Thermodynamics ch (9) (1)

Chapter 9 Solutions Engineering and Chemical Thermodynamics 2e

Milo Koretsky Wyatt Tenhaeff

School of Chemical, Biological, and Environmental Engineering

Oregon State University

[email protected]

2

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

2D

A B

yK

y y P

3

9.2 Some products can form. Since 0o

rxng , K < 1. But some products form

4

9.3

,298orxng = -500 J/mol and KT = 4

5

9.4 It would not change

6

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

7

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

8

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

9

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

10

9.9 1

A

KP

11

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

1 2 2

0.1252

0.25B

A

xK

x

2

0.252

0.125C

B

xK

x

12

9.11 It does not change

13

9.12 No

14

9.13 Largest 0.1 m CaCl2

0.1 m NaCl 0.1 m scrose smallest

15

9.15 C on the Ga site C on the As site

16

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

17

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

18

9.17 The reaction is H2O (l) + ½ O2 (g) H2O2 (l) The Gibbs energy of reaction is calculated as follows

grxn,298º gf ,298

º H2O2

gf ,298º

H2O

Therefore,

gf ,298º

H2O2

grxn,298º gf ,298

º H2O

From Appendix A.3.2:

gf ,298º

H2O 237.14

kJ

mol

gf ,298º

H2O2

116.8 kJ

mol

237.14

kJ

mol

gf ,298º

H2O2

120.34 kJ

mol

19

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 , ST , and G :

2222

00 IIIIII hnhnhnHHH

Assuming we start with 1 mole of I2 and no I, we have: 2In 1

2In

Therefore,

2 2 2

12 1 2

2I I I I IH h h h h h

Applying the definition of the heat of formation gives: IfhH ,2

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

2222

0IImixIIII snssnsnS

So

2 2 2

12 ln ln

2I I I I I IS s s R n y n y

or

2 2,2 ln lnf I I I I IS s R n y n y

: We can find the change in Gibb’s energy from its definition:

STHG

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

20

To find the equilibrium conversion, we must locate the minimum on the plot of G. This occurs when 23.0 At this extent of reaction 0.37Iy

20.63Iy

Thermodynamic Quantities as a Function of Extent of Reaction

-20000

0

20000

40000

60000

80000

100000

120000

140000

160000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Extent of Reaction ()

(J)

H

S

G

21

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

322 NHH2

3N

2

1

At 298 K,

mol

J 1045.16 3º

298,rxng

mol

J 1011.46 3º

298,rxnh

Therefore,

K298 exp 16.45103 J/mol 8.314 J/mol K 298 K

764.76

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: 9357.2A , 00209.0B , 0C , 33050D Substituting numerical values into Equation 9.24 and evaluating, we obtain

41000 106.5 K

Therefore,

grxn,1000º 8.314 1000 ln 5.6104 62250

J

mol

Since the Gibbs energies of formation of N2 and H2 are zero at 1000 K,

grxn,1000º gf ,1000

º NH3

gf ,1000º

NH3

62.25 kJ

mol

22

9.20 First, calculate the equilibrium constant. From data in Appendix A.3

mol

J 1051.88 3º

298,rxng

mol

J 101.126 3º

298,rxnh

Use Equation 9.20:

K1000 K298.15 exphrxn

º

R

1

T2

1

T1

K1000 exp 88.51103

8.314 298.15

exp

126.1103

8.314

1

1000 1

298.15

11000 K

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

11

184 HCy

11104 HCy

11

102Hy

The gas is assumed ideal; therefore,

4 10

4 8 2

1 C H

C H H

yK P

y y

11

10

11

1

11bar 51 1

Solving: 818.0 Therefore,

23

mol 818.0818.01104

HCn

mol 182.0818.0184

HCn

49.484

104 HC

HC

n

n

24

9.21 (a) The reaction is SO2 (g) + ½ O2 (g) SO3 (g) From Table A.3.2.

Species kJ/mol º298,fh kJ/mol º

298,fg

SO2 -296.813 -300.1 O2 0 0

SO3 -395.77 -371.02 From these data:

mol

J 1092.70 3º

298,rxng

mol

J 1096.98 3º

298,rxnh

Use Equation 9.21:

K973 K298 exphrxn

º

R

1

T2

1

T1

K973 exp 70.92103

8.314 298

exp

98.96103

8.314

1

973 1

298

5.2973 K

(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:

2/76.5

12

SOy 2/76.5

76.32 Ny

2/76.5

2/12

Oy

2/76.53

SOy

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

25

K 1

P1/2

ySO3 ySO3 yO2 1/2

2.5 1

1 bar 1/2

5.76 / 2

15.76 / 2

1 / 25.76 / 2

1/2

Solving, we get 481.0 The compositions are calculated using the extent of reaction 094.0

2SOy

138.02Oy

681.02Ny

087.03SOy

(c) The energy balance around the reactor is qhh inout

Since the reaction is isothermal, the change in enthalpy comes from the enthalpy of reaction. Therefore,

qhrxn

mol

kJ 96.98481.0

mol

kJ 6.47q

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

2/1KP increases, which is equal to

26

ySO3

ySO3 yO2 1/2

5.76 / 2

15.76 / 2

1 / 25.76 / 2

1/2

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.

27

9.22 The reaction is SO2 (g) + ½ O2 (g) SO3 (g)

The variation of ºrxnh is described by the following equation

hrxnº hrxn,298

º R A T 298 B

2T 2 2982 C

3T 3 2983 D

1

T 1

298

The A, B, C, and D parameters are calculated with data from Table A.2.2:

Species v A 310B 610C 510D SO2 -1 5.699 0.801 0 -1.015 O2 -1/2 3.639 0.506 0 -0.227

SO3 1 8.06 1.056 0 -2.028

A 1 5.699 1

23.639 1 8.06 0.5415

B 1 0.801103 1

20.506103 1 1.056103 2106

C 0

D 1 1.015105 1

20.227105 1 2.028105 8.995104

From Problem 9.21:

mol

J 1096.98 3º

298,rxnh

Therefore, at 700 ºC:

hrxnº 98.96103

J

mol

8.314

J

mol

0.5415 T 298 2106

2T 2 2982

8.995104 1

T 1

298

Substituting the above expression, we get

62 2

3

4

2

2 100.5415 298 298

J J 298.96 10 8.314

mol mol 1 18.995 10

298ln

T T

Td K

dT RT

28

Integrating:

67.27ln298

15.973 K

K

At 298 K

12298 1066.2 K

Therefore, 56.215.973 K

29

9.23 (a) The reaction is 3 H2 (g) + C6H6 (g) C6H12 (g) From Table A.3.2:


298,fg

H2 (1) 0 0 C6H6 (2) 82.98 129.75 C6H12 (3) -123.22 31.78

From these data:

mol

J 100.98 3º

298,rxng

mol

J 102.206 3º

298,rxnh

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:

K538 K298 exphrxn

º

R

1

T2

1

298.15

K538 exp 98.0103

8.314 298.15

exp

206.2103

8.314

1

538.15 1

298.15

44.11538 K


311

3101

y

311

12

y

3113

y

The gas is assumed ideal. Therefore,

30

3

33

1 2

1 yK

P y y

11.44 1

5 bar 3

113

1113

103113

3

Solving, we get 999.0 Therefore, the final compositions are 875.01 y 02 y 125.03 y

(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:

33

3

3101

311

PK

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.

31

9.24 (a) First, obtain an expression for the equilibrium constant. From data in Appendix A.3:

mol

J 106.241 3º

298,rxng

mol

J 105.293 3º

298,rxnh

Use Equation 9.20 (assume the heat of reaction is constant):

KT K298.15 exphrxn

º

R

1

T 1

298.15

KT exp241.6103

8.314 298.15

exp

293.5103

8.314

1

T 1

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:

3

14SiCly

3

4HCly

3

222Hy

Now calculate the required equilibrium constant needed to obtain the given conversion. For the specified conversion, 75.0 Therefore, 0667.0

4SiCly

8.0HCly

133.02Hy

If the gas is assumed ideal,

K PyHCl 4

ySiCl4 yH2 2

32

K 0.001 bar 0.8 4

0.0667 0.133 2

347.0K Now we can solve for the required temperature:

0.347 exp241.6103

8.314 298.15

exp

293.5103

8.314

1

T 1

298.15

K 1605T (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.

33

9.25 (a) The reaction is gClHCClgHC 242242 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:

3

11y

33y

3

22y

For 90% conversion, 9.0 . Therefore, 0476.01 y

524.02 y

429.03 y

Now calculate the equilibrium constant:

0476.0524.0

429.0bar 1 1K

2.17K From Appendix A.3:

mol

J 1029.141 3º

298,rxng

mol

J 1026.182 3º

298,rxnh

To calculate the minimum temperature we can use Equation 9.21.

