Michigan State University
DEPARTMENT OF CHEMICAL ENGINEERING AND MATERIALS SCIENCE
CHE 201: Material and Energy Balances, Section 001
Fall, 2021
Written Homework Problem Set 4; 115 points
Due Date: Friday, October 1st by 5:00 PM EST on Crowdmark
1. (40 pts.) Ethanol is produced by the hydration (addition of water) of ethylene, and some of the
product is converted to diethyl ether in a side reaction. A feed to a reactor contains 2.5 times as
much steam as ethylene.
a. (10 pts.) Write out the correctly balanced chemical equations for the two reactions
occurring in the reactor.
b. (30 pts.) If the fractional conversion of ethylene is 0.65 and the yield is 30%, find the
composition of the product gas. (*Hint: you will need to choose an initial feed basis)
Use an atomic species approach when solving this problem to receive full credit.
A)
Cztty
2CzHs
5- pts
for
,
nie
not
moi
cette
tho
OH
→
GHSOH
( cztts)z0
each reaction
b)
hi
→
Hzo
t
Hzo
t
( properly
☒
balanced)
-
rig
ri ,
Mol
Cztty
not the
cztls-OHrismollGHs-koweknowiriz-2.tn#
n'
5
pts
1
diagram
5hr01
/ setup
laltomic species)
DOF
#Unkn_
ato-zmspe.ie
#
-
6
Cñ
,
,
the
we
②
's ,ñ•)
Choose
100
+
Ñz=
hi
=
+2.5
=
,
yield ,
molls
}
)
spts
=0
.
can
=
also
10.0
use
:
.
a
basis
Choosingusing
and
'
65
molls
yield
N
proyl
e
hhfprl
h
t
stat
appropr:#
Mfo
71.4m¥
fc2.eu
things
)
not feed
rii-m.tn#--28r6zg-.rishis
we
,
Ñi
Ñr=
use
can
,
2. 5Ñ
28.6
therefore
0.65
,
Ñz
few
a
Goo
basis
Ñ,
=
for
info
our
a
vi. =
we
,
ñz= 2.5N ,
use
can
since
③
C§-qµ
n
100m¥
②
3
(c) HD )
rising ,
>
#otheo
-
}
fracti
o
nal
cow
of
✓
g- pts
30%
→
←
w§,¥¥¥y
yield
-0.3
2 all
✗
Ctu
limiting
0.3
can
C?
§y¥y§µ
getup
of
( 28.6)
2
=
his
(
2
up
balances :
17.74) +4 rig
2
Kitts)z0
zany ,
+
1015.43)
SP
"
25.8m¥
molls
m¥
=
is
%,
+
5.43
=
.
maximum
so
atomic
10.0)
rxtntcons
product
7.74
yyg.gg , yay,
y , , , .o,
t
5
,
(7.7-4)
"
Ñy
goneYanks
atomy
set
now
↳
,
n'
=
A. g- =
g- pts
limiting
reactant ,
desired
of
amount
Therefore:
we
nmgt.saesdfsedrpdrod.PT?'nfYFardx-n@
=
is
vis
Now that
calculate
Yczthe
=
his
=
-
621m04s tho
rig
are
known ,
composition of
ñztÑ÷-ñstn•
=
we
can
the exit stream :
t.tt#z7.-uTs-.y3 =°
Ipt
62.1
Ipt
Ytho-ioF.tt#l+s-.Es ¥3
Yarkon
Davis)zo
=
?I?¥
,
=
1-
,
-
Ipt
nuts .u5
0.12-0.73-0.09
=
0.0-6
t.pt
2. (40 pts) In the Wacker process, ethylene (C2H4) is oxidized by oxygen gas (O2) to produce
acetaldehyde (C2H4O). However, acetic acid is also produced (C2H4O2) as a side product. The
selectivity for acetaldehyde over acetic acid with the reactor you are using is 12, both for overall
and single-pass. The product stream is then fed to one condenser that removes all of the acetic
acid, followed by another condenser that removes all of the acetaldehyde. The remaining
components comprise a vapor stream, 10.0% of which is purged to the outside and the remainder
of which is recycled back to a mixing point prior to the reactor entrance. Your boss wants a
85.0% overall yield of acetaldehyde from ethylene and is willing to allow a molar flow rate of
oxygen in the feed stream that is 15 times larger than the molar flow rate of ethylene in the feed
stream.
a. (5 pts) Write balanced chemical reactions for this process.
b. (10 pts) Draw and label a diagram for this process.
c. (10 pts) Determine the composition of the purge stream.
