CHAPTER 9

CHEMICAL EQUILIBRIA

9.1 (a) False, Equilibrium is dynamic. At equilibrium, the concentrations of

reactants and products will not change, but the reaction will continue to

proceed in both directions.

(b) False. Equilibrium reactions are affected by the presence of both

products and reactants.

(c) False. The value of the equilibrium constant is not affected by the

amounts of reactants or products added as long as the temperature is

constant.

(d) True.

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 1 2 3 4

time

[PCl3] = [Cl2]

[PCl5]

9.3

9.5 (a) C2

[COCl][Cl] ;[CO][Cl ]

K = (b) 2

C2 2

[HBr] ;[H ][Br ]

K = (c) 2 2

2 2C 2 3

2 2

[SO ] [H O][H S] [O ]

K =

SM-235

9.7 (a) Because the volume is the same, the number of moles of is larger

in the second experiment. (b) Because is a constant and the

denominator is larger in the second case, the numerator must also be

larger; so the concentration of is larger in the second case.

(c) Although is the same, will be different, a result

seen by solving for in each case. (d) Because is a constant,

is the same. (e) Because is the same, its

reciprocal must be the same.

2O

CK

2O

32 3[O ] /[O ]2

2 2

2 3[O ]/[O ]

CK CK

32 3[O ] /[O ] 3

2 3[O ] /[O ]

9.9 2

2 2

[HI][H ][I ]CK =

2

3 3

2

3 3

2

3 3

(0.0137)for condition 1, 48.8(6.47 10 ) (0.594 10 )

(0.0169)for condition 2, 48.9(3.84 10 ) (1.52 10 )

(0.0100)for condition 3, 48.9(1.43 10 ) (1.43 10 )

C

C

C

K

K

K

− −

− −

− −

= =× ×

= =× ×

= =× ×

9.11 (a) 2

3O

1P

(b) 2

7H OP

(c) 2

2 3

NO NO

N O

P PP

9.13 To answer these questions, we will first calculate G∆ ° for each reaction

and then use that value in the expression ln .G RT K∆ ° = −

(a) 2 2 22 H (g) O (g) 2 H O(g)+ ⎯⎯→

SM-236

1r f 2

r

p

1 1

80

2( (H O, g) 2( 228.57 kJ mol ) 457.14 kJln or

ln

457140 Jln 184.42(8.314 J K mol ) (298.15 K)

1 10

G GG RT K

GK

RT

K

K

−

− −

∆ ° = ∆ ° = − ⋅ = −∆ ° = −

∆ °= −

−= − = +

⋅ ⋅

= ×

(b) 2 22 CO(g) O (g) 2 CO (g)+ ⎯⎯→

r f 2 f1 1

1 1

2 (CO , g) [2 (CO, g)]

2( 394.36 kJ mol ) [2( 137.17 kJ mol )]514.38 kJ

514 380 Jln 207.5(8.314 J K mol ) (298.15 K)

G G G

K

− −

− −

∆ ° = × ∆ ° − × ∆ °

= − ⋅ − − ⋅= −

−= − = +

⋅ ⋅

901 10K = ×

(c) 3 2CaCO (s) CaO(s) CO (g)⎯⎯→ +

r f f 2 f 31 1

1 1

(CaO, s) (CO , g) [ (CaCO (s)]

( 604.03 kJ mol ) ( 394.36 kJ mol ) [ 1128.8 kJ mol ]130.4 kJ

130 400 Jln 52.6(8.314 J K mol ) (298 K)

G G G G

K

1− − −

− −

∆ ° = ∆ ° + ∆ ° − ∆ °

= − ⋅ + − ⋅ − − ⋅= +

+= − = −

⋅ ⋅

231 10K −= ×

9.15 r(a) lnG RT∆ ° = − K

11 1 (8.314 J K mol ) (1200 K) ln 6.8 19 kJ mol− − −= − ⋅ ⋅ = − ⋅

(b) 1 1r

1

ln (8.314 J K mol ) (298 K) ln 1.1 10

68 kJ mol

G RT K 12− − −

−

∆ ° = − = − ⋅ ⋅ ×

= + ⋅

9.17 First we must calculate K for the reaction, which can be done using data

from Appendix 2A:

1 1r f2 (NO, g) 2 86.55 kJ mol 173.1 kJ molG G − −∆ ° = × ∆ ° = × ⋅ = ⋅

SM-237

1

1 1

ln

ln

173100 J molln 69.9(8.314 J K mol ) (298 K)

G RT KGK

RT

K−

− −

∆ ° = −∆ °

= −

+ ⋅= − = −

⋅ ⋅

314 10K −= ×

Because Q > K, the reaction will tend to proceed to produce reactants.

9.19 The free energy at a specific set of conditions is given by

r r

r2

r2

1 1

21 1

-1 1

lnln ln

[I]ln ln[I ]

(8.314 J K mol ) (1200 K) ln (6.8)(0.98)(8.314 J K mol ) (1200 K) ln(0.13)

8.3 x 10 kJ mol

G G RT QG RT K RT Q

G RT K RT

− −

− −

−

∆ = ∆ ° +∆ = − +

∆ = − +

= − ⋅ ⋅

+ ⋅ ⋅

= ⋅

Because is positive, the reaction will be spontaneous to produce IrG∆ 2.

9.21 The free energy at a specific set of conditions is given by

r r

r2

3r 3

2 21 1

21 1

3

1

lnln ln

[NH ]ln ln[N ][H ]

(8.314 J K mol ) (400 K) ln (41)(21)(8.314 J K mol ) (400 K) ln

(4.2) (1.8)27 kJ mol

G G RT QG RT K RT Q

G RT K RT

− −

− −

−

∆ = ∆ ° +∆ = − +

∆ = − +

= − ⋅ ⋅

+ ⋅ ⋅

= − ⋅

Because is negative, the reaction will proceed to form products. rG∆

9.23 (a) 2

2NO Cl 2

2NOCl

1.8 10P P

KP

−= = ×

SM-238

(3 2)2 4

12.027 K

12.027 K 12.027 K 1.8 10 4.3 10500 K

n

C

n

C

TK K

K KT

∆

−∆− −

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= = × =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

×

(b) 2CO 167K P= =

(1)12.027 K 12.027 K 167 1.87

1073 K

n

CK KT

∆ ⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

=

9.25 For the reaction written as 2 2 3N (g) 3 H (g) 2 NH (g) Eq.1+ →

3

2 2

2NH

3N H

41P

KP P

= =

(a) For the reaction written as 3 2 22 NH (g) N (g) 3 H (g) Eq. 2→ +

2 2

3

3N H

2NH

P PK

P=

This is Eq. 1

1 1 0.02441K

= =

(b) For the reaction written as 312 2 32 2N (g) H (g) NH (g) Eq. 3+ →

3

2 2

NHEq. 3 Eq.11/ 2 3/ 2

N H

41 6.4P

K KP P

= = = =

Note that 1/ 21Eq. 3 Eq.12Eq. 3 Eq.1and thus .K K= =

(c) For the reaction written as 2 2 32 N (g) 6 H (g) 4 NH (g) Eq. 4+ →

3

2 2

4NH 2 2

Eq. 4 Eq.12 6N H

41 1.7 10P

K KP P

= = = = × 3

Note that 2Eq. 3 Eq.1Eq. 4 2 Eq.1and thus .K K= =

9.27 2 2H (g) I (g) 2 HI(g)+ → 160CK =

2

2 2

[HI][H ][I ]CK =

SM-239

3 2

32

(2.21 10 )160[H ](1.46 10 )

−

−

×=

×

3 2

2 3

5 -2

(2.21 10 )[H ](160) (1.46 10 )

[H ] 2.1 10 mol L

−

−

−

×=

×

= × ⋅ 1

9.29 3 2

5

PCl Cl

PCl

P PK

P=

3

3

3

PCl

PCl

PCl

(5.43)25

1.18(25) (1.18) 5.4

5.435.4 bar

P

P

P

=

= =

=

9.31 (a) 2 2

2

160HI

H I

pKp p

= =⋅

2(0.10) 0.50(0.20) (0.10)

Q = =

(b) the system is not at equilibrium ,Q K≠ ∴

(c) Because more products will be formed. ,Q K<

9.33 (a) 2

632

2 2

[SO ] 1.7 10[SO ] [O ]CK = = ×

24

23 4

1.0 100.500

6.91.20 10 5.0 10

0.500 0.500

CQ

−

− −

⎡ ⎤×⎢ ⎥⎣ ⎦= =

⎡ ⎤ ⎡ ⎤× ×⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

(b) Because more products will tend to form, which will result

in the formation of more

,C CQ K<

3SO .