298

11

314.8

1026.182

15.298314.8

1029.141exp

2.17ln

3

3 T

K 1131T

34

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

3

1

1 2

3

1 2

3 3

K P

We calculate the fugacity coefficients using

expi

a Pb

RT RT

The following table was created with the above equation

Species C2H4 (1) 1.00 Cl2 (2) 0.996

C2H4Cl2 (3) 0.977 Therefore,

3

2996.0

3

100.1

3977.0

bar 302.17 1

996.0

35

9.26 The equilibrium constant can be expressed as follows

K

SO2ySO2

P

fSO2

4

fCaO

fCaO

fCaSO4

fCaSO4

3

fCa

fCa

We can assume that the solids form distinct phases and are immiscible with each other. Furthermore, the gas is assumed ideal. Therefore,

42SOpK

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

ln K hrxn,298

º

R

1

T

C

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 1173.15 0.000852 0.00533 8.07E-10 -20.9376 960 1233.15 0.000811 0.0253 4.10E-07 -14.7078 1000 1273.15 0.000785 0.0547 8.95E-06 -11.6236 1040 1313.15 0.000762 0.11 1.46E-04 -8.8291 1080 1353.15 0.000739 0.206 1.80E-03 -6.31952 1120 1393.15 0.000718 0.317 1.01E-02 -4.59541

y = -121600x + 83.40R2 = 0.991

-25

-20

-15

-10

-5

00.0007 0.00075 0.0008 0.00085 0.0009

ln K

1/T (K-1)

ln K vs. 1/T

36

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

mol

kJ 1011º

298,rxnh

Now we can use the following expression to obtain º298,rxng at each temperature.

K expgrxn,298

º

R 298 K

exp

hrxn,298º

R

1

T 1

298

T (K) grxn,298

(kJ mol-1) 1173.15 805.6 1233.15 802.6 1273.15 802.6 1313.15 802.9 1353.15 803.5 1393.15 805.6

Calculate the average

mol

kJ 8.803º

298,rxng

37

9.27 (a) From data in Appendix A.3:

mol

J 1072.50 3º

298,rxng

Therefore,

K exp50.72103

8.314 298

1.29109

(b) From data in Appendix A.3:

mol

J 1081.74 3º

298,rxnh

Calculate K with Equation 9.21:

K 1.29109 exp74.81103

8.314

1

1000 1

298

2.04

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

1

22Hy

1

14CHy

Note: The moles of carbon are not considered when developing these expressions because it is in the solid phase.

The expression for the equilibrium constant:

K P4 2

1 1

2.04 0.01 bar 4 2

1 1

Solve for :

38

99.0 The amount of H2: mol 98.12

2 Hn

(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

K

P 4 2

1 1

larger. Correspondingly, the equilibrium conversion must be larger.

39

9.28 (a) Use the following equation

K xi i fi

fio

vi

which for ideal solutions reduces to

K xi vi

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

xi 1

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

sat yBP xBPBsat

xA 1/ 3 0.1 atm

0.1 atm 1

3 xB

2 / 3 0.1atm 0.5 atm

2

15

Therefore,

15

82Ax

Now calculate the equilibrium constant

K 8 /15 1/ 3 2 4.8

(b) Multiply the total number of moles by the liquid mole fractions:

moles 7.66

moles 7.266

moles 7.166

2

B

A

A

n

n

n

40

(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, moles 7002

2 AAtotalA nnn

(d) Equate the liquid and vapor fugacities of species B.

satBBBB PxPy

Hence,

B yBP

xBPBsat

2 / 3 0.1 atm 66.7 mol

700mol 66.7 mol

0.5 atm

1.53

(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.

41

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

grxn,298º 113.5103

J

mol

hrxn,298º 164.94103

J

mol

The A, B, C, and D parameters are calculated with data from Table A.2.2:

Species v A 310B 610C 510D CH4 -1 1.702 9.081 -2.164 0 H2O

-2 3.470 1.45 0 0.121 CO2 1 5.457 1.045 0 -1.157 H2 4 3.249 0.422 0 0.083

9.811A 39.248 10B

62.164 10C 51.067 10D Using Equation 9.24, we get 773 0.0503K

(Alternatively if we assume º

rxnh = const, we can use Equation 9.20:

K2 K1 exphrxn

º

R

1

T2

1

T1

K773 exp113.5103

8.314 298.15

exp

164.94103

8.314

1

773.15 1

298.15

00739.0773 K

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

26

14

CHy

262 COy

42

26

252

OHy

26

42 Hy

Since the temperature is 500 ºC and the pressure is 1 bar, the gas can be assumed ideal. Therefore,

K P2yCO2 yH2 4

yCH4 yH2O 2

0.00739 1 bar 2

6 2

46 2

4

16 2

526 2

2

Solving: 0.576 Therefore,

24 0.576 2.30 molHn

(Alternatively, solving if we assume º

rxnh = const):

423.0 Therefore, nH2

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.

43

9.30 (a) The fugacity can be found as follows

low

vi

P

Pi P

fRTdPv

low

ln

For the EOS given in the problem statement

'1

ii BP

RTv

Therefore,

lowilow

P

Pi

low

vi PPB

P

PdPB

PP

f

low

'' ln

1ln

Let Plow go to zero and simplify:

iv eBi

'P

For H2S and SO2

H2Sv e

BH2S' P

SO2

v eBSO2

' P

(b) The equilibrium constant can be calculated at 500 ºC (773.15 K) with the following equation

lnK773.15

K298.15

hrxn,298º

RT 21C T 298 dT

298.15 K

773.15 K

K773.15 expgrxn,298

º

RT

exp

hrxn,298º

RT 21C T 298 dT

298.15 K

773.15 K

From data in Appendix A.3

mol

J 440,90º

298,rxng

44

mol

J 830,145º

298,rxnh

Substituting these and other values from the problem statement into the expression for the equilibrium constant provides 824.015.773 K

(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:

4

12SOy

4

232SOy

4

22OHy

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

K yH2OH2OP 2

ySO2SO2

P yH2SH2SP 2 yH2O 2

ySO2 yH2S 2

H2O 2

SO2 H2S 2 P1

(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 OH 2

.

From the steam tables:

kg

kJ 6.3373h

Kkg

kJ 5965.6s (500 ºC, 10 MPa)

kg

kJ5.1726g

mol

J6.31102g

kg

kJ 0.3489ºh

Kkg

kJ 8977.9ºs (500 ºC, 10 kPa)

kg

kJ4.4163ºg

mol

J8.75002ºg

We can calculate the fugacity from the following equation

45

low

v

P

fRTgg lnº

Substituting values, we obtain

MPa 25.92

vOHf

Therefore, H2O 0.925

Now solve the following equation for the extent of reaction.

0.824

24

2

14

3 24

2

0.925 2

exp 4.4109 10106

exp 2.2109 10106

2 100 bar 1

915.0

(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.

0.824

2 0.95 4 0.95

2

1 0.954 0.95

3 2 0.95 40.95

2

P

1105

1

bar 221P (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.

46

(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.

47

9.31 A sketch of the process follows:

1 mol/s H2

2 mol/s CO

1 mol/s CO2

P = 1.5 bar

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) hf,298[kJ/mol] gf,298[kJ/mol]

H2 0 0 CO2 -393.51 -394.36 H2O -241.82 -228.57 CO -110.53 -137.17

From these data:

grxn,298 =28.62 10

hrxn,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:

48

Species Initial number of

moles/s (n)

Number of moles/s being produced

()

Moles/s of species as a function of

nH2 1 -1 1 - nCO2 1 -1 1 - nH2O 0 1 nCO 2 1 2 + nT 4 0 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 Mole fraction yH2 0.15 yCO2 0.15 yH2O 0.1 yCO 0.6 yi 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

49

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:

1

2 2

, 1

A A

C C

T

n n

n n

n

1

1

2

1

A

C

y

y

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

2

222 2

1 2

221( ) 12

4( ) 1 1 11

1

iv v Ci

i A

PPy P

K y Py P

P

2 24 1 12

12 mole A reacted .

Since ½ mole of A reacted, and 2 moles of C are formed per mole of A consumed, 1 moleCn

(b) Looking at the expression for K given above, we can write:

2

2

4

1

K

P

Recall that we can also write the equilibrium constant as rxng

RTK e

(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 ).

50

9.33

, ,

, ,

(a)

C3H8O (g) ⇌ C3H6 (g) + H2O (g)

K298 = ∆

grxn = ∆ ∆ ∆ 62.76 228.57 163.08

K = . 3

(b)

From the Antoine equation

0.027 barsataP 0.031 barsat

bP

3 4

3.62

3 41.14

From the equilibrium constant expression:

33 0.67

51

3 2.11

(c)

0.009

0.013

0.977

52

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 rxnh is constant).

