Chapter 13 - Section A - Mathcad Solutions

Note: For the following problems the variable kelvin is used for the SI
unit of absolute temperature so as not to conflict with the variable K
used for the equilibrium constant

13.4 H2(g) + CO2(g) = H2O(g) + CO(g)

Q=
¦ Q i = 1  1  1  1 = 0 n0 = 1  1 = 2
i
1H H
By Eq. (13.5). yH = yCO = yH2O = yCO =
2 2 2 2

By Eq. (A) and with data from Example 13.13 at 1000 K:

T  1000˜ kelvin

G H  ¨
§ 1  H · ˜ (395790)˜ J  H ˜ (192420  200240)˜ J 
© 2 ¹ mol 2 mol
§ 1  H ˜ ln § 1  H ·  2˜ H ˜ ln§ H · ·
 R˜ T˜ ¨ 2˜ ¨ ¨
© 2 © 2 ¹ 2 © 2 ¹¹
Guess: H e  0.5

Given
d

G H e = 0˜
J

H e  Find H e He 0.45308
dH e mol

H  0.3  0.31  0.6

2.082

2.084
G H
5
10
2.086

2.088
0.2 0.3 0.4 0.5 0.6
H

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 13.5 (a) H2(g) + CO2(g) = H2O(g) + CO(g)

Q=
¦ Q i = 1  1  1  1 = 0 n0 = 1  1 = 2
i
1H H
By Eq. (13.5). yH = yCO = yH2O = yCO =
2 2 2 2

By Eq. (A) and with data from Example 13.13 at 1100 K:

T  1100˜ kelvin

G H  ¨
§ 1  H · ˜ ( 395960) ˜ J  H ˜ ( 187000  209110) ˜ J 
© 2 ¹ mol 2 mol
§
 R˜ T˜ ¨ 2˜
1  H §
˜ ln ¨
1  H ·  2˜ ˜ ln§ · ·
H H
¨
© 2 © 2 ¹ 2 © 2 ¹¹

Guess: H e  0.5

Given
d

G H e = 0˜
J
H e  Find H e  He 0.502 Ans.
dH e mol

H  0.35  0.36  0.65

2.102

2.103

2.104
G H
5
10
2.105

2.106

2.107
0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65
H

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 (b) H2(g) + CO2(g) = H2O(g) + CO(g)

Q=
¦ Q i = 1  1  1  1 = 0 n0 = 1  1 = 2
i
1H H
By Eq. (13.5), yH = yCO = yH2O = yCO =
2 2 2 2

By Eq. (A) and with data from Example 13.13 at 1200 K:

T  1200˜ kelvin

G H  ¨
§ 1  H · ˜ (396020)˜ J  H ˜ (181380  217830)˜ J 
© 2 ¹ mol 2 mol
§
 R˜ T˜ ¨ 2˜
1  H §
˜ ln ¨
1  H ·  2˜ ˜ ln§ · ·
H H
¨
© 2 © 2 ¹ 2 © 2 ¹¹
Guess: H e  0.1

Given
d

G H e = 0˜
J
H e  Find H e He 0.53988 Ans.
dH e mol

H  0.4  0.41  0.7

2.121

2.122

2.123

G H
5 2.124
10

2.125

2.126

2.127
0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7
H

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 (c) H2(g) + CO2(g) = H2O(g) + CO(g)

Q=
¦ Q i = 1  1  1  1 = 0 n0 = 1  1 = 2
i
1H H
By Eq, (13.5), yH = yCO = yH2O = yCO =
2 2 2 2

By Eq. (A) and with data from Example 13.13 at 1300 K:

T  1300˜ kelvin

G H  ¨
§ 1  H · ˜ ( 396080) ˜ J  H ˜ ( 175720  226530) ˜ J 
© 2 ¹ mol 2 mol
§ 1  H ˜ ln § 1  H ·  2˜ H ˜ ln§ H · ·
 R˜ T˜ ¨ 2˜ ¨ ¨
© 2 © 2 ¹ 2 © 2 ¹¹

Guess: H e  0.6

Given
d

G H e = 0˜
J
H e  Find H e He 0.57088 Ans.
dH e mol

H  0.4  0.41  0.7

2.14

2.142

G H
5
2.144
10

2.146

2.148
0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7
H

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 13.6 H2(g) + CO2(g) = H2O(g) + CO(g)

Q=
¦ Q i = 1  1  1  1 = 0 n0 = 1  1 = 2
i
1H H
By Eq, (13.5), yH = yCO = yH2O = yCO =
2 2 2 2
With data from Example 13.13, the following vectors represent values for
Parts (a) through (d):

§ 1000 · § 3130 ·
¨ ¨
1100 ¸ 150 ¸ J
T ¨ ˜ kelvin 'G  ¨ ˜
¨ 1200 ¸ ¨ 3190 ¸ mol
¨ ¨
© 1300 ¹ © 6170 ¹
Combining Eqs. (13.5), (13.11a), and (13.28) gives

§ H ·˜§ H ·
¨ ¨ 2
© 2¹ © 2¹ =
H § 'G ·
= K = exp ¨
§ 1  H · ˜ § 1  H · 1  H 2 © R˜ T ¹
¨ ¨
© 2 ¹© 2 ¹

 o § 0.4531 ·
o ¨
[
§ 'G ·
exp ¨ H
[
H ¨ 0.5021 ¸ Ans.
© R˜ T ¹ 1[ ¨ 0.5399 ¸
¨
© 0.5709 ¹

13.11 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g)

Q = 1 n0 = 6 T  773.15˜ kelvin T0  298.15˜ kelvin

J J
'H298  114408˜ 'G298  75948˜
mol mol

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 The following vectors represent the species of the reaction in the order in
which they appear:
§ 4 · § 3.156 · § 0.623 · § 0.151 ·
¨ ¨ ¨ ¨
1 3.639 ¸ 0.506 ¸  3 ¨ 0.227 ¸ ˜ 105
Q ¨ ¸ A ¨ B ¨ ˜ 10 D
¨2 ¸ ¨ 3.470 ¸ ¨ 1.450 ¸ ¨ 0.121 ¸
¨ ¨ ¨ ¨
©2 ¹ © 4.442 ¹ © 0.089 ¹ © 0.344 ¹
end  rowsA
() i  1  end

'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'D 
¦ Qi˜ Di
i i i
5 4
'A 0.439 'B 8 u 10 'C  0 'D 8.23 u 10

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

4 J
'G 1.267 u 10
mol

K  exp ¨
§ 'G · K 7.18041
© R˜ T ¹
5  4˜ H
By Eq. (13.5) yHCl =
6H
1H 2˜ H 2˜ H
yO2 = yH2O = yCl2 =
6H 6H 6H

Apply Eq. (13.28); H  0.5 (guess)

4
§ 2˜ H · ˜ § 6  H · = 2˜ K H  Find H H
Given ¨ ¨ 0.793
© 5  4˜ H ¹ © 1  H ¹
5  4˜ H 1H 2˜ H 2˜ H
yHCl  yO2  yH2O  yCl2 
6H 6H 6H 6H

yHCl 0.3508 yO2 0.0397 yH2O 0.3048 yCl2 0.3048 Ans.

488
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 13.12 N2(g) + C2H2(g) = 2HCN(g) Q= 0 n0 = 2
This is the reaction of Pb. 4.21(x). From the answers for Pbs. 4.21(x),
4.22(x), and 13.7(x), find the following values:

J J
'H298  42720˜ 'G298  39430˜
mol mol
3 5
'A  0.060 'B  0.173˜ 10 'C  0 'D  0.191˜ 10

T  923.15˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 3.242 u 10
4 J
K  exp ¨
§ 'G · K 0.01464
mol © R˜ T ¹
By Eq. (13.5),
1H 1H 2e
yN2 = yC2H4 = yHCN = = H
2 2 2

By Eq. (13.28), H  0.5 (guess)

2
§ 2˜ H · = K H  Find H H
Given ¨ 0.057
©1  H¹
1H 1H
yN2  yC2H4  yHCN  H
2 2
yN2 0.4715 yC2H4 0.4715 yHCN 0.057 Ans.
Given the assumption of ideal gases, P has no effect on the equilibrium
composition.

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 13.13 CH3CHO(g) + H2(g) = C2H5OH(g) Q = 1 n0 = 2.5
This is the reaction of Pb. 4.21(r). From the answers for Pbs. 4.21(r),
4.22(r), and 13.7(r), find the following values:

J J
'H298  68910˜ 'G298  39630˜
mol mol
3 6 5
'A  1.424 'B  1.601˜ 10 'C  0.156˜ 10 'D  0.083˜ 10

T  623.15˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 6.787 u 10
3 J
K  exp ¨
§ 'G · K 3.7064
mol © R˜ T ¹
1H 1.5  H H
By Eq. (13.5), yCH3CHO = yH2 = yC2H5OH =
2.5  H 2.5  H 2.5  H

By Eq. (13.28), H  0.5 (guess)

H ˜ 2.5  H
Given = 3˜ K H  Find H H 0.818
1  H ˜ 1.5  H
1H 1.5  H H
yCH3CHO 
2.5  H yH2  yC2H5OH 
2.5  H 2.5  H

yCH3CHO 0.108 yH2 0.4053 yC2H5OH 0.4867 Ans.