SM-240

9.35 1

1.90 g HI 0.0148 mol HI127.91 g mol−

=⋅

2 22 HI(g) H (g) I⎯⎯→ +

0.0172 mol 2 x− x x

2

32 2

0.0148 mol 0.0172 mol 20.0012 mol

0.0012 0.0012 0.00122.00 2.00 2.00 0.0066 or 6.6 10

0.0148 0.01482.00 2.00

C

xx

K −

= −=

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦= = =

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

×

9.37 4 2 24 2 21

4 2 2

25.0 g NH (NH CO ) 0.320 mol NH (NH CO )78.07 g mol NH (NH CO )− =

⋅

4221

24

3 2 3

4

24 42 8

3 2

0.0174 g CO 3.95 10 mol CO44.01g mol CO2 mol NH are formed per mol of CO , so mol NH 2 3.95 10

7.90 10

7.90 10 3.95 10[NH ] [CO ] 1.58 100.250 0.250CK

−−

−

−

− −−

= ×⋅

= × ×

= ×

⎛ ⎞ ⎛ ⎞× ×= = = ×⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

9.39 (a) The balanced equation is: 2Cl (g) Cl(g)�

The initial concentration of 122

0.0020 mol ClCl (g) is 0.0010 mol L2.0 L

−= ⋅

Concentration → 2 C 1(mol L )−⋅ 2Cl (g) l(g)

initial 0.0010 0

change x− 2 x+

equilibrium 0.0010 x− 2 x+

SM-241

2 27

C2

2 7

2 7 10

7 7 2

7 5

6 6

[Cl] (2 ) 1.2 10[Cl ] (0.0010 )

4 (1.2 10 ) (0.0010 )4 (1.2 10 ) (1.2 10 ) 0

(1.2 10 ) (1.2 10 ) 4(4) ( 1.2 10 )2.4

(1.2 10 ) 4.4 108

5.5 10 or 5.5 10

xKx

x xx x

x

x

x

−

−

− −

− −

− −

− −

= = = ×−

= × −

+ × − × =

− × ± × − − ×=

− × ± ×=

= − × + ×

10−

1−

1

The negative answer is not meaningful, so we choose

The concentration of is essentially

unchanged because The concentration of

atoms is

65.5 10 mol L .x −= + × ⋅ 2Cl

60.0010 5.5 10 0.0010.−− × ≅

Cl 6 52 (5.5 10 ) 1.1 10 mol L .− −× × = × ⋅ − The percentage

decomposition of is given by 2Cl

65.5 10 100 0.55%0.0010

−×× =

(b) The balanced equation is: 2F (g) 2 F(g)�

The problem is worked in an identical fashion to (a) but the equilibrium

constant is now 41.2 10 .−×

The initial concentration of 122

0.0020 mol FF (g) is 0.0010 mol L2.0 L

−= ⋅

Concentration 1(mol L )−⋅ 2F (g) → 2 F(g)

initial 0.0010 0

change x− 2 x+

equilibrium 0.0010 x− 2 x+

SM-242

2 24

22 4

2 4 7

4 4 2

4 3

4 4

[F] (2 ) 1.2 10[F ] (0.0010 )

4 (1.2 10 ) (0.0010 )4 (1.2 10 ) (1.2 10 ) 0

(1.2 10 ) (1.2 10 ) 4(4) ( 1.2 10 )2.4

(1.2 10 ) 1.4 108

1.9 10 or 1.6 10

CxK

x

x xx x

x

x

x

−

−

− −

− −

− −

− −

= = = ×−

= × −

+ × − × =

− × ± × − − ×=

− × ± ×=

= − × + ×

7−

1−

1−

1−

The negative answer is not meaningful, so we choose

The concentration of is

The concentration of F atoms is

The percentage decomposition of

is given by

41.6 10 mol L .x −= + × ⋅ 2F

4 40.0010 1.6 10 8 10 mol L .− −− × = × ⋅

4 42 (1.6 10 ) 3.2 10 mol L .− −× × = × ⋅ 2F

41.6 10 100 16%0.0010

−×× =

(c) is more stable. This can be seen even without the aid of the

calculation from the larger equilibrium constant for the dissociation for

compared to

2Cl

2F

2Cl .

9.41 Concentration (bar) 2 HBr(g) → 2 2H (g) Br (g)+

initial 31.2 10−× 0 0

change 2 x− x+ x+

final 31.2 10 2 x−× − x+ x+

2 2

2H Br

HBr

p pK

p⋅

=

SM-243

2 2

211

3 2 3 2

211

3 2

63

6 3

6 6 3

8

8

( ) ( )7.7 10(1.2 10 2 ) (1.2 10 2 )

7.7 10(1.2 10 2 )

8.8 10(1.2 10 2 )

(8.8 10 ) (1.2 10 2 )2(8.8 10 ) (8.8 10 ) (1.2 10 )1.1 10

1.1 10 bar; the pressure ofH Br

x x xx x

xx

xx

x xx xxp p

−− −

−−

−−

− −

− − −

−

−

× = =× − × −

× =× −

= ×× −

= × × −

+ × = × ×

≅ ×

= = ×

2 2

HBr is essentially unaffected

by the formation of Br and H .

The percentage decomposition is given by 8

33

2 (1.1 10 bar) 100 1.8 10 %1.2 10 bar

−−

−

×× = ×

×

9.43 (a) Concentration of initially 5PCl

51

5 1

1.0 g PCl208.22 g mol PCl

0.019 mol L0.250 L

−−

⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ⋅

Concentration + 1(mol L )−⋅ 5PCl (g) → 3PCl (g) 2Cl

initial 0.019 0 0

change x− x+ x+

final 0.019 x− x+ x+2

2 2

5

[PCl ][Cl ] ( ) ( )[PCl ] (0.019 ) (0.019 )C

x x xKx x

= = =− −

22

2 2

2 2

1.1 10(0.019 )

(1.1 10 ) (0.019 )(1.1 10 ) 2.1 10 0

−

−

− −

= ×−

= × −

+ × − × =

xx

x xx x 4

SM-244

2 2 2

2

(1.1 10 ) (1.1 10 ) (4)(1)( 2.1 10 )2.1

(1.1 10 ) 0.031 0.010 or 0.0212.1

− −

−

− × ± × − − ×=

− × ±= = + −

x

x

4−

The negative root is not meaningful, so we choose 10.010 mol L .x −= ⋅ 1 1

3 2 5[PCl ] [Cl ] 0.010 mol L ; [PCl ] 0.009 mol L .− −= = ⋅ = ⋅

(b) The percentage decomposition is given by

0.010 100 53%0.019

× =

9.45 Starting concentration of 13

0.400 molNH 0.200 mol L2.00 L

−= = ⋅

Concentration 1(mol L )−⋅ → 4NH HS(s) 3 2NH (g) + H S(g)

initial — 0.200 0

change — x+ x+

final — 0.200 x+ x+

3 2

4

2 4

2 4

[NH ][H S] (0.200 ) ( )

