Chapter 4 Homework Our procedure is therefore to take 5 points

PROPRIETARY MATERIAL

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τ2.256=TτT0

⋅:= T 1202.8 K=

Ans.

T 1705.4degF=

4.3 Assume air at the

g

iven conditions an ideal

g

as. Basis of calculation is 1 second

.

P 1 atm⋅:= T0122 degF⋅:= V 250 ft3

⋅:= T 932 degF⋅:=

Convert given values to SI units V 7.079 m3

=

T T 32degF−( ) 273.15K+:= T0T032degF−

()

273.15K+:=

T 773.15 K=T0323.15 K=

nPV⋅

RT

0

⋅

:= n 266.985 mol=

For air: A 3.355:= B 0.575 10 3−

⋅:= C 0.0:= D 0.016−105

⋅:=

∆H R ICPH T0T,A,B,C,D,

()

⋅:=

τ3:= (guess) Given

QnR⋅AT

0

⋅τ1−

()

⋅B

2T02

⋅τ

21−

()

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

C

3T03

⋅τ

31−

()

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

τFind τ

()

:= τ 2.652=TτT0

⋅:= T 1413.8 K=Ans.

τFind τ

()

:=

77

PROPRIETARY MATERIAL

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⎢

⎥

⎢

⎥

Given T2T1

P2

P1

⎛

⎜

⎝

⎞

⎠

R

CP

⋅=CPFind CP

()

:= CP56.95 J

mol K⋅

=Ans.

∆Hncalc RT

n

⋅

1.092 ln Pc

bar

⎛

⎜

⎝

⎞

⎠1.013−

⎛

⎜

⎝

⎞

⎠

⋅

0.930 Trn

−

⋅:= ∆Hncalc 30.108 kJ

mol

=Ans.

∆H 13.707 kJ

mol

=Qn∆H⋅:=

Q 3.469 103

×BTU=Ans.

∆H R ICPH T0T,A,B,C,D,

()

⋅:=

∆H 9.441 104

×J

mol

=Qn∆H⋅:= Q 9.4315 106

×kJ=Ans.

78

PROPRIETARY MATERIAL

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To compare with the value listed in Table B.2, calculate the % error.

%error

∆Hncalc ∆Hn

−

∆Hn

:= %error 3.464 %=

PROPRIETARY MATERIAL

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The ln P vs. 1/T relation over a short range is very nearly linear. Our

procedure is therefore to take 5 points, including the point at the

temperature of interest and two points on either side, and to do a linear

least-squares fit, from which the required derivative in Eq. (4.11) can be

found. Temperatures are in rankines, pressures in psia, volumes in cu

ft/lbm, and enthalpies in Btu/lbm. The molar mass M of tetrafluoroethane is

102.04. The factor 5.4039 converts energy units from (psia)(cu ft) to Btu.

4.10

The values calculated with Eq. (4.13) are within 2% of the handbook values.

%error

0.072

0.052−

0.814−

1.781

⎛

⎜

⎝

⎞

⎟

⎠

%=

∆H2

26.429

31.549

33.847

32.242

⎛

⎜

⎝

⎞

⎟

⎠

kJ

mol

=

Ans.

∆H2calc

26.448

31.533

33.571

32.816

⎛

⎜

⎝

⎞

⎟

⎠

kJ

mol

=

%error

∆H2calc ∆H2

−

∆H2

→

⎯

⎯⎯⎯⎯

⎯

:=

Eq. (4.13)

∆H2calc ∆H1

1T

r2

−

1T

r1

−

⎛

⎜

⎝

⎞

⎠

0.38

⋅

⎡

⎢

⎣

⎤

⎥

⎦

→

⎯

⎯⎯⎯⎯⎯⎯⎯

⎯

:=

∆H2

26.429

31.549

33.847

32.242

⎛

⎜

⎝

⎞

⎟

⎠

kJ

mol

=Tr1

0.658

0.673

0.628

0.631

⎛

⎜

⎝

⎞

⎟

⎠

=

∆H1∆Hn

:=∆H2∆H25 M⋅

()

