Kotz & Treichel: Chapter 20

4)       a) True

b) False. "Exothermic reactions are always spontaneous if DS° _sys is positive."

c) False. "Reactions with positive DH° _rxn and DS° _rxn are spontaneous at high temperature".

d) True

e) False. "When the equilibrium constant of a reaction is less than 1, the DG° _rxn is greater than zero."

9)       a) MgCl₂ has the higher entropy. (Both are crystalline solids, but more complex materials tend to have greater
                entropy than simpler materials.)

b) (CH₃)₂NH₂ (g) has the greater entropy since it is a gas.

c) Hg (l) has the greater entropy since it is a liquid.

11)    a) NH₄Cl(s) ® NH₄Cl (aq)

DS° = 169.9 J K^-1mol^-1 - 94.6 J K^-1mol^-1 = 75.3 J K^-1mol^-1
 

b) C₂H₅OH (l) ® C₂H₅OH (g)

DS° = 282.7 J K^-1mol^-1 - 160.7 J K^-1mol^-1 = 122.0 J K^-1mol^-1

c) Br₂ (l) ® Br₂ (g)

DS° = 245.46 J K^-1mol^-1 - 152.2 J K^-1mol^-1 = 93.26 J K^-1mol^-1

17)    a) ½ I₂ (s) + ½ Cl₂ (g) ® ICl (g)

DS° = 247.55 J K^-1mol^-1 - [0.5(116.14 J K^-1mol^-1) + 0.5(223.07 J K^-1mol^-1)] = 77.95 J K^-1mol^-1

b) C (s, graphite) + ½ O₂ (g) + Cl₂ (g) ® COCl₂ (g)

DS° = 283.53 J K^-1mol^-1 - [5.74 J K^-1mol^-1 + 0.5(205.14 J K^-1mol^-1) + 223.07 J K^-1mol^-1] 
        = -47.85 J K^-1mol^-1

c) Ca (s) + C (s, graphite) + 1½ O₂ (g) ® CaCO₃ (s)

DS° = 92.9 J K^-1mol^-1 - [41.42 J K^-1mol^-1 + 5.74 J K^-1mol^-1 + 1.5(205.14 J K^-1mol^-1)]
       = -262.0 J K^-1mol^-1

21) H₂O (l) ® H₂(g) + ½ O₂ (g)

This reaction should have a positive DS as well as a positive DH (since it is highly endothermic). These values can be quantified as follows:

DS° = [130.68 J K^-1mol^-1 + 0.5(205.14 J K^-1mol^-1)] - 69.96 J K^-1mol^-1 =  +163.29 J K^-1mol^-1

DH° = +285.8 kJ mol^-1

Consequently, this reaction is expected to be spontaneous only at high temperature (~ 1500ºC).
 
 

25)     a) Mg (s) + 2 H₂O (l) ® Mg(OH)₂ (s) + H₂ (g)
 

DS_system = [63.18 J K^-1mol^-1 + 130.68 J K^-1mol^-1] - [32.68 J K^-1mol^-1 + 2(69.91 J K^-1 mol^-1)]

= + 21.36 J K^-1mol^-1
 

 

DS_universe = DS_system + DS_surroundings = 1205 J K^-1mol^-1
 

29) In the following problems, T = 298 K.
 

a) Mg (s) + O₂ (g) + H₂ (g) ® Mg(OH)₂ (s)

DHº_f = -924.54 kJ/mol

DSº = 63.18 J K^-1mol^-1 - [32.68 J K^-1mol^-1 + 130.68 J K^-1mol^-1 + 205.14 J K^-1mol^-1]

        = -305.32 J K^-1mol^-1
 

DGº_f = DHº_f - TDS = -833.55 kJ/mol
 

b) ½ N₂ (g) + ½ O₂ (g) + ½ Cl₂ (g) ® NOCl (g)

DHº_f = + 51.71 kJ/mol

DSº = 261.8 J K^-1mol^-1 - [0.5(191.6 J K^-1mol^-1) + 0.5(223.07 J K^-1mol^-1) + 
0.5(205.14 J K^-1mol^-1)]  = -48.11 J K^-1mol^-1
 