The reaction scheme is given by:

5 12 2 2C H + 5 H O 5 CO + 11 H

(1) (2) (3) (4) The feed has molar compositions 1 1 molen and 2 10 molen . Writing the mole balances for

the four individual species lets us find the overall molar change as a function of :

1

2

3

4

1

10 5

5

11

n

n

n

n

11 10vTn

Since we know that 0.98 (from the problem statement), we can compute the gas-phase mole fractions for each component:

1

2

3

4

0.02

5.1

4.9

10.78

n

n

n

n

1

2

3

4

0.000962

0.245

0.236

0.518

y

y

y

y

20.8 molesvTn 0.5 barP

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

5 11 5 11 10

3 4 3 45 5

1 2 1 2

( ) ( ) ( ) ( )0.00060

( )( ) ( )( )T

y P y P y y PK

y P y P y y

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:

53

νi Species ,298 kJ/mol f Kg ,298 kJ/mol f Kh

- 1 C5H12 - 8.37 -146.54 + 5 CO - 137.17 -110.53 - 5 H2O - 228.57 -241.82

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 gas-phase composition above): ,298 ,298 465.37 kJ/molrxn K f K

i

g g , and

,298 ,298 802.99 kJ/molrxn K f Ki

h h

,298 82298 exp 2.66 10rxng

KRT

.

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:

,298

298

1 1ln

298rxnT

hK

K R T

.

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

802,990 J/mol 1 1

180.48.314 J/mol K 298T

Solving for the unknown T, we find that the reaction temperature must be 672 400T K 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 A B C D E C5H12 2.464 45.351x10-3 -14.111 x10-6 0 0 H2O 3.470 1.45 x10-3 0 0.121 x105 0 CO 3.376 0.557 x10-3 0 -0.031 x105 0

54

(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 12 2 2C H + 5 H O 5 CO + 11 H

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).

One possible scheme to separate a mixed reactor effluent containing pentane, water, hydrogen and carbon monoxide vapors.

Reactor effluent

C5H12 (liq)

H2O (liq)

Cool to T<36°C

Cryogenic Distillation

Liquid-Liquid Separation

H2

CO (b.p. -191.5°C)

(b.p. -252.5°C)

55

9.35 3 2g s g gSiClH Si HCl H

Species Initial number of

moles/s (n) Number of moles/s being produced (v)

Moles/s of species as a function of

3SiClHn 1 -1 1 -

HCln 0 1

2Hn 0 1 vTn 1 1 1+

3

2

1

1

1

SiClH

HCl H

y

y y

So the equilibrium constant can be written:

2

3

2

21HCl H

SiClH

y Py P PK

y P

Using the thermochemical data:

grxn,298 =24 10

hrxn,298 = 49.6 10

We find K

KT =K298 exp∆°

13.43

/

11 /

K P

K P

56

9.36 (a) At equilibrium, how much H2 is produced per mole of steam fed?

⇌ (1) (2) (3)

1 11

1

__________

1

∆ ° 91,4008.314298

9.5 10

ln∆ 1 1

298131,2908.314

11000

1298

37.2

1.36

.

.⟹ 2.72 2.72

2.723.72

⟹ 0.86

0.86

∆ ,° ∆ ,

°

H2O -241.82 -228.57

H2 0 0 CO -110.53 -137.17

131.29 kJ/mol

91.4 kJ/mol

∆ ° ∆ ,°

∆ ° 131.29

∆ ° 91.4

57

(b) If the flow rate of steam into the reactor is 5 mol/s, how much heat needs to be added or removed to keep the system at 727 C?

Δ5

0.86 131.29 561

58

9.37 (a) For every kg of magnesium placed in the reactor, how many kg of Ti will be produced?

2 mol Mg 0.4 mol Ti 24.3 2 mol Mg 0.4 mol 47.9

1000 Mg .

. 394 0.39

48.6 g Mg = 19.2 g Ti mTi = 0.39 kg (b) Calculate the approximate pressure needed. You can assume ideal gas and ideal liquid behavior.

∆ 560008.3141000

0.0012

∆ 2 298 642 56

0.0012 0.40.6

0.4

0.6 0.0012374

59

9.38 (a) The reaction is 2 4 2 2 5( ) ( ) ( )C H g H O g C H OH g

From Table A.3.2:


298,fg

2 4 ( )C H g (1) 52.26 68.15

2 ( )H O g (2) -241.82 -228.57

2 5 ( )C H OH g (3) -234.96 -168.39

From these data:

º 3,298

J8.0 10

molrxng

º 3,298

J45.4 10

molrxnh

First we calculate the equilibrium constant

º

355 2982

1 1exp 1.3

298.15rxnh

K KR T


1

1

2y

2

1

2y

3 2y


3

1 2

yK

Py y

60

Solving, we get 0.34 (b) The conversion would increase (c) At 10 bar, some of the water in the feed would condense

61

9.39 The reaction is 3 6 3 6C H O propylene oxide C H O acetone

Let the subscript “1” represent propylene oxide and “2” represent acetone. Since the pressure is 1 bar, the expression for the equilibrium constant is

11

22

x

xK

Using the two-suffix Margules equation we obtain the following expression

K x2 exp

Ax12

RT

x1 expAx2

2

RT

x2

1 x2 exp

A

RT12x2

From Appendix A.3:

mol

J 1075.128 3º

298,rxng

Therefore,

22107.3 K

2

2

222 21exp1

107.3 xRT

A

x

x

Solving: 12 x

02 x

62

9.40 (a) First, calculate the number of independent chemical reactions using Equation 9.44: 2012232 smR The two independent reactions: (2)butenecis1 butene-1 (Equation 1)

(3)butenetrans1 butene-1 (Equation 2) From Table A.3.1:

Species kJ/mol º298,fg

1-butene (1) 71.34 cis-butene (2) 65.90

trans-butene (3) 63.01 Using the data in the above table, calculate the equilibrium constants

K1 exp 65.90 71.34 0.008314 298

8.99

K2 exp 63.01 71.34 0.008314 298

28.85

Using 1 mole of species 1 as the basis results in the following expressions: 211 1 y 12 y 23 y

Assuming ideal gas behavior, we obtain two equations for two unknowns:

1

21 y

yK

1

32 y

yK

21

1

199.8

21

2

185.28

Solving, we get 231.01

743.02

63

Therefore, 026.01 y

231.02 y

743.03 y

(b) For 1000 K, we need to scale the equilibrium constants appropriately.

K1,1000 K1,298 exphrxn,298

º

R

1

T 1

298

K2,1000 K2,298 exphrxn,298

º

R

1

T 1

298

From Table A.3.1:

Species kJ/mol º298,fh

1-butene (1) -0.13 cis-butene (2) -6.99

trans-butene (3) -11.18 Using these data, calculate the equilibrium constants

K1,1000 8.99 exp 6.99 0.13

0.008314

1

1000 1

298

1.29

K2,1000 28.85 exp 11.18 0.13

0.008314

1

1000 1

298

1.26

Using 1 mole of species 1 as the basis results in the following expressions 211 1 y 12 y 23 y


1

21 y

yK

1

32 y

yK

21

1

129.1

21

2

126.1

64


355.02 Therefore, 282.01 y

363.02 y

355.03 y

65

9.41 (a) First, calculate the number of independent chemical reactions using Equation 9.44: 2012232 smR The two independent reactions: (2)OHC1 OHC 8383 (Equation 1)

(3)OHC1 OHC 8383 (Equation 2)

Calculate the change in Gibbs energy for the reactions:

mol

kJ 1.48º

500,1 rxng

mol

kJ 9.55º

500,2 rxng

Now find the equilibrium constants:

K1 exp 48.1

0.008314 500

1.06106

K2 exp 55.9

0.008314 500

6.92106

Using 1 mole of species 1 as the basis results in the following expressions: 211 1 y 12 y 23 y


1

21 y

yK

1

32 y

yK

21

16

11006.1

21

26

11092.6


867.02

66

Therefore, 01 y

133.02 y

867.03 y

(b) First, write down the and b matrices.

