Lecture 1C: Examples of Chemical Equilibrium Calculation

Lecture 1C: Examples of Chemical Equilibrium Calculation
Reaction System: 2 SO 2 + O2
= 2 SO3
o
o
Conditions: T = 600 C = 873 K
P = 1 atm
Stoichiometric
Feed
No Nitrogen
Calculate equilibrium conversion of SO 2 . Conversion of 99% is desired.
Species Names
SO
O2
SO
2
3
S = 3
No.
Assume ideal gas mixture.
K = K
K
y
=
y
∏
Now
n
y1
=
n
K y
=
1
tot
⎛
⎜
⎝
P
P
y
j
o
s
⎞
∑
j = 1
⎟
⎠
υ
j
1
2
3
υ = y
2 − 2X
=
3 − X
j
K
y
−2
1
Stoich. Coeff.
∑υ
= −1
⎛
⎜
⎝
y
j
P
P
−1
2
y
2 ( 3 − X )
2 ( 1 − X ) ( 1 − X )
2 ( 3 − X ) ( 3 − X )
4
4 X
2
2
o
⎞
⎟
⎠
y
− 2
−1
−1
−2
3
2
= K
=
1 −
=
3 −
y
X
X
y
y
⎛
⎜
⎝
P
2
3
2
1 y2
P
∆G
o
⎞
⎟
⎠
k cal
mol
f
∆H
− 71.
7
0
− 88.
6
⎛
⎜ y
⎜=
2
⎜ y
⎝
y
( )
( ) 3
2
3 − X
1 − X
where = equilibrium
extent
X X
= e
=
x
3
SO
2
2
SO
3
y
2X
=
3 − X
∑ n
1
O
2
f
⎞
⎟
⎟
⎠
jo
n
jo
− 70.
9
0
− 94.
4
= 3n
tot
n
j
= n
= ∑ n
2 − 2X
1 − 1X
0 + 2X
j
jo
+ υ X
j
= 3 − x
EECE 503 February, 2009
(A)

From (A)
x
( )
( ) ⎟ 2
3 − x ⎛ P ⎞
= K
⎜
3 873
1 − x P
Calculate K 298 at 298 K
∆ o
K
G r
298
=
= e
⎝
o
⎠
( − ) × ( − 71.
7)
+ ( −1)
× ( 0)
+ 2 × ( − 88.
6)
2 = − 33
o
∆Gr
−
RT
o
= e
Calculate ∆ H r , K873
∆
H o
r
=
−33,
800
−
1.
987 × 298
= e
57.
08
= 6 . 17 × 10
( − ) × ( − 70.
9)
+ ( 1)
× ( 0)
+ 2 × ( − 94.
4)
2 = − 47
24
2
. 0
. 8
k cal
mol
k cal
mol
Assume for simplicity (in order to find a first estimate for equilibrium conditions) that
o
∆ H ≈ const ≈ ∆H
r
Then
d ln
K ∆ H
=
dT RT
⎛ K
l n
⎜
⎝ K
K
T
T
298
= K
o
r
2
⎞ ∆H
⎟ =
⎠ R
298
e
r
; and at T = T = 298,
K = K
o
r
⎛
⎜
⎝
1
298
o
∆H
r ⎛ 1 1 ⎞
⎜ − ⎟
R ⎝ 298 T ⎠
1
−
T
47,
000 ⎛ 1
1.
987 ⎝ 298
57.
08
K 873 = e − ⎜ − ⎟ =
K 873 = 121.
78
⎞
⎟
⎠
o
1 ⎞
873 ⎠
e
4.
80
Note the dramatic drop in K with temperature due to the exothermocity of the reaction
( ∆H r < 0)
. While equilibrium would have been all the way to the right at 298 K we cannot
operate at such conditions because the rate is too low.
298
(B)

Let us see what are the equilibrium limitations at 873 K.
Solve by trial and error equation (B) for equilibrium extent X.
2 ( 3 − X )
3 ( 1 − X )
X ⎛ o
= K 873
⎜
⎝
P
P
⎞
⎟
⎠
−1
P
where K 873 = 121.
78;
= 1
P
Rearrange to use Newton-Raphson procedure
o
( ) ( ) ( ) 3
2
x = 0 = X 3 − X − 121.
78 1 − X
( ) ( ) ( ) 2
2
φ x = 2X 3 − X − X + 3 × 121.
78 1 X
n = X n
φ ( X n )
D φ ( X )
φ
D −
X
+ 1 (C)
n
Using a starting guess of X = 0.
5 the Newton Raphson algorithm (C) yields:
Iteration No n = 0
Extent 0.5
X = X e = 0.
777
( mol)
1
0.656
Conversion of SO 2 is:
nSO
− n
2,
0 SO2
2 −
xSO
=
=
2 nSO2
0
x = 0.
777