# If the rate constant is #3.5 xx 10^(-3) "s"^(-1)# for a first order decay of reactant #A#, how long does it take the reaction to get to #24%# completion?

##
#A)# #"285.7 s"#

#B)# #"407.7 s"#

#C)# #7.84 xx 10^(-5) "s"#

#D)# #4.08 xx 10^(-4) "s"#

#E)# #"78.41 s"#

##### 1 Answer

I got about

A first-order reaction

#A -> B#

follows the rate law:

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

For an infinitesimally small time

#r(t) = k[A] = -(d[A])/(dt) = (d[B])/(dt)#

Rearranging, we get:

#-kdt = 1/([A])d[A]#

What we can do is "integrate" from time

In simple words, this describes the evolution of time as compared to how the concentration changes, in a "smooth" manner (rather than "stepwise").

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

#-(k*t - k*0) = ln[A] - ln[A]_0#

Therefore, we have the integrated rate law for first-order kinetics:

#bb(ln[A] = -kt + ln[A]_0)#

Manipulating this rate law, we can determine at what time this reaction is at

This means:

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

#ln\frac([A])([A]_0) = -kt#

#ln\frac(0.76cancel([A]_0))(cancel([A]_0)) = -kt#

#ln(0.76) = -kt#

Therefore:

#color(blue)(t) = -ln(0.76)/k#

#= -ln(0.76)/(3.5 xx 10^(-3) "s"^(-1))#

#=# #color(blue)("78.41 s")#

This is closest to **answer E**.