Question #f0318
1 Answer
Feb 7, 2018
The first-order integrated rate law is:
#ln[A] = -kt + ln[A]_0# where
#[A]# is the current concentration of#A# ,#[A]_0# is for time zero, and#k# is the rate constant (in what units?).
We know the concentration at the start and at
#ln(3.55) = -k("4.78 s") + ln(4.16)#
Thus,
#color(blue)(k) = ln(3.55/4.16)/(-"4.78 s") = color(blue)(3.32 xx 10^(-2) "s"^(-1))#
The half-life is simply when the concentration has halved. Therefore:
#ln(1/2[A]_0) = -kt_"1/2" + ln[A]_0#
#ln(1/2[A]_0) - ln[A]_0 = -kt_"1/2"#
#ln((1/2[A]_0)/([A]_0)) = -kt_"1/2"#
#=> t_"1/2" = (ln(1/2))/(-k) = (ln2)/k#
And as a result,
#color(blue)(t_"1/2") = (ln2)/(3.32 xx 10^(-2) "s"^(-1)) = color(blue)("20.9 s")#