If the pressure is reduced to 1 bar,
H ˜ 2.5  H
Given = 1˜ K H  Find H H 0.633
1  H ˜ 1.5  H
1H
yCH3CHO  1.5  H H
2.5  H yH2  yC2H5OH 
2.5  H 2.5  H

yCH3CHO 0.1968 yH2 0.4645 yC2H5OH 0.3387 Ans.

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 13.14 C6H5CH:CH2(g) + H2(g) = C6H5.C2H5(g) Q = 1 n0 = 2.5
This is the REVERSE reaction of Pb. 4.21(y). From the answers for Pbs.
4.21(y), 4.22(y), and 13.7(y) WITH OPPOSITE SIGNS, find the following
values:

J J
'H298  117440˜ 'G298  83010˜
mol mol
3 6 5
'A  4.175 'B  4.766˜ 10 'C  1.814˜ 10 'D  0.083˜ 10

T  923.15˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 2.398 u 10
3 J
K  exp ¨
§ 'G · K 1.36672
mol © R˜ T ¹
1H
By Eq. (13.5), yC6H5CHCH2 =
2.5  H
1.5  H H
yH2 = yC6H5C2H5 =
2.5  H 2.5  H

By Eq. (13.28), H  0.5 (guess)

H ˜ 2.5  H
Given = 1.0133˜ K H  Find H H 0.418
1  H ˜ 1.5  H
1H 1.5  H H
yC6H5CHCH2  yH2  yC6H5C2H5 
2.5  H 2.5  H 2.5  H

yC6H5CHCH2 0.2794 yH2 0.5196 yC6H5C2H5 0.201 Ans.

491
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 13.15 Basis: 1 mole of gas entering, containing 0.15 mol SO2, 0.20 mol O2,
and 0.65 mol N2.
SO2 + 0.5O2 = SO3 Q = 0.5 n0 = 1
By Eq. (13.5),
0.15  H 0.20  0.5˜ H H
ySO2 = yO2 = ySO3 =
1  0.5˜ H 1  0.5˜ H 1  0.5˜ H

From data in Table C.4,

J J
'H298  98890˜ 'G298  70866˜
mol mol
The following vectors represent the species of the reaction in the order
in which they appear:
§¨ 1 · §¨ 5.699 · §¨ 0.801 · §¨ 1.015 ·
3 5
Q  ¨ 0.5 ¸ A  ¨ 3.639 ¸ B  ¨ 0.506 ¸ ˜ 10 D ¨ 0.227 ¸ ˜ 10
¨ 1 ¨ 8.060 ¨ 1.056 ¨ 2.028
© ¹ © ¹ © ¹ © ¹
end  rowsA
() i  1  end

'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'D 
¦ Qi˜ Di
i i i
6 4
'A 0.5415 'B 2 u 10 'C  0 'D 8.995 u 10

T  753.15˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 2.804 u 10
4 J
K  exp ¨
§ 'G · K 88.03675
mol © R˜ T ¹
By Eq. (13.28), H  0.1 (guess)

H ˜ 1  0.5˜ H
0.5
Given = K H  Find H H 0.1455
0.15  H ˜ 0.2  0.5˜ H 0.5

492
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 By Eq. (13.4), nSO3 = H = 0.1455
By Eq. (4.18), 
'H753  'H298  R˜ IDCPH T0 'T ' A ' B ' C  D

J J
'H753 98353 Q  H'
˜ H753 Q 14314 Ans.
mol mol

13.16 C3H8(g) = C2H4(g) + CH4(g) Q= 1

Basis: 1 mole C3H8 feed. By Eq. (13.4) nC3H8 = 1  H

n0  nC3H8 1  1  H
Fractional conversion of C3H8 = = = H
n0 1
1H H H
By Eq. (13.5), yC3H8 = yC2H4 = yCH4 =
1H 1H 1H

From data in Table C.4,

J J
'H298  82670˜ 'G298  42290˜
mol mol

The following vectors represent the species of the reaction in the order
in which they appear:

§¨ 1 · §¨ 1.213 · §¨ 28.785 · §¨ 8.824 ·

3 6
Q ¨1 ¸ A  ¨ 1.424 ¸ B  ¨ 14.394 ¸ ˜ 10 C ¨ 4.392 ¸ ˜ 10
¨1 ¨ 1.702 ¨ 9.081 ¨ 2.164
© ¹ © ¹ © ¹ © ¹
end  rowsA
() i  1  end

'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'C 
¦ Qi˜Ci
i i i
3 6
'A 1.913 'B 5.31 u 10 'C 2.268 u 10 'D  0

(a) T  625˜ kelvin T0  298.15˜ kelvin

493
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 'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 2187.9
J
K  exp ¨
§ 'G · K 1.52356
mol © R˜ T ¹
By Eq. (13.28), H  0.5 (guess)
2
H
Given = K H  Find H
1  H ˜ 1  H
H 0.777 This value of epsilon IS the fractional conversion. Ans.

2
H
(b) H  0.85 K K 2.604
1  H ˜ 1  H
J
'G  R˜ T˜ ln ( K) 'G 4972.3 Ans.
mol
The problem now is to find the T which generates this value.
It is not difficult to find T by trial. This leads to the value:
T = 646.8 K Ans.

13.17 C2H6(g) = H2(g) + C2H4(g) Q= 1

Basis: 1 mole entering C2H6 + 0.5 mol H2O.

n0 = 1.5 By Eq. (13.5),

1H H H
yC2H6 = yH = yC2H4 =
1.5  H 1.5  H 1.5  H

From data in Table C.4,

J J
'H298  136330˜ 'G298  100315˜
mol mol

The following vectors represent the species of the reaction in the order in
which they appear:

494
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 §¨ 1 · §¨ 1.131 · §¨ 19.225 ·
3
Q ¨1 ¸ A  ¨ 3.249 ¸ B  ¨ 0.422 ¸ ˜ 10
¨1 ¨ 1.424 ¨ 14.394
© ¹ © ¹ © ¹
§¨ 5.561 · §¨ 0.0 ·
6 5
C  ¨ 0.0 ¸ ˜ 10 D  ¨ 0.083 ¸ ˜ 10
¨ 4.392 ¨ 0.0
© ¹ © ¹
end  rowsA
() i  1  end

'A 
¦ Qi˜ Ai 'B  ¦ Qi˜Bi 'C 
¦ Qi˜Ci 'D 
¦ Qi˜ Di
i i i i
3 6 3
'A 3.542 'B 4.409 u 10 'C 1.169 u 10 'D 8.3 u 10

T  1100˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 5.429 u 10
3 J
K  exp ¨
§ 'G · K 1.81048
mol © R˜ T ¹

By Eq. (13.28), H  0.5 (guess)

2
H
Given = K H  Find H H 0.83505
1.5  H ˜ 1  H
By Eq. (13.4), nC2H6 = 1  H nH2 = nC2H4 = H n= 1H
1H H H
yC2H6  yH2  yC2H4 
1H 1H 1H

yC2H6 0.0899 yC2H4 0.4551 yH2 0.4551 Ans.

495
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 13.18 C2H5CH:CH2(g) = CH2:CHCH:CH2(g) + H2(g) Q= 1
(1) (2) (3)

Number the species as shown. Basis is 1 mol species 1 + x mol steam.

n0 = 1  x
1H H
By Eq. (13.5), y1 = y2 = y3 = = 0.10
1 Hx  1 Hx 
From data in Table C.4,
J J
'H298  109780˜ 'G298  79455˜
mol mol

The following vectors represent the species of the reaction in the

order in which they appear:

§¨ 1 · §¨ 1.967 · §¨ 31.630 ·
3
Q ¨1 ¸ A  ¨ 2.734 ¸ B  ¨ 26.786 ¸ ˜ 10
¨1 ¨ 3.249 ¨ 0.422
© ¹ © ¹ © ¹
§¨ 9.873 · §¨ 0.0 ·
6 5
C  ¨ 8.882 ¸ ˜ 10 D  ¨ 0.0 ¸ ˜ 10
¨ 0.0 ¨ 0.083
© ¹ © ¹
end  rows ( A) i  1  end

'A 
¦ Qi˜ Ai 'B  ¦ Qi˜Bi 'C 
¦ Qi˜Ci 'D 
¦ Qi˜ Di
i i i i

3 7 3
'A 4.016 'B 4.422 u 10 'C 9.91 u 10 'D 8.3 u 10

T  950˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

496
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 'G 4.896 u 10
3 J
K  exp ¨
§ 'G · K 0.53802
mol © R˜ T ¹

By Eq. (13.28), (0.1)˜ (0.1)˜ 1 Hx 

= K
1H
K
Since 0.10˜H1 Hx  = H H 0.843
K  0.10
H
x H1  x 6.5894
0.10

1H
(a) y1  yH2O  1  0.2  y1
1 Hx 

y1 0.0186 yH2O 0.7814 Ans.

6.5894
(b) ysteam  ysteam 0.8682 Ans.
7.5894

13.19 C4H10(g) = CH2:CHCH:CH2(g) + 2H2(g)

Q= 2
(1) (2) (3)
Number the species as shown. Basis is
n0 = 1  x
1 mol species 1 + x mol steam entering.

