CHAPTER 9

CHEMICAL EQUILIBRIA

9.2 (a) True

(b) False. Changing the rate of a reaction will not affect the value of the

equilibrium constant; it merely changes how fast one gets to equilibrium.

(c) True

(d) False. The standard reaction free energy is not 0 at equilibrium.

The reaction free energy

∆ rG

∆ rG , which is dependent upon the

concentrations of the products and reactants, is 0 at equilibrium.

9.4 Cl2 decreases from 2.15 to 2.13 bar, PCl3 decreases from 1.28 to 1.26 bar,

and PCl5 increases from 0.02 to 0.04 bar. The shapes for the curves can

only be determined accurately if the rate law for the reaction is known (see

Chapter 13).

0

0.5

1

1.5

2

2.5

Pre

ssur

e (b

ar)

Time

[Cl2]

[PCl3]

[PCl5]

247

9.6 (a) 2

2

2NO2

NO O

=P

KP P

; (b) 3 2

5

SbCl Cl

SbCl

=P P

KP

; (c) 2 4

2 2

N H2

N H

=P

KP P

9.8 All values should be the same because the same amounts of the substances

are present at equilibrium. It doesn’t matter whether we begin with

reactants or with products; the equilibrium composition will be the same if

the same amounts of materials are used. If different amounts had been

used, only (e) would be the same in the two containers. A more detailed

analysis follows:

H2(g) + Br2(g) 2 HBr(g) →

The equilibrium constant expression for this system is

2

2 2

[HBr][H ] [Br ]

=CK

In terms of the change in concentration, x, of H2 and Br2 that has come

about at equilibrium, we may write

2 2

2

[2 ] [2 ][0.05 ] [0.05 ] [0.05 ]

= =− − −C

x xKx x x

(1)

in the first container, and in terms of the change in concentration, y, of

HBr that has come about at equilibrium in the second container, we may

write

2 2

2

[0.10 ] [0.10 ]

2 2 2

− −= =

⎡ ⎤ ⎡ ⎤ ⎛ ⎞⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎣ ⎦ ⎝ ⎠

Cy yK

y y y (2)

Because KC is a constant, and because the relative amounts of starting

materials are in the ratio of their stoichiometric factors, we must have

2 2

2 2

[2 ] [0.10 ][0.05 ]

2[2 ] [0.10 ]

[0.05 ]2

−=

− ⎛ ⎞⎜ ⎟⎝ ⎠−

=− ⎛ ⎞

⎜ ⎟⎝ ⎠

x yx y

x yyx

248

Cross multiplying:

(0.05 )(0.10 )0.005 0.05 0.100.10 2

= − −= − − += −

xy x yxy yy x

x xy

This may also be seen by solving quadratic equations for Eqs. 1 and 2 for

x and y to obtain any value of KC.

(a) 2[Br ] 0.05= − x in the first container, 2[Br ] /2= y in the second

container. Because /2 0.05= −y x satisfies the conditions of Eq. 3, the

concentrations and hence the amounts of Br2 are the same in the two cases.

(b) 2[H ] 0.05 /2= − =x y , as above; hence the concentrations of H2 are

the same in both systems.

(c) Because all concentrations are the same in both cases, this ratio will

also be the same.

(d) For the same reason as in part (c), this ratio is the same in both cases.

(e) This ratio is the equilibrium constant, so it must be the same for both

systems.

(f) Because all the concentrations and amounts are the same in both cases,

and because the volumes and temperatures are the same, the total pressure

must be the same:

2 2H Br HBr( )+ +=

n n n RP

VT

9.10 3 2NH H S=K P P

For condition 1, 0.307 0.307 0.0942For condition 2, 0.364 0.258 0.0939For condition 3, 0.539 0.174 0.0938

= × == × == × =

KKK

9.12 (a) 2

33

[Cl ] [ClO ][ClO ]

− −

−

(b) [CO2]

249

(c) 3

3

[H ][CH COO ][CH COOH]

+ −

9.14 (a) 2 CH4(g) + S8(s) → 2 CS2(l) + 4 H2S(g)

r f 2 f 2

f 41 1

1

1

2 (CS , l) 4 (H S, g) [2 (CH , g)]

2(65.27 kJ mol ) 4 ( 33.56 kJ mol ) [2 ( 50.72 kJ mol )]

97.74 kJ mol

− −

−

−

∆ ° = × ∆ ° + × ∆ °− × ∆ °

= ⋅ + − ⋅

− − ⋅

= + ⋅

G G GG

ln∆ ° = −G RT K

or

1

1 1

18

ln

97 740 J molln 39.4(8.314 J K mol )(298 K)

8 10

−

− −

−

∆ °= −

+ ⋅= − = −

⋅ ⋅

= ×

GKRT

K

K

(b) CaC2(s) + 2 H2O(l) Ca(OH)→ 2(s) + C2H2(g)

r f 2 f 2 2

f 2 f 21 1

1 1

1

(Ca(OH) , s) (C H , g) [ (CaC , s) 2 (H O, l)]

( 898.49 kJ mol ) ( 209.20 kJ mol ) [( 64.9 kJ mol ) 2( 237.13 kJ mol )]

150.1 kJ mol

− −

− −

−

∆ ° = ∆ ° + ∆ °− ∆ ° + × ∆ °

= − ⋅ + + ⋅

− − ⋅ + − ⋅

= − ⋅

G G GG G

1

1 1

26

150 100 J molln 60.6(8.314 J K mol )(298 K)

2 10

−

− −

− ⋅= − = +

⋅ ⋅

= ×

K

K

(c) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l)

r f f 2 f 3

1 1

1

4 (NO, g) 6 (H O, l) [4 (NH , g)]

4(86.55 kJ mol ) 6( 237.13 kJ mol ) [4( 16.45 kJ mol )]1010.8 kJ mol

− −

−

∆ ° = ∆ ° + ∆ ° − ∆ °

= ⋅ + − ⋅ − − ⋅

= − ⋅

G G G G1−

3 1

1 1

177

1010.8 10 J molln 408(8.314 J K mol )(298 K)

10

−

− −

− × ⋅= − = +

⋅ ⋅

≅

K

K

250

(d) CO2(g) + 2 NH3(g) CO(NH→ 2)2(s) + H2O(l)

r f 2 2 f 2

f 2 f 3

1 1

1 1

1

(CO(NH ) , s) (H O, l) [ (CO , g) 2 (NH , g)]

( 197.33 kJ mol ) ( 237.13 kJ mol ) [( 394.36 kJ mol ) 2( 16.45 kJ mol )]

7.20 kJ mol

− −

− −

−

∆ ° = ∆ ° + ∆ °− ∆ ° + × ∆ °

= − ⋅ + − ⋅

− − ⋅ + − ⋅

= − ⋅

G G GG G

3

1 1

7.20 10 Jln 2.91(8.314 J K mol )(298 K)

18

− −

− ×= − = +

⋅ ⋅=

K

K

9.16 (a) r1 1

1

ln K(8.314 J K mol )(700 K) ln 5423.2 kJ mol

− −

−

∆ ° = −

= − ⋅ ⋅

= − ⋅

G RT

(b) r1 1

1

ln K(8.314 J K mol )(298 K) ln 0.302.98 kJ mol

− −

−

∆ ° = −

= − ⋅ ⋅

= + ⋅

G RT

9.18 r f 21

(CO , g)

394.36 kJ mol−∆ ° = ∆ °

= − ⋅

G G

3 1

1 1

69

ln

ln

394.36 10 J molln 159.17(8.314 J K mol )(298 K)

1.3 10

−

− −

∆ ° = −∆ °

= −

− × ⋅= − = +

⋅ ⋅

= ×

G RT KGK

RT

K

K

In practice, no K will be so precise. A better estimate would be .

Because Q < K, the reaction will tend to proceed to produce products.

691 10×

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

251

5

3 2

r r

r

Clr

Cl Cl

1 1

1 1

1

lnln ln

ln ln

(8.314 J K mol )(503 K) ln 49(1.33) (8.314 J K mol )(503 K) ln

(0.22)(0.41)5.0 kJ mol

− −

− −

−

∆ = ∆ ° +∆ = − +

∆ = − +⋅

= − ⋅ ⋅

+ ⋅ ⋅

= − ⋅

G G RT QG RT K RT Q

pG RT K RT

p p

Because r∆ is negative, the reaction will be spontaneous to form

products.

