CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

1. Metallurgical-grade silicon is purified to electronic grade for use in the semiconductor

industry by chemically separating it from impurities. The Si metal reacts with hy-
drogen chloride gas at 300◦ C in different ways to form several polychlorinated silanes
(Each of the three polychlorinated silanes is produced in a separate reaction; hydrogen
gas may or may not be a byproduct of each reaction). Trichlorosilane is liquid at room
temperature and easily separated from the other gases.

Si (A) HSiCl3 (D)

HCl (B) H2 (E)
H2 SiCl2 (C) SiCl4 (F) n4 = ?
xC4 = 0.2142
xE4 = 0.6429
n1 = ? xF 4 = 0.1429
xB1 = 1.0
1 4
3
Reactor Separator
n3 = ?
xC3 = ?
2 xD3 = ? 5
n2 = 100 kg xE3 = ?
xA2 = 1.0 xF 3 = ? n5 = ?
xD5 = 1.0

(a) If 100 kg of silicon is reacted, how much trichlorosilane is produced?

Begin this problem by writing the balanced chemical equations for each of the
reactions taking place. If you can’t figure out the reactions, the reactions are the
same as the last problem in homework set 3!

Si + 2 HCl −−→ H2 SiCl2

H2 SiCl2 + HCl −−→ HSiCl3 + H2

HSiCl3 + HCl −−→ SiCl4 + H2
To save us some writing, let’s denote each distinct species with a different letter:

A + 2 B −−→ C (1)

C + B −−→ D + E (2)
D + B −−→ F + E (3)
As I’ve advised before, when you have multiple units, a good place to start with
balances is an overall system balance. Let’s do a DOF analysis, to see if we
can attack this problem here. There are six unknowns: n1 , n4 , n5 , ξ1 , ξ2 , and
ξ3 (don’t forget that each chemical reaction has an extent of reaction associated
with it, they are unknowns too; the subscripts for each extent correspond to the

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

numbering scheme I’ve used above). We have six distinct species, so we can write
six balance equations, bringing our DOF to 0. We can solve!
The A mass balance is:
0 = xA2 n2 − ξ1 (4)
The B mass balance is:

0 = xB1 n1 − 2ξ1 − ξ2 − ξ3 (5)

The C mass balance is:

0 = −xC4 n4 + ξ1 − ξ2 (6)

The D mass balance is:

0 = −xD5 n5 + ξ2 − ξ3 (7)

The E mass balance is:

0 = −xE4 n4 + ξ2 + ξ3 (8)

The F mass balance is:

0 = −xF 4 n4 + ξ3 (9)
We were given the initial feed of Si in kg, but we wrote these equations on a mole
basis, so we need to convert our starting mass into a molar quantity. Fortunately,
this is simple, just divide using the molar mass!

n2 = 100 kg = 3.57 kmol

Now we can begin to solve our six equations! Start by solving equation (4) for ξ1 :

ξ1 = xA2 n2 = 3.57 kmol

Next, solve equation (6) for ξ2 :

ξ2 = ξ1 − xC4 n4

Solving equation (9) for ξ3 :

ξ3 = xF 4 n4
Substitute these two results into equation (8) and solve for n4 :

0 = −xE4 n4 + ξ1 − xC4 n4 + xF 4 n4

0 = n4 (xF 4 − xE4 − xC4 ) + ξ1

−ξ1
n4 = = 5.00 kmol
xF 4 − xE4 − xC4
Substitute this result into equation (9) and solve for ξ3 :

ξ3 = xF 4 n4 = 0.7145 kmol

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

We can also solve equation (6) for ξ2 :

ξ2 = ξ1 − xC4 n4 = 2.5 kmol

Solving equation (5) for n1 :

2ξ1 + ξ2 + ξ3
n1 = = 10.4 kmol
xB1
Finally, solving equation (7) for n5 :
ξ2 − ξ3
n5 = = 1.79 kmol
xD5

1.79 kmol of trichlorosilane (HSiCl3 ) is produced

(b) What is the yield of trichlorosilane?

The yield is defined as:
moles of desired product actually formed
Yield = ×100%
moles of desired product theoretically formed (no side reactions)
You already calculated the numerator in the previous part, 1.79 kmol. Now, we
have to calculate the denominator.
First, you need to ignore all reactions that CONSUME the desired product, so in
this case, we need to ignore the third reaction.
Now, we need to determine the limiting reagent, because it limits how much
product we could theoretically make. We started with 3.57 kmol of Si and on the
way to solving part (a), we found we also started with 10.4 kmol of HCl. Using
your gen chem skills, you should be able to determine that Si is the limiting
reagent!
Next, it’s just some stoichiometric calculations!
  
