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Chapter 9 Solutions Engineering and Chemical Thermodynamics 2e

Milo Koretsky Wyatt Tenhaeff School of Chemical, Biological, and Environmental Engineering Oregon State University [email protected]

9.1 You need to neglect the solid in the vapor mole fraction

K

yD 2 y A yB P

2

9.2 o Some products can form. Since g rxn  0 , K < 1. But some products form

3

9.3 o g rxn ,298 = -500 J/mol and KT = 4

4

9.4 It would not change

5

9.5 The value of the equilibrium constant does not depend on pressure; it would not change

6

9.6 (a) Yes – since there are more moles of reactant – see Example 9.5 (b) No - inspection of Appendix A3 shows that this reaction is exothermic, equilibrium conversion decreases with increasing temperature (c) No – since there are more reactants - – see Example 9.5

7

9.7 Yes, but only if it is rate limited (kinetics). Equilibrium conversion decreases for this exothermic reaction

8

9.8 (d) High pressure increases conversion, high temperature increases rate.

9

9.9

K

1 AP

10

9.10 To find the equilibrium constants we look at the concentrations at long time. Assuming ideal solutions, we get

xB 0.125  2 x A2 0.252 x 0.25 K2  C  2 xB 0.125

K1 

11

9.11 It does not change

12

9.12 No

13

9.13 Largest 0.1 m CaCl2 0.1 m NaCl 0.1 m scrose smallest

14

9.15 C on the Ga site C on the As site

15

9.16 (a) Both are approximately the same (b) Ge

16

9.16 (second one) (a) Ge (b) Ge

17

9.17 The reaction is

H2O (l) + ½ O2 (g)  H2O2 (l) The Gibbs energy of reaction is calculated as follows º grxn,298   g ºf ,298 

  g ºf ,298 

H 2O

º  grxn,298   g ºf ,298 

H 2O

H 2O2

Therefore,

 g 

º f ,298 H O 2 2

From Appendix A.3.2:

 g  º f ,298

 kJ   237.14  H 2O  mol 

 kJ   kJ     237.14   116.8   H 2O2  mol   mol   kJ   gºf ,298  H2O2  120.34  mol 

 g  º f ,298

18

9.18 We will simplify the problem by assuming the gaseous reaction mixture behaves ideally. This assumption is valid since the temperature is 1300 K and the pressure is 1 bar. We first need to obtain expressions for H , TS , and G : H  H   H 0  n I hI  n I 2 hI 2  n I0 h I 2 2

Assuming we start with 1 mole of I2 and no I, we have:

nI 2  1  

n I  2

Therefore, 1   H  2 hI  1    hI 2  hI 2  2  hI  hI 2  2   Applying the definition of the heat of formation gives: H  2h f , I

For entropy, we must also include the entropy of mixing term after the reaction has commenced:





S  n I s I  n I 2 s I 2  s mix  n I0 s I 2 2

So

   1  S   2  sI  sI 2   R nI ln yI  nI 2 ln yI2  2    





or





S   2s f , I  R nI ln yI  nI2 ln yI2    : We can find the change in Gibb’s energy from its definition: G  H  TS

Now we can create a spreadsheet that computes the desired quantities for various extents of reaction. Plotting the data in the table, we obtain 19

Thermodynamic Quantities as a Function of Extent of Reaction 160000

H 140000

S 120000

100000

(J)

80000

60000

40000

20000

G 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-20000

Extent of Reaction ()

To find the equilibrium conversion, we must locate the minimum on the plot of G. This occurs when

  0.23 At this extent of reaction yI  0.37 yI 2  0.63

20

9.19 The Gibbs energy of formation of NH3 can be calculated by considering the following reaction:

1 3 N 2  H 2  NH 3 2 2 At 298 K, º 3 J  g rxn ,298  16.45  10    mol  º 3 J  hrxn ,298  46.11  10   mol 

