A certain first-order reaction has a rate of #"0.04 M/s"# after 10 seconds and #"0.03 M/s"# after 20 seconds. What is the half-life for this reaction?
1 Answer
#t_"1/2" = "24.1 s"# .
Interesting question. The rates
#r(10) = k[A]_(10) = "0.04 M"cdot"s"^(-1)#
#r(20) = k[A]_(20) = "0.03 M"cdot"s"^(-1)# where the subscript in front of
#[A]# indicates the time in seconds. The rate constant#k# is the same for the same reaction at the same temperature.
Each would follow the same integrated rate law for first-order one-reactant reactions:
#bb(ln[A] = -kt + ln[A]_0)# where
#[A]_0# is the initial concentration of#A# .
Since the rate constants
#([A]_20)/([A]_10) = (r_20)/(r_10)#
So, using the integrated rate law for each time:
#ln[A]_10 = -10k + ln[A]_0#
#=> ln[A]_10 + 10k = ln[A]_0#
#ln[A]_20 = -20k + ln[A]_0#
#=> ln[A]_20 + 20k = ln[A]_0#
Since
#ln[A]_20 + 20k = ln[A]_10 + 10k#
#ln[A]_20 - ln[A]_10 = ln(([A]_20)/([A]_10)) = 10k - 20k#
Plug in the ratio of the rates:
#ln((r_20)/(r_10)) = -10k#
Lastly, the half-life
#bb(t_"1/2" = (ln2)/k)# ,
so
#ln((r_20)/(r_10)) = -(10ln2)/t_"1/2"#
#=> color(blue)(t_"1/2") = -(10ln2)/(ln(r_20//r_10)) = -(10ln2)/(ln(0.03//0.04))#
#=# #ulcolor(blue)("24.1 s")#