G

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

r r

r2

r2 2

1 1

21 1

1

lnln ln

[HI]ln ln[H ] [I ]

(8.314 J K mol )(700 K) ln 54(2.17) (8.314 J K mol )(700 K) ln

(0.16)(0.25)4.5 kJ mol

− −

− −

−

∆ = ∆ ° +∆ = − +

∆ = − +

= − ⋅ ⋅

+ ⋅ ⋅

= ⋅

G G RT QG RT K RT Q

G RT K RT

Because r∆ is positive, the reaction will proceed to form reactants. G

9.24 (a) 3

2 2

2SO2

SO O

3.4P

= =P

KP

C12.027 K

∆⎛ ⎞

= ⎜ ⎟⎝ ⎠

nTK K

(2 3)

212.027 K 12.027 K 3.4 2.8 101000 K

−∆ ⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

n

CK KT

×

(b) 3 2

2NH H S 9.4 10−= = ×K P P

252

(2 0)

2 412.02 K 12.027 K 9.4 10 1.5 10297 K

−∆− −⎛ ⎞⎛ ⎞= = × =⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

n

CK KT

×

9.26 For the equation written 2 2 32 SO (g) O (g) 2 SO (g)+ → Eq. 1

3

2 2

2SO 102

SO O

2.5 10= = ×P

KP P

(a) For the equation written 12 22SO (g) O (g) SO (g)3+ → Eq. 2

3

2 2

SO 10 5Eq. 2 Eq.11/ 2

SO O

2.5 10 1.6 10= = = × =P

K KP P

×

(b) For the equation written 13 2 2SO (g) SO (g) O (g)→ + 2 Eq. 3

This equation is the reverse of Eq. 2, so Eq. 3Eq. 2 Eq.1

1 1= =K

K K

2 2

3

1/ 2SO O 6

Eq. 2 10SO Eq.1

1 1 6.3 102.5 10

−= = = = ××

P PK

P K

(c) For the equation written 32 223 SO (g) O (g) 3 SO (g)3+ → Eq. 4

This equation is 3/23Eq. 4 Eq.12 Eq. 1, so× =K K

3

2 2

3SO 3/ 2 10 3/ 2 15

Eq. 2 Eq.13 3/ 2SO O

(2.5 10 ) 4.0 10= = = × =P

K KP P

×

9.28 H2(g) + Cl2(g) 2 HCl(g) → 85.1 10= ×CK

2

2 2

[HCl][H ][Cl ]

=CK

3 28

32

3 2

2 8 3

12 12

(1.45 10 )5.1 10[H ](2.45 10 )

(1.45 10 )[H ](5.1 10 )(2.45 10 )

[H ] 1.7 10 mol L

−

−

−

−

− −

×× =

×

×=

× ×

= × ⋅

253

9.30 3 2

5

SbCl Cl

SbCl

=P P

KP

2

3Cl4 (5.02 10 )

3.5 100.072

−−

×× =

P

2

43

Cl 3

(3.5 10 )(0.072) 5.0 10 bar5.02 10

−−

−

×= = ×

×P

9.32 (a) 2

33

2 2

[NH ]0.278

[N ][H ]= =CK

2

3

[0.122] 0.248[0.417][0.524]

= =CQ

(b) ≠CQ KC

C

; therefore the system is not at equilibrium.

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

9.34 (a) 2

33

2 2

[NH ]62

[N ][H ]= =CK

4 2

33 3 3

[1.12 10 ] 2.95 10[2.23 10 ][1.24 10 ]

−

− −

×= =

× ×CQ ×

C

(b) Because , ammonia will decompose to form reactants. >CQ K

9.36 2 22 21 1

1.00 g I 0.830 g I0.003 94 mol I ; 0.003 27 mol I

253.8 g mol 253.8 g mol− −= =⋅ ⋅

I2(g) 2 I(g)

0.003 94 mol − x 2 x

0.003 94 mol 0.003 27 mol− =x

0.000 67 mol; 2 0.0013 mol= =x x

2

4

0.00131.00 5.2 10

0.003 271.00

−

⎡ ⎤⎢ ⎥⎣ ⎦= =⎡ ⎤⎢ ⎥⎣ ⎦

CK ×

254

9.38 Pressure (Torr) CO(g) + H2O(g) CO2(g) + H2(g)

initial 200 200 0 0

final 20200 88− 0 88− 88 88

Note: Because pressure is directly proportional to the number of moles of

a substance, the pressure changes can be used directly in calculating the

reaction stoichiometry. Technically, to achieve the correct standard state

condition, the Torr must be converted to bar ; however,

in this case those conversion factors will cancel because there are equal

numbers of moles of gas on both sides of the equation.

1(750.1 Torr bar )−⋅

2 2

2

CO H

CO H O

88 88750.1 750.1 0.62112 112

750.1 750.1

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= = =⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

P PK

P P

9.40 (a) The balanced equation is Cl2(g) 2 Cl(g)

The initial concentration of 122

0.0050 mol ClCl (g) is 0.0025 mol L

2.0 L−= ⋅

Concentration Cl1(mol L )−⋅ 2(g) 2 Cl(g)

initial 0.0025 0

change −x +2 x

equilibrium 0.0025 − x +2 x

223

C2

2 3

2 3 6

(2 )[Cl] 1.7 10[Cl ] (0.0025 )

4 (1.7 10 )(0.0025 )4 (1.7 10 ) (4.25 10 ) 0

−

−

− −

= = = ×−

= × −

+ × − × =

xKx

x xx x

3 3 2

3 3

3 4

(1.7 10 ) (1.7 10 ) 4(4)( 4.3 10 )2 4

(1.7 10 ) 8.47 108

1.3 10 or 8.5 10

− −

− −

− −

− × ± × − − ×=

×− × ± ×

=

= − × + ×

x

x

x

6−

255

The negative answer is not meaningful, so we choose =x

The concentration of Cl48.5 10 mol L .−× ⋅ 1−2 is

The concentration of Cl atoms is

The percentage decomposition of Cl

40.0025 8.5 10−− ×

0.0017.= 42 (8.5 10 )−× ×

41.7 10−= × 1mol L .−⋅ 2 is given by

48.5 10 100 34%

0.0025

−×× =

(b) The balanced equation is Br2(g) 2 Br(g)

The initial concentration of 122

5.0 mol BBr (g) is 2.5 mol L

2.0 L−= ⋅

r

Concentration Br1(mol L )−⋅ 2(g) 2 Br(g)

initial 2.5 0

change −x +2 x

equilibrium 2.5 − x +2 x

223

C2

2 3

2 3 3

(2 )[Br] 1.7 10[Br ] (2.5 )

4 (1.7 10 )(2.5 )4 (1.7 10 ) (4.25 10 )

−

−

− −

= = = ×−

= × −

+ × − × =

xKx

x xx x 0

3 3 2

3 1

2 2

(1.7 10 ) (1.7 10 ) 4(4)( 4.25 10 )2 4

(1.7 10 ) 2.608 108

3.3 10 or 3.2 10

− −

− −

− −

− × ± × − − ×=

×− × ± ×

=

= − × + ×

x

x

x

3−

The negative answer is not meaningful, so we choose =x

The concentration of Br23.2 10 mol L .−× ⋅ 1−2 is

The concentration of Br atoms is

22.5 3.2 10 2.5.−− × ≈

22 (3.2 10 )−× × = 2 16.4 10 mol L .− −× ⋅

he percentage decomposition of Br2 is given by

23.2 10 100 1.3%

2.5

−×× =

(c) At this temperature, Cl2 and Br2 are equally stable since their

equilibrium constants are the same. If we had used the same initial

256

amounts of Cl2 and Br2 in parts (a) and (b), then the percent decomposition

would have been the same as well.