Therefore,

  16.4510 3  J/mol    764.76 K 298  exp   8.314  J/mol  K   298 K   We will use Equation 9.24 to calculate the equilibrium constant at 1000 K, but to do so, we need heat capacity data from Appendices A.2 and A.3:

A  2.9357 , B  0.00209 , C  0, D  33050 Substituting numerical values into Equation 9.24 and evaluating, we obtain

K1000  5.6  10 4 Therefore,

 J  º grxn,1000   8.314 1000 ln  5.6 10 4   62250   mol  Since the Gibbs energies of formation of N2 and H2 are zero at 1000 K, º grxn,1000   g ºf ,1000 

 g

º f ,1000



NH 3

NH 3

 kJ   62.25   mol 

21

9.20 First, calculate the equilibrium constant. From data in Appendix A.3 º 3 J  g rxn , 298  88.51  10   mol  º 3 J  hrxn ,298  126.1  10    mol 

Use Equation 9.20:

 h º  1 1  rxn K1000  K 298.15 exp     R  T2 T1       88.51103     126.110 3  1 1  K1000  exp    exp     1000 298.15  8.314  8.314  298.15  

K1000  1 Take 1 mole of C4H8 as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

1 11   10   yH 2  11   yC 4 H 8 

yC 4 H 10 

 11  

The gas is assumed ideal; therefore,





  yC4 H10  K  P 1   yC4 H8 yH 2         11     1  1  5 bar    1    10        11    11  



 

      

Solving:

  0.818 Therefore,

22

nC 4 H 10  10.818  0.818 mol nC 4 H 8  1  0.818  0.182 mol nC 4 H 10 nC 4 H 8

 4.49

23

9.21 (a) The reaction is

SO2 (g) + ½ O2 (g)  SO3 (g) From Table A.3.2. Species

h ºf ,298 kJ/mol

g ºf ,298 kJ/mol

SO2 O2 SO3

-296.813 0 -395.77

-300.1 0 -371.02

From these data: º 3 J  g rxn , 298  70.92  10   mol  º 3 J  hrxn , 298  98.96  10   mol 

Use Equation 9.21:

 h º  1 1  rxn K 973  K 298 exp     R  T2 T1       70.92 10 3     98.96 10 3  1 1  K 973  exp    exp     973 298  8.314  8.314  298  

K 973  2.5 (b) Take 1 mole of SO2 as the basis for the calculations. Create expressions for the compositions of the reactor components as functions of conversion:

1 5.76   / 2 1 / 2 yO2  5.76   / 2 y SO 2 

yN2 

3.76 5.76   / 2

y SO3 



5.76   / 2

Since the temperature is 700 ºC and the pressure is 1 bar, the gas can be assumed ideal. Therefore, 24

  1   ySO3   K  1/2 P   ySO   yO 1/2    3 2          1  5.76   / 2    2.5  1/2  1/2  1 bar    1   1  / 2        5.76   / 2  5.76   / 2   Solving, we get

  0.481 The compositions are calculated using the extent of reaction

y SO 2  0.094 yO2  0.138

y N 2  0.681 y SO3  0.087 (c) The energy balance around the reactor is

hout  hin  q Since the reaction is isothermal, the change in enthalpy comes from the enthalpy of reaction. Therefore,



 kJ     q  mol  

hrxn  0.481  98.96   kJ   q  47.6    mol 

(d) The equilibrium constant does not change with pressure; however, the term

KP1 / 2 increases, which is equal to

25

y  y y  SO3

1/2

SO3

O2

      5.76   / 2   1/2  1   1  / 2      5.76   / 2  5.76   / 2 

Therefore, the extent of reaction increases. (e) The equilibrium constant is a function of temperature only, so increasing pressure does not change the equilibrium constant. (f) An increase in pressure is justified because the extent of reaction will increase.

