# 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 **rate law**:

#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 thetimein seconds. Therate constant#k# isthe samefor 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 *same*, we can write a **ratio of the concentrations** and relate that to the ratio of the *rates*:

#([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** *first-order* reaction is given by

#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")#