9.42 Pressure (bar) 2 BrCl(g) Br2(g) + Cl2(g)

initial 31.4 10−× 0 0

change 2− x +x +x

final 31.4 10 2−× − x +x +x

2 2Br Cl2

BrCl

2

3 2 3

2

3 2

( )( )32(1.4 10 2 ) (1.4 10 2 )

32(1.4 10 2 )

− −

−

⋅=

= =× − × −

=× −

p pK

p

x x x2x x

xx

3

3

3

3

32(1.4 10 2 )

( 32)(1.4 10 2 )

2 32 ( 32)(1.4 10 )

(1 2 32) ( 32)(1.4 10 )

−

−

−

−

=× −

= × −

+ = ×

+ = ×

xx

x x

x x

x

3

4

( 32)(1.4 10 )(1 2 32)

6.4 10

−

−

×=

+

= ×

x

x

2 2

4Br Cl 6.4 10 bar = 0.64 mbar−= = ×p p

3 4 4BrCl 1.4 10 bar 2 (6.4 10 bar) 1.2 10 bar = 0.12 mbar− − −= × − × = ×p

The percentage decomposition is given by

4

3

2 (6.4 10 bar) 100 91%1.4 10 bar

−

−

×× =

×

257

9.44 (a) concentration of PCl5 initially =

51

5 1

2.0 g PCl208.22 g mol PCl

0.032 mol L0.300 L

−−

⎛ ⎞⎜ ⎟⋅⎝ ⎠ = ⋅

Concentration PCl1(mol L )−⋅ 5(g) PCl3(g) + Cl2(g)

initial 0.032 0 0

change −x +x +x

final 0.032 − x +x +x

2

3 2C

5

[PCl ][Cl ] ( )( )[PCl ] (0.032 ) (0.032 )

= = =− −

x x xKx x

2

2

2

2

0.61(0.032 )

(0.61)(0.032 )(0.61) 0.020 0

(0.61) (0.61) (4)(1)( 0.020)2 1

(0.61) 0.67 0.03 or 0.642 1

=−

= −

+ − =

− ± − −=

⋅− ±

= = + −⋅

xx

x xx x

x

x

The negative root is not meaningful, so we choose 10.03 mol L .−= ⋅x

13 2[PCl ] [Cl ] 0.03 mol L ;−= = ⋅

15[PCl ] 0.032 0.03 mol L 0.002 mol L .1− −= − ⋅ = ⋅ The solution of this

problem again points up the problems with following the significant figure

conventions for calculations of this type, because we would use only one

significant figure as imposed by the subtraction in the quadratic equation

solution. The fact that the equilibrium constant is given to two significant

figures, however, suggests that the concentrations could be determined to

that level of accuracy. Plugging the answers back into the equilibrium

expression gives a value of 0.45, which seems somewhat off from the

starting value of 0.61. The closest agreement to the equilibrium expression

comes from using 13 2[PCl ] [Cl ] 0.0305 mol L−= = ⋅ and

258

15[PCl ] 0.0015 mol L−= ⋅ ; this gives an equilibrium constant of 0.62.

Rounding these off to two significant figures gives

13 2[PCl ] [Cl ] 0.030 mol L−= = ⋅ and 5[PCl ] 0.0015= , which in the

equilibrium expression produces a value of 0.60. The percentage

decomposition is given by

0.030 100% 94%0.032

× =

9.46 Starting concentration of 13

0.200 molNH 0.100 mol L2.00 L

−= = ⋅

Concentration NH1(mol L )−⋅ 4HS(s) NH3(g) + H2S(g)

initial — 0.100 0

change — +x +x

final — 0.100 + x +x

C 3 2

4

2 4

2 4

[NH ][H S] (0.100 )( )

1.6 10 (0.100 )( )0.100 1.6 10 0

( 0.100) ( 0.100) (4)(1)( 1.6 10 )2 1

0.100 0.1031 0.002 or 0.1022 1

−

−

−

= = +

× = +

+ − × =

− + ± + − − ×=

⋅− ±

= = + −⋅

K x x

x xx x

x

x

The negative root is not meaningful, so we choose 3 12 10 mol L .− −= × ⋅x

1 33

1

3 12

[NH ] 0.100 mol L 2 10 mol L

0.102 mol L[H S] 2 10 mol L

1− − −

−

− −

= + ⋅ + × ⋅

= ⋅

= × ⋅

Alternatively, we could have assumed that x << 0.100, in which case

40.100 1.6 10−= ×x 3or 1.6 10 .−= ×x

9.48 The initial concentrations of PCl3 and Cl2 are calculated as follows:

259

1

3 2

1

0.200 mol 0.600 mol[PCl ] 0.0250 mol L ; [Cl ]8.00 L 8.00 L

0.0750 mol L

−

−

= = ⋅ =

= ⋅

Concentrations PCl1(mol L )−⋅ 5 PCl3(g) + Cl2(g)

initial 0 0.0250 0.0750

change +x −x −x

final +x 0.0250 − x 0.0750 − x

3 2C

5

[PCl ][Cl ] (0.0250 )(0.0750 ) 33.3[PCl ] ( )

− −= = =

+x xK

x

2

2

0.100 0.001875 33.333.40 0.001875 0

− + =

− + =

x x xx x

2

5

33.40 ( 33.40) (4)(1)(0.001 875) 33.40 33.399 89(2)(1) 2

5.5 10 or 33.4−

+ ± − − + ±= =

= + × +

x

The root +33.4 has no physical meaning because it is greater than the

starting concentrations of PCl3 and Cl2, so it can be discarded.

5 15 3[PCl ] 5.5 10 mol L ; [PCl ] 0.0250 mol L− −= × ⋅ = ⋅ 1−

1−

2[Cl ] 0.0750 mol L 5.5 10 mol L

5 15.5 10 mol L 0.0249 mol L ;− −− × ⋅ = ⋅

1 5 1− − −= ⋅ − × ⋅ 10.0749 mol L .−= ⋅ Note that

the normal conventions concerning significant figures were ignored in

order to obtain a meaningful answer.

9.50 The initial concentrations of N2 and O2 are

1

2 2

1

0.0140 mol 0.240 mol[N ] 0.001 40 mol L ; [O ]10.0 L 10.0 L

0.0214 mol L

−

−

= = ⋅ =

= ⋅

Concentrations (m L⋅ N1ol )−2(g) + O2(g) 2 NO(g)

initial 0.001 40 0.0214 0

change −x −x + 2x

final 0.001 40 − x 0.0214 − x + 2x

260

22

2 2

(2 )[NO][N ][O ] (0.001 40 )(0.0214 )

= =− −C

xKx x

25

25

2 5

2 5 2

2 5 2 7

2 7 10

(2 )1.00 10(0.001 40 )(0.0214 )

41.00 100.0228 3.0 10

4 (1.00 10 )( 0.0228 3.0 10 )4 1.00 10 2.28 10 3.0 104 2.28 10 3.0 10 0

−

−−

− −

− −

− −

× =− −

× =− + ×

= × − + ×

= × − × + ×

+ × − × =

xx xx

x xx x xx x xx x

5

10−

7 7 22.28 10 (2.28 10 ) (4)(4)( 3.0 10 )

(2)(4)

− −− × ± × − − ×=x

10−

7 5

6 62.28 10 6.93 10 8.6 10 or 8.7 108

− −− −− × ± ×

= = ×x − ×

The negative root can be discarded because it has no physical meaning.

[NO] 2= =x 6 52(8.6 10 ) 1.7 10 mol L ;1− −× = × ⋅ − the concentrations of N2

and O2 remain essentially unchanged at 10.001 39 mol L−⋅ and

, respectively. 10.0214 mol L−⋅

9.52 The initial concentrations of N2 and H2 are

12 2

0.20 mol[N ] [H ] 0.0080 mol L .25.0 L

−= = = ⋅

At equilibrium, 5.0 % of the N2 had reacted, so 95.0 % of the N2 remains:

1 12[N ] (0.950)(0.0080 mol L ) 0.0076 mol L− −= ⋅ = ⋅

If 5.0 % reacted, then

32 3

2

2 mol NH0.050 0.200 mol N 0.020 mol NH

mol N× × = formed.

The concentration of 4 13

0.020 molNH formed 8.0 10 mol L .25.0 L

− −= = × ⋅

The amount of

22 2

2

3 mol HH reacted 0.050 0.200 mol N 0.030 mol H

mol N= × × = 2 used.

261

Concentration of H2 pr

esent at equilibrium

10.200 mol 0.030 mol 0.0068 mol L .25.0 L

−−= = ⋅

2 4 2

3[NH ] (8.0 10 ) 2.7−×

= = =K 23 3

2 2

10[N ][H ] (0.0076)(0.0068)

×C

Note: The volume of the system is not used because we are given

pressures and K.

Pressures (bar) N2(g) + 3 H2(g) 2 NH3(g)

9.54

initial 0.025 0.015 0

change −x 3− x 2+ x

final 0.025 − x 0.015 3− x 2+ x

3NH3

N H .0= =K

P P2 2

3 0.036(0 25 )(0.015 3 )

=− −

xx x

explicitly will lead to a high-order equation, so first check to

see if the assumption that

2 2(2 )P

Solving this

can be use plify the math: d to sim3 0.015<<x

2

3

2 9

0.036(0.025)(0.015)4 3.04 10−

=

= ×x 5

(2 )

2.8 10−= ×

x

x

Comparing x to 0.015, we see that the approximation was justified.

At equilibrium, ; the pressures of

N2 and H2 remain essentially unchanged.

