What is the half-life of a first-order reaction at #40^@ "C"# if its half-life at #25^@ "C"# was #"400 s"# and the activation energy is #"80 kJ/mol"#?

1 Answer
Jan 28, 2017

#t_"1/2"^((2)) = "85.25 s"#


Notice how you're given the half-life (for one temperature), a second temperature, and the activation energy. The key to doing this problem is recognizing that:

  • the half-life for a first-order reaction is related to its rate constant.
  • the rate constant changes at different temperatures.

Go here for a derivation of the half-life of a first-order reaction. You should find that:

#t_"1/2" = ln2/k#

Therefore, if we label each rate constant, we have:

#k_1 = ln2/(t_"1/2"^((1)))##" "" "" "##k_2 = ln2/(t_"1/2"^((2)))#

Recall that the activation energy can be found in the Arrhenius equation:

#bb(k = Ae^(-E_a"/"RT))#

where:

  • #A# is the frequency factor, i.e. it is proportional to the number of collisions occurring over time.
  • #E_a# is the activation energy in #"kJ/mol"#.
  • #R = "0.008314472 kJ/mol"cdot"K"# is the universal gas constant. Make sure you get the units correct on this!
  • #T# is the temperature in #"K"# (not #""^@ "C"#).

Now, we can derive the Arrhenius equation in its two-point form. Given:

#k_2 = Ae^(-E_a"/"RT_2)##" "" "" "##k_1 = Ae^(-E_a"/"RT_1)#

we can divide these:

#k_2/k_1 = e^(-E_a"/"RT_2)/e^(-E_a"/"RT_1)#

Take the #ln# of both sides:

#color(green)(ln(k_2/k_1)) = ln(e^(-E_a"/"RT_2)/e^(-E_a"/"RT_1))#

#= ln(e^(-E_a"/"RT_2)) - ln(e^(-E_a"/"RT_1))#

#= -E_a/(RT_2) - (-E_a/(RT_1))#

#= color(green)(-E_a/R[1/(T_2) - 1/(T_1)])#

Now if we plug in the rate constants in terms of the half-lives, we have:

#ln((cancel(ln2)"/"t_"1/2"^((2)))/(cancel(ln2)"/"t_"1/2"^((1)))) = -E_a/R[1/(T_2) - 1/(T_1)]#

This gives us a new expression relating the half-lives to the temperature:

#=> color(green)(ln((t_"1/2"^((1)))/(t_"1/2"^((2)))) = -E_a/R[1/(T_2) - 1/(T_1)])#

Now, we can solve for the new half-life, #t_"1/2"^((2))#, at the new temperature, #40^@ "C"#. First, convert the temperatures to #"K"#:

#T_1 = 25 + 273.15 = "298.15 K"#

#T_2 = 40 + 273.15 = "313.15 K"#

Finally, plug in and solve. We should recall that #ln(a/b) = -ln(b/a)#, so the negative cancels out if we flip the #ln# argument.

#=> ln((t_"1/2"^((2)))/(t_"1/2"^((1)))) = E_a/R[1/(T_2) - 1/(T_1)]#

#=> ln((t_"1/2"^((2)))/("400 s")) = ("80 kJ/mol")/("0.008314472 kJ/mol"cdot"K")[1/("313.15 K") - 1/("298.15 K")]#

#= ("9621.78 K") (-1.607xx10^(-4) "K"^(-1))#

#= -1.546#

Now, exponentiate both sides to get:

#(t_"1/2"^((2)))/("400 s") = e^(-1.546)#

#=> color(blue)(t_"1/2"^((2))) = ("400 s")(e^(-1.546))#

#=# #color(blue)("85.25 s")#

This should make sense, physically. From the Arrhenius equation, the higher #T_2# is, the more negative the #[1/T_2 - 1/T_1]# term, which means the larger the right hand side of the equation is.

The larger the right hand side gets, the larger #k_2# is, relative to #k_1# (i.e. if #ln(k_2/k_1)# is very large, #k_2 ">>" k_1#). Therefore, higher temperatures means larger rate constants.

Furthermore, the rate constant is proportional to the rate of reaction #r(t)# in the rate law. Therefore...

The higher the rate constant, the faster the reaction, and thus the shorter its half-life should be.