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?

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")`