26

9.22 The reaction is

SO2 (g) + ½ O2 (g)  SO3 (g) º is described by the following equation The variation of hrxn

 1 B 2 C 3 1  º º hrxn  hrxn,298  R A T  298  T  2982   T  2983   D       T 298  2 3  The A, B, C, and D parameters are calculated with data from Table A.2.2: Species SO2 O2 SO3

v -1 -1/2 1

B  10 3 0.801 0.506 1.056

A 5.699 3.639 8.06

C  10 6 0 0 0

D  10 5 -1.015 -0.227 -2.028

1 3.639 18.06  0.5415 2 1 B  1 0.801103    0.506 10 3  11.056 10 3   2 10 6 2 C  0 1 D  1 1.015 10 5   0.227 10 5  1 2.02810 5   8.99510 4 2 A  1 5.699  

From Problem 9.21: º 3 J  hrxn ,298  98.96  10   mol 

Therefore, at 700 ºC:

  2 10 6 2 0.5415 T  298  T  2982       J    J  2 º   8.314  hrxn  98.96 10 3      mol    mol   1  4 1   8.99510   T  298  

Substituting the above expression, we get  2  106 2 T  2982 0.5415 T  298    2  J    J  98.96  103    8.314     1  1  mol    mol     8.995 104     d ln K  T 298    dT RT 2





27





    

Integrating: ln

K 973.15  27.67 K 298

At 298 K K 298  2.66  1012 Therefore, K 973.15  2.56

28

9.23 (a) The reaction is

3 H2 (g) + C6H6 (g)  C6H12 (g) From Table A.3.2: Species

h ºf ,298 kJ/mol

g ºf ,298 kJ/mol

H2 (1) C6H6 (2) C6H12 (3)

0 82.98 -123.22

0 129.75 31.78

From these data: º 3 J  g rxn , 298  98.0  10   mol  º 3 J  hrxn ,298  206.2  10   mol 

We do not need to calculate the composition at the exit of the first reactor to determine the equilibrium composition at the exit of the second reactor. For the second reactor, Equation 9.21 gives:

 h º  1 1  rxn K 538  K 298 exp      R  T2 298.15      98.0 10 3      206.2 10 3  1 1  K 538  exp    exp     538.15 298.15  8.314  8.314  298.15   K 538  11.44 Take 1 mole of C6H6 as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

y1  y3 

10  3 11  3

y2 

 11  3

The gas is assumed ideal. Therefore,

29

1 11  3

K

1 P3

  y3     3   y1   y2  

         1   11 3   11.44  3 3  5 bar    1   10  3        11 3  11 3   Solving, we get

  0.999 Therefore, the final compositions are y1  0.875

y3  0.125

y2  0

(b) The purpose of the first reactor is to increase the speed the reaction process. At higher temperatures, more collisions occur with sufficient energy to overcome the activation energy of the reaction. However, as the temperature increases, the conversion decreases. The second reactor increases the conversion. (c) Less product will be obtained if the pressure decreases. This becomes clear be rearranging the expression for equilibrium:

K  P3 

 11  3 3

1   10  3 3

Therefore, as the left-hand side decreases due to lower pressures, the right hand-side decreases correspondingly – resulting in a lower extent of reaction. (d) Addition of inert will cause the yield to decrease. The inert molecules interfere with collisions between reactant molecules. In other words, the reactant molecules collide less frequently. Dilution is not recommended.

30

9.24 (a) First, obtain an expression for the equilibrium constant. From data in Appendix A.3: º 3 J  g rxn ,298  241.6  10   mol  º 3 J  hrxn ,298  293.5  10   mol 

Use Equation 9.20 (assume the heat of reaction is constant): º  hrxn 1 1  KT  K 298.15 exp      R  T 298.15   241.6 103   293.510 3  1 1  KT  exp   exp      8.314  298.15   8.314  T 298.15 

Take 1 mole of SiCl4 and 2 moles of H2 as the basis for the calculations. Create expressions for the gas-phase components as functions of conversion:

1 3 2  2 yH 2  3 y SiCl 4 

y HCl 

4 3

Now calculate the required equilibrium constant needed to obtain the given conversion. For the specified conversion,

