Example 8.1
Consider the reaction: C 2 H 4 (g) + H 2 O(g) → C 2 H 5 OH(g). If an equimolar mixture of
ethylene and water vapor is fed to a reactor which is maintained at 500 K and 40 bar
determine the degree of conversion, assuming that the reaction mixture behaves like an ideal
gas. Assume the following ideal gas specific heat data: C p ig = a + bT + cT2 + dT3 + eT–2
(J/mol); T(K).
a
bx103
cx106
dx109
ex10-5
C2H4
20.691
205.346
– 99.793
18.825
-
H2O
4.196
154.565
– 81.076
16.813
-
C 2 H 5 OH
28.850
12.055
-
-
1.006
Species
From standard tables ΔH0 R,298 ≈ -52.7 KJ; ΔG0 R,298 = 14.5 KJ
T
0
Now ∆H T0 =
∆H 298
+
∫ ∆C dT
0
P
---------------------- (A)
298
∆CP0 = ∑ α i CP0 ,i = ∆a + ∆bT + ∆CT 2 + ∆dT 3 + ∆e / T 2 ------------- (B)
Where:
∆a =∑ α i ai , ∆b =∑ α i bi , etc
For example Δa = 20.691 – 4.196 -28.850 = -12.355
Δb = (205.346 – 154.565 – 12.055) x 10-3 = 3.8726 x 10-2
Similarly
Δc = -1.8717 x 10-5 ; Δd = 2.012 x 10-9 ; Δe = -1.006 x 105
Putting B in A and integrating A we get:
∆H T0 =
−50.944( KJ ) − 12.355T +
3.8726 × 10−2 2 1.8717 × 10−5 3
T −
T +
2
3
2.012 × 10−9 4
T + 1.006 × 105 T
4
By Vant Hoff equation:
…………..(C)
d (∆GT0 RT )
∆H T0
= −
dT
RT 2
0
∆GT0 ∆G298
∆H T0
−
=
−∫
dT
RT R(298)
RT 2
298
T
∴
------------------------------------------------- D
We already know ∆H T0 from (C); putting C in D and integrating we obtain:
∆GT0 =
−50.944 + 12.355T ln T −
+
1.006 ×105
+ 56.681T
2T
3.8726 ×10−2 2 1.8718 ×10−5 3 2.012 ×10−9 4
T +
T −
T
2
6
12
Putting T=500k, ∆GT0 = 11.43kJ
Example 8.2
Consider the following reaction: A(g) + B(g) = C(g) + 3D(g)
CH 4 + H 2 O → CO + 3H 2
Intially the following number of moles are introduced in the reactor. Obtian the mole fraction
expressions in terms of reaction coordinate.
n0, A = 2 mol, n0,B = 1 mol, n0,C = 1 mol n0,D = 4 mol
α ∑ α i = -i = -1-1+1+3 = 2
no =
∑n
io
= 2+ 1 +1 + 4 = 8
i
y=
i
ni nio + α iξ
=
n no + αξ
∴∴ y A =
2 −ξ
;
8 + 2ξ
yB =
1− ξ
;
8 + 2ξ
yC =
1+ ξ
;
8 + 2ξ
yH 2 =
4+ξ
8 + 2ξ
Example 8.3
Consider the following simultaneous reactions. Express the reaction mixture composition as
function of the reaction co-ordinates. All reactants and products are gaseous.
A + B = C + 3D
..(1)
A + 2B = E + 4D
..(2)
Initial number of moles:
n0, A = 2 mol;
n0,B = 3 mol
Let the reaction co-ordinates for each reaction be ξ1 and ξ 2 respectively.
