Question

How do you determine the temperature at which the reaction undergoes a thermal runaway?

A heat sink is removing heat from a `"1-L"` reaction vessel closed to mass transfer (but not energy transfer) at a rate of `"400 kJ/min"`, and the enthalpy of this zero-order reaction is `"540 kJ/mol"`.

Given the following data:

`ul(T(""^@ "C")" "" "k" "" "" "" ")`
`25" "" "1.286 xx 10^(-3)`
`35" "" "1.984 xx 10^(-3)`
`45" "" "2.792 xx 10^(-3)`
`55" "" "4.109 xx 10^(-3)`

and knowing that a thermal runaway is when the reaction gets out of control and too much heat is absorbed (that's what the heat sink tries to prevent!)...

a) Find the maximum initial rate at which the thermal runaway occurs.
b) Find the rate constant at the maximum temperature at which the reaction undergoes a thermal runaway.
c) Determine the maximum temperature at which the reaction undergoes a thermal runaway (which is bad! It means it gets overly hot!).

Answer 1

The hints are very helpful, actually. If you know how the units are going to work out, this is just a fancy way of using the Arrhenius equation.

I got `90.01^@ "C"` using the hottest tabled temperature for `T_1`, and `90.49^@ "C"` using the coldest tabled temperature for `T_1` in the Arrhenius equation.

So, a good average maximum temperature to one sig fig (from your rate of heat removal) is `color(blue)(90^@ "C")`.

---

DISCLAIMER: LONG ANSWER!

Basically, the idea is that if you have an exothermic reaction in a mechanically-closed system (constant number of system particles, but not insulated), heat will be released from the system into the surroundings (heat sink).

Your heat sink (which could just be a huge water bath let's say), will allow excess heat to flow into it (because the temperature gradient will direct heat towards colder regions), and is supposed to establish a maximum reaction temperature at thermal equilibrium.

Fortunately, you are already given the rate of heat removal.

1. If rate of heat production `>=` the rate of heat removal, then thermal runaway occurs...
2. As the temperature of the heat sink increases, higher temperatures are allowed in the system, increasing the likelihood of a thermal runaway due to higher reaction rates reached (almost without limit).

Therefore, we do NOT want the rate of heat removal to be higher than `"400 kJ/min"`, which we assume is the maximum rate the heat sink can manage.

`1)`

Knowing the enthalpy of reaction (endothermic with respect to the heat sink), we can calculate the maximum rate of reaction in `"M/s"` we can achieve, above which the thermal runaway occurs:

`(400 cancel"kJ")/cancel"min" xx "1 mol"/(540 cancel"kJ") xx 1/"1.00 L tank" xx cancel"1 min"/"60 s"`

`=` `"0.0123 M/s"`

(haha, the number is `0.0123456789cdots` What a sense of humor your instructor has...)

`2)` We are told the reaction is zero-order, so this requires no calculations. The rate law is therefore:

`r(t) = kcancel([A]^0)^(1) = k`

And so, the rate you JUST calculated IS the rate constant at this maximum temperature:

`color(blue)(k = "0.0123 M/s")`

`3)` And therefore, since we know the rate constant at the maximum temperature, we must solve for this temperature using an Arrhenius plot based on the table given to us.

However, we don't know the activation energy yet, so the useful form of the equation right now is:

`ln stackrel("rate constant(s)")overbrace(k) = overbrace(-E_a/R)^"slope" 1/T + overbrace(ln beta)^"y-int."`

In Excel, we graph `lnk` vs. `1"/"T` to obtain the activation energy (remember to put temperature in `"K"`...), the quantity that does not change with the other form of the Arrhenius equation we'll need (with two temperatures).

Since the slope is `-3744.6`, we have that

`E_a = -Rcdot"slope"`

`= -"8.314472 J/mol"cdot"K" cdot -"3744.6 K" = "31134.4 J/mol"`,

which is physically reasonable (should be around 10 - 100 kJ/mol for common reactions).

Finally, solve for the maximum temperature, `T_2`! The other form of the Arrhenius equation we now need is:

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

which we can now use `E_a` for! I used the hottest tabled temperature provided just in case the maximum temperature is sensitive to the temperature range.

`ln("0.0123 M/s"/"0.004109 M/s") = -"31134.4 J/mol"/("8.314472 J/mol"cdot"K")[1/T_2 - 1/"328.15 K"]`

`ln(3.005) = -"3744.6 K"[1/T_2 - 1/"328.15 K"]`

`ln(3.005)/(-"3744.6 K") + 1/"328.15 K" = 1/T_2`

Therefore, the maximum temperature before thermal runaway occurs is:

`color(blue)(T_2) = [ln(3.005)/(-"3744.6 K") + 1/"328.15 K"]^(-1)`

`=` `"363.16 K"`

`=` `color(blue)(90.01^@ "C")`

And if you chose the coldest tabled temperature for `T_1`, then `T_2 = 90.49^@ "C"`.

Huh, can't even boil water in this reaction. Well then.