1 kmol H2 SiCl2 1 kmol HSiCl3
3.57 kmol Si = 3.57 kmol HSiCl3
1 kmol Si 1 kmol H2 SiCl2
Therefore, the yield is:
1.79
Yield = × 100%
3.57
The yield of HSiCl3 is 50%.

(c) In the reactor, what is the selectivity of trichlorosilane relative to silicon chloride?
The selectivity is defined as:
moles of desired product formed
Selectivity =
moles of undesired product formed

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

In this case, it appears the desired product is trichlorosilane (species D in our

notation), and the undesired product is silicon chloride (species F in our notation).
Be careful, it is asking for the selectivity in the reactor, so we have use variables
around the reactor. Therefore, we can write:
xD3 n3
Selectivity =
x F 3 n3
We never solved for these directly, but we don’t need to! The moles of each
species is the same as the moles leaving the separator, because there is no reaction
occurring within the separator, so:
xD3 n3 xD5 n5
Selectivity = =
xF 3 n 3 xF 4 n 4

The selectivity is 2.5.

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

2. Sulfur dioxide may be converted to SO3 , which is useful in the production of sulfuric
acid. A gas stream of 10 mol% SO2 , 9.0 mol% O2 , and the balance nitrogen gas is
passed through a two-stage reaction process. The fractional conversion of SO2 to SO3
in the single pass through the first reactor is 75%, and in one pass through the second
reactor, the fractional conversion is 65%. To boost the overall conversion of the entire
process to 95%, some of the exit gas from the second reactor is recycled back to the
inlet of the second reactor.
Sketch the block flow diagram for the above process. If 100 gmol/hr is fed to the
first reactor, what is the molar flow rate and composition of this recycle stream? If
all reactions take place at 500◦ C and 1 atm, what is the volumetric flow rate of the
recycle stream? (You may assume the mixture is an ideal gas.)

SO2 (A) SO3 (C)

O2 (B) N2 (D)

ṅ1 = 100 gmol/hr ṅ2 = ? ṅ3 = ? ṅ4 = ? ṅ5 = ?

xA1 = 0.10 xA2 = ? xA3 = ? xA4 = ? xA5 = ?
xB1 = 0.09 xB2 = ? xB3 = ? xB4 = ? xB5 = ?
xD1 = 0.81 xC2 = ? xC3 = ? xC4 = ? xC5 = ?
xD2 = ? xD3 = ? xD4 = ? xD5 = ?
Reactor Reactor
1 2
1 2 3 4 5

ṅ6 = ?
6 xA6 = ?
xB6 = ?
xC6 = ?
xD6 = ?

Start all reaction problems by writing the balanced chemical equation. In this case,
sulfur dioxide is oxidized to sulfur trioxide via the following stoichiometry:

2 SO2 + O2 −−→ 2 SO3

To save me some typing (and you some writing):

2 A + B −−→ 2 C

By now, you should know that a DOF analysis is a good first step. If you are careful,
you will find that the DOF around both the overall system and around reactor 1 is zero.
Let’s start by solving around the overall system. There are six unknowns: ṅ5 , xA5 , xB5 ,

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

xC5 , xD5 , and ξoverall . Since we are told that each reactor and the overall system have
distinct conversions, they must also have different extent of reactions. Remember, each
distinct reactive system must have its own extent! There are six equations we can write,
4 component balances, 1 mole fraction equation, and one conversion specification.
The A mass balance is:

0 = xA1 ṅ1 − xA5 ṅ5 − 2ξoverall (1)

Remember, A is consumed in the reaction, so you must use a negative sign in front of
ξoverall . The “2” comes from the stoichiometric coefficient.
The B mass balance is:

0 = xB1 ṅ1 − xB5 ṅ5 − ξoverall (2)

The C mass balance is:

0 = −xC5 ṅ5 + 2ξoverall (3)
The D mass balance is:
0 = xD1 ṅ1 − xD5 ṅ5 (4)
The mole fraction equation is:

xA5 + xB5 + xC5 + xD5 = 1 (5)

The conversion specification is:

xA5 ṅ5
1− = 0.95
xA1 ṅ1
0.05xA1 ṅ1 = xA5 ṅ5 (6)
First, solve equation (4) for xD5 ṅ5 (this is the molar flow rate of D in stream 5):

xD5 ṅ5 = 81 mol

Next, solve equation (6) for xA5 ṅ5 :

xA5 ṅ5 = 0.5 mol

Then, solve equation (1) for ξoverall :