→

⎯

⎯⎯

⎯

:=Tr2

25 273.15+()K

Tc

:=Tr1

Tn

Tc

→

⎯

:=

M

72.150

86.177

78.114

82.145

⎛

⎜

⎝

⎞

⎟

⎠

gm

mol

:=∆H25

366.3

366.1

433.3

392.5

⎛

⎜

⎝

⎞

⎟

⎠

J

gm

:=Tn

36.0

68.7

80.0

80.7

⎛

⎜

⎝

⎞

⎟

⎠

273.15+

⎡

⎢

⎣

⎤

⎥

⎦

K:=

∆Hn

25.79

28.85

30.72

29.97

⎛

⎜

⎝

⎞

⎟

⎠

kJ

mol

:=Pc

33.70

30.25

48.98

43.50

⎛

⎜

⎝

⎞

⎟

⎠

bar:=

Tc

469.7

507.6

562.2

560.4

⎛

⎜

⎝

⎞

⎟

⎠

K:=

b)

80

PROPRIETARY MATERIAL

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∆H 85.817=85.834()

(c) ∆H 81.034=81.136()

(d) ∆H 76.007=75.902()

(e) ∆H 69.863=69.969()

4.11

M

119.377

32.042

153.822

⎛

⎜

⎝

⎞

⎟

⎠

gm

mol

⋅:= Tc

536.4

512.6

556.4

⎛

⎜

⎝

⎞

⎟

⎠

K⋅:= Pc

54.72

80.97

45.60

⎛

⎜

⎝

⎞

⎟

⎠

bar⋅:= Tn

334.3

337.9

349.8

⎛

⎜

⎝

⎞

⎟

⎠

K⋅:=

∆Hexp is the given

value at the normal

boiling point.

∆H is the value at

0 degC. Tr1

273.15K

Tc

→

⎯

:= Tr2

Tn

Tc

→

⎯

:=

(a) T 459.67 5+:= ∆V 1.934 0.012−:= i15..:=

Data: P

18.787

21.162

23.767

26.617

29.726

⎛

⎜

⎝

⎞

⎟

⎠

:= t

5−

0

5

10

15

⎛

⎜

⎝

⎞

⎟

⎠

:= xi

1

ti459.67+

:= yiln Pi

()

:=

slope slope x y,():= slope 4952−=

dPdT

P−()

3

T2slope⋅:= dPdT 0.545=

∆HT∆V⋅dPdT⋅

5.4039

:= ∆H 90.078=Ans.

The remaining parts of the problem are worked in exactly the same

way. All answers are as follows, with the Table 9.1 value in ( ):

(a) ∆H 90.078=90.111()

(b)

81

PROPRIETARY MATERIAL

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Pr0.022=Pr

P

Pc

:=Tr0.648=Tr

Tn

Tc

:=

∆Hn29.1 kJ

mol

:=P 1atm:=Tn329.4K:=Vc209 cm3

mol

⋅:=

Zc0.233:=Pc47.01bar:=Tc508.2K:=ω 0.307:=

Acetone4.12

PCE

0.34

8.72

0.96−

⎛

⎜

⎝

⎞

⎟

⎠

%=

∆Hn

247.7

1195.3

192.3

⎛

⎜

⎝

⎞

⎟

⎠

J

gm

=

PCE

0.77−

4.03−

0.52−

⎛

⎜

⎝

⎞

⎟

⎠

%=

∆Hn

245

1055.2

193.2

⎛

⎜

⎝

⎞

⎟

⎠

J

gm

=

This is the % error

PCE

∆Hn∆Hexp

−

∆Hexp

100⋅%

⎛

⎜

⎝

⎞

⎠

→

⎯

⎯⎯⎯⎯⎯⎯⎯

⎯

:=

∆Hn∆H1T

r2

−

1T

r1

−

⎛

⎜

⎝

⎞

⎠

0.38

⋅

⎡

⎢

⎣

⎤

⎥

⎦

→

⎯

⎯⎯⎯⎯⎯⎯

⎯

:=

(a) By Eq. (4.13)