  0.75 Therefore, y SiCl 4  0.0667

y HCl  0.8 y H 2  0.133 If the gas is assumed ideal, 4   yHCl     KP 2  y    SiCl4   yH 2  

31

4   0.8   K   0.001 bar   2   0.0667  0.133  K  0.347

Now we can solve for the required temperature:

 241.6 10 3   293.510 3  1 1  0.347  exp   exp      8.314  298.15   8.314  T 298.15  T  1605 K (b)

i. Decreasing the pressure will reduce the minimum temperature because the pressure appears in the numerator. As the pressure is decreased, the required equilibrium constant decreases. ii. Diluting the feed stream will decrease the required minimum temperature. For this proposed ratio, the mole fraction of hydrogen essentially remains constant at one despite the reaction. Therefore, the equilibrium constant becomes equal in magnitude to the pressure, which is less than the equilibrium constant in Part (a). Therefore, the temperature can be lower.

32

9.25 (a) The reaction is C 2 H 4 g   Cl 2  g   C 2 H 4 Cl 2 g 

Let C2H4 be species 1, Cl2 be species 2, and C2H4Cl2 be species 3. Create expressions for the compositions of the reaction components as functions of conversion:

1 3 2  y2  3  y1 

y3 

 3

For 90% conversion,   0.9 . Therefore, y1  0.0476 y 2  0.524 y3  0.429

Now calculate the equilibrium constant:

  0.429 K  1 bar 1    0.5240.0476  K  17.2 From Appendix A.3: º 3 J  g rxn , 298  141.29  10   mol  º 3 J  hrxn , 298  182.26  10   mol 

To calculate the minimum temperature we can use Equation 9.21. ln

17.2  141.29  10 3   exp  8.314 298.15   



182.26  10 3 8.314

1  1  T  298   

T  1131 K

33

(b) The equilibrium constant remains the same, but the fugacity terms in the expression for the equilibrium constant are different. The expression is

    3   3     1  KP   1    2    1  3     2  3      

     

We calculate the fugacity coefficients using



i  exp  b  

a  P   RT  RT 

The following table was created with the above equation Species C2H4 (1) Cl2 (2) C2H4Cl2 (3)

 1.00 0.996 0.977

Therefore,       0.977    3   17.2  30 bar 1    1   2   0.996   1.00  3   3     

  0.996

34

9.26 The equilibrium constant can be expressed as follows 4

 ˆ y P   fˆ   SO2 SO2   CaO   fSO2   fCaO  K 3  fˆ   fˆ  CaSO4    Ca  f   CaSO4   fCa  We can assume that the solids form distinct phases and are immiscible with each other. Furthermore, the gas is assumed ideal. Therefore, 4 K  p SO 2

We also have another relationship involving K assuming the heat of reaction is independent of temperature:

ln K 

º 1 hrxn,298  C T  R

Below, we have plotted ln K vs. 1/T with data given in the problem statement. T (ºC)

T (K)

1/T (K)

pSO2 (bar)

K

ln(K)

900 960 1000 1040 1080 1120

1173.15 1233.15 1273.15 1313.15 1353.15 1393.15

0.000852 0.000811 0.000785 0.000762 0.000739 0.000718

0.00533 0.0253 0.0547 0.11 0.206 0.317

8.07E-10 4.10E-07 8.95E-06 1.46E-04 1.80E-03 1.01E-02

-20.9376 -14.7078 -11.6236 -8.8291 -6.31952 -4.59541

ln K vs. 1/T 0 0.0007

0.00075

0.0008

0.00085

ln K

-5 -10

y = -121600x + 83.40 R2 = 0.991

-15 -20 -25 1/T (K-1)

35

0.0009

We can see there is a systematic deviation from the linear fit, indicating the enthalpy of reaction is not constant. From the slope, we obtain  kJ  º hrxn ,298  1011   mol  º Now we can use the following expression to obtain g rxn , 298 at each temperature.