1H H
By Eq. (13.5), y1 = y2 = = 0.12
1  x  2˜ H 1  x  2˜ H

y3 = 2˜ y2 = 0.24
From data in Table C.4,
J J
'H298  235030˜ 'G298  166365˜
mol mol
The following vectors represent the species of the reaction in the order in
which they appear:
§¨ 1 · §¨ 1.935 · §¨ 36.915 ·
3
Q ¨1 ¸ A  ¨ 2.734 ¸ B  ¨ 26.786 ¸ ˜ 10
¨2 ¨ 3.249 ¨ 0.422
© ¹ © ¹ © ¹

497
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 §¨ 11.402 · §¨ 0.0 ·
6 5
C  ¨ 8.882 ¸ ˜ 10 D  ¨ 0.0 ¸ ˜ 10
¨ 0.0 ¨ 0.083
© ¹ © ¹
end  rows ( A) i  1  end

'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'C 
¦ Qi˜Ci 'D 
¦ Qi˜ Di
i i i i
3 6 4
'A 7.297 'B 9.285 u 10 'C 2.52 u 10 'D 1.66 u 10

T  925˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 9.242 u 10
3 J
K  exp ¨
§ 'G · K 0.30066
mol © R˜ T ¹
( 0.12) ˜ ( 0.24) ˜ 1  x  2˜ H
2
By Eq. (13.28), = K
1H
K
Because 0.12˜H1  x  2˜ H = H
2
K  ( 0.24)
H
x  1  2˜ H x 4.3151 H 0.839
0.12

1H
(a) y1  yH2O  1  0.36  y1
1  x  2˜ H

y1 0.023 yH2O 0.617 Ans.

4.3151
(b) ysteam  ysteam 0.812 Ans.
5.3151

498
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 13.20 1/2N2(g) + 3/2H2(g) = NH3(g) Q = 1
Basis: 1/2 mol N2, 3/2 mol H2 feed n0 = 2
This is the reaction of Pb. 4.21(a) with all stoichiometric coefficients divided
by two. From the answers to Pbs. 4.21(a), 4.22(a), and 13.7(a) ALL
DIVIDED BY 2, find the following values:

J J
'H298  46110˜ 'G298  16450˜
mol mol
3 5
'A  2.9355 'B  2.0905˜ 10 'C  0 'D  0.3305˜ 10
(a) T  300˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 1.627 u 10
4 J
K  exp ¨
§ 'G · K 679.57
mol © R˜ T ¹
P 1 P0  1

From Pb. 13.9 for ideal gases:

 0.5
§
H  1  ¨ 1  1.299˜ K˜
P·
H 0.9664
© P0 ¹
H
yNH3  yNH3 0.9349 Ans.
2H

(b) For yNH3 = 0.5 by the preceding equation

2
H Solving the next-to-last equation for K with P = P0 gives:
3
2
§ 1 · 1
¨
1  H¹
K © K 6.1586
1.299

499
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 Find by trial the value of T for which this is correct. It turns out to be

T = 399.5˜ kelvin Ans.

(c) For P = 100, the preceding equation becomes

2
§ 1 · 1
¨
K
©1  H¹ K 0.06159
129.9
Another solution by trial for T yields T = 577.6˜ kelvin Ans.
(d) Eq. (13.27) applies, and requires fugacity coefficients, which can be
evaluated by the generalized second-virial correlation. Since iteration
will be necessary, we assume a starting T of 583 K for which:

T  583kelvin P  100bar

For NH3(1): Tc1  405.7kelvin Pc1  112.8bar Z 1  0.253

T P
Tr1  Tr1 1.437 Pr1  Pr1 0.887
Tc1 Pc1

For N2(2): Tc2  126.2kelvin Pc2  34.0bar Z 2  0.038

583 100
Tr2  Tr2 4.62 Pr2  Pr2 2.941
126.2 34.0
For H2(3), estimate critical constants using Eqns. (3.58) and (3.59)

Tc3  §
43.6 · kelvin Tc3 42.806 K
¨ 21.8
¨1  ¸
T
¨ 2.016 T
© kelvin ¹ Tr3  Tr3 13.62
Tc3

20.5
Pc3  ˜ bar Pc3 19.757 bar
44.2
1 P
T Pr3  Pr3 5.061
2.016 Pc3
kelvin

Z3  0

500
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 Therefore, i  1  3

§ PHIBTr1 ZPr1  1 · §¨ 0.924 ·

¨
I  ¨ PHIBTr2 ZPr2  2 ¸ I ¨ 1.034 ¸
¨ ¨ 1.029
© PHIBTr3 ZPr3  3 ¹ © ¹

§¨ 1 · Qi
Q  ¨ 0.5 ¸ – Ii 1.184
¨ 1.5 i
© ¹
The expression used for K in Part (c) now becomes:
2
§ 1 · 1
¨
1  H¹
K © K 0.07292
§ 129.9 ·
¨
© 1.184 ¹
Another solution by trial for T yields T = 568.6˜ K Ans.

Of course, the INITIAL assumption made for T was not so close to the
final T as is shown here, and several trials were in fact made, but not
shown here. The trials are made by simply changing numbers in the
given expressions, without reproducing them.

13.21 CO(g) + 2H2(g) = CH3OH(g) Q = 2

Basis: 1 mol CO, 2 mol H2 feed n0 = 3
From the data of Table C.4,

J J
'H298  90135˜ 'G298  24791˜
mol mol

This is the reaction of Ex. 4.6, Pg. 142 from which:

3 6 5
'A  7.663 'B  10.815˜ 10 'C  3.45˜ 10 'D  0.135˜ 10

(a) T  300˜ kelvin T0  298.15˜ kelvin

501
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 'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 2.439 u 10
4 J § 'G ·
K  exp ¨ K 1.762 u 10
4
mol © R˜ T ¹
P 1 P0  1

By Eq. (13.5), with the species numbered in the order in which they appear
in the reaction,
1H 2  2˜ H H
y1 = y2 = y3 =
3  2˜ H 3  2˜ H 3  2˜ H

By Eq. (13.28), H  0.8 (guess)

H ˜ 3  2˜ H
2 2
Given
§ P · ˜K
= ¨ H  Find H H 0.9752
4˜ 1  H
3 © P0 ¹
H
y3  y3 0.9291 Ans.
3  2˜ H
(b) y3  0.5 By the preceding equation

3˜ y3
H H 0.75
2˜ y3  1

Solution of the equilibrium equation for K gives

H ˜ 3  2˜ H
2
K K 27
4˜ 1  H
3

Find by trial the value of T for which this is correct. It turns out to be:

T = 364.47˜ kelvin Ans.

502
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 (c) For P = 100 bar, the preceding equation becomes

H ˜ 3  2˜ H
2
2 3
K ˜ 100 K 2.7 u 10
4˜ 1  H
3

Another solution by trial for T yields T = 516.48˜ kelvin Ans.

(d) Eq. (13.27) applies, and requires fugacity coefficients. Since iteration
will be necessary, assume a starting T of 528 K, for which:
T  528kelvin P  100bar

For CO(1): Tc1  132.9kelvin Pc1  34.99bar Z 1  0.048

T P
Tr1  Tr1 3.973 Pr1  Pr1 2.858
Tc1 Pc1

For CH3OH(3): Tc3  512.6kelvin Pc3  80.97bar Z 3  0.564

T P
Tr  Tr 1.03 Pr  Pr 1.235
Tc3 Pc3

By Eq. (11.67) and data from Tables E.15 & E.16.

Z3
I 3  0.6206˜ 0.9763 I3 0.612

For H2(2), the reduced temperature is so large that it may be

assumed ideal: I 

Therefore: i  1  3
§ PHIBTr1 ZPr1  1 · §¨ 1.032 ·
¨
I ¨ 1.0 ¸ I ¨ 1 ¸
¨ ¨ 0.612
© 0.612 ¹ © ¹

§¨ 1 ·
Qi
Q  ¨ 2 ¸
¨1
– Ii 0.5933
© ¹ i

503
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 The expression used for K in Part (c) now becomes:

H ˜ 3  2˜ H
2
2 3
K ˜ 100 ˜ 0.593 K 1.6011 u 10
4˜ 1  H
3

Another solution by trial for T yields: T = 528.7˜ kelvin Ans.

13.22 CaCO3(s) = CaO(s) + CO2(g)

Each species exists PURE as an individual phase, for which the activity is
f/f0. For the two species existing as solid phases, f and f0 are for practical
purposes the same, and the activity is unity. If the pure CO2 is assumed
an ideal gas at 1(atm), then for CO2 the activity is f/f0 = P/P0 = P (in bar).
As a result, Eq. (13.10) becomes K = P = 1.0133, and we must find the T
for which K has this value.
From the data of Table C.4,
J J
'H298  178321˜ 'G298  130401˜
mol mol
The following vectors represent the species of the reaction in the order in
which they appear:
§¨ 1 · §¨ 12.572 · §¨ 2.637 · §¨ 3.120 ·
3 5
Q ¨1 ¸ A  ¨ 6.104 ¸ B  ¨ 0.443 ¸ ˜ 10 D  ¨ 1.047 ¸ ˜ 10
¨1 ¨ 5.457 ¨ 1.045 ¨ 1.157
© ¹ © ¹ © ¹ © ¹
i  1  3 'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'D 
¦ Qi˜ Di
i i i
3 4
'A 1.011 'B 1.149 u 10 'C  0 'D 9.16 u 10

T  1151.83˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

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 'G 126.324
J
K  exp ¨
§ 'G · K 1.0133
mol © R˜ T ¹
Thus T = 1151.83˜ kelvin Ans.

