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Chapter 13 - Section B - Non-Numerical Solutions

13.1 (a) 4NH3(g)+ 5O2(g) 4NO(g)+ 6H2O(g)

=ii = 4 5 + 4 + 6 = 1 n0 =

i0

= 2 + 5 = 7

By Eq. (13.5),

yNH3 =2 4

7 + yO2 =

5 5

7 + yNO =

4

7 + yH2O =

6

7 +

(b) 2H2S(g)+ 3O2(g) 2H2O(g)+ 2SO2(g)

=i

i = 2 3 + 2 + 2 = 1 n0 =i0

= 3 + 5 = 8

By Eq. (13.5),

yH2S =3 2

8 yO2 =

5 3

8 yH2O =

2

8 ySO2 =

2

8

(c) 6NO2(g)+ 8NH3(g) 7N2(g)+ 12H2O(g)

=i

i = 6 8 + 7 + 12 = 5 n0 =i0

= 3 + 4 + 1 = 8

By Eq. (13.5),

yNO2 =3 6

8 + 5yNH3 =

4 8

8 + 5yN2 =

1 + 7

8 + 5yH2O =

12

8 + 5

13.2 C2H4(g)+12 O2(g) (CH2)2O(g) (1)

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

The stoichiometric numbers i,j are as follows:

i = C2H4 O2 (CH2)2O CO2 H2O

j j

1 1 12

1 0 0 12

2 1 3 0 2 2 0

n0 =i0

= 2 + 3 = 5

By Eq. (13.7),

yC2H4 =2 1 2

5 121

yO2 =3 1

21 32

5 121

y(CH2)2O =1

5 121

yCO2 =22

5 121

yH2O =22

5 121

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13.3 CO2(g)+ 3H2(g) CH3OH(g)+ H2O(g) (1)

CO2(g)+ H2(g) CO(g)+ H2O(g) (2)

The stoichiometric numbers i,j are as follows:

i = CO2 H2 CH3OH CO H2O

j j

1 1 3 1 0 1 2

2 1 1 0 1 1 0

n0 =i0

= 2 + 5 + 1 = 8

By Eq. (13.7),

yCO2 =2 1 2

8 21yH2 =

5 31 2

8 21yCH3OH =

1

8 21yCO =

1 + 2

8 21yH2O =

1 + 2

8 21

13.7 The equation for G, appearing just above Eq. (13.18) is:

G = H0 T

T0(H0 G

0)+ R

TT0

CP

RdT RT

TT0

CP

R

dT

T

To calculate values ofG, one combines this equation with Eqs. (4.19) and (13.19), and evaluates

parameters. In each case the value ofH0 = H298 is tabulated in the solution to Pb. 4.21. In

addition, the values ofA, B, C, and D are given in the solutions to Pb. 4.22. The required

values ofG0 = G298 in J mol

1 are:

(a) 32,900; (f) 2,919,124; (i) 113,245; (n) 173,100; (r) 39,630; (t) 79,455; (u) 166,365;(x) 39,430; (y) 83,010

13.8 The relation ofKy to P and K is given by Eq. (13.28), which may be concisely written:

Ky =

P

P

K

(a) Differentiate this equation with respect to T and combine with Eq. (13.14):Ky

T

P

=

P

P

dK

dT=

Ky

K

dK

dT= Ky

dln K

dT=

KyH

RT2

Substitute into the given equation for (e/T)P :

e

T

P

=Ky

RT2

de

dKyH

(b) The derivative ofKy with respect to P is:Ky

P

T

=

P

P

11

P K= K

P

P

P

P

11

P =Ky

P

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Substitute into the given equation for (e/P)T:e

P

T

=Ky

P

de

dKy()

(c) With Ky i (yi )i , ln Ky = i

i ln yi . Differentiation then yields:

1

Ky

dKy

de=i

i

yi

dyi

de(A)

Because yi = ni/n,dyi

de=

1

n

dni

de

ni

n2

dn

de=

1

n

dni

de yi

dn

de

But ni = ni0 + ie and n = n0 + e

Whence,dni

de= i and

dn

de=

Therefore,dyi

de=

i yi

n0 + e

Substitution into Eq. (A) gives

1

Ky

dKy

de=

i

i

yi

i yi

n0 + e

=

1

n0 + e

i

2i

yi i

=1

n0 + e

mi=1

2i

yi i

mk=1

k

In this equation, both Ky and n0 + e (= n) are positive. It remains to show that the summationterm is positive. Ifm = 2, this term becomes

21

y1 1(1 + 2)+

22

y2 2(1 + 2) =

(y21 y12)2

y1y2

where the expression on the right is obtained by straight-forward algebraic manipulation. One

can proceed by induction to find the general result, which is

mi=1

2i

yi i

mk=1

k

=

mi

mk

(yki yik)2

yiyk(i < k)

All quantities in the sum are of course positive.