 g º   h º 1  rxn,298 rxn,298  1 K  exp   exp      T 298  R  R  298 K    T (K) 1173.15 1233.15 1273.15 1313.15 1353.15 1393.15

grxn,298 (kJ mol-1) 805.6 802.6 802.6 802.9 803.5 805.6

Calculate the average  kJ  º g rxn , 298  803.8   mol 

36

9.27 (a) From data in Appendix A.3: º 3 J  g rxn ,298  50.72  10   mol 

Therefore,

 50.72 10 3    1.29 109 K  exp   8.314  298  (b) From data in Appendix A.3: º 3 J  hrxn , 298  74.81  10   mol 

Calculate K with Equation 9.21:

 74.8110 3  1 1  K  1.29 10  exp      2.04  8.314  1000 298  9

(c) Create expressions for gas compositions in terms of conversion:

2 1 yCH 4  1 1 Note: The moles of carbon are not considered when developing these expressions because it is in the solid phase. yH 2 

The expression for the equilibrium constant:

  4 2 K  P   1   1      4 2 2.04   0.01 bar     1   1    Solve for  :

37

  0.99 The amount of H2: n H 2  2  1.98 mol

(d) We run at 1000 K because it increases the reaction rate and equilibrium conversion. (e) Pressure appears in the numerator of the expression for the equilibrium constant. Decreasing the pressure makes

 4 2 K    P  1   1    larger. Correspondingly, the equilibrium conversion must be larger.

38

9.28 (a) Use the following equation v

 xi i fi  i K   o   fi  which for ideal solutions reduces to

K    xi 

vi

To find the mole fractions of A and A2, we apply the following constraint:

 x 1 i

Find the mole fractions of A and B by equating liquid and vapor fugacities.

yA P  x A PAsat yB P  xB PBsat 1 / 3  0.1 atm   1  2 / 3  0.1atm   2 xA  xB  3 15 0.1 atm 0.5 atm Therefore,

x A2 

8 15

Now calculate the equilibrium constant

K

8 /15  4.8 2 1 / 3

(b) Multiply the total number of moles by the liquid mole fractions: n A  166.7 moles n A2  266.7 moles n B  66.7 moles

39

(c) If the dimerization reaction does not occur, then for every mole of A2 calculated in Part (b), two moles of A actually exist. Therefore, n A total  n A  2n A2  700 moles

(d) Equate the liquid and vapor fugacities of species B. y B P  x B  B PBsat

Hence,

B 

yB P  2 / 3  0.1 atm    1.53 sat   xB PB 66.7 mol    0.5 atm   700 mol  66.7 mol 

(e) Both the model containing the dimerization reaction and the colleague’s model represent the data. There is no reason to exclude one or the other; they are different ways of viewing reality. Since the colleague’s model will always predict more moles of A in the liquid phase than the total number of moles, the model can only predict positive deviations from ideality. However, solvation reactions can be used to predict negative deviations.

40

9.29 (a) First, calculate the equilibrium constant. From data in Appendix A.3:

 J  º grxn,298  113.5 10 3   mol   J  º hrxn,298  164.94 103   mol  The A, B, C, and D parameters are calculated with data from Table A.2.2: Species CH4 H2O CO2 H2

v -1 -2 1 4

B  10 3 9.081 1.45 1.045 0.422

A 1.702 3.470 5.457 3.249

C  10 6 -2.164 0 0 0

D  10 5 0 0.121 -1.157 0.083

A  9.811 B  9.248 103 C  2.164 106 D  1.067 105

Using Equation 9.24, we get K 773  0.0503 º (Alternatively if we assume hrxn = const, we can use Equation 9.20:

 h º  1 1  rxn K 2  K1 exp     R  T2 T1    113.5103   164.94 103  1 1  K 773  exp    exp     773.15 298.15  8.314  8.314  298.15   K 773  0.00739

Take 1 mole of CH4 as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

yCH 4 

1 6  2

yCO 2 

41

 6  2

yH 2O 

5  2 6  2

yH 2 

4 6  2

Since the temperature is 500 ºC and the pressure is 1 bar, the gas can be assumed ideal. Therefore, 4   y y     CO H 2 2  K  P2  2  y    CH 4   yH 2O  