Although a number of trials were required to reach this result, only the
final trial is shown. A handbook value for this temperature is 1171 K.

13.23 NH4Cl(s) = NH3(g) + HCl(g)

The NH4Cl exists PURE as a solid phase, for which the activity is f/f0.
Since f and f0 are for practical purposes the same, the activity is unity. If
the equimolar mixture of NH3 and HCl is assumed an ideal gas mixture at
1.5 bar, then with f0 = 1 bar the activity of each gas species is its partial
pressure, (0.5)(1.5) = 0.75. As a result, Eq. (13.10) becomes K =
(0.75)(0.75) = 0.5625 , and we must find the T for which K has this value.
From the given data and the data of Table C.4,
J J
'H298  176013˜ 'G298  91121˜
mol mol

The following vectors represent the species of the reaction in the order in
which they appear:

§¨ 1 · §¨ 5.939 · §¨ 16.105 · §¨ 0.0 ·

3 5
Q ¨1 ¸ A  ¨ 3.578 ¸ B  ¨ 3.020 ¸ ˜ 10 D  ¨ 0.186 ¸ ˜ 10
¨1 ¨ 3.156 ¨ 0.623 ¨ 0.151
© ¹ © ¹ © ¹ © ¹
i  1  3 'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'D 
¦ Qi˜ Di
i i i
3
'A 0.795 'B 0.012462 'C  0 'D 3.5 u 10

T  623.97˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

505
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 'G 2.986 u 10
3 J
K  exp ¨
§ 'G · K 0.5624
mol © R˜ T ¹
Thus T = 623.97˜ K Ans.

Although a number of trials were required to reach this result, only the
final trial is shown.

13.25 NO(g) + (1/2)O2(g) = NO2(g) Q = 0.5

yNO2 yNO2
= = K T  298.15˜ kelvin
yNO˜ yO2
0.5 0.5
yNO˜ (0.21)

J
From the data of Table C.4, 'G298  35240˜
mol
§ 'G298 ·
K  exp ¨ K 1.493 u 10
6
© R˜ T ¹
 12 6
yNO  10 yNO2  10 (guesses)

0.5 6
Given yNO2 = (0.21) ˜ K˜ yNO yNO2  yNO = 5˜ 10

§ yNO ·  12
¨  FindyNO  yNO2 yNO 7.307 u 10
© y NO2 ¹
6
This is about 7˜ 10 ppm (a negligible concentration) Ans.

13.26 C2H4(g) + (1/2)O2(g) = <(CH2)2>O(g) Q = 0.5

See Example 13.9, Pg. 508-510 From Table C.4,
J J
'H298  105140˜ 'G298  81470˜
mol mol
Basis: 1 mol C2H4 entering reactor.
Moles O2 entering: nO2  1.25˜ 0.5
79
Moles N2 entering: nN2  nO2˜
21

506
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 n0  1  nO2  nN2 n0 3.976
Index the product species with the numbers:
1 = ethylene
2 = oxygen
3 = ethylene oxide
4 = nitrogen
The numbers of moles in the product stream are given by Eq. (13.5).

For the product stream, data from Table C.1:

Guess: H  0.8

§ 1H · § 1.424 · § 14.394 ·

¨ ¨ ¨
nO2  0.5˜ H 3.639 ¸ 0.506 ¸ 10 3
n H  ¨ ¸ A ¨ B ¨ ˜
¨ H ¸ ¨ 0.385 ¸ ¨ 23.463 ¸ kelvin
¨ ¨ ¨
© nN2 ¹ © 3.280 ¹ © 0.593 ¹

§ 4.392 · § 0.0 · § 1 ·
¨ ¨ ¨
0.0 ¸ 10 6 0.227 ¸ 5 0.5 ¸
C ¨ D ¨ Q ¨
2
˜ ˜ 10 ˜ kelvin
¨ 9.296 ¸ kelvin2 ¨ 0.0 ¸ ¨ 1 ¸
¨ ¨ ¨
© 0.0 ¹ © 0.040 ¹ © 0 ¹
i  1  4 A H 
¦ n H i˜Ai B H 
¦ nH i˜ Bi
i i

C H 
¦ nH i˜ Ci D H 
¦ n H i˜Di
i i
n H Qi
y H 
n0  0.5˜ H
K H 
– yH i K H 15.947
i
The energy balance for the adiabatic reactor is:

'H298  'HP = 0 For the second term, we combine Eqs. (4.3) & (4.7).

The three equations together provide the energy balance.

507
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 For the equilibrium state, apply a combination of Eqs. (13.11a) &
(13.18).The reaction considered here is that of Pb. 4.21(g), for which the
following values are given in Pb. 4.23(g):
3 6
10 10
'A  3.629 'B  8.816˜ 'C  4.904˜
kelvin 2
kelvin
5 2
'D  0.114˜ 10 ˜ kelvin T0  298.15˜ kelvin

Guess: W 3

ª
idcph  « 'A˜WT0˜   1 
'B 2 2
˜WT0 ˜
º
 1  »
2
« 'C 3 3 'D § W  1 · »
« ˜W
T0 ˜ 
1  ˜¨ »
¬ 3 T0 © W ¹ ¼

ln W  « B˜'
idcps  'A˜'
ª
T0  ª C˜ T0 
2 º ˜ §¨ W  1 · º ˜ W  1
« » © 2 ¹»
« «  'D » »
« « W˜T 2 » »
¬ ¬ 0  ¼ ¼

idcph 130.182 kelvin idcps 0.417

Given

B H
ª
'H298 = R˜ « A H ˜WT0˜   1 
2 2
˜WT0 ˜
º
 1  » 
2
« C H D H § W  1 · »
«
3 3
˜WT0 ˜ 
1  ˜¨ »
¬ 3 T 0 © W ¹ ¼
ª§ 'H298  'G298 'H298 · 1 º
K H = exp «¨   idcps  ˜ idcph»
¬© R˜ T0 R˜WT0˜ ¹ T 0˜ W ¼

§H · §H · § 0.88244 ·
¨  Find HW ¨ ¨
©W ¹ ©W ¹ © 3.18374 ¹

508
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 § 0.0333 ·
¨
0.052 ¸
y(0.88244) ¨ Ans.
¨ 0.2496 ¸
¨
© 0.6651 ¹
T  W ˜ T0 T 949.23 kelvin Ans.

13.27 CH4(g) = C(s) + 2H2(g) Q= 1 (gases only)

The carbon exists PURE as an individual phase, for which the activity is
unity. Thus we leave it out of consideration.

From the data of Table C.4,

J J
'H298  74520˜ 'G298  50460˜
mol mol
The following vectors represent the species of the reaction in the order in
which they appear:

§¨ 1 · §¨ 1.702 · §¨ 9.081 ·
3
Q ¨1 ¸ A  ¨ 1.771 ¸ B  ¨ 0.771 ¸ ˜ 10
¨2 ¨ 3.249 ¨ 0.422
© ¹ © ¹ © ¹
§¨ 2.164 · §¨ 0.0 ·
6 5
i  1  3 C  ¨ 0.0 ¸ ˜ 10 D  ¨ 0.867 ¸ ˜ 10
¨ 0.0 ¨ 0.083
© ¹ © ¹

'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'C 
¦ Qi˜Ci 'D 
¦ Qi˜ Di
i i i i
3 6 4
'A 6.567 'B 7.466 u 10 'C 2.164 u 10 'D 7.01 u 10

T  923.15˜ kelvin T0  298.15˜ kelvin

509
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 'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 1.109 u 10
4 J
K  exp ¨
§ 'G · K 4.2392
mol © R˜ T ¹
1H 2˜ H
By Eq. (13.5), n0 = 1 yCH4 = yH2 =
1H 1H

(a) By Eq. (13.28),

2˜ H 2 4˜ H
2
= = K
1  H ˜ 1  H 1H
2

K
H H 0.7173 (fraction decomposed)
4K

1H 2˜ H
yCH4  yH2  yCH4 0.1646
1H 1H
Ans.
yH2 0.8354

(b) For a feed of 1 mol CH4 and 1 mol N2, n0 = 2

By Eq. (13.28), H  .8 (guess)

2˜ H 2
Given = K H  Find H
2  H ˜ 1  H
H 0.7893 (fraction decomposed)

1H 2˜ H
yCH4  yH2  yN2  1  yCH4  yH2
2H 2H

yH2 0.5659 yCH4 0.0756 yN2 0.3585 Ans.

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 13.28 1/2N2(g) + 1/2O2(g) = NO(g) Q= 0 (1)

This is the reaction of Pb. 4.21(n) with all stoichiometric coefficients

divided by two. From the answers to Pbs. 4.21(n), 4.22(n), and 13.7(n) ALL
DIVIDED BY 2, find the following values:

J J
'H298  90250˜ 'G298  86550˜
mol mol
3 5
'A  0.0725 'B  0.0795˜ 10 'C  0 'D  0.1075˜ 10

T  2000˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 6.501 u 10
4 J
K1  exp ¨
§ 'G · K1 0.02004
mol © R˜ T ¹
1/2N2(g) + O2(g) = NO2(g) Q = 0.5 (2)

From the data of Table C.4,

J J
'H298  33180˜ 'G298  51310˜
mol mol

The following vectors represent the species of the reaction in the order in
which they appear:

§¨ 0.5 · §¨ 3.280 · §¨ 0.593 · §¨ 0.040 ·

3 5
Q  ¨ 1 ¸ A  ¨ 3.639 ¸ B  ¨ 0.506 ¸ ˜ 10 D  ¨ 0.227 ¸ ˜ 10
¨ 1 ¨ 4.982 ¨ 1.195 ¨ 0.792
© ¹ © ¹ © ¹ © ¹

i  1  3 'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'D 
¦ Qi˜ Di
i i i
4 4
'A 0.297 'B 3.925 u 10 'C  0 'D 5.85 u 10

T  2000˜ kelvin T0  298.15˜ kelvin

511
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 'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 1.592 u 10
5 J
K2  exp ¨
§ 'G · K2 6.9373 u 10
5
mol © R˜ T ¹

With the assumption of ideal gases, we apply Eq. (13.28):

yNO yNO
(1) = = K1
yN2 0.5
˜ yO2
0.5 0.5
(0.7) ˜ (0.05)
0.5

0.5 0.5 3
yNO  K1˜ (0.7) ˜ (0.05) yNO 3.74962 u 10 Ans.

