# 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

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 thefrequency factor, i.e. it is proportional to the number of collisions occurring over time.#E_a# is theactivation energyin#"kJ/mol"# .#R = "0.008314472 kJ/mol"cdot"K"# is theuniversal gas constant. Make sure you get the units correct on this!#T# is thetemperaturein#"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

#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 = 25 + 273.15 = "298.15 K"#

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

Finally, plug in and solve. We should recall that

#=> 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

The larger the right hand side gets, the larger **Therefore, higher temperatures means larger rate constants.**

Furthermore, the rate constant is proportional to the rate of reaction

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