DGº_f = DHº_f - TDS = + 66.05 kJ/mol
 

c) 2 Na (s) + C (s, graphite) + 1½ O₂ (g) ® Na₂CO₃ (s)

DHº_f = -1130.68 kJ/mol

DSº = 134.98 J K^-1mol^-1 - [2(51.21 J K^-1mol^-1) + 5.74 J K^-1mol^-1 + 1.5(205.14 J K^-1mol^-1)]

        = -280.89 J K^-1mol^-1
 

DGº_f = DHº_f - TDS = -1,047 kJ/mol
 

33)   a) HgS (s) + O₂ (g) ® Hg (l) + SO₂ (g)

DG° _rxn = [DG° _f (Hg) + DG° _f (SO₂)] - [DG° _f (HgS)+ DG° _f (O₂)]

DG° _rxn = [0 kJ/mol + (-300.19 kJ/mol)] - [(-50.6 kJ/mol)+ 0 kJ/mol] = -249.6 kJ/mol
 

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

DG° _rxn = [2 DG° _f (H₂O) + 2 DG° _f (SO₂)] - [2 DG° _f (H₂S)+ 3 DG° _f (O₂)]
 

DG° _rxn = [2(-228.57 kJ/mol) + 2(-300.19 kJ/mol)] - [2(-33.56 kJ/mol)+ 3(0 kJ/mol)] 
                = -990.38 kJ/mol

c) SiCl₄ (g) + 2 Mg (s) ® 2 MgCl₂ (s) + Si (s)

 
DG° _rxn = [2 DG° _f (MgCl₂) + DG° _f (Si)] - [DG° _f (SiCl₄)+ 2 DG° _f (Mg)]
 

DG° _rxn = [2(-591.79 kJ/mol) + 0 kJ/mol] - [-616.98 kJ/mol + 2(0 kJ/mol)] 
                = -566.6 kJ/mol
 
 

DG° _rxn = [DG° _f (TiCl₄)] - [DG° _f (TiCl₂)+ DG° _f (Cl₂)]

-272.8 kJ/mol = [-737.2 kJ/mol] - [DG° _f (TiCl₂) + 0 kJ/mol]

DG° _f (TiCl₂) = -464.4 kJ/mol

39) C (s, graphite) + ½ O₂ (g) + 2 H₂ (g) ® CH₃OH (l)
 

DG° = -RT ln K
 

43)    a) 2 C (s, graphite) + H₂ (g) ® C₂H₂ (g)

DS° = 200.94 J K^-1mol^-1 - [2(5.74 J K^-1mol^-1) + 130.68 J K^-1mol^-1] = 58.78 J K^-1mol^-1
 

b) 2 C (s, graphite) + 2 H₂ (g) ® C₂H₄ (g)

DS° = 219.56 J K^-1mol^-1 - [2(5.74 J K^-1mol^-1) + 2(130.68 J K^-1mol^-1)] = -53.28 J K^-1mol^-1

c) 2 C (s, graphite) + 3 H₂ (g) ® C₂H₆ (g)

DS° = 229.6 J K^-1mol^-1 - [2(5.74 J K^-1mol^-1) + 3(130.68 J K^-1mol^-1)] = -173.92 J K^-1mol^-1

DH° _rxn = [4 DH° _f (Ag) + DH° _f (O₂)] - [2 DH° _f (Ag₂O)] = + 62.10 kJ/mol

DS° _rxn = [4 S° (Ag) + S° (O₂)] - [2 S° (Ag₂O)] = + 132.7 J K^-1 mol^-1

Since both the changes in enthalpy and the change in entropy for this reaction are positive, the reaction will become spontaneous (or, "product-favored") at high temperature. At 25° C, we have,

DG° _rxn = DH° _rxn - T DS° _rxn = + 22.45 kJ/mol (not spontaneous)

To find out where the reaction becomes spontaneous, we set DG° _rxn (the equilibrium condition) and solve for T. We make the assumption that DH° _rxn and DS° _rxn are independent of temperature, which is a good approximation in most cases.