3

5 5NH 2 2.8 10 bar 5.6 10 bar− −= × × = ×P

262

9.56 Concentrations

CH1(mol L )−⋅ 3COOH + C2H5OH CH3COOC2H5 + H2O

initial 0.024 0.059 0 0.015

change −x −x +x +x

final 0.024 − x 0.059 − x +x 0.015 + x

3 2 5 2

3 2 52

2

2

2

2 2

2

2

[CH COOC H ][H O] ( )(0.015 )[CH COOH][C H OH] (0.024 )(0.059 )

0.0150.083 0.0014

0.0154.00.083 0.001 42

4.0 0.332 0.005 68 0.0153.0 0.347 0.005 68 0

( 0.347) ( 0.347) (4

+= =

− −

+=

− +

+=

− +

− + = +

− + =

− − ± − −=

Cx xK

x x

x xx x

x xx x

x x x xx x

x)(3.0)(0.005 68) 0.347 0.228

(2)(3.0) 6.00.0958 or 0.0198

+ ±=

=x

The root 0.0958 is meaningless because it is larger than the initial

concentration of acetic acid and ethanol, so the value 0.0198 is chosen.

The equilibrium concentration of the product ester is, therefore,

. The numbers can be confirmed by placing them into the

equilibrium expression:

10.0198 mol L−⋅

3 2 5 2

3 2 5

[CH COOC H ][H O] (0.0198)(0.015 0.0198) 4.1[CH COOH][C H OH] (0.024 0.0198)(0.059 0.0198)

+= =

− −CK =

This is reasonably good agreement, given the nature of the calculation.

Given that the KC value is reported to only two significant figures, the best

report of the concentration of ester will be 0.020 1mol L−⋅ .

9.58 5

3 2

PCl

PCl Cl

=P

KP P

2

24

Cl

1.3 103.5 10(9.56)

×× =

P

263

2

24

Cl 4

1.3 10 3.9 10 bar(9.56)(3.5 10 )

−×= = ×

×P

9.60 We use the reaction stoichiometry to calculate the amounts of substances

present at equilibrium:

Amounts (mol) CO(g) + H2O(g) CO2(g) + H2(g)

initial 1.000 1.000 0 0

change −x −x +x +x

final 1.000 − x 1.000 − x 0.665 +x

(a) Because x = 0.665 mol, there will be (1.000 0.665− ) mol = 0.335 mol

CO; 0.335 mol H2O; 0.665 mol H2. The concentrations are easy to

calculate because V = 10.00 L:

1 12 2 2[CO] [H O] 0.0335 mol L ; [CO ] [H ] 0.0665 mol L− −= = ⋅ = = ⋅

(b) 2

2 22

2

[CO ][H ] (0.0665) 3.94[CO][H O] (0.0335)

= = =CK

9.62 The initial concentration of 12

0.100 molH S 0.0100 mol L10.0 L

−= = ⋅

The final concentration of 12

0.0285 molH 0.002 85 mol L10.0 L

−= = ⋅

Concentrations 2 H1(mol L )−⋅ 2S(g) 2 H2(g) + S2(g)

initial 0.0100 0 0

change 2− x 2+ x +x

final 0.0100 2− x 0.002 85 +x

Thus, 1 12 0.002 85 mol L or 0.001 42 mol L− −= ⋅ = ⋅x x

At equilibrium:

264

1 12

12

12

24

2

[H S] 0.0100 mol L 0.002 85 mol L 0.0072 mol L

[H ] 0.002 85 mol L

[S ] 0.001 42 mol L

(0.002 85) (0.001 42) 2.22 10(0.0072)

1− − −

−

−

−

= ⋅ − ⋅ = ⋅

= ⋅

= ⋅

= = ×CK

9.64 The initial concentration of 15

0.865 molPCl 1.73 mol L0.500 L

−= = ⋅

Concentrations (m L PCl1ol )−⋅ 5(g) PCl3(g) + Cl2(g)

initial 1.73 0 0

change −x +x +x

final 1.73 − x +x +x

3 2

5

2

2

[PCl ][Cl ][PCl ]

( )( )1.80(1.73 ) (1.73 )

(1.80)(1.73 )

=

= =− −

− =

CK

x x xx x

x x

2

2

1.80 3.114 0

1.80 (1.80) (4)(1 3.114) 1.80 3.96(2)(1) 2

1.08 or 2.88

+ − =

− ± − − − ±= =

= + −

x x

x

x

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

concentrations of PCl3 and Cl2 are, therefore, 1.08 1mol L−⋅ at equilibrium,

and the concentration of PCl5 is 1.73 1mol L−⋅ 11.08 mol L−− ⋅

. These numbers can be checked by substituting back into

the equilibrium expression:

10.65 mol L−= ⋅

3 2

52

[PCl ][Cl ][PCl ]

(1.08) 1.79(0.65)

=

=

CK

which compares well to KC (1.80).

265

9.66 We use the ideal gas relationship to find the initial concentration of HCl(g)

at 25 C:°

1 1

1 atm1.00 bar (4.00 L)1.013 25 bar

0.176 mol(0.082 06 L atm K mol )(273 K)− −

=

⎛ ⎞×⎜ ⎟

⎝ ⎠= =⋅ ⋅ ⋅

PVnRT

n

10.176 mol[HCl] 0.0147 mol L12.00 L

−= = ⋅

2 HCl(g) + I1Concentrations (mol L )−⋅ 2(s) 2 HI(g) + Cl2(g)

initial 0.0147 — 0 0

change 2− x — 2+ x +x

final 0.147 2− x — 2+ x +x

22

2

234

2

[HI] [Cl ][HCl]

(2 ) ( )1.6 10(0.0147 2 )

−

=

× =−

CK

x xx

Because the equilibrium constant is very small, we will assume that x <<

0.0147:

3

342

41.6 10(0.0147)

−× =x

34 2

3 (1.6 10 )(0.0147)4

−×=x

34 2

133(1.6 10 )(0.0147) 2.1 10

4

−−×

= =x ×

At equilibrium:

13 13[HI] 2 2.1 10 4.2 10− −= × × = ×

132[Cl ] 2.1 10−= ×

1[HCl] 0.0147 mol L−= ⋅

266

9.68 initial [NH3] 1

1

25.6 g17.03 g mol

0.301 mol L5.00 L

−−

⎛ ⎞⎜ ⎟⋅⎝ ⎠= = ⋅

2 NH1Concentrations (mol L )−⋅ 3(g) N2(g) + 3 H2(g)

initial 0.301 0 0

change 2− x +x 3+ x

final 0.301 2− x +x 3+ x

32 2

23

3

2

4

2

[N ][H ][NH ]

( )(3 ) 0.395(0.301 2 )

27 0.395(0.301 2 )

=

=−

=−

CK

x xx

xx

4

2

2

27 0.395(0.301 2 )

3 3 0.628(0.301 2 )

=−

=−

xx

xx

2

2

2

3 3 (0.628)(0.301 2 ) 0.189 1.265.20 1.26 0.189 0

1.26 (1.26) (4)(5.20)( 0.189) 1.26 2.34(2)(5.20) 10.4

0.105 or 0.396

= − = −

+ − =

− ± − − − ±= =

= + −

x x xx x

x

x

The negative root is discarded because it is not physically meaningful.

Thus, at equilibrium we should have

1 12 2 3[N ] 0.105 mol L ; [H ] 0.315 mol L ; [NH ] 0.091 mol L .− −= ⋅ = ⋅ = 1−⋅

9.70 2 HCl(g) → + 2H (g) 2Cl (g)

2 2H Cl 342

HCl

3.2 10−= = ×P P

KP

(Eq. 1)

At equilibrium, . (Eq. 2) 2 2H Cl HCl 3.0 bar= + + =TotalP P P P

267

And, since the atomic ratio , Cl:H is 1:3

2

2

HCl Cl

HCl H

2moles Cl 1moles H 3 2

+= =

+

n nn n

2 2

2 2

2 2

HCl H HCl Cl

H HCl Cl

H HCl Cl

2 3( 2 )

3

at , constant so

3 (Eq. 3)

+ = +

= +

∝

= +gas gas

n n n n

n n n

P n T V

P P P

We now have three equations in three unknowns, so we can rearrange and

substitute to find each partial pressure.

Substituting equation 3 into equation 2 gives

2Cl HCl4 2 3.0 b= + =TotalP P P ar

2HCl Cl

3.0 bar 22

= −P P

Substitution back into equation 3 gives

2 2H C

3.0 bar 2

= +P P l

Using these two expressions for partial pressures in equation 1 gives

2 2

2

Cl Cl34

2

Cl

3.0 2 3.2 103.0 2

2

−

⎛ ⎞+ ⋅⎜ ⎟⎝ ⎠= =⎛ ⎞−⎜ ⎟⎝ ⎠

P PK

P×

Since the equilibrium constant is so small, we would expect the partial

pressure of chlorine to be very low at equilibrium, so we can neglect it to

simplify this expression.