    4 4      2   6  2   6  2   0.00739  1 bar    1   5  2 2        6  2  6  2   Solving:

  0.576 Therefore, nH 2  4  0.576   2.30 mol º (Alternatively, solving if we assume hrxn = const):

  0.423 Therefore,

nH 2  4  0.423 1.69 mol These values are significantly different) (b) At the process conditions, a higher conversion can never be achieved. Equilibrium is the upper limit. However, low conversions may be achieved. Equilibrium does not say anything about the kinetics. Kinetic considerations may result in lower conversions. (c) Increasing the pressure does not make sense. In order to increase conversion, you should decrease the pressure since the pressure term appears in the numerator of the expression for the equilibrium constant.

42

9.30 (a) The fugacity can be found as follows P

 f iv     v dP RT ln i  Plow    Plow For the EOS given in the problem statement 1  vi  RT   Bi'  P  Therefore,  fv  P 1 ' ln i      Bi  Plow    Plow  P

 dP  ln P   B ' P  P  i low P   low   

Let Plow go to zero and simplify: ' i

iv  e B P For H2S and SO2

 Hv S  e

BH' 2 S P

2

v SO e

' BSO P 2

2

(b) The equilibrium constant can be calculated at 500 ºC (773.15 K) with the following equation º K  773.15 K hrxn,298 1 C T  298 dT ln  773.15    2   K 298.15  298.15 K RT º º 773.15 K hrxn,298   grxn,298  1 C T  298 dT  exp K 773.15  exp     2   RT  298.15 K RT 

From data in Appendix A.3  J  º g rxn , 298  90,440   mol 

43

 J  º hrxn , 298  145,830   mol  Substituting these and other values from the problem statement into the expression for the equilibrium constant provides K 773.15  0.824 (c) Consider a basis of 3 moles of H2S and 1 mole of SO2. Create expressions for the mole fractions of the gaseous species as functions of extent of reaction:

1 4  2 yH 2O  4 

y SO2 

y SO2 

3  2 4 

The equation for the equilibrium constant can be written as follows:

K

y

SO2

y



H 2O H 2O

P

2

2

SO P   yH S H S 2

2

y     P   y   y      

2

2

H 2O

H 2O

2

2

SO2

H 2S

SO2

2

P 1

H2S

(We are approximating the fugacity coefficients by the pure species fugacity coefficients.) We have expressions or values for all of the variables in the above expression except  H 2 O . From the steam tables:

 kJ  hˆ  3373.6    kg   kJ   gˆ  1726.5    kg 

 kJ  sˆ  6.5965    kg  K   J  g  31102.6   mol 

 kJ  hˆº  3489.0    kg   kJ   gˆ º  4163.4    kg 

 kJ  sˆº  9.8977    kg  K   J  g º  75002.8   mol 

We can calculate the fugacity from the following equation

44

(500 ºC, 10 MPa)

(500 ºC, 10 kPa)

 fv    g  g º  RT ln   Plow    Substituting values, we obtain f Hv O  9.25 MPa 2

Therefore,

 H O  0.925 2

Now solve the following equation for the extent of reaction. 2

 2  2   0.925   4   0.824  2  1   3 2  exp  4.4 109 10 10 6  exp  2.2 10 9 10 10 6            4    4   





2

100 bar 

1

  0.915 (d) By Le Chatelier’s principle, you would want to increase the pressure. (e) Assume that the fugacity coefficients are equal to unity and solve the following equation. 2

 2  0.95    1  P   4  0.95  0.824  2  5  1 0.95  3 2  0.95   110      4  0.95  4  0.95 

P  221 bar

(f) Since the reaction is exothermic, you would want to lower the temperature as low as possible in order to maximize conversion. However, in a real reactor it is better to change pressure because you want to keep the temperature high for kinetic reasons.

