This is a problem where we're interested in the concept of energy balance on a chemical
reactor, so we have an adiabatic reactor, a simple A goes to B reaction and the conversion
is 50 percent in the reactor, and the notation we're using H is enthalpy and this is enthalpy
per mole. So the question is which one of these answers is the correct energy balance
for this adiabatic reactor.
So because this is a steady state reactor then any derivative
with respect to time in the energy balance is zero, there's no work in the reactor so
it's just a plug flow reactor, there's no heat transfer because it's adiabatic so the
only terms left are the energy flow in in the form of enthalpy and the energy flow out
so our energy balance really reduces to the enthalpy in equals the enthalpy out and to
be general we would talk about the total enthalpy but then we can take a basis of one mole of
A entering the reactor and the enthalpy into this reactor then would just be the enthalpy
of A per mole and at the inlet conditions, whatever temperature it's coming into the
reactor, the enthalpy leaving then, since it's 50 percent conversion we have a half
a mole of A leaving, but it's at the outlet conditions, different temperature, and we
have a half of a mole of B leaving again at the outlet conditions, this is then, because
it's 50 percent conversion, this is the complete energy balance for this system, enthalpy of
A in is equal to the enthalpy of A out at the outlet conditions times point five plus
the enthalpy of B out at outlet conditions and this is kilojoules per mole for example
times point five, so this means the correct answer is D for this system, and the question
that arises is how did we account for the heat of reaction in this type of energy balance
and the easiest way to understand that is to look at some more specific numbers so let's
suppose that A comes in at 50 degrees C and let's assume that A and B leave at 300 degrees
C, so that happens to be the adiabatic temperature we'd get from the reaction and so the enthalpy
of A in is the enthalpy of A at 50 degrees C, enthalpy of A out is the enthalpy of A
at 300 degrees C which we could write as the enthalpy of A at 50 degrees C plus the heat
capacity of A times the temperature difference, 300 minus 50, and we could do the same thing
for B, so I've written this for B also, and then we're just going to substitute back in
to the first law for this specific example, enthalpy of A at 50 degrees C is what's entering
our reactor and then we have point five enthalpy of A at 50 plus point five heat capacity of
A times this difference and then similar terms for enthalpy of B at 50 degrees C plus point
five heat capacity of B again multiplied by 250, so we can simplify this and we're going
to do two things we're going to combine these two heat capacity terms because otherwise
they're similar and we're going to bring this to the other side because we can see we're
going to simplify, we bring this to the other side it because minus H A at 50 degrees C
and we have here plus a half of H A so left side then is zero, this becomes minus point
five times H A at 50 degrees C and then we're going to add enthalpy of B at 50 degrees C
and then we're going to combine these two terms, point 5 times 250, that that's 125
heat capacity of A plus the heat capacity of B, so I've just factored out this point
5 and the important thing to notice is this term enthalpy of B at 50 minus enthalpy of
A is just the heat of reaction at 50 degrees C so the energy balance that might look more
familiar is the heat of reaction at 50 degrees C and then the heat capacity term, but notice
this follows directly when we start with the first law and just do an enthalpy balance,
we don't have to introduce a separate term in this expression anywhere for heat of reaction
for this reaction, instead heat of reaction drops out naturally when we do the enthalpy
balance, heat of reaction per mole since only half a mole reacts then the point five term
ends up here.