(2) P0  1 P  200

0.5
= §¨
yNO2 yNO2 P·
= ˜ K2
yN2 0.5˜ yO2 0.5
(0.7) ˜ (0.05) © P0 ¹

0.5
yNO2  §¨
P· 0.5 5
˜ K2˜ (0.7) ˜ (0.05) yNO2 4.104 u 10 Ans.
© P0 ¹

13.29 2H2S(g) + SO2(g) = 3S(s) + 2H2O(g)

The sulfur exists PURE as a solid phase, for which the activity is f/f0. Since
f and f0 are for practical purposes the same, the activity is unity, and it is
omitted from the equilibrium equation. Thus for the gases only,

Q = 1
From the given data and the data of Table C.4,
J J
'H298  145546˜ 'G298  89830˜
mol mol

512
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 The following vectors represent the species of the reaction in the order in
which they appear:

§ 2 · § 3.931 · § 1.490 · § 0.232 ·

¨ ¨ ¨ ¨
1 5.699 ¸ 0.801 ¸  3 ¨ 1.015 ¸ ˜ 105
Q ¨ ¸ A ¨ B ¨ ˜ 10 D
¨3 ¸ ¨ 4.114 ¸ ¨ 1.728 ¸ ¨ 0.783 ¸
¨ ¨ ¨ ¨
©2 ¹ © 3.470 ¹ © 1.450 ¹ © 0.121 ¹

i  1  4 'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'D 
¦ Qi˜ Di
i i i
3 4
'A 5.721 'B 6.065 u 10 'C  0 'D 6.28 u 10

T  723.15˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 1.538 u 10
4 J
K  exp ¨
§ 'G · K 12.9169
mol © R˜ T ¹
By Eq. (13.5), gases only: n0 = 3 (basis)

2  2˜ H 1H 2˜ H
yH2S = ySO2 = yH2O =
3H 3H 3H

By Eq. (13.28), H  0.5 (guess)

2˜ H 2˜ 3  H
Given = 8˜ K H  Find H H 0.767
2  2˜ H 2˜ 1  H
Percent conversion of reactants = PC

ni0  ni HQ i˜
PC = ˜ 100 = ˜ 100 [By Eq. (13.4)]
ni0 ni0

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 Since the reactants are present in the stoichiometric proportions, for each
reactant,
ni0 = Q i Whence PC  H ˜ 100 PC 76.667 Ans.

13.30 N2O4(g) = 2NO2(g)

Q= 1
(a) (b)
Data from Tables C.4 and C.1 provide the following values:

J J
'H298  57200˜ 'G298  5080˜
mol mol

T0  298.15˜ kelvin T  350˜ kelvin

3 5
'A  1.696 'B  0.133˜ 10 'C  0 'D  1.203˜ 10

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 3.968 u 10
3 J
K  exp ¨
§ 'G · K 3.911
mol © R˜ T ¹

Basis: 1 mol species (a) initially. Then

1
1H 2˜ H 2˜ H 2 § P · ˜K
ya = yb = = ¨
1H 1H 1  H ˜ 1  H © P0 ¹

K
(a) P 5 P0  1 H H 0.4044
4˜ P  K
1H
ya  ya 0.4241 Ans.
1H
K
(b) P 1 P0  1 H H 0.7031
4˜ P  K

514
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 By Eq. (4.18), at 350 K:


'H  'H298  R˜ IDCPH T0 'T ' A ' B ' C  D 'H 56984
J
mol
This is Q per mol of reaction, which is

'H  0.7031  0.4044 'H 0.299

J
Whence Q  'H˜ 'H Q 17021 Ans.
mol

13.31 By Eq. (13.32),

xB˜ J B 1  xA ˜ J B
K= =
xA˜ J A xA˜ J A


ln J a = 0.1˜ xB
2

ln J b = 0.1˜ xA
2
Whence

§ 1  xA · exp 0.1˜ xA
K= ¨ ˜
2

=
1  xA
˜ exp ª¬0.1˜ xA  xB º¼
2 2

© xA ¹ exp 0.1˜ xB 2
 xA

1  xA § 'G ·
K= ˜ exp ª¬0.1˜ 2˜ xA  1 º¼ K = exp ¨
xA © R˜ T ¹
J
'G  1000˜ T  298.15˜ kelvin
mol

xA  .5 (guess)

1  xA § 'G ·
Given ˜ exp ª¬0.1˜ 2˜ xA  1 º¼ = exp ¨ xA  FindxA
xA © R˜ T ¹
xA 0.3955 Ans.

For an ideal solution, the exponential term is unity:

1  xA § 'G ·
Given = exp ¨ xA  FindxA xA 0.4005
xA © R˜ T ¹
This result is high by 0.0050. Ans.

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 13.32 H2O(g) + CO(g) = H2(g) + CO2(g) Q= 0
From the the data of Table C.4,
J J
'H298  41166˜ 'G298  28618˜
mol mol

T0  298.15˜ kelvin T  800˜ kelvin

3 5
'A  1.860 'B  0.540˜ 10 'C  0 'D  1.164˜ 10

'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 9.668 u 10
3 J
K  exp ¨
§ 'G · K 4.27837
mol © R˜ T ¹
(a) No. Since Q= 0 , at low pressures P has no effect

(b) No. K decreases with increasing T. (The standard heat of reaction is

negative.).

(c) Basis: 1 mol CO, 1 mol H2, w mol H2O feed.

From the problem statement,

nCO
= 0.02
nCO  nH2  nCO2

By Eq. (13.4), nCO = 1  H nH2 = 1  H NCO2 = H

1H 1H 0.96

= = 0.02 H H 0.941
1  H H1 H  2H 1.02

Let z = w/2 = moles H2O/mole "Water gas".

By Eq. (13.5),
wH 2˜Hz 1H 1H
yH2O = = yCO = yH2 =
2w 2  2˜ z 2  2˜ z 2  2˜ z

516
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 H
yCO2 = By Eq. (13.28) z 2 (guess)
2  2˜ z

Given H ˜ 1  H
= K z  Find()
z z 4.1 Ans.
2˜Hz  ˜ 1  H

(d) 2CO(g) = CO2(g) + C(s) Q = 1 (gases)

Data from Tables C.4 and C.1:
J J
'H298  172459˜ 'G298  120021˜
mol mol

3 5
'A  0.476 'B  0.702˜ 10 'C  0 'D  1.962˜ 10

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 3.074 u 10
4 J
K  exp ¨
§ 'G · K 101.7
mol © R˜ T ¹
By Eq. (13.28), gases only, with P = P0 = 1 bar
yCO2
= K = 101.7 for the reaction AT EQUILIBRIUM.
yCO 2

If the ACTUAL value of this ratio is GREATER than this value, the
reaction tries to shift left to reduce the ratio. But if no carbon is present, no
reaction is possible, and certainly no carbon is formed. The actual value of
the ratio in the equilibrium mixture of Part (c) is
H 1H
yCO2  yCO 
2  2˜ z 2  2˜ z
3
yCO2 0.092 yCO 5.767 u 10

yCO2 3
RATIO  RATIO 2.775 u 10
yCO 2
No carbon can deposit from the equilibrium mixture.