45

(g) The addition of inerts would lower the conversion. This becomes apparent mathematically by examining the expressions for the mole fractions in the equilibrium constant equation. However, it can also be explained physically. The inert molecules block the collisions between gaseous reactant molecules (more gaseous reactant molecules than gaseous product molecules), so fewer reactions occur, and the overall conversion is less.

46

9.31 A sketch of the process follows:

P = 1.5 bar

1 mol/s H2 2 mol/s CO  1 mol/s CO2

T=?K

0.40 mol/s H2O H2  CO CO2

We first have to determine the chemical reaction occurring. No other species are involved in the reaction, so let’s look at the possibilities. To make H2O we need to react H2 with one or both of the oxygen containing components. Reacting CO with H2 to produce H2O would also produce C, which is not an option. Therefore, the reaction for our system must be: H2(g) + CO2(g)  H2O(g) + CO(g) Next, from table A.3.2 Species (all gas)

hf,298[kJ/mol]

gf,298[kJ/mol]

H2 CO2 H2O CO

0 -393.51 -241.82 -110.53

0 -394.36 -228.57 -137.17

From these data: grxn,298 = 28.62

10

hrxn,298 = 41.16

10



Use Equation 9.21: ∆°

KT =K298 exp KT =exp

. .

exp

. .

Next, let’s examine the number of moles of each species as a function of the extent of a reaction:

47

Species

Initial number of moles/s (n)

nH2 nCO2 nH2O nCO nT

1 1 0 2 4

Number of moles/s being produced () -1 -1 1 1 0

Moles/s of species as a function of  1- 1-  2+ 4

We know what nH2O at the outlet (at equilibrium) is, so we know . 0.4

Now we can look at our mole fractions: Species yH2 yCO2 yH2O yCO yi

Mole fraction 0.15 0.15 0.1 0.6 1

Next, setting up the equilibrium constant given our known information… .

K=

.

.

2.67

.

Now set sides of our equilibrium constant equation equal to each other and solve for T. 2.67 = exp

. .

exp

. .

T = 1213 K

48

9.32 (a) First, write out the expressions for the moles of each species as a function of the extent of reaction, ξ. From these, we calculate the gas-phase mole fractions:

nA  nA    1   nC  nC  2  2 , nT  1  

1  1  2 yC  1  yA 

Now apply Equation 9.27 (we are told to assume an ideal gas phase): 2

K  4    yi  i v

i

4 1   2   12 2

 2  2 2  1   P 2 2  P  ( yC P) 12 2   v    P  ( y A P)1  1   1   1    1   2 P 1    

  12 mole A reacted . Since ½ mole of A reacted, and 2 moles of C are formed per mole of A consumed, nC  1 mole (b) Looking at the expression for K given above, we can write: K 4 2  P 1  2

Recall that we can also write the equilibrium constant as K  e

 g rxn RT

(a constant at any T).

Since K is constant at any given temperature, and the expression in the right hand side is greater than one for all physically valid conversions (what is the domain of  ?), a lower P will generate a higher conversion (or ).

49

9.33

,

,

,

,

(a)

C3H8O (g) ⇌ C3H6 (g) + H2O (g) ∆

K298 = grxn = ∆ K=

∆

∆

62.76

3

.

(b)



From the Antoine equation

Pbsat  0.031 bar

Pasat  0.027 bar 3

4

3

4

3.62 1.14

From the equilibrium constant expression: 3

3

0.67

50

228.57

163.08

3 (c) 0.009 0.013

0.977

51

2.11

9.34 You wish to generate H2 through a gas-phase reaction of pentane with steam. The feed is 1 mole C5H12 per 10 moles of water, at 0.5 bar. You need 98% conversion.  (a) Calculate the minimum temperature to achieve 98% conversion (assuming hrxn is constant).

