What is the time it takes for a first-order decay to leave the reactant at #1/8# of its original concentration if the reaction's rate constant was #3.30 xx 10^(-3) "s"^(-1)#? I put in #"1680 s"# but it was incorrect.

#t_"1/8" = 6.30 xx 10^2 " s"#

The time you entered was in fact the time it takes to experience #8# first-order half-lives, not #3#.

A first-order, one-reactant reaction has a rate law of:

#r(t) = k[A] = -(Delta[A])/(Deltat)#

where #r# is the initial rate of reaction, #k# is the rate constant, and #[A]# is the molar concentration of reactant #A#.

For an infinitesimally small change in time, #Delta[A] -> d[A]# and #Deltat -> dt#. Separation of variables eventually gives the integrated rate law.

#-int_(0)^(t) kdt = int_([A]_0)^([A]) 1/([A])d[A]#

#-kt = ln[A] - ln[A]_0#

#bb(barul|stackrel(" ")(" "ln[A] = -kt + ln[A]_0" ")|)#

From the integrated rate law, a half-life #t_"1/2"# passes by when the initial concentration halves, i.e. #[A]_0 -> 1/2[A]_0#. So...

#ln(1/2[A]_0) - ln[A]_0 = -kt_"1/2"#

#ln((1/2cancel([A]_0))/(cancel([A]_0))) = -kt_"1/2"#

#-ln2 = -kt_"1/2"#

Therefore, the first-order half-life is given by:

#barul|stackrel(" ")(" "bb(t_"1/2" = (ln2)/k)" ")|#

In one half-life, this much time would pass:

#t_"1/2" = (ln2)/(3.30 xx 10^(-3) "s"^(-1))#

#=# #2.10 xx 10^(2) " s"#

Since each time a half-life passes, the concentration halves, halving 3 times gives #1/8# the initial concentration.

So when the concentration becomes #1/8# of what it was, three half-lives passed by:

#color(blue)(t_"1/8") = 3t_"1/2" = ulcolor(blue)(6.30 xx 10^2 " s")#