183

183

183

1

8

3

b

Equation 9.45 gives:

1

8

3

183

183

183

321 nnn

Therefore, 3333 321 nnn

8888 321 nnn

1111 321 nnn

Now, use Equation 9.47 to obtain three more equations:

47.3 0.008314 500 ln n1

n1 n2 n3 3C 8H 1O 0

95.4 0.008314 500 ln n2

n1 n2 n3 3C 8H 1O 0

103.2 0.008314 500 ln n3

n1 n2 n3 3C 8H 1O 0

Simultaneously solving the six equations, we get

867.0

133.0

0

3

2

1

n

n

n

67

Therefore,

867.0

133.0

0

3

2

1

y

y

y

68

9.42 (a) First, create expressions for the reaction components in terms of conversion:

3

1Ay

3

2Cy

3

22By

Assuming ideal gas behavior, the expression for the equilibrium constant is

2

21

BA

C

yy

yPK

At 473.15 K and 1 bar: 25.0Cy

3/1 and 115.473 K

At 573.15 K and 1 bar: 539.0Cy

637.0 and 05.2015.573 K

To find the equilibrium constant at 250 ºC (523.15 K) we can use Equation 9.20 if we assume the enthalpy of reaction is constant:

lnK573.15

K473.15

hrxnº

R

1

573.15 1

473.15

Determine the enthalpy of reaction:

hrxnº 67.6103 J

mol

Now calculate the equilibrium constant at 250 ºC:

K523.15 K473.15 exphrxn

º

R

1

523.15 1

473.15

5.16

69

The conversion can be calculated using the following expression

5.16 2 bar 1

23

2

13

2 23

2

574.0 Substituting this value into the expressions for the compositions, we obtain: 176.0Ay

351.0By

473.0Cy

(b)

i. Increasing the temperature increases the equilibrium constant because the reaction is endothermic. Thus, the conversion will increase, as well as the reaction rate.

ii. Increasing the pressure will also increase the conversion because the pressure appears in the denominator of the expression for the equilibrium constant.

iii. Adding an inert will decrease the conversion because the inert molecules hinder collisions between reactant molecules. There are more reactant molecules than product molecules.

70

9.43 First, calculate the equilibrium constant. From data in Appendix A.3:

grxn,298º 28.62103

J

mol

hrxn,298º 41.16103

J

mol

Since we are calculating the compositions at 1000 K, we won’t assume that the heat of reaction is independent of temperature. At 298.15 K K298.15 1.03105

Using Equation 9.24 and data from Appendix A.2, we get

46.11000 K

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

8

62

OHy

8

12

Hy

8

1 COy

82

COy

The gas is assumed ideal; therefore

K P0yH2 yCO2 yH2O yCO

1.46 1 bar 0 1 6 1

Solving: 807.0 Therefore, 024.0COy 226.0

2Hy

649.02

OHy 101.02COy

71

9.44 Let Species 1 = O Species 2 = O2 Assuming the gas is ideal. Therefore, the equilibrium expression is

K Py1

2

y2

Calculate the equilibrium constant required for 10% O:

K 1 bar 0.10 2

0.90

011.0K From Table A.3.2:


298,fg

1 249.17 231.74 2 0 0

Equation 9.15:

K298 exp 2 231.74

0.008314 298

5.711082

Assuming the heat of reaction is constant, we obtain the following expression:

K K298 exphrxn,298

º

R

1

T 1

298

0.011 5.711082 exp498.34

0.008314

1

T 1

298

Solve for T: K3211T To lower the minimum temperature, decrease pressure.

72

9.45 The reaction is 2 2 2N O NO

From data in Appendix A.3:

grxn,298º 173.2103

J

mol

hrxn,298º 180.58103

J

mol

Since we are assuming the enthalpy of reaction is constant, we can use Equation 9.20 to find the temperature dependence of K:

KT K298.15 exphrxn

º

R

1

T2

1

298.15

KT exp173.2103

8.314 298.15

exp

180.58103

8.314

1

T 1

298.15

The following table has been created using the above equation.

T (K) K

1100 81022.5

1250 71058.5

1500 51001.1

1750 51099.7

2000 41077.3

2250 31026.1

2500 31031.3

2750 31029.7

3000 21041.1 Take 1 mole of air (79% N2, 21% O2) as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

1

79.02

Ny

1

2NOy

73

1

21.02

Oy


K P0 yNO yN2 yO2

As an example, the conversion is calculated below at 1100 K:

5.22108 1 bar 0 2 2

0.21 0.79

Solving:

51065.4 Therefore,

5103.9 NOy

This has been repeated for 1100 to 3000 K. The following plot is obtained.

Composition of NO From Air as a Function of Temperature

0.00E+00

1.00E-02

2.00E-02

3.00E-02

4.00E-02

5.00E-02

1000 1500 2000 2500 3000

Temperature (K)

y NO (

mo

le f

ract

ion

)

74

9.46 The reaction is 2 2 22 2N O NO

From Appendix A.3:

grxn,298º 102.52103

J

mol

hrxn,298º 66.2103

J

mol

Assuming the enthalpy of reaction is constant, we can use Equation 9.21 to find K:

K3000 K298.15 exphrxn

º

R

1

3000 1

298.15

K3000 exp102.52103

8.314 298.15

exp

66.2103

8.314

1

3000 1

298.15

K3000 3.05108

Take 1 mole of air (79% N2, 21% O2) as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

1

79.02Ny

1

2NOy

1

221.02Oy


K P1yNO2

yN2 yO2

For 1 bar:

3.05108 1 bar 1

21

2

0.2121

20.79

1

75

51063.1 Assume ideal gas behavior at 500 bar since the temperature is 3000 K:

3.05108 500 bar 1

21

2

0.2121

20.79

1

41064.3 Therefore,

510266.32

NOy (1 bar)

410283.72

NOy (500 bar)

76

9.47 (a) The subscript “1” refers to the first reaction and “2” refers to second. We will calculate the equilibrium constants for both reactions at 3000 K without assuming the heats of reaction are constant. From Appendix A.3:

grxn,298º

1173.2103

J

mol

grxn,298

º 2 51.26103

J

mol

hrxn,298º

1180.58103

J

mol

hrxn,298

º 2 33.1103

J

mol

From Appendix A.2: A1 1.45, B1 0.000159, C1 0, D1 21500 A2 0.297, B2 0.0003925, C1 0, D1 58500 Using Equation 9.24, we find 0154.01 K

000142.02 K Take 1 mole of air as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

2

21

5.01

5.079.02

Ny

2

1

5.01

2

NOy

2

21

5.01

21.02

Oy

2

2

5.012

NOy

Assuming ideal gas behavior, we obtain the following expressions for the equilibrium constants:

0.0154 P0

21

1 0.52

2

0.791 0.52

10.52

0.211 2

1 0.52

0.000142 P0.5

2

10.52

0.791 0.52

1 0.52

0.50.211 2

10.52

77

Solving simultaneously: 0234.01

000517.02 Therefore, 0468.0NOy

000517.02NOy

186.02Oy

766.02Ny

(b) The subscript “1” refers to the first reaction and “2” refers to second. We will calculate the equilibrium constants for both reactions at 3000 K without assuming the heat of reactions are constant. From Appendix A.3:

grxn,298º

1173.2103

J

mol

grxn,298

º 2 35.34103

J

mol

hrxn,298º

1180.58103

J

mol

hrxn,298

º 2 57.19103

J

mol

From Appendix A.2: A1 1.45, B1 0.000159, C1 0, D1 21500 A2 0.2245, B2 0.000313, C1 0, D1 69250 Using Equation 9.24, we find 0154.01 K

00114.02 K Take 1 mole of air as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

yN2 0.791

1 0.52

yNO 21 2

10.52

yO2 0.211 0.52

1 0.52

yNO2 2

1 0.52

78


0.0154 P0

21 2

1 0.52

2

0.791

1 0.52

0.211 0.52

1 0.52

0.00114 P0.5

2

1 0.52

0.211 0.52

10.52

0.521 2

1 0.52


00515.02 Therefore, 047.0NOy

000518.02NOy

186.02Oy

766.02Ny

(c) The conversions and compositions in parts A and B are equal.