53) CH₃OH (l) ® CH₄ (g) + ½ O₂ (g)

a) DS° _rxn = [S° (CH₄) + ½ S° (O₂)] - [S° (CH₃OH)] = + 162.03 J K^-1 mol^-1

The positive entropy change is expected, since a liquid usually has lower entropy than a gas.
 

b) DH° _rxn = [DH° _f (CH₄) + ½ DH° _f (O₂)] - [DH° _f (CH₃OH)] = + 163.85 kJ/mol

DG° _rxn = DH° _rxn - T DS° _rxn = + 115.6 kJ/mol (not spontaneous)
 

c) The temperature at which the reaction becomes spontaneous is:
 

 

59)    a) H₂O (l) ® H₂ (g) + O₂ (g)

DH° _rxn: positive         DS° _rxn: positive         DG° _rxn: positive

b) explosion of dynamite:

        DH° _rxn: negative         DS° _rxn: positive         DG° _rxn: negative

c) combustion of gasoline:

        DH° _rxn: negative         DS° _rxn: positive         DG° _rxn: negative
 

Reaction 1:     AgCl (s) « Ag⁺ (aq) + Cl^- (aq)             K₁ = K_sp = 1.8 ´ 10^-10

Reaction 2:     Ag⁺ (aq) + 2 Cl^- (aq) « AgCl₂^- (aq)     K₂ = K_f = 2.5 ´ 10⁵
 

Sum: AgCl (s) + Cl^- (aq) « AgCl₂^- (aq)                     K_sum = K₁K₂ = 4.5´ 10^-5
 

The free energy changes for the above reactions are:

DG° _rxn1 = [DG° _f (Ag⁺) + DG° _f (Cl^-)] - [DG° _f (AgCl)] =

                = [77.1 kJ/mol + (-131.2 kJ/mol)] - [-109.79 kJ/mol] = +55.7 kJ/mol
 

DG° _rxn2 = [DG° _f (AgCl₂^-)] - [DG° _f (Ag⁺) + 2 DG° _f (Cl^-)] =

                = [-215.4 kJ/mol] - [77.1 kJ/mol + 2(-131.2 kJ/mol)] = -30.1 kJ/mol
 

DG° _sum = [DG° _f (AgCl₂^-)] - [DG° _f (AgCl) + DG° _f (Cl^-)] =

                = [-215.4 kJ/mol] - [-109.79 + (-131.2 kJ/mol)] = +25.59 kJ/mol
 

 

This result is similar to that obtained using the equilibrium constants of the two "component" reactions.

64) C₆H₆ (l) + H₂O (l) ® C₆H₅OH (s) + H₂ (g)

DS° = [S° (C₆H₅OH) + S° (H₂)] - [S° (C₆H₆) + S° (H₂O)] = + 162.03 J K^-1 mol^-1

DS° = [144.0 J K^-1 mol^-1 + 130.68 J K^-1 mol^-1] - [69.91 J K^-1 mol^-1 + 172.8 J K^-1 mol^-1]
        = + 31.97 J K^-1 mol^-1

DH° _rxn = [DH° _f (C₆H₅OH) + DH° _f (H₂)] - [DH° _f (C₆H₆) + DH° _f (H₂O)]

DH° _rxn = [-165.1 kJ/mol + 0 kJ/mol] - [49.03 kJ/mol + (-237.13 kJ/mol)] = +23.0 kJ/mol
 

DG° _rxn = DH° _rxn - T DS° _rxn = + 13.5 kJ/mol

This reaction is not spontaneous (i.e., it is reactant favored) at standard temperature. It is an entropy-driven process.

 
b) The oxygen atom is sp³ hybridized. Therefore, the ideal VSEPR bond angle for the C-O-H- bond would be 109.5° .

c) The hybridization of the carbon atoms in phenol is sp².
 

d) This is a typical weak acid problem. The reaction is:

C₆H₅OH (aq) + H₂O (l) ® C₆H₅O^- (aq) + H₃O⁺ (aq)

	C6H5OH (M)	C6H5O-  (M)	H3O+ (M)
Initial	0.012	0	0
D	-x	x	x
Final	0.012-x	x	x