Question

A first-order reaction has a half-life of 22.9 s . how long does it take for the concentration of the reactant in the reaction to fall to one-sixteenth of its initial value

Answer 1

The half-life of a first order reaction is in how much τtime, the concentration of the reactant halves. After one half-life, the concentration of the reactant becomes half its initial value. After another half-life, itsconcentration becomes (M: initial concentration):
`((1)/(2) *M)* (1)/(2) = (1)/(4) * M`
In general, after n half-lifes, the concentration of the reactant becomes:
`( (1)/(2) ^ n)*M`
Since 1/16=
`((1)/(2)) ^ 4` , we have that 4 half-lifes need to pass so that the concentration of the reactant becomes 1/16 of the initial one (=1/16*M). Hence, 4*22.9 sec=91.6 seconds is the required time.