13.9 12

N2(g)+32

H2(g) NH3(g)

For the given reaction, = 1, and for the given amounts of reactants, n0 = 2.

By Eq. (13.5), yN2 =

12(1 e)

2 eyH2 =

32(1 e)

2 eyNH3 =

e

2 e

By Eq. (13.28),yNH3

y1/2N2

y3/2H2

=e(2 e)

[ 12(1 e)]1/2[

32(1 e)]3/2

= KP

P

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Whence,e(2 e)

(1 e)2=

1

2

1/2 3

2

3/2K

P

P = 1.299K

P

P

This may be written: re2 2 re + (r 1) = 0

where, r 1 + 1.299KP

P

The roots of the quadratic are: e = 1 1

r1/2= 1 r1/2

Because e < 1, e = 1 r1/2, e = 1

1 + 1.299K

P

P

1/2

13.10 The reactions are written:

Mary: 2NH3 + 3NO 3H2O +52

N2 (A)

Paul: 4NH3 + 6NO 6H2O+ 5N2 (B)

Peter: 3H2O +52

N2 2NH3 + 3NO (C)

Each applied Eqs. (13.11b) and (13.25), here written:

ln K= G/RT and K= (P )i

( fi)i

For reaction (A), GA = 3GfH2O

2GfNH3 3GfNO

For Marys reaction = 12 , and:

KA = (P)

12

f3fH2Of

5/2fN2

f2fNH3f3fNO

and ln KA =GART

For Pauls reaction = 1, and

KB = (P)1

f6fH2Of5fN2

f4fNH3f6fNO

and ln KB =2GART

For Peters reaction = 1

2 , and:

KC = (P)

12

f2fNH3f3fNO

f3fH2Of

5/2fN2

and ln KC =GA

RT

In each case the two equations are combined:

Mary: (P )12

f3fH2Of

5/2fN2

f2fNH3f3fNO

= expGART

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Paul: (P )1f6fH2O

f5fN2f4fNH3

f6fNO

=

exp

GART

2

Taking the square root yields Marys equation.

Peter: (P )12

f

2

fNH3f

3

fNO

f3fH2Of

5/2fN2

=

exp GART

1

Taking the reciprocal yields Marys equation.

13.24 Formation reactions: 12

N2 +32

H2 NH3 (1)

12

N2 +12

O2 NO (2)

12

N2 + O2 NO2 (3)

H2 +12

O2 H2O (4)

Combine Eq. (3) with Eq. (1) and with Eq. (2) to eliminate N2:

NO2 +32

H2 NH3 +O2 (5)

NO2 12

O2 +NO (6)

The set now comprises Eqs. (4), (5), and (6); combine Eq. (4) with Eq. (5) to eliminate H2:

NO2 +32

H2O NH3 + 134

O2 (7)

Equations (6) and (7) represent a set of independent reactions for which r= 2. Other equivalent sets

of two reactions may be obtained by different combination procedures. By the phase rule,

F= 2 + N r s = 2 1 + 5 2 0 F= 4

13.35 (a) Equation (13.28) here becomes:yB

yA=

P

P

0K= K

Whence,yB

1 yB= K(T)

(b) The preceding equation indicates that the equilibrium composition depends on temperature only.

However, application of the phase rule, Eq. (13.36), yields:

F= 2 + 2 1 1 = 2

This result means in general for single-reaction equilibrium between two species A and B that

two degrees of freedom exist, and that pressure as well as temperature must be specified to fix the

equilibrium state of the system. However, here, the specification that the gases are ideal removes

the pressure dependence, which in the general case appears through the is.

13.36 For the isomerization reaction in the gas phase at low pressure, assume ideal gases. Equation (13.28)

then becomes:

yB

yA=

P

P

0K= K whence

1 yA

yA= K(T)

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Assume that vapor/liquid phase equilibrium can be represented by Raoults law, because of the low

pressure and the similarity of the species:

xAP

sat

A(T) =

yAP and(

1xA

)P

sat

B(T) = (

1yA

)P

(a) Application of Eq. (13.36) yields: F= 2 + N r= 2 2 + 2 1 = 1

(b) Given T, the reaction-equilibriuum equation allows solution for yA. The two phase-equilibrium

equations can then be solved for xA and P . The equilibrium state therefore depends solely on T.