2

2

2

Cl34

2

34 34Cl

HCl H

3.0 2 3.2 103.0

2(1.5)(3.2 10 ) 4.8 10 bar

1.5 bar

−

− −

⎛ ⎞ ⋅⎜ ⎟⎝ ⎠≈ = ×⎛ ⎞⎜ ⎟⎝ ⎠

= × = ×

= =

PK

P

P P

268

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

pressure will shift the equilibrium to the left, increasing the partial

pressure of .

2CO

4CH

(b) According to Le Chatelier’s principle, a decrease in the partial

pressure of will shift the equilibrium to the left, decreasing the

partial pressure of CO

4CH

2.

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

unaffected by any change in concentration.

(d) According to Le Chatelier’s principle, a decrease in the concentration

of H2O will shift the equilibrium to the right, increasing the concentration

of CO2.

9.74 The questions can all be answered qualitatively using Le Chatelier’s

principle:

(a) Adding a reactant will promote the formation of products; the amount

of HI should increase.

(b) I2 is solid and already present in excess in the original equilibrium.

Adding more solid will not affect the equilibrium so the amount of Cl2 will

not change.

(c) Removing a product will shift the reaction toward the formation of

more products; the amount of Cl2 should increase.

(d) Removing a product will shift the reaction toward the formation of

more products; the amount of HCl should decrease.

(e) The equilibrium constant will be unaffected by changes in the

concentrations of any of the species present.

(f) Removing the reactant HCl will cause the reaction to shift toward the

production of more reactants; the amount of I2 should increase.

(g) As in (e), the equilibrium constant will be unaffected by the changes

to the system.

269

9.76 Increasing the total pressure on the system by decreasing its volume will

shift the equilibrium toward the side of the reaction with fewer numbers of

moles of gaseous components. If the total number of moles of gas is the

same on the product and reactant sides of the balanced chemical equation,

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

distribution of species present. (a) decrease in the amount of NO2;

(b) increase in the amount and concentration of NO; (c) decrease in the

amount of HI; (d) decrease in the amount of SO2; (e) increase in the

amount and concentration of NO2. (Note that it is not possible to predict

whether or not the concentration will increase, decrease, or remain the

same in cases where the amount of the substance is decreasing. More

specific information about the extent of the changes in the amount and the

volume is needed in those cases.)

9.78 (a) The partial pressure of SO3 will decrease when the partial pressure of

SO2 is decreased. According to Le Chatelier’s principle, a decrease in the

partial pressure of a reactant will shift the equilibrium to the left,

decreasing the partial pressure of the products, in this case SO3.

(b) When the partial pressure of SO2 increases, the partial pressure of O2

will decrease. According to Le Chaltelier’s principle, an increase in the

partial pressure of a reactant shifts the equilibrium toward products,

decreasing the partial pressure of the other reactant, O2.

9.80 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 reactions given, raising

the temperature should favor products in (a) and reactants in (b) and (c).

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

answer this qualitative question. Because the equilibrium constant is larger

at lower temperatures 24 10(4.0 10 at 298 K vs. 2.5 10× × at 500 K, see

270

, more products will be present at the lower

temperature. Thus we expect more SO

Table 9.2, page 334)

3 to be present at 25ºC than at 500

K, assuming that no other changes occur to the system (the volume is

fixed and no reactants or products are added or removed from the vessel).

9.84 To answer this question, we must calculate Q:

2 2

2 2

[ClF] (0.92) 15[Cl ][F ] (0.18)(0.31)

= =Q =

Because Q ≠ K, the system is not at equilibrium, and because Q < K, the

reaction will proceed to produce more products so ClF will tend to form.

9.86 (a) 122HgO(s) Hg(l) O (g)+

r f1

r1

r1

r 221 1 1 11

r 21 1

1 1r

[ (HgO, s)]

[ 90.83 kJ mol ]

90.83 kJ mol(Hg, l) (O , g) [ (HgO, s)]

76.02 J K mol 205.14 J K mol

[70.29 J K mol ]108.30 J K mol

−

−

− − − −

− −

− −

∆ ° = − ∆ °

∆ ° = − − ⋅

∆ ° = + ⋅

∆ ° = ° + × ° − °

∆ ° = ⋅ ⋅ + × ⋅ ⋅

− ⋅ ⋅

∆ ° = ⋅ ⋅

H H

H

HS S S S

S

S

At 298 K:

1 1r(298 K)

1

r(298 K)

90.83 kJ (298 K)(108.30 J K )/(1000 J kJ )

58.56 kJ molln

− −

−

∆ ° = − ⋅ ⋅

= ⋅∆ ° = −

G

G RT K

r(298 K)

1

ln

58 560 J 23.6(8.314 J K )(298 K)−

∆ °= −

= − = −⋅

GK

RT

116 10−= ×K

At 373 K:

1 1

r(373 K)

1

90.83 kJ (373 K)(108.3 J K )/(1000 J kJ )

50.4 kJ mol

− −

−

∆ ° = − ⋅ ⋅

= ⋅

G

271

1

8

50 400 Jln 16.3(8.314 J K )(373 K)

8 10

−

−

= − = −⋅

= ×

K

K

(b) propene (C3H6, g) → cyclopropane (C3H6, g)

r f f1 1

r1

r

r1 1 1 1

r1 1

r

(cyclopropane, g) [ (propene, g)]

53.30 kJ mol [(20.42 kJ mol )]

32.88 kJ mol(cyclopropane, g) [ (propene, g)]

237.4 J K mol [266.6 J K mol ]

29.2 J K mol

− −

−

− − − −

− −

∆ ° = ∆ ° − ∆ °

∆ ° = ⋅ − ⋅

∆ ° = + ⋅∆ ° = ° − °

∆ ° = ⋅ ⋅ − ⋅ ⋅

∆ ° = − ⋅ ⋅

H H H

H

HS S S

S

S

At 298 K:

1r(298 K)

1 1

1

r(298 K)

32.88 kJ mol

(298 K)( 29.2 J K mol )/(1000 J kJ )41.58 kJ mol

ln

−

− − −

−

∆ ° = ⋅

− − ⋅ ⋅ ⋅

= ⋅∆ ° = −

G

G RT K

1

r(298 K)

1

ln

41 580 J 16.8(8.314 J K )(298 K)−

∆ °= −

= − = −⋅

GK

RT

85 10−= ×K

At 373 K:

1r(373 K)

1 1

1

32.88 kJ mol

(373 K)( 29.2 J K mol )/(1000 J kJ )43.8 kJ mol

−

− − −

−

∆ ° = ⋅

− − ⋅ ⋅ ⋅

= ⋅

G1

1

7

43 800 Jln 14.1(8.314 J K )(373 K)

7 10

−

−

= − = −⋅

= ×

K

K

9.88 We can think of the vaporization of a liquid as an equilibrium reaction,

e.g., for substance A, A( ) A( )l g . This reaction has an equilibrium

constant expression A=K P . Since is the pressure of the gas above a AP

272

liquid, it is the equilibrium vapor pressure of that liquid. Therefore, the

vapor pressure varies with temperature in the same way that this

equilibrium constant varies with temperature.

The van’t Hoff equation, 2

1 1

1 1ln⎛ ⎞∆

= −⎜⎝ ⎠

rK HK R T T2

⎟ , (see 16, page 354),

gives the temperature dependence of K. To make this equation fit the

special case of vapor pressure all we need to do is substitute for K

and ∆ for∆

P

vapH rH .

2

1 1

1 1ln∆ ⎛ ⎞

= −⎜ ⎟⎝ ⎠

vapHPP R T T2

This relationship is known as the Clausius-Clapeyron equation (see

equation 4, page 285). Since 0∆ >vapH , vapor pressure increases with

temperature.

9.90 Recall ln and 10−∆ = − = ∆ − ∆ = pKr r rG RT K H T S K

and 2

1 1

1 1ln⎛ ⎞∆

= −⎜⎝ ⎠

rK HK R T T2

⎟ , the van’t Hoff equation (see equation 16,

page 354).

Rearranging and plugging in values,

13.8330 1 1

215.136

1

1 2

4 1

1

10 8.314 J K molln ln1 1101 1

293 K 303 K

(3.000)(7.381 10 J mol )221.4 kJ mol

− −

−

−

−

⎛ ⎞⎜ ⎟⎛ ⎞ ⎛ ⎞ ⋅ ⋅

∆ = = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠ ⎜ ⎟−⎜ ⎟−⎜ ⎟ ⎝ ⎠⎝ ⎠= × ⋅

= ⋅

rK RHK

T T

−

3 11 1 15.136

1 1

ln

221.4 10 J mol(8.314 J K mol ) ln(10 )293 K

465.9 J K mol

−− − −

− −

∆∆ = +

× ⋅= ⋅ ⋅ +

= ⋅ ⋅

rr

HS R K

T

273

The sign on is positive because of the randomization that occurs as

deuterons dissociate and redistribute throughout the liquid.