517
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 13.33 CO(g) + 2H2(g) = CH3OH(g) Q = 2 (1)

This is the reaction of Pb. 13.21, where the following parameter values are
given:
J J
'H298  90135˜ 'G298  24791˜
mol mol

T  550˜ kelvin T0  298.15˜ kelvin

3 6 5
'A  7.663 'B  10.815˜ 10 'C  3.45˜ 10 'D  0.135˜ 10

'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 3.339 u 10
4 J
K1  exp ¨
§ 'G · K1 6.749 u 10
4
mol © R˜ T ¹
H2(g) + CO2(g) = CO(g) + H2O(g) Q= 0 (2)

From the the data of Table C.4,

J J
'H298  41166˜ 'G298  28618˜
mol mol

T  550˜ kelvin T0  298.15˜ kelvin

The following vectors represent the species of the reaction in the order in
which they appear:
§ 1 · § 3.249 · § 0.422 · § 0.083 ·
¨ ¨ ¨ ¨
1 5.457 ¸ 1.045 ¸  3 ¨ 1.157 ¸ ˜ 105
Q ¨ ¸ A ¨ B ¨ ˜ 10 D
¨1 ¸ ¨ 3.376 ¸ ¨ 0.557 ¸ ¨ 0.031 ¸
¨ ¨ ¨ ¨
©1 ¹ © 3.470 ¹ © 1.450 ¹ © 0.121 ¹

i  1  4 'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'D 
¦ Qi˜ Di
i i i
4 5
'A 1.86 'B 5.4 u 10 'C  0 'D 1.164 u 10
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 'G  'H298 
T0
T

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 1.856 u 10
4 J
K2  exp ¨
§ 'G · K2 0.01726
mol © R˜ T ¹

Basis: 1 mole of feed gas containing 0.75 mol H2,

0.15 mol CO, 0.05 mol CO2, and 0.05 mol N2.
Stoichiometric numbers, Q i.j
i= H2 CO CO2 CH3OH H2O
_______________________________________________
j
1 -2 -1 0 1 0
2 -1 1 -1 0 1

By Eq. (13.7)

0.75 H2˜ H 1  2 0.15 HH 1  2

yH2 = yCO =
1  2˜ H 1 1  2˜ H 1

0.05  H 2 H1 H2
yCO2 = yCH3OH = yH2O =
1  2˜ H 1 1  2˜ H 1 1  2˜ H 1

P  100 P0  1
By Eq. (13.40), H 1  0.1 H 2  0.1 (guesses)
Given


H 1˜ 1  2˜ H 1
2
= §¨
P·
2
˜ K1
0.75 H2˜ H 1  2
2

˜ 0.15 HH 1  2 © P0 ¹

0.15 HH 1  2 ˜H2

= K2
§¨ H 1 ·
 Find H 1  H 2
0.75 H2˜ H 1  
2 ˜ 0.05  H 2 ¨ H2
© ¹

519
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 3
H1 0.1186 H2 8.8812 u 10

0.75 H2˜ H 1  2 0.15 HH 1  2

yH2  yCO 
1  2˜ H 1 1  2˜ H 1

0.05  H 2 H1 H2
yCO2  yCH3OH  yH2O 
1  2˜ H 1 1  2˜ H 1 1  2˜ H 1

yN2  1  yH2  yCO  yCO2  yCH3OH  yH2O

yH2 0.6606 yCO 0.0528 yCO2 0.0539

Ans.
yCH3OH 0.1555 yH2O 0.0116 yN2 0.0655

13.34 CH4(g) + H2O(g) = CO(g) + 3H2(g) Q= 2 (1)

From the the data of Table C.4,

J J
'H298  205813˜ 'G298  141863˜
mol mol
The following vectors represent the species of the reaction in the order in
which they appear:

§ 1 · § 1.702 · § 9.081 ·
¨ ¨ ¨
1 3.470 ¸ 1.450 ¸  3
Q ¨ ¸ A ¨ B ¨ ˜ 10
¨1 ¸ ¨ 3.376 ¸ ¨ 0.557 ¸
¨ ¨ ¨
©3 ¹ © 3.249 ¹ © 0.422 ¹
§ 2.164 · § 0.0 ·
¨ ¨
0.0 ¸  6 0.121 ¸ 5
C ¨ ˜ 10 D ¨ ˜ 10 i  1  4
¨ 0.0 ¸ ¨ 0.031 ¸
¨ ¨
© 0.0 ¹ © 0.083 ¹
'A 
¦ Qi˜ Ai 'B 
¦ Qi˜Bi 'C 
¦ Qi˜Ci 'D 
¦ Qi˜ Di
i i i i

3 6 3
'A 7.951 'B 8.708 u 10 'C 2.164 u 10 'D 9.7 u 10
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 T  1300˜ kelvin T0  298.15˜ kelvin

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 1.031 u 10
5 J
K1  exp ¨
§ 'G · K1 13845
mol © R˜ T ¹

H2O(g) + CO(g) = H2(g) + CO2(g) Q = 0 (2)

This is the reaction of Pb. 13.32, where parameter values are given:

J J
'H298  41166˜ 'G298  28618˜
mol mol

3 5
'A  1.860 'B  0.540˜ 10 'C  0.0 'D  1.164˜ 10

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

'G 5.892 u 10
3 J
K2  exp ¨
§ 'G · K2 0.5798
mol © R˜ T ¹

(a) No. Primary reaction (1) shifts left with increasing P.

(b) No. Primary reaction (1) shifts left with increasing T.

(c) The value of K1 is so large compared with the value of K2 that for all
practical purposes reaction (1) may be considered to go to
completion. With a feed equimolar in CH4 and H2O, no H2O then
remains for reaction (2). In this event the ratio, moles H2/moles CO
is very nearly equal to 3.0.

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 (d) With H2O present in an amount greater than the stoichiometric
ratio, reaction (2) becomes important. However, reaction (1) for all
practical purposes still goes to completion, and may be considered
to provide the feed for reaction (2). On the basis of 1 mol CH4 and
2 mol H2O initially, what is left as feed for reaction (2) is: 1 mol
H2O, 1 mol CO, and 3 mol H2; n0 = 5. Thus, for reaction (2) at
equilibrium by Eq. (13.5):

1H H 3H
yCO = yH2O = yCO2 = yH2 =
5 5 5
By Eq. (13.28), H  0.5 (guess)

H ˜ 3  H
Given = K2 H  Find H H 0.1375
1  H 2

yH2 3H
Ratio = Ratio  Ratio 3.638 Ans.
yCO 1H

(e) One practical way is to add CO2 to the feed. Some H2 then reacts
with the CO2 by reaction (2) to form additional CO and to lower the
H2/CO ratio.

(f) 2CO(g) = CO2(g) + C(s) Q = 1 (gases)

This reaction is considered in the preceding problem, Part (d), from
which we get the necessary parameter values:

J J
'H298  172459˜ 'G298  120021˜
mol mol

For T = 1300 K, T  1300˜ kelvin T0  298.15˜ kelvin

3 5
'A  0.476 'B  0.702˜ 10 'C  0.0 'D  1.962˜ 10

'G  'H298 
T
T0

˜ 'H298  'G298 


 R˜ IDCPH T0 'T ' A ' B ' C  D 

 R˜ T˜ IDCPS T0 'T ' A ' B ' C  D

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 'G 5.673 u 10
4 J
K  exp ¨
§ 'G · K 5.255685 u 10
3
mol © R˜ T ¹

As explained in Problem 13.32(d), the question of carbon deposition

depends on:
yCO2
RATIO =
yCO 2
When for ACTUAL compositions the value of this ratio is greater than the
equilibrium value as given by K, there can be no carbon deposition. Thus
in Part (c), where the CO2 mole fraction approaches zero, there is danger
of carbon deposition. However, in Part (d) there can be no carbon
deposition, because Ratio > K:

H˜5
Ratio  Ratio 0.924
1  H 2

13.37 Formation reactions:

C + 2H2 = CH4

H2 + (1/2)O2 = H2O

C + (1/2)O2 = CO

C + O2 = CO2

Elimination first of C and then of O2 leads to a pair of reactions:

CH4 + H2O = CO + 3H2 (1)
CO + H2O = CO2 + H2 (2)

There are alternative equivalent pairs, but for these:

Stoichiometric numbers, Q i.j

i= CH4 H2O CO CO2 H2 Qj

_________________________________________________
j
1 -1 -1 1 0 3 2
2 0 -1 -1 1 1 0
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 For initial amounts: 2 mol CH4 and 3 mol H2O, n0 = 5, and by Eq. (13.7):
2  H1 3 HH 1  2 H1  H2
yCH4 = yH2O = yCO =
5  2˜ H 1 5  2˜ H 1 5  2˜ H 1
H2 3˜HH 1  2
yCO2 = yH2 =
5  2˜ H 1 5  2˜ H 1

By Eq. (13.40), with P = P0 = 1 bar

yCO˜ yH2
3
yCO2˜ yH2
= k1 = k2
yCH4˜ yH2O yCO˜ yH2O

From the data given in Example 13.14,

J J
'G1  27540˜ 'G2  3130˜ T  1000˜ kelvin
mol mol

§ 'G1 · § 'G2 ·
K1  exp ¨ K2  exp ¨
© R˜ T ¹ © R˜ T ¹
K1 27.453 K2 1.457

H 1  1.5 H2  1 (guesses)

Given
H 1  H 2 ˜ 3˜HH 1  2 3 = K1
2  H 1 ˜ 3 HH 1  2 ˜ 5  2˜ H 1 2


H 2˜ 3˜HH 1  2
= K2
H 1  H 2 ˜ 3 HH 1  2

§¨ H 1 ·
 Find H 1  H 2 H1 1.8304 H2 0.3211
¨ H2
© ¹

2  H1 3 HH 1  2 H1  H2
yCH4  yH2O  yCO 
5  2˜ H 1 5  2˜ H 1 5  2˜ H 1

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 H2 3˜HH 1  2
yCO2  yH2 
5 2˜ H 1 5 2˜ H 1

yCH4 0.0196 yH2O 0.098 yCO 0.1743

yCO2 0.0371 yH2 0.6711

These results are in agreement with those of Example 13.14.