Tr2

0.623

0.659

0.629

⎛

⎜

⎝

⎞

⎟

⎠

=

Tr1

0.509

0.533

0.491

⎛

⎜

⎝

⎞

⎟

⎠

=

∆Hexp

246.9

1099.5

194.2

⎛

⎜

⎝

⎞

⎟

⎠

J

gm

⋅:=∆H

270.9

1189.5

217.8

⎛

⎜

⎝

⎞

⎟

⎠

J

gm

⋅:=

82

PROPRIETARY MATERIAL

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⎜

⎯

⎢

⎥

⎢

⎥

⎯

C 228.060:=B 2756.22:=A 14.3145:=

∆V 2.602 104

×cm3

mol

=∆VVV

sat

−:=

∆HT∆V⋅B

TC+()

2

⋅e

AB

TC+()

−

⎡

⎢

⎣⎤

⎥

⎦

⋅=

gives

Psat e

AB

TC+

−

=

with Antoine’s Equation

∆HT∆V⋅

T

Psat

d

⋅=

Combining the Clapyeron equation (4.11)

Vsat 70.917 cm3

mol

=

Eq. (3.72)

Vsat VcZc

1T

r

−

()

2

7

⋅:=

Liquid Volume

V 2.609 104

×cm3

mol

=VZR⋅Tn

⋅

P

:=

(Pg. 102)

Z 0.965=Z1B

0

Pr

Tr

⋅+ωB1

⋅Pr

Tr

⋅+:=

Eq. (3.66)

B10.924−=B10.139 0.172

Tr4.2

−:=

Eq. (3.65)

B00.762−=

B00.083 0.422

Tr1.6

−:=

Vapor Volume

Generalized Correlations to estimate volumes

83

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∆Hcalc Tn∆V⋅B

Tn273.15K−

KC+

⎛

⎜

⎝

⎞

⎠

2e

AB

Tn273.15K−

KC+

⎛

⎜

⎝

⎞

⎠

−

⎡

⎢

⎣

⎤

⎥

⎦

⋅kPa

K

⎡

⎢

⎣

⎤

⎥

⎦

⋅:=

∆Hcalc 29.662 kJ

mol

=Ans. %error

∆Hcalc ∆Hn

−

∆Hn

:= %error 1.9 %=

Acetic Acid 37.6 58.7%

Acetonitrile 31.3 3.5%

Benzene 30.8 0.2%

iso-Butane 21.2 -0.7%

n-Butane 22.4 0.0%

1-Butanol 43.5 0.6%

n-Pentane 25.9 0.2%

Phenol 45.9 -0.6%

1-Pro

p

anol 41.9 1.1%

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p

2-Propanol 40.5 1.7%

Toluene 33.3 0.5%

Water 41.5 2.0%

o-Xylene 36.7 1.2%

m-Xylene 36.2 1.4%

PROPRIETARY MATERIAL

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Q 1.372 103

×kW=Ans.

(b) Benzene: ∆Hv28.273 kJ

mol

=∆H 55.296 kJ

mol

=Q 1.536 103kW⋅=

(c) Toluene ∆Hv30.625 kJ

mol

=∆H 65.586 kJ

mol

=Q 1.822 103kW⋅=

4.15 Benzene Tc562.2K:= Pc48.98bar:= Tn353.2K:=

T1sat 451.7K:= T2sat 358.7K:= Cp162 J

mol K⋅

⋅:=

CPV T() A

V

BV

KT⋅+ CV

K2T2

⋅+

⎛

⎜

⎝

⎞

⎠

R⋅:=

P 3bar:= Tsat 368.0K:= T1300K:= T2500K:=

∆H

T1

Tsat

TCPL T()

⌠

⎮

⌡d∆Hv

+

Tsat

T2

TCPV T()

⌠

⎮

⌡d+:= ∆H 49.38 kJ

mol

=

n 100 kmol

hr

:= Qn∆H⋅:=

86

PROPRIETARY MATERIAL

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∆Hv∆Hn

1T

r

−

1T

rn

−

⎛

⎜

⎝

⎞

⎠

0.38

⋅:= ∆Hv6.638 kJ

mol

=

∆Hf227480 J

mol

⋅:= ∆H298 ∆Hf∆Hv

−:= ∆H298 220.8 kJ

mol

=Ans.

