Raw milk sours in about 4 hours at #28°C# but in about 48 hours at #5°C#. What is the activation energy (in #(kJ)/(mol)#) for the souring of milk?

1 Answer
Truong-Son N.
Sep 18, 2017

I got an activation energy of around #"75.25 kJ/mol"#. This is reasonable, as ordinary reactions' activation energies are usually around #"25 - 100 kJ/mol"#.

The activation energy is essentially independent of temperature, and typical reactions follow Arrhenius behavior:

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

where:

  • #k# is the rate constant at temperature #T#.
  • #A# is the frequency factor, which is implicitly temperature-dependent, but we assume it is not, within this small temperature range.
  • #E_a# is the activation energy for the process.

Thus, we define two trials, one with temperature #T_1# and one with temperature #T_2#, having the same activation energy and frequency factor:

#k_1 = Ae^(-E_a//RT_1)#

#k_2 = Ae^(-E_a//RT_2)#

Therefore:

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

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

#= e^(-E_a/R[1/T_2 - 1/T_1])#

The activation energy can then be gotten once we know the ratio of the rate constants:

#ln(k_2/k_1) = -E_a/R[1/T_2 - 1/T_1]#

#-># This is known as the natural log form of the Arrhenius equation. Solving for the activation energy gives:

#E_a = -Rln(k_2/k_1) cdot [1/T_2 - 1/T_1]^(-1)#

Now the problem is, we don't know what the order of this reaction is. The main reaction that occurs is the conversion of lactose to lactic acid:

#"C"_12"H"_22"O"_11(aq) + "H"_2"O"(l) stackrel("multiple steps"" ")(->) 4 "C"_3"H"_6"O"_3(aq)#

I'm not going to go too into the specifics, but it seems that it is effectively modeled as a first-order process according to this paper, and that's what we need to know.

A first-order process has a half-life (which is derived in your General chemistry textbook) of:

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

And so, the rate constant is proportional to the half-life, which is also temperature-dependent:

#k_1 = (ln2)/t_("1/2",(1))#

#k_2 = (ln2)/t_("1/2",(2))#

Thus, #k_2/k_1 = t_("1/2",(1))/(t_("1/2",(2))#. Lastly, we choose our variables:

  • #T_1 = 28^@ "C"# #-># Convert to #"K"#!
  • #T_2 = 5^@ "C"# #-># Convert to #"K"#!
  • #t_("1/2",(1))# is for the process at #T_1#
  • #t_("1/2",(2))# is for the process at #T_2#

So, the activation energy is around:

#color(blue)(E_a) = -Rln(t_("1/2",(1))/(t_("1/2",(2)))) cdot [1/T_2 - 1/T_1]^(-1)#

#= -("0.008314472 kJ/mol"cdot"K")ln("4 hours"/"48 hours") cdot [1/(5^@ "C" + "273.15 K") - 1/(28^@ "C" + "273.15 K")]^(-1)#

#=# #color(blue)ul("75.25 kJ/mol")#

Related questions
Impact of this question
15166 views around the world
You can reuse this answer
Creative Commons License