1.6 10 (0.200 ) ( )0.200 1.6 10 0

( 0.200) ( 0.200) (4) (1) ( 1.6 10 )2.1

0.200 0.2016 0.0008 or 0.20082.1

CK x x

x xx x

x

x

−

−

−

= = +

× = +

+ − × =

− + ± + − − ×=

− ±= = + −

The negative root is not meaningful, so we choose 4 18 10 mol Lx − −= × ⋅

(note that in order to get this number we have had to ignore our normal

significant figure conventions). 1 4 1

3[NH ] 0.200 mol L 8 10 mol L 0.200 mol L 1− − −= + ⋅ + × ⋅ = ⋅ −

4 12[H S] 8 10 mol L− −= × ⋅

Alternatively, we could have assumed that 4 40.2, the 0.200 1.6 10 , 8.0 10 .x x x− −<< = × = ×

SM-245

9.47 The initial concentrations of and are calculated as follows: 5PCl 3PCl

1 15 3

0.200 mol 0.600 mol[PCl ] 0.0500 mol L ; [PCl ] 0.150 mol L4.00 L 4.00 L

− −= = ⋅ = = ⋅

Concentrations 1(mol L )−⋅ 5PCl → 3 2PCl (g) Cl (g)+

initial 0.0500 0.150 0

change x− x+ x+

final 0.0500 – x 0.150 + x x+

3 2

5

2

2

2

[PCl ][Cl ] (0.150 ) ( ) 33.3[PCl ] (0.0500 )

0.150 (33.3) (0.0500 )33.45 1.665 0

33.45 (33.45) (4) (1) ( 1.665) 33.4 33.6 0.0497(2) (1) 2

Cx xK

x

x x xx x

x

+= = =

−

+ = −

+ − =

− ± − − − ±= = =

1

1

The negative root has no physical meaning and so it can be discarded. 1 1 4

51 1

31

2

[PCl ] 0.0500 mol L 0.497 mol L 3 10 mol L

[PCl ] 0.150 mol L 0.0497 mol L 0.200 mol L

[Cl ] 0.0497 mol L

− − −

− −

−

= ⋅ − ⋅ = ×

= ⋅ + ⋅ = ⋅

= ⋅

−

−

⋅

9.49 The initial concentrations of and are equal at 0.114

because the vessel has a volume of 1.00 L.

2N 2O 1mol L−⋅

Concentrations 1(mol L )−⋅ 2N (g) + → 2 2O (g) NO(g)

initial 0.114 0.114 0

change x− x− 2 x+

final 0.114 – x 0.114 – x 2 x+

SM-246

2 2

22 2

25

2

25

2

[NO] (2 ) (2 )[N ][O ] (0.114 ) (0.114 ) (0.114 )

(2 )1.00 10(0.114 )

(2 )1.00 10(0.114 )

Cx xK

2

x x x

xx

xx

−

−

= = =− − −

× =−

× =−

3

3

4

4

4

(2 )3.16 10(0.114 )

2 (3.16 10 ) (0.114 )2.00316 3.60 10

1.8 10[NO] 2 2 1.8 10 3.6 10 ;

xx

x xx

xx

−

−

−

−

4− −

× =−

= × −

= ×

= ×

= = × × = × the concentrations of and

remain essentially unchanged at 0.114

2N 2O

1mol L .−⋅

9.51 The initial concentrations of and are: 2H 2I

1 12 2

0.400 mol 1.60 mol[H ] 0.133 mol L ; [I ] 0.533 mol L3.00 L 3.00 L

− −= = ⋅ = = ⋅

Concentrations 1(mol L )−⋅ 2H (g) + 2I (g) 2 HI(g)

initial 0.133 0.533 0

change x− x− 2 x+

final 0.133 x− 0.533 x− 2 x+

At equilibrium, 60.0% of the had reacted, so 40.0% of the

remains:

2H 2H

1 1

1 1

1

(0.400)(0.133 mol L ) 0.133 mol L0.133 mol L (0.400) (0.133 mol L )0.080 mol L

xxx

− −

− −

−

⋅ = ⋅ −

= ⋅ − ⋅

= ⋅

at equilibrium: [ 1 12H ] 0.133 mol L 0.080 mol L 0.053 mol L 1− − −= ⋅ − ⋅ = ⋅

SM-247

1 12

1 1

2 2

2 2

[I ] 0.533 mol L 0.080 mol L 0.453 mol L

[HI] 2 0.080 mol L 0.16 mol L[HI] 0.16 1.1

[H ][I ] (0.053) (0.453)CK

− −

− −

= ⋅ − ⋅ = ⋅

= × ⋅ = ⋅

= = =

1−

9.53 Initial concentrations of CO and are given by 2O

13 1

21

2 4 12

0.28 g CO28.01g mol CO

[CO] 5.0 10 mol L2.0 L

0.032 g O32.00 g mol O

[O ] 5.0 10 mol L2.0 L

−− −

−− −

⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ×

⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ×

⋅

⋅

Concentration 1(mol L )−⋅ 2 CO(g) + → 2O (g) 22 CO (g)

initial 35.0 10−× 45.0 10−× 0

change 2 x− x− 2 x+

final 35.0 10 2 x−× − 45.0 10 x−× − 2 x+ 2

22

22

3 2 4

2

2 5

[CO ][CO] [O ]

(2 )(5.0 10 2 ) (5.0 10 )

4(4 0.020 2.5 10 ) (5.0 10 )

CK

xx x

x4x x x

− −

− −

=

=× − × −

=− + × × −

2

3 2 5 8

2 3 2 5

2 3 2 5 8

3 2 5 8

5

5 12

40.664 0.022 3.5 10 1.25 10

4 (0.66) ( 4 0.022 3.5 10 1.25 10 )6.06 4 0.022 3.5 10 1.25 100 4 6.04 3.5 10 1.25 10

4.3 10[CO ] 8.6 10 mol L ; [CO] 4.9 10

xx x x

x x x xx x x x

x x xx

− −

− −

− −

− −

−

− − −

=− + − × + ×

= − + − × + ×

= − + − × + ×

= − − − × + ×

= ×

= × ⋅ = × 3 1

4 12

mol L

[O ] 4.6 10 mol L

−

− −

⋅

= × ⋅

8

SM-248

9.55 Concentrations

+ + 1(mol L )−⋅ 3CH COOH 2 5 3 2 5C H OH CH COOC H→ 2H O

initial 0.32 6.30 0 0

change x− x− x+ x+

final 0.32 x− 6.30 x− x+ x+2

3 2 5 22

3 2 5

2

2

2 2

2

2

[CH COOC H ][H O] ( ) ( )[CH COOH][C H OH] (0.32 ) (6.30 ) 6.62 2.02

4.06.62 2.02

4.0 26.48 8.083.0 26.48 8.08 0

( 26.48) ( 26.48) (4) (3.0) (8.08) 26.48 24.58(2) (3.0) 6.0

8.51

Cx x xKx x x x

xx x

x x xx x

x

x

= = =− − − +

=− +

− + =

− + =

− − ± − − + ±= =

= or 0.317

The root 8.51 is meaningless because it is larger than the concentration of

acetic acid and ethanol, so the value 0.317 is chosen. The equilibrium

concentration of the product ester is, therefore, 0.317 1mol L−⋅ . The

numbers can be confirmed by placing them into the equilibrium

expression:

3 2 5 2

3 2 5

[CH COOC H ][H O] (0.317) (0.317) 5.6[CH COOH][C H OH] (0.320 0.317) (6.300 0.317)CK = =

− −=

Note: This number does not appear to agree well with the given value of

= 4.0. CK

If 0.316 is used, the agreement is better, giving a quotient of 4.1. If 0.315

is used, the quotient is 3.3. The better answer is thus 0.316 The

discrepancy is caused by rounding errors in places that are really beyond

the accuracy of the measurement. Given that is only given to two

significant figures, the best report of the concentration of ester would be

1mol L .−⋅

CK

SM-249

0.32 , even though this value will not satisfy the equilibrium

expression as well as 0.316

1mol L−⋅1mol L .−⋅

9.57 2

2 2

[BrCl][Cl ][Br ]CK =

2

22

12

(0.145)0.031(0.495)[Br ]

(0.145)[Br ] 1.4 mol L(0.495) (0.031)

−

=

= = ⋅

9.59 We can calculate changes according to the reaction stoichiometry:

Amount (mol) CO(g) + 23 H (g) → 4CH (g) + 2H O(g)

initial 2.00 3.00 0 0

change x− 3 x− x+ x+

final 2.00 x− 3.00 3 x− 0.478 x+

According to the stoichiometry, 0.478 mol = x; therefore, at equilibrium,

there are 2.00 mol – 0.478 mol = 1.52 mol CO, 3.00 – 3(0.478 mol) = 1.57

mol and 0.478 mol To employ the equilibrium expression, we

need either concentrations or pressures; because is given, we will

choose to express these as concentrations. This calculation is easy because

V = 10.0 L:

2H , 2H O.