79

9.48 (a) First, calculate the equilibrium constant for the reaction. Data from Table A.3.2:


298,fg

N2O4 (1) 9.079 97.79 NO2 (2) 33.1 51.26

At 298 K:

mol

kJ 4.73º

298,rxng

Calculate the equilibrium constant with Equation 9.15:

148.02980.008314

73.4exp298

K

Express the moles of reaction components as functions of : 11n

22 n Therefore,

1

11y

1

22y

Assuming ideal gas behavior:

1

22

y

yPK

1

1

1

2

bar 1148.0

2


80

Use to obtain the compositions 682.01 y

318.02 y (b) The equilibrium constant is the same as in Part (a). Express the moles of reaction components as functions of : 11n

22 n

1In (moles of inert) Therefore,

2

11y

2

22y

Assuming ideal gas behavior:

1

22

y

yPK

2

1

2

2

bar 1148.0

2

Solving, we get 25.0 Use to obtain the compositions 333.01 y

222.02 y

444.0Iy

81

9.49 First we have:

dT

RTgd

dT

Kd rxn /ln º (I)

From Equation 9.18: (II)

2

ºln

RT

h

dT

Kd rxn

Combine Equation (I) and (II) to yield

2

ºº /

RT

h

dT

RTgd rxnrxn

From the problem statement,

145log5.77570001º

TTRRT

grxn

Now differentiate the expression above,

10ln

5.77570001/2

º

TTRdT

RTgd rxn

10ln

5.7757000122

º

TTRRT

hrxn

75700010ln

5.7º Thrxn

At 298 K:

mol

J 1056.7 5º

298,rxnh

82

9.50 The subscript “1” refers to reaction 1 and “2” refers to reaction 2. We will calculate the equilibrium constants for both reactions at 800 K without assuming the heat of reactions are constant. From Appendix A.3:

grxn,298º

1142.12103

J

mol

grxn,298

º 2 28.62103

J

mol

hrxn,298º

1 206.1103

J

mol

hrxn,298

º 2 41.16103

J

mol

From Appendix A.2:

951.71 A , 008708.01 B , 61 10164.2 C , 97001 D

86.12 A , 00054.02 B , 01 C , 1164001 D Using Equation 9.24, we find 03.0800,1 K

29.4800,2 K

Create expressions for the compositions of the reaction components as functions of conversion:

1

125

14

CHy 1

2125

32

Hy

1

2125

42

OHy 1

2125

COy

1

22 25

COy


0.03 P2

1 2

5 21

31 2

5 21

3

11

5 21

41 2

521

4.29 P0

2

5 21

31 2

5 21

1 2

5 21

41 2

5 21


83

515.02 Therefore, 062.0

4CHy

461.02

OHy

378.02Hy

016.0COy

083.02COy

These compositions are equal to those in Example 9.19.

84

9.51 First, calculate the equilibrium constant. From data in Appendix A.3:

mol

J 1016.31 3º

298,rxng

mol

J 1038.34 3º

298,rxnh

We won’t assume the heat of reaction is independent of temperature. At 298.15 K

6

3

15.298 1047.3314.815.298

1016.31exp

K

Using Equation 9.24 and data from Appendix A.2, we obtain

021.0800 K

Take 1 mole of C2H2 and 1 mole of N2 as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of conversion:

2

122

HCy

2

2HCNy

2

12

Ny

The gas mixture is assumed ideal; therefore

K P0 yHCN yC2H2 yN2

0.021 1 bar 0

22

2

12

12

Solving: 0676.0 Now calculate the compositions 466.0

22HCy

85

466.02Ny

0676.0HCNy

86

9.52 We can assume ideal behavior, and since Cu and Cu2O are stable in given the temperature range, we obtain the following relationship

2

1

OpK

KpO

12

To calculate K over the temperature range, we can use the expression for Gibbs energy of formation given in the problem statement.

grxnº 2 gf ,Cu2O

º 3.40105 14.24T lnT 248T

K expgrxn

º

RT

exp

3.40105 14.24T lnT 248T RT

The following table has been created

T K P (bar) 300 3.09E+50 3.24E-51 400 8.00E+35 1.25E-36 500 1.54E+27 6.47E-28 600 2.54E+21 3.94E-22 700 1.95E+17 5.12E-18 800 1.65E+14 6.05E-15 900 6.90E+11 1.45E-12 1000 8.79E+09 1.14E-10 1100 2.51E+08 3.98E-09 1200 1.32E+07 7.59E-08 1300 1.10E+06 9.11E-07

We can also plot the data.

Partial Pressure of O2 as a Function of

Temperature

1.00E-511.00E-441.00E-371.00E-301.00E-231.00E-161.00E-091.00E-02

100 300 500 700 900 1100 1300 1500

Temperature (K)

PO

2 (b

ar)

87

9.53 (a) Assume that the data given in the problem statement is for a pressure of 1 bar. If we also assume the enthalpy of reaction is independent of temperature, we can use Equation 9.20:

121

2 11ln

TTR

h

K

K orxn

Substitute numerical values to solve the enthalpy of reaction:

hrxno

8.314 J/mol K ln 1.12103

1.98105

11079.15 K

1905.15 K

1.88105 J

mol

(b) First, calculate the equilibrium constant at 1000 ºC using Equation 9.20:

K1273.15 1.98105 exp 1.88105

8.314

1

1273.15 1

905.15

0271.015.1273 K

Since the pressure is 1 bar, we can assume the vapor phase is ideal. Also assume the fugacities of the solids in the mixture equal their respective pure solid fugacities. The expression for the equilibrium constant reduces to

K

yHClP1 bar

2

yH2P

1 bar

yHCl2

yH2

P

Now, we need to create expressions for the mole fractions containing the extent of reaction. If we take one mole of H2 and one mole of CrCl2 as the basis, the equations for the number of moles of each species are 1

2CrCln Crn

12Hn 2HCln

The expressions for the vapor mole fractions are

88

1

12Hy

1

2HCly

Substitute these expressions into the expression for the equilibrium constant and solve for :

0.0271

21

2

11

1 bar 4 2

1 1 1 bar

mol 0820.0 For every mole of hydrogen fed, 0.082 moles of Cr(s) are produced. (c) We can rewrite the expression for the extent of reaction as

K

P nHCl

2

ntotalvap nH2

4 2

1 1

Since pressure is in the denominator of the left-hand side, we want to reduce pressure to increase extent of reaction. The reduction in pressure can also be reasoned by employing Le Châtelier’s principle. There are more product gas molecules, so as you reduce pressure, the product side will be favored.

89

9.54 We need a second reaction. The two reactions can be written 2 4 2 2 5( ) ( ) ( )C H g H O g C H OH g Reaction 1

2 4 2 2 2( ) 4 ( ) 2 ( ) 6 ( )C H g H O g CO g H g Reaction 2

From Table A.3.2:


298,fg

2 4 ( )C H g (1) 52.26 68.15

2 ( )H O g (2) -241.82 -228.57

2 5 ( )C H OH g (3) -234.96 -168.39

2 ( )CO g (4) -393.51 -394.36

2 ( )H g (5) 0 0

From these data, for Reaction 1, we ger:

º 3,298

J8.0 10

molrxng

º 3,298

J45.4 10

molrxnh

From these data, for Reaction 2, we ger:

º 3,298

J57.4 10

molrxng

º 3,298

J128 10

molrxnh

First we calculate the equilibrium constant for Reaction 1

º

355 2982

1 1exp 1.3

298.15rxnh

K KR T

And for Reaction 2:

º

7355 298

2

1 1exp 3.5 10

298.15rxnh

K KR T


90

1 21

1 2

1

2 3y

1 22

1 2

1 4

2 3y

13

1 22 3y

24

1 2

2

2 3y


31

1 2

yK

Py y

2 6

34 52

1 2

y yK P

y y

Solving, we get

1

2

0.290

0.031

91

9.55 We are given the Gibbs energies of formation for several isomers of nonane. Rank the isomers in order of highest- to lowest-expected concentration. Determine the equilibrium composition of a mixture of the three most “important” isomers at 1000K. (a) For isomerization reactions of the form 1 2I I , we can write:

2 1, ,rxn formation I formation Ig g g

Therefore, the most stable isomers will be those that have the lowest Gibbs energy of formation (can you explain why?) Ranking the species in order from lowest to highest Gibbs energy of formation,

Isomer Species fg (kcal/mol) Expected Rank

4-methyloctane (B)

161.90 Highest

nonane (A)

162.15 :

4-ethylheptane (C)

164.22 :

2,2,3-trimethylhexane (D)

168.13 :

3,3-diethylpentane (E)

171.79 :

2,2,3,4-tetramethylpentane (F)

175.62 Lowest

(b) Determine the equilibrium composition mole fractions of a mixture of the three top species (highlighted in gray above). Consider the parallel isomerization reactions of nonane (A) to 4-methyloctane (B) and 4-ethyl-heptane (C). Each isomer will be in equilibrium with nonane, as shown at right. The isomers can interchange with one other by first going through nonane, then isomerizing to the other species. We can write the following equilibrium expressions for the system:

1K

A B 1By P

K Ay P

B

A

y

y

nonane(A)

4-methyloctane(B)

4-ethylheptane(C)

K1 K2

nonane(A)

4-methyloctane(B)

4-ethylheptane(C)

K1 K2

92

2K

A C 2Cy P

K Ay P

C

A

y

y

Set a basis of starting with one mole of pure A, so that 1 mole, and 0 moleA B Cn n n . Then,

1 2

1

2

1

1 (constant)A

B T

C

n

n n

n

, 1 2

1

2

1A

B

C

y

y

y

Now we need to calculate the values of K1 and K2. Can you guess at this point which isomer will have the highest mole fraction?

