Exam # ____
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Signature: ____________________________________________
General Chemistry 201
May 12, 2009
Exam 4 (150 pts)
Read all questions before you start. Show all work and explain your answers to receive full credit.
If you do not follow directions, don't expect full credit for your answers. Report all numerical answers to the proper number of significant
figures. By signing your signature above you agree that you have worked alone and neither give nor received help from any source.
There is zero tolerance for dishonesty here. Keep your eyes on your own paper at all times
1
IA
1
18
VIIIA
1
H
2
13
14
15
16
17
He
IIIA IVA
VA
VIA VIIA 4.0026
5
6
7
8
9
10
B
C
N
O
F
Ne
10.811 12.0112 14.0067 15.9994 18.9984 20.179
2
1.00797 IIA
2
3
4
Li
Be
6.939 9.0122
3
11
12
Na
Mg
3
22.9898 24.305 IIIB
4
IVB
5
VB
6
VIB
7
VIIB
8
9
VIIIB
10
4
19
20
21
22
23
24
25
26
27
28
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
39.102 40.08 44.956 47.90 50.942 51.996 54.9380 55.847 58.9332 58.71
5
37
Rb
85.47
6
7
38
39
40
41
42
Sr
Y
Zr
Nb
Mo
87.62 88.905 91.22 92.906 95.94
43
Tc
[99]
55
56
57 *
72
73
74
75
Cs
Ba
La
Hf
Ta
W
Re
132.905 137.34 138.91 178.49 180.948 183.85 186.2
87
Fr
[223]
*
Lanthanide
Series
‡
Actinide
Series
88
Ra
[226]
89 ‡
Ac
[227]
104
Ku
[261]
58
Ce
140.115
90
Th
232.0381
11
IB
29
Cu
63.54
13
14
15
16
17
18
12
Al
Si
P
S
Cl
Ar
IIB 26.9815 28.086 30.9738 32.064 35.453 39.948
30
Zn
65.37
31
Ga
65.37
32
33
34
35
36
Ge
As
Se
Br
Kr
72.59 74.9216 78.96 79.909 83.80
44
45
46
47
48
49
50
51
52
53
54
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
101.07 102.905 106.4 107.870 112.40 114.82 118.69 121.75 127.60 126.904 131.30
76
Os
190.2
77
78
79
80
81
82
83
84
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
192.2 195.09 197.0 200.59 204.37 207.19 208.980 [210]
105
106
107
108
109
110
111
112
[262]
[263]
[262]
[265]
[268]
[269]
[272]
[277]
59
60
61
62
63
Pr
Nd
Pm
Sm
Eu
140.9077 144.24
(145) 150.368 151.965
91
92
93
94
95
Pa
U
Np
Pu
Am
231.0359 238.0289 237.048 [244]
[260]
Thermodynamics
Universe = surroundings + system
State Function (X) where X = E, H, S or G
ΔXrxn = Σ n ΔX°prod - Σ n ΔX°react
w = -P ΔV
ΔE = q + w
ΔH = ΔE + P ΔV
ΔH = qp = m CsΔT
ΔS°univ = ΔS°surr + ΔS°sys
ΔS°surr = - ΔH°sys / T
ΔG = ΔH - TΔS
ΔG = ΔG° + RT ln Q
ΔG° = - RT ln Keq
Keq = exp {-ΔG° /RT}
ln Keq = (ΔS°/ R) - (ΔH°/ RT)
Temperature Conversion
°C = 5/9 (__°F - 32)
°F = (9/5 _°C) + 32)
64
Gd
157.25
96
Cm
[247]
65
Tb
158.9254
97
Bk
[247]
85
At
[210]
86
Rn
[222]
66
67
68
69
70
71
Dy
Ho
Er
Tm
Yb
Lu
162.50 164.9303 167.26 168.9342 173.04 174.967
98
99
100
101
102
103
Cf
Es
Fm
Md
No
Lr
[251]
[252]
[257]
[258]
[259]
[260]
Cell Potential, ΔG and Keq
ΔG = - nFE
ΔG° = - nFE°
E° = (0.0591 / n ) log Keq
E°cell = E°red(cathode) - E°red(anode)
E°cell = E°red + E°ox
Cell concentration and the Nernst equation
E° = (RT/ nF) ln Keq
E° = (0.0592/ n) log Keq
E = E° - (RT/ nF) ln Q
E = E° - (0.0592 / n) log Q
2H2O (l) + 2e-  H2 (g) + 2OH- (aq)
O2(g) + 4H+ (aq) 4 e-  2H2O (l)
Constants
R = 8.314 J / mol•K
F = 96,500 C / mol e1 V = 1 J/ C
E° = - 0.83 V
E° = +1.23V
(4 pts each) Choose the one best answer.
1
To determine the temperature at which carbon dioxide sublimes at 1 atm, what condition must be met ?
a) Keq = 0
b) ΔG° = 0
c) Q = 0
d) ΔS = 0
2
When a seal is broken for an ice-cold pack, NH4Cl(s) is mixed with water and the resulting aqueous solution becomes colder. Under
normal room temperature conditions, what is ΔG° for the dissolution process of NH4Cl?
a) ΔG < 0
3
4
5
6
b) ΔG = 0
c) ΔG > 0
If H < 0 and S < 0 the reaction will ?
a) will never proceed spontaneously to the right.
c) proceed spontaneously to the right only at high temperatures.
d) Not enough info to determine
b) proceed spontaneously to the right only at low temperatures.
d) proceed spontaneously to the right at all temperatures.
Which is always true for a reaction to be spontaneous.
a) ΔSuniv > 0
b) ΔHsys > 0
c) ΔGsys > 0
d) All are true
Which state function is based on an absolute scale?
a) Free Energy
b) Entropy
c) Enthalpy
d) Electromotive Force
Which of the following is true for an endothermic reaction?
a) Heat is considered a reactant
b) The sign of ΔHrxn is positive
d) All are true
c) The process absorbs heat
7
Use the Standard Reduction table to rank the following metals in order of increasing oxidizing agent. Cd Ag Zr
a) Ag - Zr - Cd
b) Ag - Cd - Zr
c) Cd - Zr - Ag
d) Zr - Cd - Ag
8
In the oxidation-reduction of water, how many electrons are exchanged? 2H2O (g)  2H2 (g) + O2 (g)
a) One
b) Two
c) Four
d) None
9
In an electrolytic cell, oxidation occurs at the __ and is assign the __ electrode.
a) anode, (+)
b) anode, (-)
c) cathode, (+)
10
d) cathode, (-)
Which one atom has the largest change in its oxidation state in the following reaction?
Fe2S3(s) + 12 HNO3(aq)  2 Fe(NO3)3(aq) + 3S(s) + 6NO2(g) + 6 H2O(l)
a) S
b)
Fe
c) N
d) All the same
11
(11 pts) Consider
[RhCl6]2- + 3e-  Rh(s) + 6Cl- (aq)
Cl2 (s) + 2e-  2Cl- (aq)
E° =
0.44 V
E° =
1.358 V
i) From the data shown below, write the net reaction for the formation reaction of [RhCl6]2- (aq).
Remember that a formation reaction is based on elements in their neutral state forming a metal complex.
ii) Calculate Kf for [RhCl6]2- (aq)
and
iii) ΔG° (kJ)
NetRxn : 3Cl2(g) + 6 Cl− + 2 Rh(s)
ΔG° = - n ℑ E°cell = - 6 mol e - *
 −ΔG 