Estimate ∆Hv using Riedel equation (4.12) and Watson correction (4.13)

Trn

Tn

Tc

:= Trn 0.628=Tr2sat

T2sat

Tc

:= Tr2sat 0.638=

∆Hn

1.092 ln Pc

bar

⎛

⎜

⎝

⎞

⎠1.013−

⎛

⎜

⎝

⎞

⎠

⋅

0.930 Trn

−R⋅Tn

⋅:= ∆Hn30.588 kJ

mol

=

∆Hv∆Hn

1T

r2sat

−

1T

rn

−

⎛

⎜

⎝

⎞

⎠

0.38

⋅:= ∆Hv30.28 kJ

mol

=

Assume the throttling process is adiabatic and isenthalpic.

Guess vapor fraction (x): x 0.5:=

Given CpT1sat T2sat

−

()

⋅x∆Hv

⋅=x Find x():= x 0.498=Ans.

87

PROPRIETARY MATERIAL

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⎜

or dQ CPdT⋅RdT⋅− RdT⋅

1δ−

+=

which reduces to dQ CPdT⋅δ

1δ− R⋅dT⋅+=

or dQ CP

R

δ

1δ−

+

⎛

⎜

⎝

⎞

⎠R⋅dT⋅=(C)

Since CP is linear in T, the mean heat capacity is the value of

CP at the arithmetic mean temperature. Thus Tam 675:=

(b) For 1,3-butadiene: ∆H298 88.5 kJ

mol

=

(c) For ethylbenzene: ∆H298 12.3−kJ

mol

⋅=

(d) For n-hexane: ∆H298 198.6−kJ

mol

⋅=

(e) For styrene: ∆H298 103.9 kJ

mol

⋅=

88

PROPRIETARY MATERIAL

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Parts (a) – (d) can be worked exactly as Example 4.7. However, with

Mathcad capable of doing the iteration, it is simpler to proceed differently.

∆H298 2 241818−()⋅2 393509−()⋅+ 52510−[]

J

mol

⋅:=

C2H4 + 3O2 = 2CO2 + 2H2O(g)4.19

Comparison is on the basis of equal numbers of C atoms.

Ans.

∆H298 3−=770,012 J⋅,∆H298 3770012−=

∆H298 6 393509−()⋅6 241818−()⋅+ 41950−()−:=

For the combustion of 1-hexene:

C6H12(g) + 9O2(g) = 6CO2(g) + 6H2O(g)

Ans.

∆H298 4−=058,910 J⋅,

For 6 MeOH:

∆H298 676485−=

Ans.

P211.45 bar=P2P1

T2

T1

⎛

⎜

⎝

⎞

⎠

δ

δ1−

⋅:=P11 bar⋅:=

Ans.

Q 6477.5 J

mol

=QCPm

R

δ

1δ−

+

⎛

⎜

⎝

⎞

⎠R⋅T2T1

−

()

⋅:=

δ1.55:=T1400 K⋅:=T2950 K⋅:=

Integrate (C):

CPm R 3.85 0.57 10 3−

⋅Tam

⋅+

()

⋅:=

89

PROPRIETARY MATERIAL

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τFind τ

()

:= τ 8.497=TT

0τ⋅:= T 2533.5 K=Ans.

(b) nO20.75=nn214.107=T 2198.6 K⋅=Ans.

(c) nO21.5=nn216.929=T 1950.9 K⋅=Ans.

(d) nO23.0=nn222.571=T 1609.2 K⋅=Ans.

