Serie 3 Soluciones Copia

November 20, 2018 | Author: Roy Roman | Category: Chemical Equilibrium, Combustion, Gases, Methane, Chemical Reactions
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13.6

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



=



i

=

1  1  1  1

0

=

n0

=

11

=

2

yCO

=

i

By Eq, (13.5),

yH

2

=

yCO

2

1

=

yH2O

2

=

 2

With data from Example 13.13, the following vectors represent values for Parts (a) through (d):

 1000  1100  T    kelvin  1200 

  3130   150  J G     3190  mol

 1300 

 6170 

Combining Eqs. (13.5), (13.11a), and (13.28) gives

        2   2   1      1      2     2  



=

2

1  

     G   exp      R T 

2

=



exp

=

 G   R T   0.4531  0.5021    0.5399 

    1

Ans.

 0.5709  13.11 4HCl(g) + O2(g) = 2H2O(g) + 2Cl(g)



=

1

n0

=

H298   114408

6 J mol

T



773.15 kelvin

G298   75948

487

T0 J

mol



298.15 kelvin

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

 4 

 3.156 

1    

A 

2   2  

 3.639   3.470 

B 

 4.442 

  A i

  0.151  

 0.506   10 3  1.450 

D 

 0.344 

 0.089 

B 

i

  B i

i

D 

i

  D i

i

B  8  10 T T0

i

i

5

A  0.439 G  H298 

 0.227   105  0.121 

i  1  end

end  rows ( A)

A 

 0.623 

4

C  0

D  8.23  10

 H298  G298 

 R IDCPH T0 T  A  B  C  D   R  T  IDCPS T0 T  A  B  C  D J

4

G  1.267  10

K  exp

mol

 G    R T 

K  7.18041

By Eq. (13.5)

yO2

=

yHCl

1

=

yH2O

6

=

5  4  6 2  6

  0.5

Apply Eq. (13.28);

yCl2

=

2  6

(guess)

4

Given

yHCl 

  2      6     5  4    1    5  4  6

yHCl   0.3508

yO2  yO2   0.0397

=

2 K 

1 6

   Find

yH2O 

yH2O   0.3048 488

2  6

   0.793

yCl2 

yCl2   0.3048

2  6 Ans.



13.12 N2(g) + C2H2(g) = 2HCN(g)

=

n0

0

=

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

H298   42720

G298   39430

mol

3

A  0.060

B  0.173 10

T  923.15 kelvin

G  H298 

T T0

J mol 5

C  0

D  0.191 10

T0  298.15 kelvin

 H298  G298 

 R IDCPH T0 T  A  B  C  D   R  T  IDCPS T0 T  A  B  C  D 4

G  3.242  10

J

K  exp

mol

 G    R T 

K  0.01464

By Eq. (13.5),

y N2

=

1

yC2H4

2

By Eq. (13.28),

y N2 

1 2

y N2   0.4715

1

  0.5

  2     1   

Given

=

2

yHCN

=

2e 2

=



(guess)

2 =

   Find



yC2H4 

1 2

yC2H4   0.4715

   0.057

yHCN   yHCN   0.057

Ans.

Given the assumption of ideal gases, P has no effect on the equilibrium composition.

489

13.13



CH3CHO(g) + H2(g) = C2H5OH(g)

=

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

H298   68910

G298   39630

mol

3

G  H298 

mol

6

A  1.424 B  1.601 10 T  623.15 kelvin

J 5

C  0.156 10

D  0.083 10

T0  298.15 kelvin

T T0

 H298  G298 

 R IDCPH T0 T  A  B  C  D   R  T  IDCPS T0 T  A  B  C  D

 G    R T 

J

3

G  6.787  10

K  exp

mol

By Eq. (13.5), yCH3CHO

=

  2.5   1    1.5  

yCH3CHO 

2.5  

=

yCH3CHO   0.108

yH2 

=

1.5   2.5  

yC2H5OH

=

 2.5  

(guess)

3 K 

1 2.5  

yH2

  0.5

By Eq. (13.28),

Given

1

K  3.7064

1.5   2.5  

yH2   0.4053

   Find 

yC2H5OH 

   0.818  2.5  

yC2H5OH   0.4867 Ans.