CK

1 12 4

12

4 23 3

2

[CO] 0.152 mol L ; [H ] 0.157 mol L ; [CH ] 0.0478 mol L ;

[H O] 0.0478 mol L[CH ][H O] (0.0478) (0.0478) 3.88[CO][H ] (0.152) (0.157)CK

− −

−

= ⋅ = ⋅ =

= ⋅

= = =

1−⋅

9.61 First, we calculate the initial concentrations of each species:

12

1 12 3

0.100 mol[SO ] [NO] 0.0200 mol L ;5.00 L

0.200 mol 0.150 mol[NO ] 0.0400 mol L ; [SO ] 0.0300 mol L5.00 L 5.00 L

−

− −

= = = ⋅

= = ⋅ = = ⋅

SM-250

We can use these values to calculate Q in order to see which direction the

reactions will go:

(0.0200) (0.0300) 0.75.(0.0200) (0.0400)

Q = = Because Q < the reaction will proceed

to produce more products.

,CK

Concentration 1(mol L )−⋅

+ NO(g) + 2SO (g) 2NO (g) → 3SO (g)

initial 0.0200 0.0400 0.0200 0.0300

change x− x− x+ x+

final 0.0200 x− 0.0400 x− 0.0200 x+ 0.0300 x+

3

2 2

2

2

[NO][SO ][SO ][NO ](0.0200 ) (0.0300 )85.0(0.0200 ) (0.0400 )

0.0500 0.000 6000.0600 0.000 800

CK

x xx x

x xx x

=

+ +=

− −

+ +=

− +

2 2

2 2

2

2

85.0( 0.0600 0.000 800) 0.0500 0.000 60085.0 5.10 0.0680 0.0500 0.000 60084.0 5.15 0.0674 0

( 5.15) ( 5.15) (4) (84.0) (0.0674) 5.15 1.97(2) (84.0) 168

x x x xx x x xx x

x

− + = + +

− + = + +

− + =

− − ± − − + ±= =

0.0424x = + or 0.0189+

The root 0.0424 is not meaningful because it is larger than the

concentration of . The root of choice is therefore 0.0189. 2NO

At equilibrium: 1 1

21 1

21 1

1 13

[SO ] 0.0200 mol L 0.0189 mol L 0.0011 mol L

[NO ] 0.0400 mol L 0.0189 mol L 0.0211 mol L

[NO] 0.0200 mol L 0.0189 mol L 0.0389 mol L[SO ] 0.0300 mol L 0.0189 mol L 0.0489 mol L

1

1

1

1

− − −

− − −

− − −

− − −

= ⋅ − ⋅ = ⋅

⋅= ⋅ − ⋅ =

= ⋅ + ⋅ = ⋅

= ⋅ + ⋅ = ⋅

To check, we can put these numbers back into the equilibrium constant

expression:

SM-251

3

2 2

[NO][SO ][SO ][NO ]

(0.0389) (0.0489) 82.0(0.0011) (0.0211)

CK =

=

Compared to = 85.0, this is reasonably good agreement given the

nature of the calculation. We can check to see, by trial and error, if a better

answer could be obtained. Because the value is low for the

concentrations we calculated, we can choose to alter x slightly so that this

ratio becomes larger. If we let x = 0.0190, the concentrations of NO and

are increased to 0.0390 and 0.0490, and the concentrations of

and are decreased to 0.0010 and 0.0200 (the stoichiometry of the

reaction is maintained by calculating the concentrations in this fashion).

Then the quotient becomes 91.0, which is further from the value for

than the original answer. So, although the agreement is not the best with

the numbers we obtained, it is the best possible, given the limitation on the

number of significant figures we are allowed to use in the calculation.

CK

CK

3SO 2SO

2NO

CK

9.63 (a) The initial concentrations are:

1 15 3

12

1.50 mol 3.00 mol[PCl ] 3.00 mol L ; [PCl ] 6.00 mol L ;0.500 L 0.500 L

0.500 mol[Cl ] 1.00 mol L0.500 L

− −

−

= = ⋅ = = ⋅

= = ⋅

First calculate : Q

5

3 2

[PCl ] 3.00 0.500[PCl ][Cl ] (6.00) (1.00)

Q = = =

Because the reaction is not at equilibrium. ,Q K≠

(b) Because Q < the reaction will proceed to form products. ,CK

(c)

Concentrations 1(mol L )−⋅ + 3PCl (g) 2Cl (g) → 5PCl (g)

initial 6.00 1.00 3.00

SM-252

change x− x− x+

final 6.00 x− 1.00 x− 3.00 x+

5

3 2

2

2

2

2

2

[PCl ][PCl ][Cl ]

3.00 3.000.56(6.00 ) (1.00 ) 7 6.00

(0.56) ( 7 6.00) 3.000.56 3.92 3.36 3.000.56 4.92 0.36 0

( 4.92) ( 4.92) (4) (0.56) (0.36) 4.92 4.48(2) (0.56) 1.12

CK

x xx x x x

x x xx x xx x

x

=

+ += =

− − − +

− + = +

− + = +

− + =

− − ± − − + ±= =

x = 9.2 or 0.07

Because the root 9.2 is larger than the amount of or available, it

is physically meaningless and can be discarded. Thus, x = 0.071

giving

3PCl 2Cl

1mol L ,−⋅

1 15

1 13

1 12

[PCl ] 3.00 mol L 0.07 mol L 3.07 mol L

[PCl ] 6.00 mol L 0.07 mol L 5.93 mol L

[Cl ] 1.00 mol L 0.07 mol L 0.93 mol L

− −

− −

− −

= ⋅ + ⋅ =

= ⋅ − ⋅ =

= ⋅ − ⋅ = ⋅

1

1

1

−

−

−

⋅

⋅

The number can be checked by substituting them back into the equilibrium

constant expression:

5

3 2

?

[PCl ][PCl ][Cl ]

(3.07) 0.56(5.93) (0.93)

0.56 0.56

CK =

=

=�

9.65 Pressures (bar) 2 HCl(g) + → 2H (g) 2Cl (g)

initial 0.22 0 0

change 2 x− x+ x+

final 0.22 2 x− x+ x+

SM-253

2 2H Cl2

HCl

342

( ) ( )3.2 10(0.22 2 )

P PK

Px x

x−

=

× =−

Because the equilibrium constant is small, assume that x << 0.22 2

342

2 34

34 2

18

3.2 10(0.22)

(3.2 10 ) (0.22)

(3.2 10 ) (0.22)

3.9 10

x

x

x

x

−

−

−

−

× =

= ×

= ×

= ± ×

2

The negative root is not physically meaningful and can be discarded. x is

small compared to 0.22, so the initial assumption was valid. The pressures

at equilibrium are

2 2

18HCl H Cl0.22 bar; =3.9 10 barP P P −= = ×

The values can be checked by substituting them into the equilibrium

expression:

2 2

18 1834?