Index the product species with the numbers:

1 = oxygen

2 = carbon dioxide

3 = water (g)

4 = nitrogen

(a) For the product species, no excess air:

n

0

2

11.286

⎛

⎜

⎝

⎞

⎟

⎠

:= A

3.639

5.457

3.470

3.280

⎛

⎜

⎝

⎞

⎟

⎠

:= B

0.506

1.045

1.450

0.593

⎛

⎜

⎝

⎞

⎟

⎠

10 3−

K

⋅:= D

0.227−

1.157−

0.121

0.040

⎛

⎜

⎝

⎞

⎟

⎠

105

⋅K2

:=

i14..:= A

i

niAi

⋅

()

∑

:= B

i

niBi

⋅

()

∑

:= D

i

niDi

⋅

()

∑

:=

A 54.872=B 0.012 1

K

=D 1.621−105

×K2

=

For the products, ∆HPR

T0

T

CP

R

⌠

⎮

⌡

d⋅=T0298.15K:=

The integral is given by Eq. (4.7). Moreover, by an energy balance,

∆H298 ∆HP

+0=

τ2:= (guess)

Given ∆H298

−RAT

0

⋅τ1−

()

⋅B

2T02

⋅τ

21−

()

⋅+ D

T0

τ1−

τ

⎛

⎜

⎝

⎞

⎠

⋅+

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

90

PROPRIETARY MATERIAL

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Ans.

T 2282.5 K K=TT

0K⋅τ⋅:=τ 7.656=τ Find τ

()

:=

∆H298

−∆Hair

−RAT

0

⋅τ1−

()

⋅B

2T02

⋅τ

21−

()

⋅+

D

T0

τ1−

τ

⎛

⎜

⎝

⎞

⎠

⋅+

…

⎡

⎢

⎣

⎤

⎥

⎦

⋅=

Given

(guess)

τ2:=

D 1.735−105

×K2

=B 0.016 1

K

=A 78.84=

D

i

niDi

⋅

()

∑

:=

B

i

niBi

⋅

()

∑

:=

A

i

niAi

⋅

()

∑

:=

D

0.227−

1.157−

0.121

0.040

⎛

⎜

⎝

⎞

⎟

⎠

105

⋅K2

⋅:=

B

0.506

1.045

1.450

0.593

⎛

⎜

⎝

⎞

⎟

⎠

10 3−

K

⋅:=A

3.639

5.457

3.470

3.280

⎛

⎜

⎝

⎞

⎟

⎠

:=

n

1.5

2

16.929

⎛

⎜

⎝

⎞

⎟

⎠

:=

∆H298 ∆Hair

+∆HP

+0=

The energy balance here gives:

∆Hair 309399−J

mol

=

∆Hair 21.429 8.314⋅3.65606⋅298.15 773.15−()⋅J

mol

⋅:=

MCPH 773.15 298.15,3.355,0.575 10 3−

⋅, 0.0,0.016−105

⋅,

()

3.65606=

For one mole of air:

∆Hair MCPH 298.15 773.15−()⋅=

∆Hair ∆H298

+∆HP

+0=

(e) 50% xs air preheated to 500 degC. For this process,

91

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4.20 n-C5H12 + 8O2 = 5CO2 + 6H2O(l)

By Eq. (4.15) with data from Table C.4:

∆H298 5 393509−()⋅6 285830−()⋅+ 146760−()−:=

∆H298 3−=535,765 J⋅, Ans.

(b) -905,468 J

(c) -71,660 J

(d) -61,980 J

(e) -367,582 J

(f) -2,732,016 J

(g) -105,140 J

(h) -38,292 J

(i) 164,647 J

(j) -48,969 J

(k) -149,728 J

(l) -1,036,036 J

(m) 207,436 J

(o) 178,321 J

(p) -132,439 J

(q) -44,370 J

(r) -68,910 J

(s) -492,640 J

(t) 109,780 J

(u) 235,030 J

(v) -132,038 J

(w) -1,807,968 J

(x) 42,720 J

(y) 117,440 J

(z) 175,305 J

92

PROPRIETARY MATERIAL

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4.22 The solution to each of these problems is exactly like that shown in

Example 4.6. In each case the value of ∆Ho

298 is calculated in Problem

4.21. Results are given in the following table. In the first column the

letter in ( ) indicates the part of problem 4.21 appropriate to the ∆Ho

298

value.

