in this problem we want to design a
packed bed reactor for a gas-phase reaction
hydrogenating acetylene to ethylene and we are given reaction data,
those obtained in a differential reactor. So a
differential reactor means that the conversion is small
so that we can assume the concentrations
are essentially uniform throughout the reactor bed and this makes analysis
the rate data easier. The reactor has 25 grams of
catalyst and we're interested in what's the ordered the reaction with
respect to each of the reactants and then interested in determining the rate constant.
Notice that we only need to determine the orders of reaction to the nearest
half order, and the idea is to simplify the calculations so you can see the
procedure, and then we want to use that rate law to
determine how much catalyst we we need for 70 percent conversion, so
no longer a differential reactor, much bigger reactor we are processing
a thousand liters per minute at 10 bar in four hundred
Kelvin at the same temperature.
What we are gonna do first is determine the
order of reaction from the rate data given here,
so we'll determine alpha first. So we will look at the data where the pressure
of A
we have three values for pressure of A
partial pressure of hydrogen is kept constant, and notice these three rates are
almost the same, namely this reaction is
zero-order in the pressure of A,
acetylene. And we are going to do the same thing now for
pressure of hydrogen, so we are going look at data
though it should not matter once we determine it's zero order but let's look at the data where
the pressure if A is constant and the pressure B changes
and just by inspection you notice as we go from
2 to 4 bar, the
rate doubles. We go from two to six bar,
it triples, so beta is approximately
one, first-order. We can use now these results to determine
the rate constant. So we look at the
mass balance for a plug flow reactor to change
in the flow rate of the product
since that's where we're measuring, the flow rate,
is going to be plus a rate constant,
pressure pf hydrogen to the first power
and independent of the pressure of acetylene.
And what we're going to do is assume that
we can make the approximation because it's a differential reactor,
that these derivatives can be
determined from difference values, so finite
rather than infinitesimal and so let's
pick conditions to use these values. Let's use the
case where the rate's the highest, so the pressure
of A is 2 bar and the pressure
of hygroen is six bar.
Notice now the rate is 0.0182.
And so we're going to substitute in,
because this is the change in the flow rate of
our product, there's no product in our feed,
so 0.0182
and we can calculate this
approximation to a derivative and now this is equal to our rate constant
times the pressure of hydrogen.
Now, the pressure of hydrogen where we made this measurement
is six bar, so that means we can solve this equation
for k, so this term of
course is equal to this term. Now notice the
the units of the rate constant, to be consistent with the rate expression in
terms of grams of catalyst
and bar and mole because our final
flow rate calculated from our mass balance is moles per minute.
Now we are going to use this rate constant to
model the larger reactor. For the larger reactor we're going to write a mass balance
for hydrogen in terms of rate constant
and the pressure of hydrogen. This is a minus sign, since
hydrogen is reacting. I'll write it in terms of the mole fraction
and the total pressure of hydrogen. So first I need to determine the
mole fraction of hydrogen. I am going to write this in terms of flow rates,
so it's the flow rate of hydrogen over the total flow rate, which is
hydrogen, flow rate acetylene, the
flow rate of ethylene. So we have the reaction
A plus hydrogen goes to ethylene,
and we can write flow rate of hydrogen
equals the flow rate of acetylene because
entering the reactor, the inlet
flow rate of hydgrogen is equal to the
inlet flow rate of acetylene and the stoichiometry is 1
to 1. Since no ethylene
enters the reactor, I can write the flow rate of
ethylene in terms of the flow rate of hydrogen, so the inlet
flow rate of hydrogen, molar flow rate minus the flow rate of hydrogen
any given point, that difference is the flow rate of
ethylene. So I am going to substitute these values into my equation
for the mole fraction of hydrogen and so the mole fraction of hydrogen
then can be written just in terms of molar flow rate of hydrogen,
and the inlet molar flow rate of hydrogen. So I am going to substitute
now back into my mass balance for the
mole fraction of hydrogen and then I can solve this
differential equation analytically so to
solve this equation I'm going to separate variables and put
to terms of the flow rate of hydrogen on the left side.
So I end up with this then I am going to integrate from the inlet of the reactor,
W equals 0, up to the any point in the reactor,
the weight of the catalyst, and then integrate from the
flow rate of hydrogen at the inlet
to the flow rate of hydrogen at any point. So I'm going to
just rearrange this and then I will integrate,
of course the right side because k is a constant
and pressure's a constant I can integrate that directly.
I can integrate the left side which is the flow rate of hydrogen
evaluated between the two limits, and then
inlet flow rate of hydrogen times the
log of the flow rate of hydrogen evaluated between the two limits.
Now since the conversion is seventy percent, that means
the outlet flow rate of hydrogen is only
30 percent of the inlet flow rate, 70 percent
reacted, 30 percent is unreacted, I can make this substitution
in my more balance equation.
So simplify the equation, we know the rate constant and the pressure.
What we need is the inlet molar
flow rate of hydrogen, so the inlet molar flow rate of hydrogen
is one 1/2 of the total
inlet flow rate, and so what we need now is the
total inlet flow rate, and we can get that from the ideal gas law
pressure, the volumetric flow rate over RT
you can substitute the values in so total flow rate
is three hundred and then the flow rate of hydrogen entering
is 1/2 of 300 or 150.
So now we just need to substitute the values in to
this equation and we can solve for W.
So 236 kilograms which
means probably 240 kilograms in terms of significant figures,
this is the amount of catalyst to get seventy percent conversion
for the inlet flow rates we're given in the larger reactor