2

34 34

18HCl H Cl

(3.9 10 ) (3.9 10 ) 3.2 10(0.22)

3.1 10 3.2 100.22 bar; 3.9 10 barP P P

− −−

− −√

−

× ×= ×

× = ×

= = = ×

The numbers agree very well for a calculation of this type.

9.67 (a) To determine on which side of the equilibrium position the conditions

lie, we will calculate Q :

1 12

13

332 2

2

0.342 mol 0.215 mol[CO] 0.114 mol L ; [H ] 0.0717 mol L ;3.00 L 3.00 L

0.125 mol[CH OH] 0.0417 mol L3.00 L

[CH OH] 0.0417 71.1 10[CO][H ] (0.114) (0.0717)

Q

− −

−

= = ⋅ = =

= = ⋅

= = = ×

⋅

SM-254

Because Q > the reaction will proceed to produce more of the

reactants, which means that the concentration of methanol will decrease.

,CK

(b)

Concentrations 1(mol L )−⋅ CO(g) + 22 H (g) → 3CH OH(g)

initial 0.114 0.0717 0.0417

change x+ 2 x+ x−

final 0.0114 x+ 0.0717 2 x+ 0.0417 x−

22

0.0417 1.1 10(0.0114 ) (0.0717 2 )C

xKx x

−−= =

+ +×

3−

−

2 2

3 2 2

2 3 3 2 4 6

2 3 3 2

0.0417 (1.1 10 ) (0.114 ) (0.0717 2 )(1.2 10 1.1 10 ) (4 0.287 5.14 10 )4.4 10 8.0 10 4.0 10 6.2 10

0 4.4 10 8.0 10 1.00 0.0417

x x xx x x

x x xx x x

−

− −

− − −

− −

− = × + +

= × + × + + ×

= × + × + × + ×

= × + × + −

This equation can be solved approximately, simply by inspection: it is

clear that the x term will be very much larger than the 3x and the 2x terms,

because their coefficients are very small compared to 1.00. This leads to a

prediction that 10.0417 mol Lx −= ⋅

1mol L ;

to within the accuracy of the data.

Essentially all of the will react, so that

0.156

3CH OH

1 1[CO] 0.114 mol L 0.0417 mol L− −= ⋅ + ⋅ = −⋅

1 12[H ] 0.0717 mol L 2 (0.0417 mol L ) 0.155 mol L .1− − −= ⋅ + ⋅ = ⋅ The

mathematical situation is odd in that clearly a 3[CH OH] 0= will not

satisfy the equilibrium constant. Knowing that the methanol concentration

is very small compared to the CO and concentrations, we can now

back-calculate to get a concentration value that will satisfy the equilibrium

expression:

2H

22 1.1 10

(0.156) (0.155)CyK −= = ×

4

2 2(1.1 10 ) (0.156) (0.155) 2.6 10y − −= × = ×

SM-255

Alternatively, the cubic equation can be solved with the aid of a graphing

calculator like the one supplied on the CD accompanying this book.

9.69 Since reactants are strongly favored it is easier to push the reaction as far

to the left as possible then start from new initial conditions.

Pressures (bar) 2 HCl(g) + → 2H (g) 2Cl (g)

original 2.0 1.0 3.0

new initial 4.0 0 2.0

change 2 x− x+ x+

final 4.0 2 x− x+ 2.0 x+

2 2H Cl2

HCl

342 2

33

-2 -1 -1

33

33

( ) (2.0 ) ( ) (2.0)3.2 10(4.0 2 ) (4.0) 16

2.6 10 barnRT (1 mole)(8.314 10 L bar K mol )(298K)V= =

p 2.6 10 bar9.7 10 L

P PK

Px x x x

xx

−

−

−

=

+× = ≈ =

−

= ×

××

= ×

9.71 (a) According to Le Chatelier’s principle, an increase in the partial

pressure of will result in creation of reactants, which will decrease

the partial pressure.

2CO

2H

(b) According to Le Chatelier’s principle, if the CO pressure is reduced,

the reaction will shift to form more CO, which will decrease the pressure

of 2CO .

(c) According to Le Chatelier’s principle, if the concentration of CO is

increased, the reaction will proceed to form more products, which will

result in a higher pressure of 2H .

(d) The equilibrium constant for the reaction is unchanged, because it is

unaffected by any change in concentration.

SM-256

9.73 (a) According to Le Chatelier’s principle, increasing the concentration of

NO will cause the reaction to form reactants in order to reduce the

concentration of NO; the amount of water will decrease.

(b) For the same reason as in (a), the amount of will increase. 2O

(c) According to Le Chatelier’s principle, removing water will cause the

reaction to shift toward products, resulting in the formation of more NO.

(d) According to Le Chatelier’s principle, removing a reactant will cause

the reaction to shift in the direction to replace the removed substance; the

amount of should increase. 3NH

(e) According to Le Chatelier’s principle, adding ammonia will shift the

reaction to the right, but the equilibrium constant, which is a constant, will

not be affected.

(f) According to Le Chatelier’s principle, removing NO will cause the

formation of more products; the amount of will decrease. 3NH

(g) According to Le Chatelier’s principle, adding reactants will promote

the formation of products; the amount of oxygen will decrease.

9.75 As per Le Chatelier’s principle, whether increasing the pressure on a

reaction will affect the distribution of species within an equilibrium

mixture of gases depends largely upon the difference in the number of

moles of gases between the reactant and product sides of the equation. If

there is a net increase in the amount of gas, then applying pressure will

shift the reaction toward reactants in order to remove the stress applied by

increasing the pressure. Similarly, if there is a net decrease in the amount

of gas, applying pressure will cause the formation of products. If the

number of moles of gas is the same on the product and reactant side, then

changing the pressure will have little or no effect on the equilibrium

distribution of species present. Using this information, we can apply it to

the specific reactions given. The answers are: (a) reactants;

(b) reactants; (c) reactants; (d) no change (there is the same number

of moles of gas on both sides of the equation); (e) reactants

SM-257

9.77 (a) If the pressure of NO (a product) is increased, the reaction will shift to

form more reactants; the pressure of should increase. 3NH

(b) If the pressure of (a reactant) is decreased, then the reaction will

shift to form more reactants; the pressure of should increase.

3NH

2O

9.79 If a reaction is exothermic, raising the temperature will tend to shift the

reaction toward reactants, whereas if the reaction is endothermic, a shift

toward products will be observed. For the specific examples given, (a) and

(b) are endothermic (the values for (b) can be calculated, but we know that

it requires energy to break an X—X bond, so those processes will all be

endothermic) and raising the temperature should favor the formation of

products; (c) and (d) are exothermic and raising the temperature should

favor the formation of reactants.

9.81 Even though numbers are given, we do not need to do a calculation to

answer this qualitative question. Because the equilibrium constant for the

formation of ammonia is smaller at the higher temperature, raising the

temperature will favor the formation of reactants. Less ammonia will be

present at higher temperature, assuming no other changes occur to the

system (i.e., the volume does not change, no reactants or products are

added or removed from the container, etc.).

9.83 To answer this question we must calculate Q:

2 2

33 3

2 2

[NH ] (0.500) 0.0104[N ][H ] (3.00) (2.00)

Q = = =

Because the system is not at equilibrium and because the

reaction will proceed to produce more products.