yi =
=
∴ yA
j
A
B
C
D
E
α j =Σα i,j
1
-1
-1
1
3
0
2
2
-2
-2
0
4
1
2
nio + ∑ j α i , jξ j
no + ∑ j α jξ j
;
2 − ξ1 − ξ 2
=
; yB
5 + 2ξ1 + 2ξ 2
n o = 2+3 = 5
3 − ξ1 − 2ξ 2
=
; yC
5 + 2ξ1 + 2ξ 2
ξ1
5 + 2ξ1 + 2ξ 2
3ξ1 + 4ξ 2
=
; yE
5 + 2ξ1 + 2ξ 2
yD
=
ξ2
5 + 2ξ1 + 2ξ 2
Example 8.4
Consider the reaction : C 2 H 4 (g) + H 2 O(g) → C 2 H 5 OH(g). If an equimolar mixture of
ethylene and water vapor is fed to a reactor which is maintained at 500 K and 40 bar
determine the equilibrium constant, assuming that the reaction mixture behaves like an ideal
gas. Assume the following ideal gas specific heat data: C p ig = a + bT + cT2 + dT3 + eT–2
(J/mol); T(K).
a
bx103
cx106
dx109
ex10-5
C2H4
20.691
205.346
– 99.793
18.825
-
H2O
4.196
154.565
– 81.076
16.813
-
C 2 H 5 OH
28.850
12.055
-
-
1.006
Species
From standard tables ΔH0 R,298 ≈ -52.7 KJ; ΔG0 R,298 = 14.5 KJ
T
0
Now ∆H T0 =
∆H 298
+
∫ ∆C dT
0
P
---------------------- (A)
298
∆CP0 = ∑ α i CP0 ,i = ∆a + ∆bT + ∆CT 2 + ∆dT 3 + ∆e / T 2 ------------- (B)
Where:
∆a =∑ α i ai , ∆b =∑ α i bi , etc
For example Δa = 20.691 – 4.196 -28.850 = -12.355
Δb = (205.346 – 154.565 – 12.055) x 10-3 = 3.8726 x 10-2
Similarly
Δc = -1.8717 x 10-5 ; Δd = 2.012 x 10-9 ; Δe = -1.006 x 105
Putting B in A and integrating A we get:
∆H T0 =
−50.944( KJ ) − 12.355T +
3.8726 × 10−2 2 1.8717 × 10−5 3
T −
T +
2
3
2.012 × 10−9 4
T + 1.006 × 105 T
4
By Vant Hoff equation:
…………..(C)
d (∆GT0 RT )
∆H T0
= −
dT
RT 2
0
∆GT0 ∆G298
∆H T0
−
=
−∫
dT
RT R(298)
RT 2
298
T
∴
------------------------------------------------- D
We already know ∆H T0 from (C); putting C in D and integrating we obtain:
∆GT0 =
−50.944 + 12.355T ln T −
+
3.8726 ×10−2 2 1.8718 ×10−5 3 2.012 ×10−9 4
T +
T −
T
2
6
12
1.006 ×105
+ 56.681T
2T
Putting T=500k, ∆GT0 = 11.43kJ
Hence K 500 = exp [ -1143/8.314 x 1000] = 0.064
Example 8.5
Consider the reaction: C 2 H 4 (g) + H 2 O(g) → C 2 H 5 OH(g). If an equimolar mixture of
ethylene and water vapor is fed to a reactor which is maintained at 500 K and 40 bar
determine the degree of conversion, assuming that the reaction mixture behaves like an ideal
gas. Assume the following ideal gas specific heat data: C p ig = a + bT + cT2 + dT3 + eT–2
(J/mol); T(K).