T/K ∆A 103 ∆B 106 ∆C 10-5 ∆D IDCPH/J ∆HoT/J

(a) 873.15 –5.871 4.181 0.000 -0.661 -17,575 -109,795

(b) 773.15 1.861 -3.394 0.000 2.661 4,729 -900,739

(f) 923.15 6.048 –9.779 0.000 7.972 15,635 –2,716,381

(i) 973.15 9.811 –9.248 2.106 –1.067 25,229 189,876

(j) 583.15 -9.523 11.355 -3.450 1.029 -10,949 -59,918

(l) 683.15 -0.441 0.004 0.000 -0.643 -2,416 -1,038,452

(m) 850.00 4.575 -2.323 0.000 -0.776 13,467 220,903

(t) 733.15 4.016 –4.422 0.991 0.083 7,424 117,204

(u) 750.00 7.297 -9.285 2.520 0.166 12,172 247,202

(v) 900.00 2.418 –3.647 0.991 0.235 3,534 -128,504

(w) 673.15 2.586 -4.189 0.000 1.586 4,184 –1,803,784

(x) 648.15 0.060 0.173 0.000 -0.191 125 42,845

(y) 1083.15 4.175 -4.766 1.814 0.083 12,188 129,628

(g) –3.629 8.816 -4.904 0.114

(h) –9.987 20.061 -9.296 1.178

(k) 1.704 -3.997 1.573 0.234

(z) -3.858 -1.042 0.180 0.919

93

PROPRIETARY MATERIAL

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∆HcCH4 890649−J

mol

=

Ans.

n HigherHeatingValue⋅5dollar

GJ

⋅7.985 105

×dollar

day

=

n 1.793 108

×mol

day

=nq

P

RT⋅

⋅:=

Assuming methane is an ideal gas at standard conditions:

4.24 q 150 106

⋅ft3

day

:= T6032−()

5

9K⋅273.15K+:= T 288.71 K=P 1atm:=

The higher heating value is the negative of the heat of combustion with water

as liquid product.

Calculate methane standard heat of combustion with water as liquid product:

CH4 + 2O2 –> CO2 +2H2O

Standard Heats of Formation:

∆HfCH4 74520−J

mol

:= ∆HfO2 0J

mol

:=

∆HfCO2 393509−J

mol

:= ∆HfH2Oliq 285830−J

mol

:=

∆Hc∆HfCO2 2∆HfH2Oliq

⋅+∆HfCH4

−2∆HfO2

⋅−:=

HigherHeatingValue ∆Hc

−:= ∆Hc8.906−105

×J

mol

=

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0.85 ∆HcCH4

⋅0.07 ∆HcC2H6

⋅+ 0.03 ∆HcC3H8

⋅+ 932.875−kJ

mol

=

c)

Gas b) has the highest standard heat of combustion. Ans.

.−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

N2(g)+2H2(g)+C(s)+1/2O2(g)=(NH2)2CO(s)

∆H298 ∆Hf1 ∆Hf2

+∆H+:= ∆H298 333509−J=Ans.

Calculate ethane standard heat of combustion with water as liquid product:

Standard Heats of Formation:C2H6 + 7/2O2 –> 2CO2 +3H2O

∆HfC2H6 83820−J

mol

:=

∆HcC2H6 2∆HfCO2 3∆HfH2Oliq

⋅+∆HfC2H6

−7

2∆HfO2

⋅−:=

∆HcC2H6 1560688−J

mol

=

0.95 ∆HcCH4

⋅0.02 ∆HcC2H6

⋅+ 0.02 ∆HcC3H8

⋅+ 921.714−kJ

mol

=

a)

0.90 ∆HcCH4

⋅0.05 ∆HcC2H6

⋅+ 0.03 ∆HcC3H8

⋅+ 946.194−kJ

mol

=

b)

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