13.39Phase-equilibrium equations:
y1˜ P
Ethylene oxide(1): p1 = y1˜ P = 415x
˜ 1 P  101.33˜ kPa x1 =
˜
415kPa
y2˜ P
Water(2): x2˜ Psat2 = y2˜ P Psat2  3.166˜ kPa x2 =
Psat2
(steam tables)

Ethylene glycol(3): Psat3 = 0.0 y3 = 0.0

Therefore, y2 = 1  y1 and x3 = 1  x2  x3

For the specified standard states:

(CH2)2O(g) + H2O(l) = CH2OH.CH2OH(l)

By Eq. (13.40) and the stated assumptions,

J 3˜ x3 x3
k= =
§y ˜ P · ˜ J ˜ x y1˜ x2 T  298.15kelvin
˜
¨ 1  2 2
© P0 ¹
J
Data from Table C.4: 'G298  72941˜
mol
§ 'G298 · 12
k  exp ¨ k 6.018 u 10 Ans.
© R˜ T ¹

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 So large a value of k requires either y1 or x2 to approach zero. If y1
approaches zero, y2 approaches unity, and the phase-equilibrium
expression for water(2) makes x2 = 32, which is impossible. Thus x2
must approach zero, and the phase-equilibrium equation requires y2 also
to approach zero. This means that for all practical purposes the reaction
goes to completion. For initial amounts of 3 moles of ethylene oxide and
1 mole of water, the water present is entirely reacted along with 1 mole of
the ethylene oxide. Conversion of the oxide is therefore
33.3 %.

§ 1 1 · § 50 ·
13.41 ¨ ¨
1 0 ¸ Initial
¨ 50 ¸ kmol
a) Stoichiometric coefficients: Q  ¨ numbers of n0 
¨ 1 1 ¸ moles
¨ 0 ¸ hr
¨ ¨
©0 1 ¹ ©0¹
Number of components: i  1  4 Number of reactions: j  1  2

vj 
¦ Qi  j v
§ 1 ·
¨
© 1 ¹
n0 
¦ n0i n0 100
kmol
hr
i
i
Given values: yA  0.05 yB  0.10
kmol kmol
Guess: yC  0.4 yD  0.4 H1  1 H2  1
hr hr
Given

n01 HH 1  2 n02  H 1

yA = yB =
n0 HH 1  2 n0 HH 1  2
Eqn. (13.7)
n03 HH 1  2 n04  H 2
yC = yD =
n0 HH 1  2 n0 HH 1  2

§ yC ·
¨
¨ yD ¸
¨ ¸  FindyC HyD H 1  2 H1 44.737
kmol
H2 2.632
kmol
¨ H1 ¸ hr hr
¨H
© 2¹

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 kmol
(i) nA  n01 HH 1  2 nA 2.632
hr
kmol
nB  n02  H 1 nB 5.263
hr
kmol
nC  n03 HH 1  2 nC 42.105
hr
kmol
nD  n04  H 2 nD 2.632
hr
kmol
n  nA  nB  nC  nD n 52.632 Ans.
hr

(ii) yC 0.8 yD 0.05 Ans.

§ 1 1 · § 40 ·
¨ ¨
1 2 ¸ Initial
¨ 40 ¸ kmol
b) Stoichiometric coefficients: Q  ¨ numbers of n0 
¨1 0 ¸ moles
¨ 0 ¸ hr
¨ ¨
©0 1 ¹ ©0¹
Number of components: i  1  4 Number of reactions: j  1  2

vj 
¦ Qi  j v
§ 1 ·
¨
© 2 ¹
n0 
¦ n0i n0 80
kmol
hr
i
i
Given values: yC  0.52 yD  0.04
kmol kmol
Guess: yA  0.4 yB  0.4 H1  1 H2  1
hr hr
Given

n01 HH 1  2 n02  H 1  2H 2

yA = yB =
n0  H 1  2H 2 n0  H 1  2H 2
Eqn. (13.7)
n03  H 1 n04  H 2
yC = yD =
n0  H 1  2H 2 n0  H 1  2H 2

527
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 § yA ·
¨
¨ yB ¸
¨ ¸  FindyA HyB H 1  2 H1 26
kmol
H2 2
kmol
¨ H1 ¸ hr hr
¨H
© 2¹
yA 0.24 yB 0.2

kmol
nA  n01 HH 1  2 nA 12
hr
kmol
nB  n02  H 1  2H 2 nB 10
hr Ans.
kmol
nC  n03  H 1 nC 26
hr
kmol
nD  n04  H 2 nD 2
hr

§ 1 1 · § 100 ·
¨ ¨
1 1 ¸ Initial
¨ 0 ¸ kmol
c) Stoichiometric coefficients: Q  ¨ numbers of n0 
¨1 0 ¸ moles
¨ 0 ¸ hr
¨ ¨
©0 1 ¹ © 0 ¹
Number of components: i  1  4 Number of reactions: j  1  2

vj 
¦ Qi  j v
§1 ·
¨
© 1 ¹
n0 
¦ n0i n0 100
kmol
hr
i
i
Given values: yC  0.3 yD  0.1
kmol kmol
Guess: yA  0.4 yB  0.4 H1  1 H2  1
hr hr
Given

n01 HH 1  2 n02 HH 1  2

yA = 0 yB =
n0 HH 1  2 n0 HH 1  2
Eqn. (13.7)
n03  H 1 n04  H 2
yC = yD =
n0 HH 1  2 n0 HH 1  2

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 § yA ·
¨
¨ yB ¸
¨ ¸  FindyA HyB H 1  2 H1 37.5
kmol
H2 12.5
kmol
¨ H1 ¸ hr hr
¨H
© 2¹
yA 0.4 yB 0.2

kmol
nA  n01 HH 1  2 nA 50
hr
kmol
nB  n02 HH 1  2 nB 25
hr Ans.
kmol
nC  n03  H 1 nC 37.5
hr
kmol
nD  n04  H 2 nD 12.5
hr

§ 1 1 · § 40 ·
¨ 1 1 ¨ 60
¨ ¸ Initial ¨ ¸ kmol
d)Stoichiometric coefficients: Q  ¨ 1 0 ¸ numbers of n0  ¨0¸
¨0 1 ¸ moles ¨ 0 ¸ hr
¨ ¨
©0 1 ¹ ©0¹
Number of components: i  1  5 Number of reactions: j  1  2

vj 
¦ Qi  j v
§ 1 ·
¨
©0 ¹
n0 
¦ n0i n0 100
kmol
hr
i
i
Given values: yC  0.25 yD  0.20
kmol kmol
Guess: yA  0.2 yB  0.4 yE  0.1 H1  1 H2  1
hr hr

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 Given n01 HH 1  2 n02 HH 1  2
yA = yB = Eqn. (13.7)
n0  H 1 n0  H 1

n03  H 1 n04  H 2 n05  H 2

yC = yD = yE =
n0  H 1 n0  H 1 n0  H 1

§ yA ·
¨
¨ yB ¸
¨ yE ¸  Find y  y Hy H  H1 20
kmol
H2 16
kmol
A B E 1 2
¨ ¸ hr hr
¨ H1 ¸
¨
© H2 ¹
(i) (ii)
kmol
nA  n01 HH 1  2 nA 4 yA 0.05
hr
kmol
nB  n02 HH 1  2 nB 24 yB 0.3
hr
kmol
nC  n03  H 1 nC 20 Ans. yC 0.25
hr
kmol
nD  n04  H 2 nD 16 yD 0.2
hr
kmol
nE  n05  H 2 nE 16 yE 0.2
hr

13.45 C2H4(g) + H2O(g) -> C2H5OH(g)

T0  298.15kelvin P0  1bar T  400kelvin P  2bar

J J
1 = C2H4(g) 'H0f1  52500 'G0f1  68460
mol mol
J J
2 = H2O(g) 'H0f2  241818 'G0f2  228572
mol mol
J J
3 = C2H5OH(g) 'H0f3  235100 'G0f3  168490˜
mol mol

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 kJ
'H0  '
'H0f1 ' H0f2  H0f3 'H0 45.782
mol
kJ
'G0  '
'G0f1 ' G0f2  G0f3 'G0 8.378
mol

'A  (1.424) (3.470) (3.518) 'A 1.376

3 3
'B  [(14.394) (1.450) (20.001)]10
˜ 'B 4.157 u 10
6 6
'C  [(4.392) ()
0  (6.002)]10
˜ 'C 1.61 u 10
5 4
'D  [()
0  (0.121) () ˜
0 ]10 'D 1.21 u 10

a) K0  exp §¨
'G0 ·
Eqn. (13.21) K298  K0 K298 29.366 Ans.
© R˜ T0 ¹

b) K1  exp ª«
'H0 § T0 ·º 3
˜ ¨1  » Eqn. (13.22) K1 9.07 u 10
¬ R˜ T0 © T ¹¼

1
K2  exp §¨ IDCPH T0 'T ' A ' B ' C  D  · K2 0.989 Eqn. (13.23)
T
¨
© IDCPST0 'T ' A ' B ' C  D
 ¹
K400  K0˜ K1˜ K2 Eqn. (13.20) K400 0.263 Ans.

c) Assume as a basis there is initially 1 mol of C2H4 and 1 mol of H2O

1  He 1  He He
y1 = y2 = y3 =
2  He 2  He 2  He
y3 P
Assuming ideal gas behavior = K˜
y1˜ y2 P0
He
2H e P
Substituting results in the following expression: = K400˜
1  He 1  He P0
˜
2  He 2  He

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 Solve for He using a Mathcad solve block. Guess: H e  0.5
He
2H e
Given
1  He 1  He
= K400˜
P
P0

H e  Find H e He 0.191
˜
2  He 2  He

1  He 1  He He
y1  y2  y3 
2  He 2  He 2  He

y1 0.447 y2 0.447 y3 0.105 Ans.

d) Since Q a decrease in pressure will cause a shift on the reaction

to the left and the mole fraction of ethanol will decrease.