∆ rS

From the results above, the “autoprotolysis” constant of heavy water at

25°C is

2

1 1 2

3 115.136

2 1 1

1 1ln

221.4 10 J mol 1 1ln ln10293 K 298 K8.314 J K mol

33.327

−−

− −

⎛ ⎞∆= −⎜ ⎟

⎝ ⎠⎛ ⎞× ⋅ ⎛ ⎞= −⎜ ⎟⎜ ⎟⋅ ⋅ ⎝ ⎠⎝ ⎠

= −

rK HK R T T

K +

15D, 298 K D, 298 K3.359 10 or 14.474−= × =K pK

Since for regular water < H, 298 K 14.00=pK D, 298 K 14.474=pK ,

substitution of D for H causes reactants, i.e., the associated molecules, to

be favored more in heavy water than in regular water even though we

expect D and H to be chemically equivalent. We can suggest that this

observation is another example of the kinetic isotope effect where the

zero-point energy of the O-D bond is lower than that of the O-H bond (see

the solutions to Exercises 17.55 and 17.85 for further discussion). If bond

dissociation is the rate-limiting step in the mechanism, and if only the

vibrational energy of the ground state of the water molecules is changed

appreciably, then the activation barrier for dissociation is bigger in heavy

water since the zero-point energy is lower due to the heavier mass of D.

This increase in the activation energy for bond dissociation causes the rate

of dissociation to decrease while the rate of association is not changed

appreciably. The position of equilibrium then favors the associated

molecules more than it does in the molecules.

2D O

2H O

9.92 (a) r f 31

1

[2 (O , g)]

[2(142.7 kJ mol )]285.4 kJ mol

−

−

∆ ° = − ∆ °

= − ⋅

= − ⋅

H H

274

r 2 3

1 1 1 1

1 1

3 (O , g) [2 (O , g)]

3(205.14 J K mol ) [2(238.93 J K mol )]137.56 J K mol

− − − −

− −

∆ ° = ° − °

= ⋅ ⋅ − ⋅ ⋅

= ⋅ ⋅

S S S

At 298 K:

1r(298 K)

1 1

1

285.4 kJ mol

(298 K)(137.56 J K mol )/(1000 J kJ )326.4 kJ mol

−

− − −

−

∆ ° = − ⋅

− ⋅ ⋅

= − ⋅

G1⋅

°G or directly from ∆ values f

r f 3

1

1

2 (O , g)

2(163.2 kJ mol )326.4 kJ mol

−

−

∆ ° = − ∆ °

= − ⋅

= − ⋅

G G

(b) r(298 K)

r(298 K)

1

57

ln

ln

326 390 J 132(8.314 J K )(298 K)

10

−

∆ ° = −

∆ °= −

−= − = +

⋅

=

G RT K

GK

RT

K

The equilibrium constant for the decomposition of ozone to oxygen is

extremely large, making this process extremely favorable. Note: The

presence of ozone in the upper atmosphere is largely due to kinetic factors

that inhibit the decomposition; however, in the presence of a suitable

catalyst, this process should occur extremely readily.

9.94 1 1

12 4 2

1

2.50 g 0.330 g92.02 g mol 46.01 g mol

[N O ] 0.0136 mol L ; [NO ]2.00 L 2.00 L

0.003 59 mol L

− −−

−

⎛ ⎞ ⎛⎜ ⎟ ⎜⋅ ⋅⎝ ⎠ ⎝= = ⋅ =

= ⋅

⎞⎟⎠

N1Concentration (mol L )−⋅ 2O4(g) 2 NO2(g)

initial 0.0136 0.003 59

change −x 2+ x

final 0.0136 − x 0.003 59 2+ x

275

22

2 42

3

3 2

5 3 2 2

2 2 5

[NO ][N O ]

(0.003 59 2 )4.66 100.0136

(4.66 10 )(0.0136 ) (0.003 59 2 )6.34 10 4.66 10 4 1.44 10 1.29 104 1.91 10 5.05 10 0

−

−

− − −

− −

=

+× =

−

× − = +

× − × = + × + ×

+ × − × =

K

xx

x xx x x

x x

�

5−

1

Solving by using the quadratic equation gives 31.89 10 .−= ×x

1 3 1 1

2 41 3 1

2

[N O ] 0.0136 mol L 1.89 10 mol L 0.012 mol L

[NO ] 0.003 59 mol L 2 (1.89 10 mol L ) 0.0074 mol L

− − − −

− − −

= ⋅ − × ⋅ = ⋅

= ⋅ + × ⋅ = −⋅

These numbers can be checked by substituting them into the equilibrium

expression:

23?

3 3

(0.0074) 4.66 100.012

4.6 10 4.66 10

−

− −√

= ×

× = ×

9.96 We can write an equilibrium expression for the interconversion of each

pair of isomers:

2-methyl propene cis-2-butene K1

2-methyl propene trans-2-butene K2

cis-2-butene trans-2-butene K3

We need to calculate only two of these values in order to determine the

ratios.

The K’s can be obtained from the r∆ °G ’s for each interconversion:

1 11

1 12

1 13

65.86 kJ mol 58.07 kJ mol 7.79 kJ mol

62.97 kJ mol 58.07 kJ mol 4.90 kJ mol

62.97 kJ mol 65.86 kJ mol 2.89 kJ mol

− −

− −

1

1

1

−

−

− − −

∆ ° = ⋅ − ⋅ = ⋅

∆ ° = ⋅ − ⋅ = ⋅

∆ ° = ⋅ − ⋅ = − ⋅

G

G

G

276

rr

11

1 11 1

12

2 21 1

13

3 1 1

ln ; ln

7790 J molln 3.14; 0.043(8.314 J K mol )(298 K)

4900 J molln 1.98; 0.14(8.314 J K mol )(298 K)

2890 J molln(8.314 J K mol )(298 K)

−

− −

−

− −

−

− −

∆ °∆ ° = − =

−∆ ° ⋅

= = − = − =− ⋅ ⋅

∆ ° ⋅= = − = − =− ⋅ ⋅

∆ ° − ⋅= = −− ⋅ ⋅

GG RT K K

RTG

K KRTG

K KRTG

KRT 31.17; 3.2= + =K

Each K gives a ratio of two of the isomers:

1 2 3[ -2-butene] [ -2-butene] [ -2-butene]; ;

[2-methylpropene] [2-methylpropene] [ -2-butene]= = =

cis trans transK K Kcis

We will choose to use the first two K’s to calculate the final answer:

[ -2-butene] 0.043[2-methylpropene]

=cis [ -2-butene] 0.14

[2-methylpropene]=

trans

If we let the number of moles of 2-methylpropene = 1, then there will be

0.14 mol trans-2-butene and 0.043 mol cis-2-butene. The total number of

moles will be 1 mol + 0.14 mol + 0.043 mol = 1.18 mol. The percentage

of each will be

1 mol% 2-methylpropene 100% 85%1.18 mol

0.043 mol% -2-butene 100% 4%1.18 mol

0.14 mol% -2-butene 100% 12%1.18 mol

= × =

= × =

= × =

cis

trans

(The sum of these should be 100% but varies by 1%, due to limitations in

the use of significant figures/rounding conventions.)

The relative amounts are what we would expect, based upon the free

energies of formation (the more negative or less positive value

corresponding to the thermodynamically more stable compound), which

indicate that the most stable compound is the 2-methylpropene followed

by trans-2-butene with cis-2-butene being the least stable isomer. We can

use K3 as a check of these numbers:

277

?