The reaction scheme is given by:  5 CO + 11 H 2 C5 H12 + 5 H 2 O  (1) (2) (3)

(4)

The feed has molar compositions n1  1 mole and n2  10 mole . Writing the mole balances for the four individual species lets us find the overall molar change as a function of : n1  1   n2  10  5 n3  5 n4  11 nTv  11  10 Since we know that   0.98 (from the problem statement), we can compute the gas-phase mole fractions for each component: n1  0.02

y1  0.000962

n2  5.1

y2  0.245

n3  4.9

y3  0.236

n4  10.78

y4  0.518

n  20.8 moles

P  0.5 bar

v T

Write an expression for the equilibrium constant, KT, of the mixture at the unknown temperature:

( y3 P)5 ( y4 P)11 ( y3 )5 ( y4 )11 P10 KT    0.00060 ( y1 P)( y2 P)5 ( y1 )( y2 )5 The K value above doesn’t have any temperature dependence. This will come from the enthalpy of formation in the next steps. From Appendix A.3, we can tabulate the standard Gibbs energy and enthalpy of formation for each species in the gas phase:

52

g f ,298 K  kJ/mol hf ,298 K  kJ/mol Species νi -1 C5H12 - 8.37 -146.54 +5 CO - 137.17 -110.53 - 228.57 -241.82 -5 H2O Next, calculate the Gibbs energy and heat of reaction of the overall reaction, and the equilibrium constant K for the reaction (note that this K is different from the K we computed from the gasphase composition above):   g rxn ,298 K    g f ,298 K  465.37  kJ/mol  , and i

h

 rxn ,298 K

   hf ,298 K  802.99  kJ/mol i

  g rxn  ,298 82 K 298  exp    2.66 10 .  RT  Note that the K value is for equilibrium the reference temperature, 298K, but unless we’re really lucky (hint: don’t bet on it), the reaction is probably not occurring at exactly that temperature. What does the magnitude of K298 tell you about how much pentane will react at room temperature?

We can introduce the temperature dependency by correcting the K value using Equation 9.21:

ln

 hrxn 1  KT ,298  1   .  K 298 R  T 298 

Substituting the values of KT and K298 provides an equation with only one unknown: T. 180.4 

802,990 J/mol  1 1    8.314 J/mol  K  T 298 

Solving for the unknown T, we find that the reaction temperature must be T  672 K  400C

(b) In Part (a), we made the implicit assumption that the enthalpy of reaction is not a function of temperature. While this may be justifiable for a small deviation from the reference temperature, we are operating 375°C higher than the reference. We could achieve much better accuracy by taking into account the change in the enthalpy of reaction as the temperature changes. To do this, we would need to use the (admittedly unwieldly) Equation 9.24 from the text. We would need the heat capacity constants for each species, from Appendix A.2:

Species C5H12 H2O CO

A 2.464 3.470 3.376

B 45.351x10-3 1.45 x10-3 0.557 x10-3

C -14.111 x10-6 0 0 53

D 0 0.121 x105 -0.031 x105

E 0 0 0

(c) The reaction stoichiometry shows that every mole of pentane gas reacted produces a net increase of 10 moles of gas in the reactor.  5 CO + 11 H 2 C5 H12 + 5 H 2 O  n = 6 moles n = 16 moles

Assuming the reactor has a fixed volume and temperature, this would cause a net increase in pressure as the reaction proceeds (P = nRT/V). Using LeChatelier’s Principle, a decrease in the system pressure will drive the reaction to the right, producing more H2 and CO than at higher pressures. While it may be preferable from an equilibrium standpoint to operate at low pressures, keep in mind that the reaction rates will increase dramatically with pressure (why?), so that it may be more economical to operate at different temperatures or pressures. Again, equilibrium does not determine the reaction kinetics (or process economics). (d) Separation could be achieved in many ways. The following process is one possibility. We will use liquid-liquid separation (settling) of the cooled vapor effluent to remove pentane and water from the gas stream. The remaining hydrogen and carbon monoxide can be separated by cryogenic distillation (note the large difference in boiling points between the two gases).

C5H12 (liq)

H2 (b.p. -252.5°C)

Reactor effluent

Cool to T