Calculating fraction : The rate constant of a first order reaction is `0.0231 min^(-1)` . What fraction of the reactant remains after 70 minuts ?
Strategy : If 'f' is the reaction of the reactant reacting after 70 minutes, then the fraction of the reactant remaining after 70 minutes will be 1 - f.

Explanation

According to Eq. `(4.22)` we have
`t_(f) = (2.303)/(k) log (1)/(1-f)`
or `log (((1)/(1-f)) = (kt_(f))/(2.303)`
Substituting the given data, we have
`log (((1)/(1-f))) = ((0.0231 min^(-1))(70 min))/(2.303)`
`= 0.7021`
Taking antilog of both sides, we get
`(1)/(1-f) = 5.036`
or `1-f = (1)/(5.036) = 0.1985 ~~ 0.2`
Altrnatively
Let the reaction be
`{:(a,o, "Initial concentration"),(A rarr,B, ),((a-x),x, "Concentration after 70 minutes"):}`
Thus, fraction of the reactant remained undreacted `= (a-x)//a`. According to intergrated rate law for a first order reaction
`log (((a)/(a-x)))= (kt)/(2.303) = ((0.231 min^(-1))(70min))/(2.303)`
`= 0.7021`
Taking antilog of both sides, we get
`(a)/(a-x) = 5.036`
Thus `(a-x)/(a) = (1)/(5.036) ~~ 0.2`