13.46 H2(g) + O2(g) -> H2O2(g)

kJ
'H0fH2O2  136.1064 T  298.15kelvin P  1bar
mol

J J
S0H2  130.680 S0O2  205.152
mol˜ kelvin mol˜ kelvin

J
S0H2O2  232.95
mol˜ kelvin

J
'S0fH2O2  S0H2  S0O2  S0H2O2 'S0fH2O2 102.882
mol˜ kelvin

kJ
'G0f  'H0fH2O2  T˜ 'S0fH2O2 'G0f 105.432 Ans.
mol

532
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 13.48 C3H8(g) -> C3H6(g) + H2(g) (I)
C3H8(g) -> C2H4(g) + CH4(g) (II)

T0  298.15kelvin P0  1bar T  750kelvin P  1.2bar

J J
1 = C3H8 (g) 'H0f1  104680 'G0f1  24290
mol mol

J J
2 = C3H6 (g) 'H0f2  19710 'G0f2  62205
mol mol

J J
3 = H2 (g) 'H0f3  0 'G0f3  0
mol mol
J J
4 = C2H4 (g) 'H0f4  52510 'G0f4  68460
mol mol

J J
5 = CH4 (g) 'H0f5  74520 'G0f5  50460
mol mol

Calculate equilibrium constant for reaction I:

kJ
'H0I  '
'H0f1 ' H0f2  H0f3 'H0I 124.39
mol
kJ
'G0I  '
'G0f1 ' G0f2  G0f3 'G0I 86.495
mol

'AI  (1.213) (1.637) (3.249) 'AI 3.673

3 3
'BI  [(28.785) (22.706) (0.422)]10
˜ 'BI 5.657 u 10
6 6
'CI  [(8.824) (6.915) () ˜
0 ]10 'CI 1.909 u 10
5 3
'DI  [()
0  ()
0  (0.083)]10
˜ 'DI 8.3 u 10

KI0  exp ¨
§ 'G0I ·
Eqn. (13.21) KI0 0
© R˜ T0 ¹

KI1  exp «
ª 'H0I ˜ § 1  T0 ·º 1.348 u 10
13
¨ » Eqn. (13.22) KI1
¬ R˜ T0 © T ¹¼

1
KI2  exp §¨ IDCPH T0 'T ' AI ' BI ' CI  DI  · KI2 1.714
T
¨
© IDCPST0 'T ' AI ' BI ' CI  DI
 ¹ Eqn. (13.23)

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 KI  KI0˜ KI1˜ KI2 Eqn. (13.20) KI 0.016
Calculate equilibrium constant for reaction II:
kJ
'H0II  '
'H0f1 ' H0f4  H0f5 'H0II 82.67
mol
kJ
'G0II  '
'G0f1 ' G0f4  G0f5 'G0II 42.29
mol

'AII  ( 1.213)  ( 1.424)  ( 1.702) 'AII 1.913

3 3
'BII  [ ( 28.785)  ( 14.394)  ( 9.081) ] ˜ 10 'BII 5.31 u 10
6 6
'CII  [ ( 8.824)  ( 4.392)  ( 2.164) ] ˜ 10 'CII 2.268 u 10
5
'DII  [ ( 0)  ( 0)  ( 0) ] ˜ 10 'DII 0

KII0  exp ¨
§ 'G0II · 8
© R˜ T0 ¹ Eqn. (13.21) KII0 3.897 u 10

KII1  exp «
ª 'H0II ˜ § 1  T0 ·º Eqn. (13.22) 5.322 u 10
8
¨ » KII1
¬ R˜ T0 © T ¹¼

1
KII2  exp §¨ IDCPH T0 'T ' AII ' BII ' CII  DII  KII2
· 1.028
T
¨
© IDCPS T0 'T ' AII ' BII ' CII  DII
 ¹Eqn. (13.23)
KII  KII0˜ KII1˜ KII2 Eqn. (13.20) KII 21.328

Assume an ideal gas and 1 mol of C3H8 initially.

1 HH I  II HI HI
y1 = y2 = y3 =
1 HH I  II 1 HH I  II 1 HH I  II

H II H II
y4 = y5 = Eqn. (13.7)
1 HH I  II 1 HH I  II

The equilibrium relationships are:

y2˜ y3 y4˜ y5
= KI˜ §¨
P0 ·
= KII˜ §¨
P0 ·
Eqn. (13.28)
y1 ©P¹ y1 ©P¹

534
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 Substitution yields the following equations:

§ HI ·§ HI ·
¨ ˜¨
© 1 HH I  II ¹ © 1 HH I  II ¹
= KI˜ §¨
P0 ·
§ 1 HH I  II · ©P¹
¨
© 1 HH I  II ¹

§ H II ·§ H II ·
¨ ˜¨
© 1 HH I  II ¹ © 1 HH I  II ¹
= KII˜ §¨
P0 ·
§ 1 HH I  II · ©P¹
¨
© 1 HH I  II ¹
Use a Mathcad solve block to solve these two equations for H I and H II. Note
that the equations have been rearranged to facilitate the numerical
solution.
Guess: H I  0.5 H II  0.5

Given

HI HI § P0 · § 1 HH I  II ·
˜ = KI˜ ¨ ˜¨
1 HH I  II 1 HH I  II © P ¹ 1 HH I  II
© ¹
H II H II P0 · 1 HH I  II
˜ = KII˜ §¨ ˜
1 HH I  II 1 HH I  II © P ¹ 1 HH I  II

§¨ H I ·
 Find H I  H II HI 0.026 H II 0.948
¨ H II
© ¹
1 HH I  II HI HI
y1  y2  y3 
1 HH I  II 1 HH I  II 1 HH I  II

H II H II
y4  y5 
1 HH I  II 1 HH I  II

y1 0.01298 y2 0.0132 y3 0.0132 y4 0.4803 y5 0.4803

535
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 A summary of the values for the other temperatures is given in the table below.

T= 750 K 1000 K 1250 K

y1 0.0130 0.00047 0.00006
y2 0.0132 0.034 0.0593
y3 0.0132 0.034 0.0593
y4 0.4803 0.4658 0.4407
y5 0.4803 0.4658 0.4407

13.49 n-C4H10(g) -> iso-C4H10(g)

T0  298.15kelvin P0  1bar T  425kelvin P  15bar

J J
1 = n-C4H10(g) 'H0f1  125790 'G0f1  16570
mol mol
J J
2 = iso-C4H10(g) 'H0f2  134180 'G0f2  20760
mol mol

kJ
'H0  '
'H0f1  H0f2 'H0 8.39
mol
kJ
'G0  '
'G0f1  G0f2 'G0 4.19
mol

'A  (1.935) (1.677) 'A 0.258

3 4
'B  [(36.915) (37.853)]10
˜ 'B 9.38 u 10
6 7
'C  [(11.402) (11.945)]10
˜ 'C 5.43 u 10
5
'D  [()
0  () ˜
0 ]10 'D 0

a) K0  exp §¨
'G0 ·
Eqn. (13.21) K0 5.421 Ans.
© R˜ T0 ¹

b) K1  exp ª«
'H0 § T0 ·º
˜ ¨1  » Eqn. (13.22) K1 0.364
¬ R˜ T0 © T ¹¼

1
K2  exp §¨
·
IDCPH T0 'T ' A ' B ' C  D  K2 1 Eqn. (13.23)
T
¨
©  IDCPST0 'T ' A ' B ' C  D ¹
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 Ke  K0˜ K1˜ K2 Eqn. (13.20) Ke 1.974 Ans.

Assume as a basis there is initially 1 mol of n-C4H10(g)

y1 = 1  H e y2 = H e

y2
a) Assuming ideal gas behavior = Ke
y1
He
Substitution results in the following expression: = Ke
1  H e
Solving for Ke yields the following analytical expression for He

1
He  He 0.336
1  Ke

y1  1  H e y1 0.664 y2  H e y2 0.336 Ans.

b) Assume the gas is an ideal solution. In this case Eqn. (13.27) applies.

ª § P·
Q º
« yi˜ I i = ¨
– ¬ © P0 ¹
˜K »
¼
Eqn. (13.27)
i

Substituting for yi yields:

1  H e ˜ I2 = K
H e˜ I 1
I2
This can be solved analytically for He to get: He =
I 2  Ke˜ I 1

Calculate Ii for each pure component using the PHIB function.

For n-C4H10: Z 1  0.200 Tc1  425.1kelvin Pc1  37.96bar

T P
Tr1  Tr1 1 Pr1  Pr1 0.395
Tc1 Pc1

I 1  PHIB Tr1 Z
Pr1  1 I1 0.872
For iso-C4H10: Z 2  0.181 Tc2  408.1kelvin Pc2  36.48bar
T
Tr2  Tr2 1.041 P
Tc2 Pr2  Pr2 0.411
Pc2

537
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I 2  PHIB Tr2 Z
Pr2  2 I2 0.884

I2
Solving for He yields: He  He 0.339
I 2  Ke˜ I 1

y1  1  H e y1 0.661 y2  H e y2 0.339 Ans.

The values of y1 and y2 calculated in parts a) and b) differ by less than 1%.
Therefore, the effects of vapor-phase nonidealities is here minimal.

538
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