In this example we calculate the adiabatic temperature for a reactor where we have 30%
conversions. So the idea is that we're feeding in this case 1 mol of NO, 1/2 mol of oxygen at
100 degrees C and what we want is what's the adiabatic temperature if 30% of the NO is
converted, with this being the reaction that we're interested in. So we've already calculated
the heat of reaction from heats of formation at 25 C and we're given simplified forms of
the heat capacity just to make the calculations a little easier to work through and less numbers
to deal with. So they have a temperature dependence as indicated here where the temperature in
these equations is in Kelvin.
So the first thing we do of course is to make
a diagram of the process and so the system is this adiabatic reactor, we've labeled the
feed conditions. We have NO, O_2 and NO_2 all leaving since we only have 30% conversion
and the unknown is the final temperature. So we want to setup... what's the approach
we use to solve this type problem. So what I'm going to do is draw a representation to
help visualize the solution. We have 1 mol of NO, 1/2 mol of O_2. It's at 373 K,
so I've converted that to absolute temperature, and we're going to end up with, if we're converting
30% than 70% of the NO is unreacted and likewise since the stoichiometry is, in this case,
1 mol of NO reacts with a 1/2 mol of O_2, then we have 0.35 mol of O_2. Namely when
3/10 of mol of NO reacts half of that, 0.15 mol of O_2 reacts. So 0.5 - 0.15
gives us the moles of O_2 and then we form 3/10 of a mole of NO_2. The temperature
is what we don't know and what we're going to do to solve this is just use the idea of
state functions. So what we're going to calculate first is if we cool down this feed to 298 K,
namely 25 degrees C. So if we cool down, there's going to be some enthalpy change,
delta H_1 for that process. Then we'll carry out reaction at 25 C and we're doing that
because we know the heat of reaction at 25 C. So delta H_2 is carrying out the reaction.
So we've listed the products now of reaction for 30% conversion, remember this is
30% conversion, and then we're going to take these products and heat them up to the final
temperature, we'll call that delta H_3 and our actual process we'll call this delta H_4
and delta H_4 is equal to 0 because it's adiabatic and so the state function says delta H_1 plus
delta H_2 plus delta H_3 equals the enthalpy change 4 which is equal to 0. So let's look at delta
H_1. We're going to look at the heat capacities, temperature change. So we're going to start
at 373 and we're cooling down to 298 and we have 1 mol times the heat capacity of NO
plus 1/2 mol times the heat capacity of O_2 and integrating that it gives us the enthalpy
change going from 373 to 298. So we were given the heat capacities and notice these were heat
capacities divided by the gas constant. So I've substituted in the heat capacities for
NO and for O_2 with the 1/2 mol of O_2 and then the gas constant is here times dT. So
we can simplify this and do the integration. So all I've done is add up the first term,
do the multiplication, bring the R outside. We'll do the calculation in J/mol and
then we've done an integration here with these limits so we need to substitute in the limits.
So this T final - T initial, 298 - 373, is - 75 and then we have these
values squared. When we do the calculations we end up with delta H_1 is - 3.42 kJ. Negative
because we're taking energy out to cool... Now the next term, delta H_2, is due to the
reaction. It's 30% conversion so we have 30% of the heat of reaction, that remember we
have at 25 degree C so this is - 57 kJ/mol and we're reacting 3/10 of a mol.
So this is - 17.1 kJ. Then the third term, delta H_3, is very similar to delta
H_1 with just more species so we won't go through in as much detail. So now we're integrating
from 298 up to 373. So this is going to be a positive term. Heat capacity of NO, 7/10
of a mol. Heat capacity of oxygen, 3/10, 0.35 mol and then 3/10 of a mol of NO_2
and its heat capacity. So the only difference now in doing this integration. It's not to
373 like previously but it's to our final temperature and we don't know the final temperature.
So up here again this should have been the final temperature and so we're going to end
up with an expression after we do an integration that still has the final temperature, the
adiabatic temperature, we're interested in and that's squared here divide by a 1,000
to convert it from joules to kilojoules so we can use it with the other terms and then
what we're going to do is say delta H_1 plus delta H_2 plus delta H_3 should add up to
0. So substituting in the values for delta H_1, delta H_2, delta H_3 we calculate
and just adding them up together we get this equation which, quadratic equation, we can
solve for the final temperature and so the adiabatic temperature is around 720 K for
this problem where we have 30% conversion. It's certainly higher than our starting temperature
as we would expect for an exothermic reaction and what we've done is take advantage, remember
of state function, to calculate the temperature here because adiabatic means delta H is 0
and we've just added up these three delta H terms in this calculation.