a
bx103
cx106
dx109
ex10-5
C2H4
20.691
205.346
– 99.793
18.825
-
H2O
4.196
154.565
– 81.076
16.813
-
C 2 H 5 OH
28.850
12.055
-
-
1.006
Species
From standard tables ΔH0 R,298 ≈ -52.7 KJ; ΔG0 R,298 = 14.5 KJ
T
0
Now ∆H T0 =
∆H 298
+
∫ ∆C dT
0
P
---------------------- (A)
298
∆CP0 = ∑ α i CP0 ,i = ∆a + ∆bT + ∆CT 2 + ∆dT 3 + ∆e / T 2 ------------- (B)
Where:
∆a =∑ α i ai , ∆b =∑ α i bi , etc
For example Δa = 20.691 – 4.196 -28.850 = -12.355
Δb = (205.346 – 154.565 – 12.055) x 10-3 = 3.8726 x 10-2
Similarly
Δc = -1.8717 x 10-5 ; Δd = 2.012 x 10-9 ; Δe = -1.006 x 105
Putting B in A and integrating A we get:
∆H T0 =
−50.944( KJ ) − 12.355T +
3.8726 × 10−2 2 1.8717 × 10−5 3
T −
T +
2
3
2.012 × 10−9 4
T + 1.006 × 105 T
4
By Vant Hoff equation:
d (∆GT0 RT )
∆H T0
= −
dT
RT 2
…………..(C)
0
∆GT0 ∆G298
∆H T0
−
=
−∫
dT
RT R(298)
RT 2
298
T
∴
------------------------------------------------- D
We already know ∆H T0 from (C); putting C in D and integrating we obtain:
∆GT0 =
−50.944 + 12.355T ln T −
+
3.8726 ×10−2 2 1.8718 ×10−5 3 2.012 ×10−9 4
T +
T −
T
2
6
12
1.006 ×105
+ 56.681T
2T
Putting T=500k, ∆GT0 = 11.43kJ
Hence K 500 = exp [ -1143/8.314 x 1000] = 0.064
α = Σα i = 1-1-1 = -1
∴ K = K ϕK y Pα ; K ϕ=1(since ideal gas assumption is made)
P = 40bar
Component
n i0
n(exit)
y i (exit)
C2H4
1
1-ε
(1-ε )/(2-ε)
H2O
1
1-ε
(1-ε )/(2-ε)
C2H5OH
0
ε
(ε )/(2-ε)
n t (at exit) = 2-ε
∴ K = K y = yC2 H5OH yC2 H 4 yH 2O =
ξ (2 − ξ )
{(1 − ξ ) ( 2 − ξ )}
2
=
ξ (2 −ξ )
(1 − ξ )
2
Now K = K ϕK y Pα = (1) K y P-1
∴ K y = 40 K ; K y = ξ ( 2 − ξ ) (1 − ξ )
∴Ky =
ξ (2 −ξ )
(1 − ξ )
2
2
= 40 × K 500 = 40 × 0.064 = 2.56
On solving ξ = 0.47
Thus yC2 H 4 =(1 − ξ ) ( 2 − ξ ) =0.3464 = yH 2O
yC2 H5OH= ξ
( 2 − ξ )=
0.3072
Example 8. 6
The following two independent reactions occur in the steam cracking of methane at 1000 K
and 1 bar: CH 4 (g) + H 2 O(g) → CO(g) + 3H 2 (g); and CO(g) + H 2 O(g) → CO 2 (g) + H 2 (g).
Assuming ideal gas behaviour determine the equilibrium composition of the gas leaving the
reactor if an equimolar mixture of CH 4 and H 2 O is fed to the reactor, and that at 1000K, the
equilibrium constants for the two reactions are 30 and 1.5 respectively.
Let ξ 1 and ξ 2 be the reaction co-ordinate for the two reactions, we have
Comp
n i0
n exit
y exit
CH 4
1
1 - ξ1
(1 - ξ 1 ) / 2(1 + ξ 1 )
H2O
1
1 – ξ 1 – ξ2
(1 – ξ 1 – ξ 2 ) / 2(1 + ξ 1 )
CO
0
ξ1 – ξ2
(ξ 1 – ξ 2 ) / 2(1 + ξ 1 )
CO 2
0
ξ2
ξ 2 / 2(1 + ξ 1 )
H2
0
3 ξ1 + ξ2
(3 ξ 1 + ξ 2 ) / 2(1 + ξ 1 )
Total moles at equilibrum: 2(1 + ξ 1 )
K = K ϕK y Pα (for each reaction); K ϕ = 1.0 (ideal gas assumption); P = 1 bar
 ξ1 − ξ 2 )   ( 3ξ1 + ξ 2 ) 


 2 (1 + ξ1 )   2 (1 + ξ1 ) 
Thus K1 =  (
 ( ε1 − ξ 2 )   ( 3ξ1 + ξ 2 ) 



 2 (1 + ξ1 )   2 (1 + ξ1 ) 
3
(ξ1 − ξ 2 )( 3ξ1 + ξ 2 )
=
2
4 (1 + ξ1 ) (1 − ξ1 )(1 − ξ1 − ξ 2 )
3
=
Similarly K 2
=
( 3ξ1 + ξ 2 ) ξ 2
=
1.5
(ξ1 − ξ 2 )(1 − ξ1 − ξ 2 )
30
……………………………. A
………………………………. B
A and B needs to be solved simultaneously; a simple way to do this is to
(i)
Assume ε 2 , calculate ε1 using B
(ii)
Use ξ 2 and ξ1 in A to check if K 1 = 30
(iii)
If K1 ≠ 30, assume new ε 2 and go to step 1
Using the above algorithm, one finally obtains: ξ1 = 0.7980, ξ 2 = 0.0626.