calmol

1 calmol K

161,900 162,150exp 1.13

1.987 1000K

K

calmol

2 calmol K

164,220 162,150exp 0.35

1.987 1000K

K

Now write the two equilibrium expressions together:

11

1 2 1 1

2 2 22

1 2

1

1

KK

KK

Usie the above equations to solve for the unknowns i ,

21 1 1 1

1

1K

KK

21 1 1

1

1 1K

KK

11

1 2

0.4021 B

Ky

K K

1 0.456A A By y y

22

1 2

0.1421 C

Ky

K K

So, the equilibrium gas composition is:

93

0.456

0.402

0.142

A

B

C

y

y

y

94

9.56 The following two cracking reactions for n-pentane occur in parallel: C5H12(g) ⇌ C3H6(g) C2H6(g) (I) and C5H12(g) ⇌ C4H8(g) CH4(g) (II) At 183 C and 0.5 bar, both these reactions contribute to the equilibrium composition of species in the system. For every 1.0 mole of n-pentane gas initially fed, 0.10 mol of polypropylene (C3H6) is formed at equilibrium from Reaction I. Calculate the equilibrium amount of trans-2-butene formed per mole of pentane fed. You may assume that ∆ for each reaction does not change with temperature. First we take a molar basis of 1 mole of pentane in the feed. Next we can find thermochemical data for our reactions. Next, from table A.3.1

Species (all gas) ∆ ,° ⁄ ∆ ,

° ⁄

C5H12 -146.54 -8.37 C3H6 20.43 62.76 C2H6 -84.68 -32.84

trans C4H8 -11.18 63.01 CH4 -74.81 -50.72

From these data:

∆ ,° 38.29 10

∆ ,° 20.66 10

∆ ,° 82.29 10

∆ ,° 60.55 10

Use Equation 9.21:

, ,∆ °

95

,.

.

.

.0.0193

, ,∆ °

,.

.

.

.1.14

Next if we look at the moles of each species as a function of the extent of reactions. Because we have two reactions we have two extent of reactions, 1 for Reaction I and 2 for Reaction II.

Moles of species

Initial number of moles (n)

Number of moles being

produced (v)

Moles of species as a function of

nC5H12 1 -11 - 12 1 - 1 - 2 nC3H6 0 11 1 nC2H6 0 11 1

ntrans C4H8 0 12 2 nCH4 0 12 2 nT 1 11 + 12 1 + 1 + 2

Next we look at the mole fractions…

Species Mole fraction

yC5H12 11

yC3H6 1

yC2H6 1

ytrans C4H8 1

yCH4 1

yi 1

96

Now, reading the problem statement carefully reveals that for every 1.0 mole of n-pentane fed, 0.10 moles of propylene are produced from Reaction I. The number of moles of propylene produced at equilibrium is equal to the extent of Reaction I. 0.1 Next let’s fine the equilibrium constants in terms of the extent of reactions.

, 0.5 . .

. .

, 0.0193 0.5 . .

. .

, 0.5

. .

, 1.14 0.5. .

Solving either K equation results in the same extent of reaction for Reaction II. 0.76 Now we can find the amount of trans-2-butene formed per mole of pentane fed. Since we took a molar basis of 1 mole of pentane fed, the amount of trans-2-butene formed is simply equal to the extent of reaction for Reaction II.

0.76moles

mol C H fed

97

9.57

C2H2 + N2 ⟶ 2HCN (1) (2) (3)

C H O → 2CO 5H O

(4) (5) (6) 1 mole C2H2

52

5molesO → 252molesO

0.720.21

252molesN → 47molesN

.

.

.

K

∆ (kJ/mol) ∆ (kJ/mol)

C2H2 226.88 209.24

HCN 130.63 120.2

∆ ,° 2 130.63 226.88 34.38 kJ mol

∆ ,° 2 120.2 209.24 31.16 kJ mol

K ∆ ,° / / . 3.45 10

ln∆ 10.6

K 3.46 10 . 0.14

n1 = 1

n2 = 47

n3 = 2

n4 = 12.5

nT = 60.5

98

0.14448 47

1.77 1.77 4 1.732

0.7

260.5

0.0231

99

9.58 The solution assuming º º

rxn rxnh h T is presented

ºrxnh = const

(a) First write the chemical reaction:

5 12 2 2 210 5 16C H H O CO H

Calculate the equilibrium constant. From data in Appendix A.3:

∆ ∆ C5H12 -146.54 -8.37 -1 H2O -241.82 -228.57 -10 CO2 -393.51 -394.36 5 H2O 0 0 16 rxn 597.19 322.27

º 3,298

J322.27 10

molrxng

º 3,298

J597.19 10

molrxnh

We assume the heat of reaction is independent of temperature. At 298.15 K

357

298.15

322 10exp 3.23 10

298.15 8.314K


600

298

1 1 597,190 1 1ln 121.35

298 8.314 600 298

orxnK h

K R T

4

600 1.58 10K

Take 1 mole of C5H12 and 10 mole of H2O as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of extent of reaction:

100

5 12

1

11 10C Hy

2

5

11 10COy

2

10 1

11 10H Oy

2

16

11 10Hy


2 2

5 12 2

5 16 5 16

10 10600 10 10 10

5 16

1 10 1 11 10

CO H

C H H O

y yK P P

y y

Solving: 0.56 Now calculate the compositions

5 120.0265C Hy

20.169COy

2

0.265H Oy 2

0.540Hy

(b) We need to write 2 equations:

Equation 1: 5 12 2 2 210 5 16C H H O CO H

Equation 2: 2 2 2CO H CO H O (note there are alternatives to this equation that can

be used) From above: 4

1 1.58 10K

Similarly:

º 3,298,2

J28.62 10

molrxng

º 3,298,2

J41.16 10

molrxnh

We assume the heat of reaction is independent of temperature. At 298.15 K

36

298.15,2

28.62 10exp 9.62 10

298.15 8.314K


101

2

298,2

1 1 41,160 1 1ln 8.36

298 8.314 600 298

orxnhK

K R T

2 0.0412K

Take 1 mole of C5H12 and 10 mole of H2O as the basis for the calculations. Create expressions for the compositions of the reaction components as functions of extent of reaction:

5 12

1

1

1

11 10C Hy

2

1 2

1

5

11 10COy

2

111 10COy

2

1 2

1

10 10

11 10H Oy

2

1 2

1

16

11 10Hy


2 2

5 12 2

5 16

101 10

CO H

C H H O

y yK P

y y

And

2

2 2

2H O CO

CO H

y yK

y y

Solving: 1 0.584 and 2 0.230

Now calculate the compositions

5 120.0247C Hy

20.160COy 0.014COy

2

0.260H Oy 2

0.542Hy

102

9.60 (a)

1 31 0.3 molAn

1 23 2 2 molBn

1 3 0.5 molDn

32 0.2 molFn

3 molv

Tn

yA = 0.1; yB = 2/3; yD = 1/6; yF = 1/15;

1 5D

A B

yK

Py y

2 2 2

19

B

Ky P

2

3

4

15F

A D

yK

y y

(b) Increase the pressure

103

9.61 (a)

C s H O g ⇌ H CO g

K 11

1

K

∆ ,° ∆ ,

° ∆ ,° 137.17 228.57 kJ

mol 91.4 kJ mol

K∆ ,

R 29896.52 10

∆ ,

° ∆ , ∆ , , 110.53 241.82 131.28

KK

∆R

1T

1298

1312808.314

11000

1298

53

K 1.36 K P 2.72

2.723.72

0.86

0.86molHmolH

(b) Explain why you think that this reaction runs at 50 kPa instead of at atmospheric pressure?

Smaller P, larger y2, more H2

° = 1 mol

= 1

=

=

= 1

104

(c) In solving part A, you assumed that ∆ ° constant. What do you think is the single most important reason that this assumption might not be valid?

Large T difference. cp reactants does not equal cp products.

(d) Without redoing all the calculations that you did in Part A, determine how the values of the following quantities change if you write the reaction as

2C s 2H O g ⇌ 2H 2CO g

i. K 1.86

ii. same

(e) You believe that CO2 might also form when water vapor is passed through the bed of activated carbon—write the set of equations that you would need to solve to see how much CO2 forms (as well as CO and H2). You do not need to solve this set of equations—just set them up.