ξoverall = 4.75 mol

Solving equation (2) for xB5 ṅ5 :

xB5 ṅ5 = 4.25 mol

Solving equation (3) for xC5 ṅ5 :

xC5 ṅ5 = 9.5 mol

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

The total molar flow rate of stream 5 (ṅ5 ) is the sum of the molar flow rate of its
components:
ṅ5 = 95.25 mol/hr
This now allows us to solve for the individual compositions of stream 5. These com-
positions are also exactly equal to those in streams 4 and 6 because these streams are
connected by a splitter.
xA4 = xA5 = xA6 = 0.005
xB4 = xB5 = xB6 = 0.045
xC4 = xC5 = xC6 = 0.100
xD4 = xD5 = xD6 = 0.850
Now, let’s solve around the first reactor (you should be able to prove the DOF = 0
here). There is a different conversion here, compared to the overall system, so this has
its own extent, which I’ll denote ξ1 .
The four component balances are:

0 = xA1 ṅ1 − xA2 ṅ2 − 2ξ1

0 = xB1 ṅ1 − xB2 ṅ2 − ξ1

0 = −xC2 ṅ2 + 2ξ1
0 = xD1 ṅ1 − xD2 ṅ2
The mole fraction equation is:

xA2 + xB2 + xC2 + xD2 = 1

The conversion specification is:

xA2 ṅ2
1− = 0.75
xA1 ṅ1
The algebra is very similar to the overall unit, so I’ll trust you with that. You should
be able to arrive at the following for the unknowns:

ξ1 = 3.75 mol

ṅ2 = 96.25 mol

xA2 = 0.026
xB2 = 0.055
xC2 = 0.078
xD2 = 0.841
Now, we have 3 more systems to analyze: the mixing point, reactor 2, and the splitting
point. However, a quick DOF check around these three systems reveals that DOF = 1

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

everywhere! So what do you do? You could write all the balances for all the remaining
systems and solve simultaneously, but that would be a nightmare. Instead, I suggest
picking one species and writing a balance on it for each remaining system. You should
also write any relevant specifications you may have. Then, see if those equations get
you to your goal.
I will do species A, since we are given an extra piece of information on it, specifically,
the conversion specification in reactor 2.
The A balance at the mixing point is:

0 = xA2 ṅ2 + xA6 ṅ6 − xA3 ṅ3 (7)

Note that there is no need for an extent term here because reaction does not occur at
the mixing point!
The A balance in reactor 2 is:

0 = xA3 ṅ3 − xA4 ṅ4 − 2ξ2 (8)

The conversion specification as it pertains to reactor 2 is:

0.35xA3 ṅ3 = xA4 ṅ4 (9)

Finally, the A balance at the splitting point is:

0 = xA4 ṅ4 − xA5 ṅ5 − xA6 ṅ6

which reduces to the following because the compositions around the splitter are all the
same:
0 = ṅ4 − ṅ5 − ṅ6 (10)
If you examine these four equations, there are 5 unknowns: xA3 , ṅ3 , ṅ4 , ξ2 , and ṅ6 .
We can’t directly solve this, so we can play a little algebra trick. The quantity xA3 ṅ3
comes up quite a bit, and it contains quantities we don’t exactly need right now. We
are striving for n6 ! Let’s replace xA3 ṅ3 with the placeholder term Y . So equations
(7-10) become:
0 = xA2 ṅ2 + xA6 ṅ6 − Y
0 = Y − xA4 ṅ4 − 2ξ2
0 = ṅ4 − ṅ5 − ṅ6
0.35Y = xA4 ṅ4
Now, we only have 4 unknowns to solve for: Y , ṅ4 , ξ2 , and ṅ6 . We also have 4 equations,
so we can do this!
Solving for Y by combining the first, third, and fourth equations:

Y = 3.08 mol

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CHEG 112 - Introduction to Chemical Engineering Homework 8 Solutions

Then, solving for n4 :

n4 = 205 mol/hr
Which gives us the recycle flow rate:

n6 = 109.75 mol/hr

Rearranging the ideal gas law for the volumetric flow rate:
ṅRT
V̇ =
P
Substituting in the correct constants and the total molar flow rate, the volumetric flow
rate of the recycle stream is:
V̇ = 6970 L/hr