,Q K≠ ,Q K<

9.85 Because we want the equilibrium constant at two temperatures, we will

need to calculate and rH∆ ° rS∆ ° for each reaction:

SM-258

(a) 4 3NH Cl NH (g) HCl(g)→ +

r f 3 f f 4(NH , g) (HCl, g) (NH Cl, s)H H H H∆ ° = ∆ ° + ∆ ° − ∆ °

1 1r ( 46.11 kJ mol ) ( 92.31 kJ mol ) ( 314.43 kJ mol )H 1− − −∆ ° = − ⋅ + − ⋅ − − ⋅

1r 176.01 kJ molH −∆ ° = ⋅

r 3 4(NH , g) (HCl, g) (NH Cl, s)S S S S∆ ° = ° + ° − °

1 1 1 1 1 1r 192.45 J K mol 186.91 J K mol 94.6 J K molS − − − − − −∆ ° = ⋅ ⋅ + ⋅ ⋅ − ⋅ ⋅

1 1r 284.8 J K molS − −∆ ° = ⋅ ⋅

r rG H T S∆ ° = ∆ ° − ∆ °r

At 298 K:

1 1r(298 K)

1

r(298 K)

176.01 kJ (298 K) (284.8 J K )/(1000 J kJ )

91.14 kJ molln

G

G RT K

− −

−

∆ ° = − ⋅ ⋅

= ⋅∆ ° = −

r(298 K)

1

ln

91140 J 36.8(8.314 J K ) (298 K)

GK

RT

−

∆ °= −

= − = −⋅

161 10

At 423 K:K −= ×

1 1r(423 K)

1

r(423 K)

176.01 kJ (423 K) (284.8 J K )/(1000 J kJ )

55.54 kJ molln

G

G RT K

− −

−

∆ ° = − ⋅ ⋅

= ⋅∆ ° = −

r(423 K)

1

ln

55 540 J 15.8(8.314 J K )(423 K)

GK

RT

−

∆ °= −

= − = −⋅

71 10K −= ×

(b) 2 2 2 2H (g) D O(l) D (g) H O(l)+ → +

SM-259

r f 2 f 21 1

r1

r

r 2 2 2 21 1 1 1

r1 1 1

(H O, l) [ (D O, l)]

( 285.83 kJ mol ) [ 294.60 kJ mol ]

8.77 kJ mol(D , g) (H O, l) [ (H , g) (D O, l)]

144.96 J K mol 69.91 J K mol

[130.68 J K mol 75.94 J K

H H H

H

HS S S S S

S

− −

−

− − − −

− − −

∆ ° = ∆ ° − ∆ °

∆ ° = − ⋅ − − ⋅

∆ ° = ⋅∆ ° = ° + ° − ° + °

∆ ° = ⋅ ⋅ + ⋅ ⋅

− ⋅ ⋅ + ⋅ 1

1 1r

mol ]8.25 J K molS

−

− −

⋅

∆ ° = ⋅ ⋅

At 298 K:

1 1 1

r(298 K)

1

8.77 kJ mol (298 K) (8.25 J K mol )/(1000 J kJ )

6.31 kJ mol

G − − −

−

∆ ° = ⋅ − ⋅ ⋅ ⋅

= ⋅

1−

K r(298 K) lnG RT∆ ° = −

r(298 K)lnG

KRT

∆ °= −

1

6310 J 2.55(8.314 J K ) (298 K)−= − = −

⋅

27.8 10

At 423 K:K −= ×

1 1 1

r(423 K)

1

8.77 kJ mol (423 K) (8.25 J K mol )/(1000 J kJ )

5.28 kJ mol

G − − −

−

∆ ° = ⋅ − ⋅ ⋅ ⋅

= ⋅

1−

1

5280 Jln 1.50(8.314 J K )(423 K)

0.22

K

K

−= − = −⋅

=

9.87 ( ) ncK RT K∆=

2

1 1

1 1ln° ⎛ ⎞∆

= − −⎜ ⎟⎝ ⎠

rK HK R T T2

( )( )

2 2

1 21 1

1 1ln∆ °

∆

⎛ ⎞ ⎛ ⎞∆⎜ ⎟ = − −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

nc r

nc

RT K HR T TRT K

SM-260

22

1 1 1

2 2

1 1 2

1 1ln ln

1 1ln ln

°

°

⎛ ⎞⎛ ⎞ ⎛∆∆ + = − −⎜ ⎟⎜ ⎟ ⎜

⎝ ⎠ ⎝⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛∆

= − − − ∆⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝⎝ ⎠

c r

c

c r

c

KT Hn

T K R T

K H Tn

K R T T

2

1

⎞⎟⎠⎞⎟⎠

T

T

9.89

ln∆ ° = − = ∆ ° − ∆ °G RT K H T S-1 -1 15

2984 -1

-1 -1 14303

4 -1

298 303

(8.3145JK mol )(298.15K) ln(9.9197 10 )

7.9933 10 Jmol(8.3145JK mol )(303.15K) ln(1.4689 10 )

8.0283 10 Jmol(298.15K) [ (303.15K

−

−

∆ ° = − ×

= ×

∆ ° = − ×

= ×∆ ° − ∆ ° = ∆ ° − ∆ ° − ∆ ° −

G

G

G G H S H2 -1

-1 -1

) ]

3.5097 10 Jmol (5.00K)70.194JK mol

∆ °

− × = ∆ °

∆ ° = −

S

SS

The sign is negative indicating that the product ions create more order in

the system probably because of their interaction with solvating water

molecules.

9.91 (a) According to Le Chatelier’s principle, adding a product should cause a

shift in the equilibrium toward the reactants side of the equation.

(b) Because there are equal numbers of moles of gas on both sides of the

equation, there will be little or no effect upon compressing the system.

(c) If the amount of is increased, this will cause the reaction to shift

toward the formation of products.

2CO

(d) Because the reaction is endothermic, raising the temperature will favor

the formation of products.

(e) If the amount of is removed, this will cause the reaction to

shift toward the formation of products.

6 12 6C H O

(f) Because water is a liquid, it is by definition present at unit

concentration, so changing the amount of water will not affect the

reaction. As long as the glucose solution is dilute, its concentration can be

considered unchanged.

SM-261

(g) Decreasing the concentration of a reactant will favor the production of

more reactants.

9.93 (a) In order to solve this problem, we will manipulate the equations with

the known so that we can combine them to give the desired overall

reaction:

’sK

First, reverse equation (1) and multiply it by 12 :

12 2 22H (g) O (g) H O(g)+ → 4 11 1/ 2

1(1.6 10 )

K −=×

(4)

Multiply equation (2) by 12 also:

12 22CO (g) CO(g) O (g)→ + (5)

1/ 2105 (1.3 10 )K −= ×

Adding equations (4) and (5) gives the desired reaction. The resultant

equilibrium constant will be the product of the K’s for (4) and (5):

2 2 2CO (g) H (g) H O(g) CO(g)+ → +1/ 210

5 11

1.3 10 2.91.6 10

K−

−

⎛ ⎞×= ⎜ ⎟×⎝ ⎠

= (3)

(b) To obtain the K value for a net equation from two (or more) others,

the K’s are multiplied, but r’sG∆ ° are added:

(1) 2 22 H O(g) 2 H (g) O (g)+� 2

11−

2

10

r

1 1

2 1

ln

(8.314 J K mol ) (1565 K) ln 1.6 103.2 10 kJ mol

G RT K− −

−

∆ ° = −

= − ⋅ ⋅ ×

= + × ⋅

(2) 22 CO (g) 2 CO(g) O (g)→ +

r

1 1

2 1

ln

(8.314 J K mol ) (1565 K) ln 1.3 103.0 10 kJ mol

G RT K− − −

−

∆ ° = −

= − ⋅ ⋅ ×

= + × ⋅

The corresponding values for (4) and (5) are

2 21

r(4) 2

2 21r(5) 2

(3.2 10 kJ) 1.6 10 kJ

(3.0 10 kJ) 1.5 10 kJ

G

G

∆ ° = − × = − ×

∆ ° = × = ×

Summing these two values will give

SM-262

2 2r(3) 1.6 10 kJ 1.5 10 kJ 10 kJ molG 1−∆ ° = − × + × = − ⋅

1

r1 1

10 000 J molln 0.8(8.314 J K mol ) (1565 K)

GKRT

−

− −

∆ ° − ⋅= − = − = +

⋅ ⋅

2K =

This value is in reasonable agreement with the one obtained in (a), given

the problems in significant figures and due to rounding errors.