One way:

C s H O g ⇌ H CO g Rxn 1 1.36

CO H O ⇌ CO H Rxn 2 0.89

Find KT for rxn 2

∆ ° 394.36 228.57 137.17 28.62

y1 =1

1

y2 = 1

y3 = 1

y4 = 1

n1 = 1

n2 =

n3 =

n4 =

nT = 1

105

K exp ∆ °104000

∆ ° 393.51 241.82 110.53 41.16

lnKK

11.66

K , 0.89

106

9.62 The following half reactions can be written, with the corresponding half-cell potentials are obtained from Table 9.1:

o 0.762 V

o 0.342 V

The overall reaction is

rxno 1.104 V

Applying Equation 9.34 to this system and assumingZn2 ,Cu2 1 gives:

rxno RT

zFln pi i

vapors

ci im i

liquids

rxn

o RT

zFln

cZn2

cCu2

(1)

or

1.104 8.314298

296, 485ln

0.5

0.1

1.12 V

Since 0, the reaction is spontaneous. To see how far this reaction goes

cZn2 c

Zn2o (2)

cCu2 c

Cu 2o (3)

At equilibrium, 0. Substituting Equations (2) and (3) into (1) and solving gives:

1 cZn2 0.6m

cCu2 1038 m

Thus, approximately all the copper in the solution is consumed.

107

9.63 (a)

(b) The following half reactions can be written, with the corresponding half-cell potentials are obtained from Table 9.1:

O 0.126 V (reduction reaction)

O 0.000 V (oxidation reaction) (c) The overall reaction is

orxn 0.126 V

Applying Equation 9.34 to this system and using the fact that Pb2+ is 1 m:

rxno RT

zFln pi i

vapors

ci im i

liquids

rxn

o RT

zF2.3log10 a

H

2 (1)

Solving Equation (1) for pH:

o10 rxnH

FpH log a

2.3RT

Thus, 0.008 V pH=2.0 0.465 V pH=10.0 (d) We can measure across an equilibrium electrochemical cell to determine pH. The values above are well within a reasonable experimental range.

Anode Cathodee- e-

cations anions

Pt Pb

108


o 0.401 V (reduction reaction)

o 0.447 V (oxidation reaction) The overall reaction is

rxno 0.848 V

Applying Equation 9.34 to this system and using the fact that Pb2+ is 1 m:

rxno RT

zFln pi i

vapors

ci im i

liquids

rxno RT

zFln

cFe2

pO2

1/2 RT

zF2.3log10 a

OH

2

(1)

By definition:

log10 aOH 14 pH 2

Taking the oxygen partial pressure to be 0.21 bar gives

0.848 8.314298

296, 485ln

0.1

0.21

8.314298

96, 485 22.3 0.99 V

Since 0, the reaction is spontaneous.

109

9.65 (a) The overall reaction is

with half reactions

o 1.358 V

o 0.828 V

(b) Neither the anode nor the cathode directly participate in the half-cell reactions

Anode(s)Cl2(g)Cl l OH l H2(g) Cathode(s)

(c) Adding together the half-cell potentials from part a gives:

rxno 2.186 V

Hence this reaction needs input of electrical energy to proceed.

110

9.66 Only two of these reactions are independent. Therefore, to be consistent, the half-cell potential of the third reaction can be obtained from the other two. We label reaction 1:

1o 0.153 V

and reaction 2,

2o 0.521 V

Applying Equation 9.35 gives:

To get the Gibbs energy for reaction 3, we add reaction 1 and 2:

The half-cell potential is given by:

3o g2

o

zF

65.1 kJ

mol

2 mol e-

mole Cu2+

96, 485

Cmole e-

0.34 V

This value is consistent with that reported in Table 9.1.

111

9.67 (a) The following half reactions can be written, with the corresponding half-cell potentials are obtained from Table 9.1:

o 1.229 V

o 0.000 V


rxno 1.229 V

Applying Equation 9.34 to this system and assumingZn2 ,Cu2 1 gives:

rxno RT

zFln pi i

vapors

ci im i

liquids

rxn

o RT

zFln pO2

1/2 pH2

(1)

1.229 8.314 298

2 96,485ln 0.21 0.4 1.25 V

The work can be obtained from as follows. Assuming complete conversion:

W * zF 241 kJ

mol H2

(b) From the Gibbs-Helmholtz relation:

If the enthalpy of reaction is constant, we can integrate to give:

At 298 K, we can calculate from rxno 1.229 V

To get the enthalpy of reaction we use values from Appendix A.3:

112

Integration gives:

so

From Equation 1 we have

0.700 8.314923

296, 485ln 0.21 0.4

0.767 V

The work can be obtained from integrating Equation 9.67. Assuming complete conversion:

W * zF 148 kJ

mol H2

113


o 0.341 V

o 0.342 V


rxno 0 V

Applying Equation 9.34 to this system:

rxno RT

zFln pi i

vapors

ci im i

liquids

RT

zFln

Cu2m c

Cu20.01m

Cu2m c

Cu20..1m

(1)

(a) AssumingZn2 ,Cu2 1 gives

8.314 298

296,485ln

0.01

0.1

0.030 V

(b) We first determine the ionic strength. Accounting for cupric and sulfate ions, the ionic strength is given by:

I 1

2zi

2 ci 1

24c

Cu2 4cSO4

2

8c

Cu2

Using the modified Debye-Huckel theory, Equation 9.38, which is valid up to about 1m, we have:

ln A zz I

1 B I

4 1.17 8cCu2

1 0.33 8cCu2

For 0.01m Cu2+, ln 1.21 For 0.1m Cu2+, ln 3.23

rxno RT

zFln pi i

vapors

ci im i

liquids

RT

zFln

Cu2m c

Cu20.01m

Cu2m c

Cu20..1m

If we assume the activities of cupric and sulfate are the same, Cu2

m

114

8.314 298

2 96,485ln

exp 1.21 0.01

exp 3.23 0.1

0.0036 V

115


o 0.000 V

o 0.222 V The overall reaction is

rxno 0.222 V

Applying Equation 9.34 to this system and taking the activities of the solids to be 1 gives:

rxno RT

zFln pi i

vapors

ci im i

liquids

rxn

o RT

zFln

H m c

HClm c

Cl

pH2

(1)

or in terms of the mean activity coefficient

rxno

RT

zFln

2cHCl2

solving gives

exp zF rxn

o RT

cHCl

0.818

116

9.70 The following set of equations and equilibrium constant expressions can be written:

K1 Cl a 2

pCl2 g (1)

K2 ClO

nCl a VO

(2)

K3 pn (3) Note from these expressions we can determine how the hole and electron densities depend on Cl2 pressure as follows: Electroneutrality gives

ClO p n

We can solve for n from Equation (2):

n K2 Cl a VO

ClO

Using Equation (1), we have:

Cl a K1pCl2 g

So

n K2 VO K1pCl2 g

ClO

At high ClO concentrations, electroneutrality gives: ClO

n

So

n K2 VO K11/4 pCl2

1/4

and from Equation (3):

p K3

n K3

K2 VO K11/4

pCl2

1/4

117

9.71 For the set of equations given in the problem, the following equilibrium constant expressions can be written:

K1 VA VB

(1)

K2 VA 2 p2

pB2

(2)

K3 VB

2n2 pB2

(3)

K4 np (4) Note that Equations (2), (3) and (4) are not independent and we only need two of these to constrain the system. In the solution presented here, we use (2) and (4), but any other 2 could be chosen. Electroneutrality gives:

VA n VB

p (5)

Region 1: low pB2

We take VA n and VB

p , so electroneutrality gives:

n VB

From Equation (1):

VA K1

VB

and from Equation (4)

p K4

n

Thus, Equation (2) gives

K2 VA

2

p2

pB2

K1K4 2

VB

2n2 pB2

From Electroneutrality:

118

VB n K1K4

K2

pB2

1/4

VA

K1

VB

K1 K2

K4

pB2

1/4

And from Equation (4):

p K4

n

K4 K2

K1

pB2

1/4

Region 2: intermediate pB2

In the region of intermediate pB2, we assume that the

concentration of electronic defects is greater than the vacancy concentration, so

We take VA n and VB


n p So from Equation (2):

n p K4

From Equation (2):

VA

K2 pB2

K4

and from Equation (1),

VB

K1

VA K1

K4

K2 pB2

Region 3: high pB2

.

We take VA n and VB


VA p

So from Equation (2):

119

VA p K2 pB2 1/ 4

From Equation (1):

VB K1

VA K1 K2 pB2 1/ 4

And from Equation (4):

n K4

p K4 K2 pB2 1/ 4

From the Brouwer diagram, we see that the material is intrinsic at intermediate pB2

. It is n- type

at low pB2and it is p-type at high pB2

.