3

0.14[ -2-butene]3.2

0.043[ -2-butene]

3.2 3.3

= = =

≅

trans VKcis

V

9.98 A 2 B + 3 C

initial 10.00 atm 0 atm 0 atm

change − +2 x +3 x x

final 10.00 − x +2 x +3 x

The equilibrium expression is 2 3×

= B C

A

P PK

P

We can also write total 10.00 2 3 10.00 4 15.76= − + + = + =P x x x x

x = 1.44 atm

PA = 8.56 atm; PB = 2.88 atm; PC = 4.32 atm

2 3

1 1

1

(2.88) (4.32) 78.18.56

ln (8.314 J K mol )(298 K)ln(78.1)10.8 kJ mol

− −

−

= =

∆ ° = − = − ⋅ ⋅

= − ⋅

K

G RT K

9.100 (a) First we calculate the initial pressure of the gas at 127ºC using the

ideal gas relationship:

=

=

PV nRTmPV RTM

1 1

1

(10.00 g)(0.082 06 L atm K mol )(298 K)(17.03 g mol )(4.00 L)

3.59 atm

− −

−

=

⋅ ⋅ ⋅=

⋅=

mRTPMV

278

1 2

1 2

23.59 atm298 K 400 K

=

=

P PT T

P

= 4.82 atm or 4.88 bar (1 atm = 1.01325 bar) 2P

From Table 9.1 we find that the equilibrium constant is 41 for the equation

2 2 3N (g) 3 H (g) 2 NH (g)+

Although this exercise technically begins with pure ammonia and allows it

to dissociate into N2(g) and H2(g), we can use the same expression.

pressure (bar) + 3 H2N (g) 2(g) 2 NH3(g)

initial 0 0 4.88

change +x 3+ x 2− x

final +x 3+ x 4.88 2− x

2

3

2

4

2

2

2

(4.88 2 ) 41( )(3 )

(4.88 2 ) 4127

4.88 2 413 3

4.88 2 3 12333.3 2 4.88 0

0.354

−=

−=

−=

− =

+ − ==

xx x

xx

xx

x xx x

x

= 4.17 bar or 4.11 atm; 3NHP

2N 0.354=P bar or 0.349 atm; bar

or 1.05 atm

2H 1.06=P

Ptotal = 4.11 atm + 0.349 atm + 1.05 atm = 5.51 atm

(b) The equilibrium constant can be calculated from ∆ °G at 400 K,

which can be obtained from ∆ °H and ∆ °S . The values are

279

1 1

1 1 1 1

1 1

1 1

1 1

1 1

2( 46.11 kJ mol ) 92.22 kJ mol2(192.45 J K mol ) [(191.61 J K mol )

3(130.68 J K mol )]198.75 J K mol

( 92.22 kJ mol )(1000 J K ) (400 K)( 198.75 J K mol )

12.72 kJ

− −

− − − −

− −

− −

− −

− −

∆ ° = − ⋅ = − ⋅

∆ ° = ⋅ ⋅ − ⋅ ⋅

+ ⋅ ⋅

= − ⋅ ⋅

∆ ° = − ⋅ ⋅

− − ⋅ ⋅

= − ⋅

HS

G

1 1mol 12720 J mol− −= − ⋅

1 1 1/ 12720 J mol / (8.314 J K mol )(400K)e e− − −−∆ ° + ⋅ ⋅ ⋅= = =G RTK 46

This is reasonably good agreement considering the logarithmic nature of

these calculations. For comparison, the ∆ °G value calculated from K = 41

at 400 K is 12.34 1kJ mol .−⋅

9.102 (a) pressure (bar) 2 AsH3(g) 2 As(s) + 3 H2(g)

initial 0.52

change 2− x 3+ x

final 0.52 2− x 3+ x

At equilibrium, total pressure = 0.64 bar.

0.64 0.52 2 3 0.52

0.12 bar= − + = +

=x x x

x

3

2

AsH 0.52 bar 2(0.12 bar) = 0.28 bar

3(0.12 bar) = 0.36 bar

= −

=H

P

P

(b) Find the number of moles of that decomposed and relate it to

the mass of As.

3AsH

1 1

33

33

3

(0.24 bar)(1.00 L)(0.0831 L bar K mol )(298 K)

9.69 10 mol AsH decomposed

2 mol As 74.9 g Asmass As = (9.69 10 mol AsH )2 mol AsH 1 mol As

0.73 g As

− −

−

−

= =⋅ ⋅ ⋅

= ×

⎛ ⎞⎛ ⎞× ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

PVnRT

280

(c) 2

3

3 3H2 2

AsH

(0.36)K 0(0.28)

= = =P

P.60

9.104 The free energy change for the reaction is calculated from the equation

ln ln ln∆ = ∆ ° + = − +G G RT Q RT K RT Q

In order to determine the range of Cl2 pressures that we need to examine,

we will first calculate the equilibrium pressures of the gases present.

Br2(g) + Cl2(g) 2 BrCl(g)

initial 1.00 bar 1.00 bar 0

change − x −x 2 x

final 1.00 − x 1.00 − x 2 x

2 2

2 2BrCl

2Br Cl

(2 )0.2(1.00 )

20.21.000.183 0.2

= =−

=−

= ≈

P xP P x

xx

x

In other words, equilibrium will be reached when BrCl 2 0.36 bar= =P x , at

the point on the graph where 0.∆ =G

-700

-600

-500

-400

-300

-200

-100

0

100

0 0.1 0.2 0.3 0.4 0.5

delta

G (k

J pe

r mol

)

Pressure of BrCl (bar)

281

Notice that the free energy change for the reaction is most negative the

farther the system is from equilibrium and that the value approaches 0 as

equilibrium is attained.

9.106 The graph is generated for ln K vs. 1/T according to the relationship

1ln ∆ ° ∆ °= − ⋅ +

H SKR T R

where ∆ °−

HR

is the slope and ∆ °SR

is the intercept.

0

4

8

12

16

20

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035

ln K

1/T (1/K)

y = -5398.4x +17.899R = 0.99213

1

1 1

5398 K

44.9 kJ mol

17.9

149 J K mol

−

− −

∆ °− = −

∆ ° = ⋅∆ °

=

∆ ° = ⋅ ⋅

HR

HSRS

Because the reaction involves breaking only the N—N bond, the enthalpy

of reaction should be an approximation of the N—N bond strength.

Notice that this value is considerably less than the average

value for N—N bonds given in Table 2.2. Therefore it is a very weak

1163 kJ mol−⋅

N—N bond.

282

9.108 (a) Since we are only using two temperatures, it really is not necessary to

make graphs in order to determine ∆ °H and ∆ °S . We can easily

substitute into the van’t Hoff equation, 2

1 1

1 1ln⎛ ⎞∆

= −⎜⎝ ⎠

rK HK R T T2

⎟ , for each

halogen to get ∆ °H , then use it to find ∆ °S from

. A spread sheet will work well. ln∆ ° = ∆ ° − ∆ = −G H T S RT K

(b) & (c) Results from our spread sheet based on the relationships given

above and data from Table 9.2 are given below. The ∆ °S values are for

the reactions X2(g) 2 X(g), so we can calculate the standard molar

entropies for the atomic species using data from Appendix 2A and the

relationship m m2 (X, g) (X , g∆ ° = ° − °S S S 2 ).

Halogen ∆ °H

1kJ mol−⋅

∆ °S 1 1J K mol− −⋅ ⋅

2(X , g)°S

1 1J K mol− −⋅ ⋅

(X, g)°S 1 1J K mol− −⋅ ⋅

Fluorine 164 6 202.78 164 12

Chlorine 256 160 223.07 192

Bromine 195 110 245.46 178

Iodine 163 152 260.69 206

.110 The general form of this type of equation is

9

r r1ln∆ ° ∆ °⎛ ⎞= − +⎜ ⎟

⎝ ⎠

H SK

R T R

which is easily derived from the relationships

r r . By inspection, we can see

that for the original expression to be valid,

∆ ° = − ∆ ° = ∆ ° − ∆ °G RT K G H T Sr rln and

rr

1 1 1

21 700, which gives

(21 700 K)(8.314 J K mol )/(1000 J kJ ) 180 kJ mol 1− − − −

− = − ∆ °

= + ⋅ ⋅ ⋅ = ⋅

HR

∆ °H

283

9.112 (a) 3 3PH BCl

1=

⋅K

P P

(b) c

c 1 1

4

( )19.2

( ) [(0.08314 L bar K mol )(333 K)]1.48 10

∆

∆ − −

=

= =⋅ ⋅ ⋅

= ×

n

n 2−

RTKK

RT

K K

(c) The initial pressure of PH3 is:

3

1 1

PH(0.0128 mol)(0.08314 L bar K mol )(333 K)

0.500 L0.7088 bar

− −⋅ ⋅= =

=

nRTPV

pressure (bar) PH3(g) + BCl3(g) PH3BCl3(s)

initial 0.7088 0

change +x +x

equilibrium 0.7088+ x x

3 3PH BCl

1119.2(0.7088 )

= = =⋅ +

KP P x x

=

2 0.7088 0.05208 0+ −x x

20.7088 (0.7088) 4(0.05208)

20.06712 (negative root isn't physically possible)

− ± +=

=

x

So 3

0.70PH 88 0.06712 0.7759 bar= + =P

and 3

3= =c PH 1

PH 1 1

0.7759 bar 0.0280 mol L(0.08314 L bar K mol )(333 K)

−− − = ⋅

⋅ ⋅

PRT

(d) K increases as T increases, so the equilibrium position is shifting

toward products with the addition of heat. The reaction is endothermic.