13.38 (a) For low pressure and a temperature of 500 K, the system is assumed to be a mixture of ideal

gases, for which Eq. (13.28) is appropriate. Therefore,

yMX

yOX= P

P

0

KI = KIyPX

yOX= P

P

0

KII = KIIyEB

yOX= P

P

0

KIII = KIII

(b) These equation equations lead to the following set:

yMX = KIyOX (1) yPX = KIIyOX (2) yEB = KIIIyOX (3)

The mole fractions must sum to unity, and therefore:

yOX + KIyOX + KIIyOX + KIIIyOX = yOX(1 + KI + KII + KIII) = 1

yOX =

1

1 + KI + KII + KIII (4)

(c) With the assumption that CP = 0 and therefore that K2 = 1, Eqs. (13.20), (13.21), and (13.22)

combine to give:

K= K0K1 = exp

G298RT0

exp

H298

RT0

1

T0

T

Whence, K= exp

H298

1

298.15

500

G298

(8.314)(298.15)

The data provided lead to the following property changes of reaction and equilibrium constants

at 500 K:

Reaction H298 G298 K

I 1,750 3,300 2.8470

II 1,040 1,000 1.2637

III 10,920 8,690 0.1778

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(d) Substitution of numerical values into Eqs. (1), (2), (3), and (4) yields the following values for the

mole fractions:

yOX = 0.1891 yMX = 0.5383 yPX = 0.2390 yEB = 0.0336

13.40 For the given flowrates, nA0 = 10 and nB0 = 15, with nA0 the limiting reactant without (II)

nA = nA0 I IInB = nB0 InC = I IInD = IIn = n0 I II

Use given values ofYC and SC/D to find I and II:

YC =I II

nA0and SC/D =

I II

II

Solve for I and II:

I =

SC/D + 1

SC/D

nA0YC =

2 + 1

2

10 0.40 = 6

II =nA0YC

SC/D=

10 0.40

2= 2

nA = 10 6 2 = 2

nB = 15 6 = 9

nC = 6 2 = 4

nD = 2 = 2

n = 17

yA = 2/17 = 0.1176

yB = 9/17 = 0.5295

yC = 4/17 = 0.2353

yD = 2/17 = 0.1176

= 1

13.42 A compound with large positive Gf has a disposition to decompose into its constituent elements.

Moreover, large positive Gf often implies large positive Hf. Thus, if any decomposition product

is a gas, high pressures can be generated in a closed system owing to temperature increases resulting

from exothermic decomposition.

13.44 By Eq. (13.12), G i

iGi and from Eq. (6.10), (G

i /P)T = V

i

G

P

T

=i

i

Gi

P

T

= i

iVi

For the ideal-gas standard state, Vi = RT/P. Therefore

G

P T

=i

iRTP = RT

P and G(P2 )G(P1 ) = RT ln P

2P1

13.47 (a) For isomers at low pressure Raoults law should apply:

P = xAPsat

A + xBPsat

B = Psat

B + xA(Psat

A Psat

B )

For the given reaction with an ideal solution in the liquid phase, Eq. (13.33) becomes:

Kl =xB

xA=

1 xA

xAfrom which xA =

1

Kl + 1

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The preceding equation now becomes,

P =

1

1

Kl + 1

P satB +

1

Kl + 1

P satA

P =

Kl

Kl + 1

P satB +

1

Kl + 1

P satA (A)

For Kl = 0 P = P satA For Kl = P = P satB

(b) Given Raoults law:

1 = xA + xB = yAP

P satA+ yB

P

P satB= P

yA

P satA+

yB

P satB

P =1

yA/Psat

A + yB/Psat

B

=P satA P

satB

yAPsat

B + yBPsat

A

=P satA P

satB

P satA + yA(Psat

B Psat

A )

For the given reaction with ideal gases in the vapor phase, Eq. (13.28) becomes:

yB

yA= Kv whence yA =

1

Kv + 1

Elimination ofyA from the preceding equation and reduction gives:

P =(Kv + 1)P satA P

satB

KvP satA + Psat

B

(B)

For Kv = 0 P = P satA For Kv = P = P satB

(c) Equations (A) and (B) must yield the same P . ThereforeKl

Kl + 1

P satB +

1

Kl + 1

P satA =

(Kv + 1)P satA Psat

B

KvP satA + Psat

B

Some algebra reduces this to:Kv

Kl=

P satB

P satA

(d) As mentioned already, the species (isomers) are chemically similar, and the low pressure favors

ideal-gas behavior.

(e) F= N+ 2 r= 2 + 2 2 1 = 1 Thus fixing T should suffice.

708