We're going to calculate the adiabatic reactor temperature for a reversible, liquid phase
reaction. So the idea is that we have some type of reactor, now it turns out because
we're calculating the adiabatic temperature it doesn't matter whether this is a plug flow
reactor or a continuous stirred tank reactor or even a closed system batch reactor, this
case we're assuming we're feeding in pure A at 300 Kelvin. We have some flow rate, what
that is doesn't matter for this. Adiabatic means Q is zero. Now we're interested in the
equilibrium mixture leaving, so we're going to run it to equilibrium and we want to know
what is the final temperature leaving the reactor.
Now we're going to make some assumptions
in order to demonstrate the idea. We're going to assume ideal solution and what that means
is then we can write down the equilibrium constant in terms of mole fractions or concentrations,
so concentration of B at equilibrium, concentration of A at equilibrium, because the reaction
is just A goes to B. Since we're interested in conversion we can write this equilibrium
constant in terms of conversion. Concentration of A would be our initial concentration that
we're feeding in times one minus our equilibrium conversion. We convert 40%, 60% remains, the
concentration of B, since we only feed A into the reactor is what fraction of A is converted
since the stoichiometry is one to one. So we're going to use this to relate fractional
conversion to equilibrium constant. We're given the equilibrium constant at one temperature,
298 Kelvin, and we're given the heat of reactions so we can calculate the equilibrium constant
at other temperatures. So what we're going to do is use the Van't Hoff equation, and
this relates equilibrium constant at one temperature to the equilibrium constant at another temperature.
So the log of K2 is at some temperature we don't know yet. K1 will be our initial, where
we have information, 298 Kelvin. This is related to the heat of reaction, divided by the gas
constant, and then one over temperature 2, one over Temperature 1, where of course temperature
must be absolute temperature. So I'm going to substitute the values we know into this
equation. So I've substituted in, what I want to solve for actually is K2 so I want to take
the exponent of both sides, because then I can solve this for the extent of reaction
and then calculate that for different values for K which corresponds of course to different
temperatures. So let me first solve for K2 as a function of temperature. Keep in mind
this equation for K, Ca0 cancels, if I solve this for equilibrium extent of reaction in
terms of the equilibrium constant it just becomes this, where remember the equilibrium
constant is a function of temperature as indicated here, so this is K at any temperature, written
for temperature T2. The other thing we need of course to finish solving the problem is
the adiabatic reactor energy balance. So let me write that down. The energy balance, we
have molar flow rate coming in to the reactor, we have heat capacity of reactant, we want
to solve at equilibrium conditions, feed temperature, so this is the energy if you like to heat
up the feed to the final temperature, since the heat capacities of A and B are the same
then I could just write everything in terms of heat capacity of A, and the reason of course
the temperature increases is I have the heat of reaction and then some fraction of that
material A flowing in is converted. This term here is moles of A reacted per time, of course
Fa0 is going to cancel out of this equation and I'm going to end up with the equation
for the equilibrium conversion in terms of the heat capacity of A equilibrium temperature,
T0 is my feed temperature and then minus the heat of reaction. So the equilibrium conversion
is just linear in temperature, I can substitute in the numbers. So now I'm about finished,
I have this relationship here, so this is one equation. My second equation is this one,
relating equilibrium conversion to equilibrium constant, and then I'm going to substitute
this equilibrium constant into here so it is a function of temperature also. We look
at what these equations look like, this is an exothermic reaction. As I raise the temperature
the equilibrium conversion goes down, very low temperatures it goes to one, and then
it decreases as I raise the temperature. That's what equation number two is going to show.
Equation number one is just a straight line, the energy balance. so this is the energy
balance and this is the equilibrium relationship. This is the adiabatic equilibrium conversion,
so here's the equilibrium conversion, here is the adiabatic temperature. If I solve these
equations, I won't go through the details here. Equilibrium adiabatic temperature 460
K, this corresponds to an equilibrium conversion of 40%. So that's just from solving the equations
simultaneously, lots of ways to do it but certainly also just plot them out in Excel
and read off the values graphically. As reaction proceeds and temperature increases of course
our equilibrium conversion decreases. If we run long enough to get to equilibrium this
is what temperature we'd reach and what equilibrium conversion we would reach.