9.95 Pressure + 5PCl (g) 3PCl (g) 2Cl (g)

initial n

change nα− nα+ nα+

final (1 )n α− nα+ nα+

2 2 2

2 2

2

( ) ( )(1 ) (1 ) (1 )

(1 ) (1 )

1

(1 ) 1 1 1

α α α αα α αα α α α

α2α α α

α α α α

= = =− − −

= − + + = +

=+

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟− + − −⎝ ⎠ ⎝ ⎠

n n n nKn n

P n n n nPn

n P PK

2 2

2 2

2 2

2

(1 )α α

α α

α α

α

α

− =

− =

= +

=+

=+

K PK K PK P K

KP K

KP K

(a) For the specific conditions 4.96K = and 0.50P = bar,

4.96 0.9530.50 4.96

α = =+

(b) For the specific conditions 4.96K = and 1.00P = bar,

4.96 0.9121.00 4.96

α = =+

SM-263

9.97 (a) If K = 1.00, then G∆ °must be equal to 0 ( ln ).G RT K∆ ° = −

(b) This can be calculated by determining the values for H∆ ° and

at 25 C.S∆ ° °

1 1

1

393.51 kJ mol [( 110.53 kJ mol ) ( 241.82 kJ mol )]41.16 kJ mol

H − −

−

∆ ° = − ⋅ − − ⋅ + − ⋅

= − ⋅

1−

1 1 1 1

1 1 1 1

1 1

130.68 J K mol 213.74 J K mol [197.67 J K mol 188.83 J K mol ]

42.08 J K mol

− − − −

− − −

− −

∆ ° = ⋅ ⋅ + ⋅ ⋅

− ⋅ ⋅ + ⋅ ⋅

= − ⋅ ⋅

S−

1

1 1 1( 41.16 kJ mol ) (1000 J kJ ) ( 42.08 J K mol ) 0

978 (or 705 C)G TT K

− − − −∆ = − ⋅ ⋅ − − ⋅ ⋅ == °

(c) CO(g) + + 2H O(g) → 2CO (g) 2H (g)

10.00 bar 10.00 bar 5.00 bar 5.00 bar

change x− x− x+ x+

net 10.00 x− 10.00 x− 5.00 x+ 5.00 x+

2.50 barx =

All pressures are equal to 7.50 bar.

(d) First, check Q to determine the direction of the reaction:

Q =(10.00) (5.00) 2.08(6.00) (4.00)

=

Because Q is greater than 1, the reaction will shift to produce reactants.

CO(g) + + 2H O(g) → 2CO (g) 2H (g)

6.00 bar 4.00 bar 10.00 bar 5.00 bar

change x+ x+ x− x−

net 6.00 x+ 4.00 x+ 10.00 x− 5.00 x−

SM-264

2 2 2

2 2

CO(g) H O(g) CO (g) H (g)

(10.00 ) (5.00 ) 1(6.00 ) (4.00 )(10.00 ) (5.00 ) (6.00 ) (4.00 )

15.00 50.0 10.00 24.025.00 26.0

1.04 bar7.04 bar; 5.04 bar; 8.96 bar; 3.96 bar

x xx xx x x x

x x x xx

xP P P P

− −=

+ +− − = + +

− + = + +=

== = = =

9.99 (a) These values are easily calculated from the relationship

ln K. For the atomic species, the free energy of the reaction will be

G R∆ ° = − T

12 of

this value because the equilibrium reactions are for the formation of two

moles of halogen atoms.

The results are

Bond Dissociation Energy G∆ °

Halogen 1(kJ mol )−⋅ 1(kJ mol )−⋅

fluorine 146 19.2

chlorine 230 47.8

bromine 181 42.8

iodine 139 5.6

(b)

0

10

20

30

40

50

60

0 50 100 150 200 250 300

bond dissociation energy (kJ⋅mol-1)

free

ene

rgy

ofr

iom

at fo

n(k

J ⋅m

ol-1

)

SM-265

There is a correlation between the bond dissociation energy and the free

energy of formation of the atomic species, but the relationship is clearly

not linear.

0

20

40

60

80

100

120

0 10 20 30 40 50 60

atomic number

free

ene

rgy

of d

isso

ciat

ion

(kJ ⋅

mol

-1)

For the heavier three halogens, there is a trend to decreasing free energy of

formation of the atoms as the element becomes heavier, but fluorine is

anomalous. The F—F bond energy is lower than expected, owing to

repulsions of the lone pairs of electrons on the adjacent F atoms because

the F—F bond distance is so short.

9.101 (a) K<1 but will increase at higher temperature because S∆ >0.

(b) K>1

(c) K>1

(d) K>1

9.103 (a) Use the relationship ln 2

1 2

1 1 .K HK R T T

⎛ ⎞∆ °= − −⎜

⎝ ⎠1⎟ The value of H∆ ° is

first obtained by using the two points given in Table 9.1.

ln 3

5

1.7 × ∆ 10 1 13.4 10 1200 1000

HR

−

−

° ⎛ ⎞= − −⎜ ⎟× ⎝ ⎠

2 11.96 10 kJ molH −∆ ° = × ⋅

SM-266

We then use this value plus one of the known equilibrium constant points

in the equation

ln1

2985 1 1

196 000 J mol 1 13.4 10 8.314 J K mol 298 1000

K −

− − −

⎛ ⎞⋅ ⎛ ⎞= − −⎜ ⎟⎜ ⎟× ⋅ ⋅ ⎝ ⎠⎝ ⎠

29298 2.6 10K −= ×

(b) Using the thermodynamic data in Appendix 2A:

2

1 1

/ 29

Br (g) 2 Br(g)

2(82.40 kJ mol ) 3.11 kJ mol 161.69 kJ mole 4.5 10G RT

GK

1− − −

−∆ ° −

→

∆ ° = ⋅ − ⋅ = ⋅

= = ×

For equilibrium constant calculations, this is reasonably good agreement

with the value obtained from part (a), especially if one considers that H∆ °

will not be perfectly constant over so large a temperature range.

(c) We will use data from Appendix 2A to calculate the vapor pressure of

bromine:

2 2

1

Br (l) Br (g)

3.11 kJ mole 0.285G /RT

GK

−

−∆ °

→

∆ ° = ⋅

= =

The vapor pressure of bromine will, therefore, be 0.285 bar or 0.289 atm.

Remember that because the standard state for the thermodynamic

quantities is 1 bar, the values in K will be derived in bar as well.

(d) 2

2 2Br(g) Br(g)29

Br (l)

4.5 100.285 bar

P PP

−× = =

2 15 15Br(g) 3.6 10 bar or 3.6 10 atmP − −= × ×

(e) Use the ideal gas law:

1 1(0.289 atm) (0.0100 mol) (0.082 06 L atm K mol ) (298 K)0.846 L or 846 mL

PV nRTV

V

− −

=

= ⋅ ⋅ ⋅=

9.105 First, we calculate the equilibrium constant for the conditions given.

SM-267

2

3

(23.72) 41.0,(3.11) (1.64)

K = =

3

3

which corresponds to the reaction written as

2 2N (g) 3 H (g) 2 NH (g)+ →

We then set up the table of anticipated changes upon introduction of the

nitrogen:

2 2N (g) 3 H (g) 2 NH (g)+ →

initial 4.68 bar 1.64 bar 23.72 bar

change x− 3 x− 2 x+

total 4.68 x− 1.64 3x− 23.72 2 x+

2

3

(23.72 2 )41.0(4.68 ) (1.64 3 )

xx x

+=

− −

The equation can be solved using a graphing calculator, other computer

software, or by trial and error. The solution is 0.176.x = The pressures of

gases are

2

2

3

N

H

NH

4.33 bar or 4.39 atm

1.11 bar or 1.12 atm

24.07 bar or 24.39 atm

P

P

P

=

=

=

50

650

1250

250 300 350

T / K

K

9.107

SM-268

9.109 (a) 2 2

2

2 4

[NO ] (2.13) 11.2[N O ] 0.405CK = = =

(b) If is added, the equilibrium will shift to produce more

The amount of will be greater than initially present, but less than the

present immediately upon making the addition. will not

be affected.