Thus:
=
yCO2 0.0174,
=
yCH 4 0.0562,
=
yH 2O 0.0388,
=
yCO 0.2045,
=
yH 2 0.6831
Example 8.7
The gas n-pentane (1) is known to isomerise into neo-pentane (2) and iso-pentane (3)
 P2 ; P2 
 P3 ; P3 
 P1 . 3 moles of
according to the following reaction scheme: P1 
o
pure n-pentane is fed into a reactor at 400 K and 0.5 atm. Compute the number of moles of
each species present at equilibrium.
o
Species
∆G 0f at 400 K (Cal/mol)
P1
9600
P2
8900
P3
8200
We use here the method of undetermined Lagrangian Multipliers.
The set of equation to be solved are:
=
AC 15,
=
AH 36
For P 1 :
 n
9600
+ ln  1
RT
 ∑ ni
 5λC 12λH
+
=
0
+
RT
 RT
For P 2 :
 n
8900
+ ln  2
RT
 ∑ ni
 5λC 12λH
+
=
0
+
RT
 RT
For P 3 :
 n  5λ 12λH
8200
+ ln  3  + C +
=
0
RT
RT
 ∑ ni  RT
15
Atomic mass balance for C: 5 ( n1 + n2 + n3 ) =
36
For H: 12 ( n1 + n2 + n3 ) =
n 1 + n 2 + n 3 = 10
Alternately:
5λ 12λH
9600
+ ln y1 + C +
=
0 ………………….
RT
RT
RT
(A)
5λ 12λH
8900
+ ln y2 + C +
=
0 …………………. (B)
RT
RT
RT
T = 400K
5λ 12λH
8200
+ ln y3 + C +
=
0 …………………. (C)
RT
RT
RT
and y 1 + y 2 + y 3 = 1 …………………. (D)
It follows from (A) – (C), y 2 / y 1 = 2.41; y 3 / y 2 = 2.41
Using eqn. (D) y 1 = 0.108, y 2 = 0.29, y 3 = 0.63
Example 8.8
Consider the liquid phase reaction: A(l) + B(l) → C(l) + D(l). At 50oC, the equilibrium
constant is 0.09. Initial number of moles, n A,0 = 1 mole; n B,0 = 1 mol Find the equilibrium
conversion. Assume ideal solution behaviour.

 ( P -1) ∑ γ iVi 
γ
∴K = ∴ ( xi γ i ) i exp 

RT






1.0
Also, γ i = 1 (ideal solution)
Hence K = π ( xi ) i
γ
x A = x B = (1-ξ)/2 ;
xC = xD = ξ / 2
∴ K = x C x D / x A x B = [ξ/(1- ξ)]2
⇒ 0.09 = [ξ/(1- ξ)]2
Thus, ξ e = 0.23
Example 8.9
Consider the following reaction: A(s) + B(g) → C(s) + D(g). Determine the equilibrium
fraction of B which reacts at 500oC if equal number of moles of A and B are introduced into
the reactor initially. The equilibrium constant for the reaction at 500oC is 2.0.
The reaction is:
A( s ) + B ( g ) → C ( s ) + D( g ) ; basis 1 mole of A & B each initially
K = a C a D a A a B
For solids: a = 1
Thus:

=
K a=
Kφ K y =
Pα ; α 0, and
=
Kφ 1
D aB
∴K =
Ky
If one assumes equimolar feed of reactants:
ξ
yB =
(1 − ξ ); yD =
∴ K = K y = 2.0 =
ξ
1− ξ
Thus 67% of B reacts.
⇒ ξ = 0.67