d. (10 pts) Determine the single-pass conversion of ethylene and single-pass yield of
acetaldehyde.
e. (5 pts) How would a change in the fraction of vapor stream that is purged impact the
overall conversion of ethylene?
a) Zczttytoz
Cztly
+
→
Oz
Czttyoz
→
b)
Ñ
-¥÷
,
(02,41-14)
ÑZ
(044+4)
2. 5- pts
Zcztty 0
( Cztluo
a.
city,
,
Gaya)
f
n' u
Ñg
( 4%0)
02,41kg
2. 5- pts
(02,41-4)
f
¥1m)
(
¥a)
his
Ccattyo)
10ptsforcorreq(G%0
diagram
2
r
c)
need
Will
Ñr
Ñ'
=
Ogg
,
,
we
of
composition
Determine
to
assume
150
mods
catty
can
do
now
Lboundaries
stream ?
→
identifying
overall
want
thats
→
overall balance %%F
i
an
pt
,
tÉ
green)
in
purge
basis
a
mots
1500
=
the
purge
composition :
Ña
Oz
,
eww
-
2
{,
-
correct
getup
{2
¥+0T
balance
"
air
Czttyoz
Czttyo
'"
=
-
{
-
'
of
NN
balance :
Au
=
1-
0
eqns
{z
balance :
rig
we
10m¥
=
of reaction
)
(extent
entry balance
=
can
yield
yield
O t
2
zpts
{,
information
use
some
0.85
=moYYµ%YmedtorFÉay
,
:
=
mainly
yield
,y÷ }
one
-
-
correct / use
setup
9
basis, the maximum amount
Gwen our chosen
of
litho
rib
Therefore :
could form
that
;
agg
¥¥f
getup
qÉs
"
I÷
riu=
can
we
purge
{,
4.25
is
equal
14
to
:
{z
=
0.59m¥
{ 2=0.59
,
solve
now
•n
ñy=
for
flow
rates
in
stream :
Mia ,czµq
=
via of
,
2
=
I¥s%%¥e%
=
=
since
=
selectivity
know
selectivity
,
10m04s
is
8. 5m04s
=
{
we also
yield
of
Zpt
-24
10
150
-
{,
-
-
{2=10-214.251-0.59
=
0.91m¥
{ 2--1500-4.25-0.51=145
Therefore:
Yeah ,9=µ¥÷
0.006
=
,
You
=
d) single
and
0.995
-
-
pass
yield
-
Single pass
-
ñ"""n÷?÷;_} setup
Ñ3,czl
Yield
acetaldehyde
pass
.
correct
( fatty
single
acetaldehyde
of
t.pt
→
catty
ethylene
of
conversion
single pass
conversion
}2pts
.
amount Cztluo prod
no side rxh
ideal
.
)
Coessumrng
Letisstartwithfq.tl#
First
,
city
we
fed
is
at
Balance
Ñz
need
,
qµy
to
to
determine how
( Ñz ,czµu )
reactor
mixing point :
=
Ñ never
+
much
Ñ
8.
catty
rig
2¥}
gown
µoÑ
mind
that
know
we
rig
a
=
:
( e. g.
10%
of vapor
stream is
purged)
he therefore
of
tie
>
Ñooczuy
port
Ñ8czH
Therefore
9
=
19m01s
:
=
,
need
we
catty
,
( 0.91m¥ )
8.
=
,
Ñzgµ
Now
9 Ña
=
cute
+8.19m¥
10m¥
=
his
determine
to
-18.19m¥
>
city
all
we
"
ne
yggfOÑ
find
go
"
Amir
leaves
{
the
catty
m
Ñ 3. Catty
purge / recycle
Ñt
=
,
cette
a
n°3
Ña
Czthe
=
Ñz,czH4
=
,
n' 3)
that
leaves
goes straight through
reactor
the
he
that
know
"
Carey
,
cute,
8.19
=
9.1
+
Ñ 8.
t
0.91
mobs
catty
,
e.
condenser
g.
Therefore
single
,
fczteu
Now
not
Yield
correct
s¥Yidd
'
↳
§rgµp%µced
µ¥¥¥f
For
Ñ3o_
the
the
fully
that
cztheo
reacted
to
¥21440
2m01
of
c~HÑ
Therefore
=
the
we
is
✗
Catty
single pass
%÷
-
=
in
that
stream
8.5m¥
=
Czthe
m°¥H4
1219
ÑG
=
of
-
know
we
,
leaves
denominator,
18.19m¥
0.5
produced
Cztheo
.
produced
n' g.
so
,
amt
numerator
Cztluo
=
yield
pass
ideal
all
6
single pass
,
Ipt
:
1%Yg.,-g
=
=
↳ For
ethylene
-
single
the
,
conv .
pass
know
fed ,
that
so
if
cztheo
form
__
rxt
18.19m¥cztkeo
yield :
0.47
-
Ipt
e)
if
is
you increased the vapor that
purged
of
amount
system
,
,
you'd
Cztly
this
have
a
leaving
higher
overall
decreasing few
overall
,
5-pts
identifying
between
purge
relationship
/ corn
.