2NO 2 4N O .

2NO

13.13 mol L−⋅ CK

(c)

Concentrations 1(mol L )−⋅ 2 4N O 22 NO

initial 0.405 3.13

change x+ 2 x−

final 0.405 x+ 3.13 2 x−

2

2

2

2

2

(3.13 2 )11.20.405

(11.2) (0.405 ) (3.13 2 )11.2 4.536 4 12.52 9.7974 23.7 5.26 0

( 23.7) ( 23.7) (4) (4) (5.26) 23.7 21.9(2) (4) 8

5.70 or 0.23

xx

x xx x x

x x

x

x

−=

+

+ = −

+ = − +

− + =

− − ± − − ±= =

=

At equilibrium 1 12 4[N O ] 0.405 mol L 0.23 mol L 0.64 mol L 1− − −= ⋅ + ⋅ = ⋅

1 12[NO ] 3.13 mol L 2(0.23 mol L ) 2.67 mol L− −= ⋅ − ⋅ = 1−⋅

These concentrations are consistent with the predictions in (b).

9.111 To find the vapor pressure, we first calculate G∆ ° for the conversion of

the liquid to the gas at 298 K, using the free energies of formation found in

the appendix:

SM-269

2 2 2 2

2 2 2 2

H O(l) H O(g) f(H O(g)) f(H O(l))

1 1

1

D O(l) D O(g) f(D O(g)) f(D O(l))

1 1

1

( 228.57 kJ mol ) [ 237.13 kJ mol ]8.56 kJ mol

( 234.54 kJ mol ) [ 243.44 kJ mol ]8.90 kJ mol

G G G

G G G

→

− −

−

→

− −

−

∆ ° = ∆ ° − ∆ °

= − ⋅ − − ⋅

= ⋅∆ ° = ∆ ° − ∆ °

= − ⋅ − − ⋅

= ⋅

The equilibrium constant for these processes is the vapor pressure of the

liquid:

2 2H O D OorK P K P= =

Using ln K, we can calculate the desired values. G R∆ ° = − T

For 2H O:

ln 1

1 1

8560 J mol 3.45(8.314 J K mol ) (298 K)

GKRT

−

− −

∆ ° ⋅= − = − = −

⋅ ⋅

bar 0.032K =

1 atm 760 Torr0.032 bar 24 Torr1.013 25 bar 1 atm

× × =

For 2D O:

ln 1

1 1

8900 J mol 3.59(8.314 J K mol ) (298 K)

GKRT

−

− −

∆ ° ⋅= − = − = −

⋅ ⋅

bar 0.028K =

1 atm 760 Torr0.028 bar 21 Torr1.013 25 bar 1 atm

× × =

The answer is that D has a lower zero point energy than H. This makes the

— “hydrogen bond” stronger than the — hydrogen

bond. Because the hydrogen bond is stronger, the intermolecular forces are

stronger, the vapor pressure is lower, and the boiling point is higher.

2D O 2D O 2H O 2H O

Potential energy curves for the O—H and O—D bonds as a function of

distance:

energy required to break O—H or O—D bond. E∆ =

SM-270

9.113 First, we derive a general relationship that relates fG∆ ° values to

nonstandard state conditions. We do this by returning to the fundamental

definition of and G∆ G∆ ° .

f m(products) m(reactants)G G G∆ ° = Σ ° − Σ °

Similarly, for conditions other than standard state, we can write

f m(products) m(reactants)G G G∆ = Σ − Σ

Because m m ln ,G G RT Q= ° +

f m i (products) m i (reactants)( ln ) ( ln )G G RT P G RT P∆ = Σ ° + − Σ ° +

But because all reactants or products in the same system refer to the same

standard state,

f m(products) m(reactants) (reactants)( lnG G G n RT Q∆ = Σ ° − Σ ° + ∆ )

) f f (reactants)( lnG G n RT Q∆ = ∆ ° + ∆

The value f , which is the nonstandard value for the conditions of 1

atm,

G∆

etc. becomes the new standard value. 11 mol L ,−⋅

(a) 1 1 1 12 22 2 2 2H (g) I (g) HI(g), 1 mol ( mol mol) 0n+ → ∆ = − + =

1 atm = 1.013 25 bar

f

1 1

1.70 kJ mol

(0) (8.314 J K mol ) (298 K) (ln 1.013 25)/(1000 J kJ )1.70 kJ mol

G1− − −

∆ = ⋅

+ ⋅ ⋅ ⋅= ⋅

(b) 1 122 2C(s) O (g) CO(g), 1 mol mol moln+ → ∆ = − = 1

2

f

1 1 1−⋅12

1

137.17 kJ mol

( ) (8.314 J K . mol ) (298 K)(ln 1.013 25)/(1000 J kJ )

137.15 kJ mol

G− −

−

∆ = − ⋅

+ ⋅

= − ⋅

(c) 1 1 1 12 22 2 2 2C(s) N (g) H (g) HCN(g), 1 mol ( mol mol)

0n+ + → ∆ = − +

=

1 Torr 31 atm 1.013 25 bar 1.333 10 bar760 Torr 1 atm

−× × = ×

SM-271

f

1 1 3

1

124.7 kJ mol

(0) (8.314 J K mol ) (298 K) (ln 1.333 10 )/(1000 J kJ )124.7 kJ mol

G1− − −

−

∆ = ⋅

+ ⋅ ⋅ × ⋅

= ⋅

−

1

(d) 2 4C(s) 2 H (g) CH (g), 1 mol 2 mol 1 moln+ → ∆ = − = −

51 Pa 10 bar−=

f

1 1 5

1

50.72 kJ mol

( 1) (8.314 J K mol ) (298 K)(ln 10 )/(1000 J kJ )22.2 kJ mol

G− − − −

−

∆ = − ⋅

+ − ⋅ ⋅ ⋅

= − ⋅

9.115 (a) (i) Increasing the amount of a reactant will push the equilibrium toward

the products. More NO2 will form. (ii) Removing a product will pull the

equilibrium toward products. More NO2 will form. (iii) Increasing total

pressure by adding an inert gas not change the relative partial pressures.

There will be no change in the amount of NO2. (b) Since there are two

moles of gas on both sides of the reaction, the volume cancels out of the

equilibrium constant expression and it is possible to use moles directly for

each component.

At equilibrium, SO2 = 2.4 moles -1.2 moles = 1.2 moles, SO3 = 1.2 moles

and NO = 1.2 moles since the reaction is 1:1:1:1. The original number of

moles of NO2 is set to x.

K = 3.00 = (1.2)(1.2)/(1.2)(x-1.2) = (1.2)/(x-1.2)

x=1.2/3.00 + 1.2

x=1.6 moles of NO2

9.117 (a) Reversing both reactions then adding them together gives the reaction

of interest.

NO + O NO2 K’=1/K=1/6.8 x 10-49=1.47 x1048

NO2 + O2 O3 + NO K”=1/K=1/5.8 x 10-34=1.72 x 1033

----------------------------

O2 + O O3 K=K’*K”=2.5 x 1081

SM-272

(b) Since the initial mixture is equimolar in the two reactants (i.e. each

partial pressure is 2.0 bar) and the equilibrium strongly favors products,

we can see that essentially all of both reactants will combine to form an

equimolar amount of ozone, or a final equilibrium partial pressure of 2.0

bar. Some very small partial pressure, x, of each reactant will be left and

must satisfy the equilibrium constant expression:

K = 2.5 x 1081 = PO3/(PO2*PO ) = 2.0 bar/(x2)

x = 2.8 x 10-41 bar = PO2 = PO and PO3 = 2.0 bar at equilibrium

SM-273