or ( 70 xx 0.0231 ) /(2.303) = log (-f) or 0.7021 = log (-f)`
` or f=` antilog `-0.7021 = 1- 0.7021 +1`
` = bar(1).2979 = 00.1985 `
Hence, fraction of the reactant left = 0.1985 or 19.85%