Summary Table: Defect concentrations for three regions of pB2

Region 1:�Low � pB2 Region 2: �Intermediate�

pB2

Region 3: High� pB2

n VB n p VA

p

p K4 K2

K1

pB2

1/ 4 p K4 p K2 pB2 1/4

n K1K4

K2

pB2

1/4 n K4 n K4 K2 pB2 1/ 4

VA K1 K2

K4

pB2

1/ 4 VA K2 pB2

K4 VA

K2 pB2 1/4

VB K1K4

K2

pB2

1 / 4 VB K1

K4

K2 pB2

VB K1 K2 pB2 1/4

120

Con

cent

rati

on

pn

n VB

p

VA

p VA

VB

n

pB2

pB2

pB2 pB2

low highintermediate

121

9.72 As in Example 9.24: K1 Bi VB (1)

B

B

V

nVK

2 (2)

K3

Bi p

Bi (3)

K4 Bi 2

pB2

(4)

K5 pn (5)

Inserting Equation 3 into Equation 4 and rearranging

Bi

K3 K4 pB2

p (6)

Equations 1, 2, and 4 give

VB K1K2

n K4 pB2

(7)

And Equation 5 Gives

p K5

n (8)

Electroneutrality yields

n Bi VB

p (9)

In region 1, we have low pB2 so we take VB

p and n Bi , and Equation (9) reduces to

n VB . Setting Equation (7) becomes:

VB n K1K2

K4

pB2

1/4 (10)

Equation (8) becomes:

122

p K5

K4

K1K2

pB2

1/4 (11)

and Equation (6) becomes:

Bi

K3

K5

K1K2 K4 pB2

1/4

In region 2, we have intermediate pB2

. We now assume the atomic defect concentration is

larger than the electronic defect concentration. Thus we have n Bi and p VB

, and

Equation E22.9 reduces to Bi VB

. Setting Equation (6) equal to Equation (7) and using Equation (8) gives

n K1K2K5

K3K4

pB2

1/2

From Equation 8, we have

p K3K4K5

K1K2

pB2

1/2

Using Equation (6), we have

Bi VB

K1K2K3

K5

In region 3, we have high pB2 so we take Bi

n and p VB , and Equation (9) reduces

to p Bi . Thus Equation (6) becomes:

Bi p K3 K4 pB2

1/4

and Equation (8) gives:

n K5

K3 K4

pB2

1/ 4

And Equation (7) gives

123

VB

K1K2 K3

K5K41/4

pB2

1/4

A summary of the defect concentrations for three regions of pB2

is shown below

Region 1:�Low � pB2 Region 2: �Intermediate� pB2

Region 3: High� pB2

n VB Bi

VB p Bi

VB K1K2

K41/ 2 pB2

1/4 Bi

K1K2K3

K5 VB

K1K2 K3

K5K41/4 pB2

1/4

Bi K1K2 K3K4

1/4

K5pB2

1/4 p K3K4 K5

K1K2pB2

1/ 2 Bi K3 K4

1/ 4 pB2

1/ 4

p K5K4

1/4

K1K2pB2

1/4 n K1K2K5

K3K4pB2

1/ 2 n K5

K3K41/4 pB2

1/ 4

From this table, the following Brouwer diagram can be constructed:

Con

cent

rati

on

p

n

n

VB

p

p

VB

n

pB2

pB2

pB2 pB2

low highintermediate

Bi

Bi

Bi

VB

124

9.73 From Example 9.23:

K1 Oi 2

pO2

(1)

K2

Oi p

Oi (2)

K3 Zni n (3)

K4 pn (4)

Electroneutrality gives:

p Zni n Oi

At high oxygen partial pressures:

p Oi

Combining Equations (1) and (2)

p K2 K1

1/4 pO2

1/4

Therefore,

n K4

K2 K11/4 pO2

1 /4

The conductivity decreases as the oxygen partial pressure increases. An increase in the oxygen partial pressure by 10 causes a decrease in n by 0.56.

125

9.74 Possible reactions include dissociative adsorption of oxygen to form an adsorbed oxygen atom, O(a):

K1 O(a) 2

pO2

(1)

and incorporation of the absorbed oxygen into a lattice site, creating two copper vacancies:

K2 VCu 2

O(a) (2)

Solving Equations (1) and (2) for oxygen partial pressure in terms of copper vacancies gives:

pO2

VCu 4

K1K22

Hence the ratio of vacancies at 3 torr to that at 760 torr (1 atm) is given by:

VCu 3 torr

VCu 760 torr

3

760

1/4

0.25

126

9.75 Electroneutrality gives

p BSi n

Since the boron concentration, BSi , is much larger than the intrinsic carrier concentration, we

have approximately:

p BSi 1015 cm-3

We can calculate the electron concentrations using the following expression: K pn (a) At 25 0C, Table 9.2 gives

K ni2 1020 cm-6

hence

n K

p105 cm-3

(b) At 100 oC, we again use Table 9.2. Now we must interpolate; however, we know the Tdependence of K as given by Equation 9.21. At 400 K

K400 ni2 1.2 10 25 cm-6

Hence

which gives

Using

gives

K373 9.6 1023 cm-6 Thus

127

n K

p 9.6 108 cm-3

(c) Electroneutrality now becomes:

p PSi BSi

n

but PSi p . Hence

n 91015 cm-3 and

p K

n1.1104 cm-3

128

9.76

K1 n Cui (1)

K2

CuSi3 p3

VSi (2)

K3

PSi n

VSi (3)

K4 pn (4)

From Table 9.2 K4 = 1020

cm-6

Since Cui =106

cm-3, K1=1016 cm-6

Since CuSi3 =102

cm-3, K2 VSi =1032 cm-12

To get the total copper, we sum together the two defects

Cui CuSi

3 1016

n

1032

p3

Using Relation (4)

p

1020

n

so

Cui

CuSi3 1016

n1028 n3

The minimum occurs when the derivative with respect to n is zero

d Cui CuSi

3 dn

1016

n2 31028n2 0

or n = 7.6 x 1010 cm-3

129

So you want a P concentration around 7.6 x 1010 cm-3. At this concentration,

Cui

CuSi3 1016

n1028 n3 1.75 105 cm-3

Thus the defect concentration is reduced by 8.25 x 105 cm-3.

9.77 (a)

K1 B(a) 2

pH2

3

pB2H6

(1)

K2 BSi

p

B(a) VSi (2)

K3 pn (3)

(b)

p n BSi

(c)

At low pB2H6, p n

p n K3

BSi

K2 B(a) VSi K3

K2 VSi K1

K3

pB2H6

1/2

pH2

3/2

At high pB2H6, p BSi

BSi p K2 B(a) VSi K2 VSi K1

pB2H6

1/4

pH2

3/4

n K3

K2 VSi K1

pH2

3/4

pB2H6

1/4

This we can make the following plot:

131

log

Con

cent

ratio

n np

p

n

low high pB2H6

pB2H6

pB2H6log

BSi

BSi

slope = 1/2

slope = 1/4

slope = -1/4

132

9.78 Consider the combustion reaction of butane:

Choose a basis of 1 mol butane. For a stoichiometric amount of air, we have,

nC4 H101 mol

nO2 6.5 mol

nN2 0.79 6.5

0.21= 24.5 mol

For 2000K and 50 bar: Formula n (Initial) y (Initial) n (Equilibrium) y (Equilibrium) Phi* C4H10 1.000 3.12e-2 2.97e-60 8.85e-62 1.021 O2 6.500 0.203 1.53e-2 4.56e-4 1.006 N2 24.500 0.766 24.494 0.731 1.007 H2O 0.000 0.000 4.991 0.149 1.004 NO 0.000 0.000 1.23e-2 3.68e-4 0.996 NO2 0.000 0.000 6.41e-6 1.91e-7 1.003 H2 0.000 0.000 8.9e-3 2.66e-4 1.005 CO 0.000 0.000 3.4e-2 1.01e-3 1.007 CO2 0.000 0.000 3.966 0.118 1.008 C H O N Lambda/RT 18.2589 2.15857 1.88732 -1.80261 For 2500K and 50 bar: Formula n (Initial) y (Initial) n (Equilibrium) y (Equilibrium) Phi* C4H10 1.000 3.12e-2 1.91e-52 5.66e-54 1.018 O2 6.500 0.203 0.129 3.84e-3 1.005 N2 24.500 0.766 24.446 0.726 1.005 H2O 0.000 0.000 4.938 0.147 1.004 NO 0.000 0.000 0.107 3.18e-3 0.995 NO2 0.000 0.000 8.25e-5 2.45e-6 0.998 H2 0.000 0.000 6.24e-2 1.85e-3 1.004 CO 0.000 0.000 0.304 9.01e-3 1.006 CO2 0.000 0.000 3.696 0.110 1.006 C H O N Lambda/RT 15.6957 1.18748 0.822376 -1.79841

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Thermodynamics ch (9) (1)