If the pressure is reduced to 1 bar,

Given

  2.5   1    1.5  

yCH3CHO 

=

1 K 

1 2.5  

yCH3CHO   0.1968

yH2 

1.5   2.5  

yH2   0.4645 490

   Find 

yC2H5OH 

   0.633  2.5  

yC2H5OH   0.3387 Ans.

nSO3

By Eq. (4.18),

H753  H298  R IDCPH T 0 T  A  B  C  D

H753   98353

13.16

   0.1455

By Eq. (13.4),

=

=

J mol



C3H8(g) = C2H4(g) + CH4(g)

=

By Eq. (13.5),

yC3H8

nC3H8

n0  nC3H8 n0

1

=

yC2H4

1

Ans.

mol

1

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

Fractional conversion of C3H8 =

J

Q  14314

Q    H753

=

=

=

1

1  1   1

 1

=

yCH4



=

 1

From data in Table C.4,

H298   82670

J

G298   42290

mol

J mol

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

 1     1    1  

 1.213  A   1.424   1.702 

end  rows ( A)

A 

  A i

i  1  end

B 

i

i

A  1.913 (a)   T

 8.824  6 C   4.392   10  2.164 

 28.785  3 B   14.394   10   9.081  

  B i

i

C 

  C

i

i

i

3

B  5.31  10

 625 kelvin

i

T0  298.15 kelvin

493

6

C  2.268  10

D  0

G  H298 

T T0

 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

mol

 G    R T 

  0.5

By Eq. (13.28),

K  1.52356

(guess)

2

 1    1  

Given

   0.777

=

   Find 



This value of epsilon IS the fractional conversion. Ans. 2

(b)

   0.85

 K   1    1  

G  R T ln( K )

G  4972.3

J mol

K  2.604

Ans.

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)

=

1

Basis: 1 mole entering C2H6 + 0.5 mol H2O.

n0

=

1.5

yC2H6

=

By Eq. (13.5),

1

yH

1.5  

=

 1.5  

yC2H4

=

 1.5  

From data in Table C.4,

H298   136330

J mol

G298   100315

J mol

The following vectors represent the species of the reaction in the order in which they appear: 494

 0.0333  y (0.88244) 

 0.052   0.2496 

Ans.

 0.6651  T    T0

13.27

T  949.23kelvin Ans.



CH4(g) = C(s) + 2H2(g)

=

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,

H298   74520

J

G 298   50460

mol

J mol

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

 1     1    2  

 1.702  A   1.771   3.249 

 9.081  3 B   0.771   10  0.422 

i  1  3

 2.164  6 C   0.0   10   0.0  

  0.0   5 D   0.867   10   0.083  

A 

   A i

i

B 

i

A  6.567

  B i

i

C 

i

i

i

3

B  7.466  10

T  923.15 kelvin

   C 6

509

D 

   D i

i

i

C  2.164  10

T0  298.15 kelvin

i

4

D  7.01  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

G  1.109  10

J mol

By Eq. (13.5),

(a)

 

K  exp

n0

=

1

yCH4

4  K 

yCH4 

   0.7173

1

yH2 

1

=

K  4.2392

1

yH2

1

2  2 1    1  

By Eq. (13.28),



 G    R T 

=

4 

Given

   0.7893 yCH4 

1 2

yH2   0.5659

2



=

yCH4   0.1646

1

Ans.

yH2   0.8354 n0

  .8 =

1

2

1

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

2  2 2    1  

2 

(fraction decomposed)

2 

By Eq. (13.28),

=

=

2

(guess)

   Find 



(fraction decomposed)

yH2 

2  2

yCH4   0.0756

510

y N2  1  yCH4  yH2 y N2   0.3585

Ans.

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