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

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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!).

A heat sink is removing heat from a

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!).

##### 1 Answer

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

So, a good **average maximum temperature** to one sig fig (from your rate of heat removal) is

**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.

- If rate of heat production
#>=# the rate of heat removal, then thermal runaway occurs... - 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

Knowing the **enthalpy of reaction** (endothermic with respect to the heat sink), we can calculate the ** maximum rate of reaction** in

*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...)

#r(t) = kcancel([A]^0)^(1) = k# And so, the rate you JUST calculated IS the

rate constantat this maximum temperature:

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

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 **activation energy** (remember to put temperature in

Since the slope is

#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,

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

which we can now use

#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

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