3. (35 pts.) Methane (CH4) reacts with Cl2 to produce methyl chloride (CH3Cl) and HCl. However
some of the CH3Cl undergoes further chlorination to form CH2Cl2 and HCl. The balanced reactions
are shown below:
CH4+ Cl2 → CH3Cl + HCl
CH3Cl + Cl2 → CH2Cl2 + HCl
A fresh feed consisting of 150 mol/h of methane with 125% excess Cl2 is joined by a recycle
stream and the mixture is fed to a reactor where it reacts to produce CH3Cl, CH2Cl2 and HCl. The
products from the reactor are sent to a separator where 75% of the unreacted CH4 and Cl2 are
recycled, while the remaining CH4, Cl2 and the products are sent to another part of the plant to be
processed. If the overall conversion is 85% and the selectivity is 3, find:
a) (25 points) the molar flow rates of HCl, CH3Cl and CH2Cl2
b) (10 pts) the single-pass conversion of CH4
sptsdiagram
To receive full credit, use the extent of reaction approach in this problem.
÷É☒Éss
4¥)
4¥¥÷¥D
¥7,9
ñ
:*
.=¥
ri.ae
✓
337.5m¥
n'
(
"
'
4
k¥:)
a)
the molar
are
ethan
he 5
flow
rates
the same
.
3
of
in
HCl ,
ctlzd.be
streams
3
we
can
use
flow
some
Ell
our
to
info
given
some
rates
Ctke
conversion
85%
→
tianya.at?s-ci-ia:rred-
{
pts
use
of
overall
eÑ
We
0.85
=
0.85
n'
-22.5 thot /h
=
5cm ,
Clz
know
also
l502goÑ5c
=
fed
is
125%
in
excess
,
zpf.wfrom
s?xngniciz--2.25l150.mot-n)(Ynm%fek-*.)- m#
337.5
excess
We
now
can
write
balances
on
OLÉ lb.0.undaro.es
CHU :
Clz
Ñ5aµµ= Ñicµ
n'
:
CHM
"
sick
n' 5
=
,
Ñiciz
=
age ,
{
-
0
-
{
)
in blue
g-pts
i
{
correct
{i
+
the
-
{z
-
,
{z
extent
of
Rlhenquatrons
ris.eu
HCl :
Cttzclz
Ñ5oµµz
:
the
to
{
We
:{
¥
3
also
=
3
we
{
'
of
sweetie "Ñow
we
2
=
that
can
values
known
-
{
,
selectivity
=
Cttzclz
t2§
=
MIN
the
Ñ5sctts#
,
our
127.5
=
,
know
ÑS
{z
in
150
=
{z
balance:
cthe
22.5m¥
+
0 +
=
plug
can
we
{,
1-
0
=
-
=3
,
so :
%_g{÷
p
31.9
we
find
know
the
9,
he
{z
flow rates
3
ÑS
Cuza
,
=
127.5-31.9
=
95.6
Ipt
moly
159.TW#1pth5,CtlzClz--
Ñsttcl
=
127.51-31.9
=
-
m¥
31.9
Ipt
-
b)
find single
to
need
we
{
setup
zpts
rect
we
£+14
Csingle
Ñ3cµu
sepor §and
galant
We
Ñ2cµy-ñs
=
I
know
also
of Ctty
conversion
:
pass)
from
separator
zpts
pass
a
that
=
balance
"
over
the
:
Ñscthe
know
that
treaty
+
Ñ5cµu= 22.5m¥
that
-_
" "
he
Ñhlcthl
"*
C problem
"
" the
statement)
Therefore :
n' zcµ
=
225
+0.75mi
,
Ñ3cµy
=
90m¥
sexy
hire
We
zp§
get
the
balWHY
ance
of
need
now
can
=
my
67.5
rizotey
from
rizane-n.am
Ñ2cµ
pi
150m¥
=
,
Ñrctiu
Therefore
£0144
:
ñaoµ ,
+
217.5
=
67.5m¥
m£
,
)=
↳ ingle pass
faey
+
which
balance
a
point
mixing
,
=
21752,29¥
0.59-a
Ipt
we
at