(e) c

c 1 1

4

( )26.2

( ) [(0.08314 L bar K mol )(343 K)]2.13 10

∆

∆ − −

=

= =⋅ ⋅ ⋅

= ×

n

n

K K RTKK

RT 2−

284

(f) has a lone pair of electrons to donate in order to form an adduct

with , which can be an electron acceptor. Therefore is a Lewis

base while is a Lewis acid.

3PH

3BCl 3PH

3BCl

9.114 (a) The values of the standard free energies of formation are

Compound 1f (kJ mol )−∆ ° ⋅G

BCl3(g) 387.969−

BBr3(g) 236.914−

BCl2Br(g) 338.417−

BClBr2(g) 287.556−

(b) A number of equilibrium reactions are possible including

2 BCl2Br(g) BCl3(g) + BClBr2(g)

2 BClBr2(g) BBr3(g) + BCl2Br(g)

2 BCl3(g) + BBr3(g) 3 BCl2Br(g)

BCl3(g) + 2 BBr3(g) 3 BClBr2(g)

(c) BCl3 because it is the most stable as determined by its f∆ °G value.

(d) No. Although the equilibrium reaction given in the exercise would

require that these two partial pressures be equal, there are other equilibria

occurring as in (b) that will cause the values to be unequal.

(e) We will solve this problem graphically rather than by writing a

computer program to solve a system of simultaneous equations. Standard

free energies of formation are only used to get equilibrium constants for

two essential reactions from −∆

=rG

RTK e and

. ∆ = ∑ ∆ −∑ ∆r prods f reacts fG n G m G

In order to solve this part of the problem, first we need to figure out the

independent reactions between the four species. If we consider BCl3 and

BBr3 as pure species in terms of the halogens atoms, then the mixed

species BClBr2 and BCl2Br can be made from the pure species by these

two reactions:

285

(1) BCl3 + 2 BBr3 3 BClBr2 1871exp 1.421

8.314 298.15⎛ ⎞= =⎜ ⎟×⎝ ⎠

K

(2) 2 BCl3 + BBr3 3 BCl2Br 22399exp 2.632

8.314 298.15⎛ ⎞= =⎜ ⎟×⎝ ⎠

K

All other reactions between the species can be derived from these two

essential reactions, so there are only two independent reactions for the

system.

Let , and 3BCl =P x

3BBr =P y then 2

3BClBr

1 2=P

Kxy

and 2

3BCl Br

2 2=P

Kx y

so , and 2

1/ 3 1/ 3 2 / 3BClBr 1=P K x y

2

1/ 3 2 / 3 1/ 3BCl Br 2=P K x y

Atom balances are determined by the initial composition of the system.

Balancing B atoms we get

3 3 3 2 2

o oBCl BBr BCl BCl Br BClBr BBr+ = + + +P P P P P P

3

2

33

3=

3

where are the initial pressures. Balancing Cl atoms

and Br atoms gives us

3 3

o oBCl BBr 1 bar= =P P

3 3 2

oBCl BCl BCl Br BClBr3 3 2= + +P P P P

3 2 2

oBBr BCl Br BClBr BBr3 2= + +P P P P

(Notice that the B balance is not an independent condition because it is

implied by the Cl and Br balances.) In terms of x and y, the Cl balance and

Br balance can be written as

Cl balance: 1/ 3 2 / 3 1/ 3 1/ 3 1/ 3 2 / 32 13 2+ +x K x y K x y

Br balance: 1/ 3 2 / 3 1/ 3 1/ 3 1/ 3 2 / 32 12 3+ + =K x y K x y y

Subtracting the second equation from the first equation, dividing the

resulting equation by y, and also using this new definition

1/ 3

⎛ ⎞≡⎜ ⎟

⎝ ⎠

x zy

we arrive at following equation in terms of z:

3 1/ 3 2 1/ 32 13 3+ − −z K z K z 0=

286

3 23 1.381 1.124 3 0+ − −z z z =

This cubic equation can be solved graphically by noting where

crosses 0. The only real root is

.

3 2( ) 3 1.381 1.124 3= + − −f z z z z

0.975=z

-25

-20

-15

-10

-5

0

5

10

15

20

25

-3 -2 -1 0 1 2 3

z

Substituting 3=x z y back into the Cl balance equation gives

3 1/ 3 2 1/ 32 1

3 2

33 2

33(0.975) 2(1.381)(0.975) (1.124)(0.975)0.461

=+ +

=+ +

=

yz K z K z

Then 3 3(0.975) (0.461) 0.428= = =x z y

And finally:

( )2

1/ 32BClBr (1.421)(0.428)(0.461) 0.506= =P

( )2

1/ 32BCl Br (2.632)(0.428) (0.461) 0.606= =P

287

The results can be summarized as:

bar 3BCl 0.428=P

bar 3BBr 0.461=P

bar 2BCl Br 0.606=P

bar 2BClBr 0.506=P

Note that in part (c) we predicted that BCl3 is the most abundant species in

the equilibrated mixture. However, this calculation reveals that it is the

least abundant species at equilibrium given the stated initial conditions.

The result is simply non-intuitive. What we need to recognize is that the

formation reactions and their corresponding values of free energy of

reaction describe equilibria between the elements B(s), Cl2(g), and Br2(l)

and the boron trihalides under standard conditions. In our mixture the

compounds are certainly not in equilibrium with their elements in their

standard states, so the prediction in part (c) fails.

Also note that in this system, B(s) and Br2(l) cannot exist at same time,

because the equilibrium constant of B(s) + 1.5 Br2(l) BBr3(g) is very

large, while the actual pressure of BBr3(g) is way below this equilibrium

constant. B(s) may exist from a microscopic amount of decomposition of

BCl3(g), so Br2(l) can not exist. This condition further prevents BCl2Br

and BClBr2 from decomposing into their elements, so only BCl3(g) may

decompose to a small extent. Clearly, this small amount of Cl2(g) is far

below standard conditions.

9.116 (a) Find Q to see if the reaction is at equilibrium.

cyclohexane (C) methylcyclohexane (M)

[M] 0.100Q 5.00 K 0.140[C] 0.0200

= = = > =

The reaction will proceed to the left to form more reactant.

288

(b) [C] [M]

initial 0.0200 0.100

change +x −x

equil 0.0200 + x 0.100 − x

2

2

[M] 0.100K 0.140[C] 0.0200 +

0.100 (0.140)(0.0200 + )1.140 9.72 10

8.53 10

−

−

−= = =

− =

= ×

= ×

xx

x xx

x

[M] = 0.100 − 0.0853 = 0.0147 = 0.015 M

[C] = 0.0200 + 0.0853 = 0.1053 M

(c) At equilibrium, [C] = 0.100 M. The total concentration of all species is

0.120 M. Therefore, [M] = 0.020 at equilibrium.

50 C

[M] 0.100K 5[C] 0.020

= = = .0

(d) As the temperature increased, K also increased resulting in the

reaction forming more products: C + heat M. Therefore, the

reaction is endothermic.

9.118 Refer to Graham’s Law of Effusion, equation 19, page 144, which states

that the ratio of the rates of effusion of two gases, both at the same

temperature, is equal to the square-root of the inverse ratio of their molar

masses.

2

2 HI

H

1

1

Rate of effusion of HRate of effusion of HI

127.9 g mol2.016 g mol

7.97

−

−

=

⋅=

⋅

=

MM

This treatment suggests that H2 effuses ~8 times faster than HI. However,

it is also important to note that the partial pressure of each gas affects the

collision frequency of the molecules with the pinhole. Therefore, an

expression that takes this dependence into account is

289

2

2

H2 H

HI H

Rate of effusion of HRate of effusion of HI

=P MP M

I

The partial pressure of hydrogen is 0.7 atm, or 0.69 bar. The partial

pressure of HI can be found from

2

2

2HI

H

1/ 2HI H (0.345 0.69) 0.49 bar

=

= ⋅ = × =

PK

P

P K P

2

2

1H2 HI

1HI H

Rate of effusion of H 0.69 bar 127.9 g molRate of effusion of HI 0.49 bar 2.016 g mol

1110 (to 1 sig fig)

−

−

⋅⎛ ⎞= = ⎜ ⎟ ⋅⎝ ⎠

==

P MP M

So hydrogen effuses ten times as fast as hydrogen iodide under these

conditions.

290