 RT 
Kf = e
 E °nF 


 RT 
Kf = e
 −173673 J 

−
 8.314(298) 
= e
= e(
 (.919V )(6 )(96485) 


8.314(298)


= e
+214.5
= e(
 2 [RhCl6 ]396,485 C
mol e
-
*
E°cell = 0.918 V
0.918 J
C
= - 531.439 kJ
) = 1.43 x 1093
+214.5
) = 1.43 x 1093
€
12
16 pts) Check mark if you agree or disagree with the following statement, then write a short convincing explanation to support or
refute the statement. No credit will be given without a statement of justification.
i) __ Agree __ Disagree:
All endothermic reactions are nonspontaneous since energy is necessary to initiate the reaction.
DISAGREE, Entropy can drive the reaction.
ii) __ Agree __ Disagree:
The following line-notation represents a voltaic cell. Cd(s) | Cd+2(aq) || Ag+(aq) | Ag(s)
AGREE, Ag+(aq) | Ag(s), E° = +0.7994 V, Cd(s) | Cd+2(aq) E° = + 0.403 V, E°cell = +.396 V, this is an voltaic cell.
iii) __ Agree __ Disagree:
The chemical reaction from a voltaic cell is said to come to a complete stops when the equilibrium constant is zero, Keq = 0.
DISAGREE, Keq does not equal zero when the reaction stops, E° does equal zero but Keq is some equilibrium constant > 0.
iv) __ Agree __ Disagree:
The boiling of water at 100 °F and 1 atm. is a spontaneous process.
DISAGREE, 100°F < 100°C. thus the temperature is lower than the normal boiling point of water, which is non spontaneous.
13
(9 pts) Another substance that may be used to standardized KMnO4 (aq) is sodium oxalate, Na2C2O4. If 0.2482 g, Na2C2O4 is dissolved in
water and titrated with 0.0296M KMnO4, what volume of KMnO4 (aq) is required to reach the equivalent point?
C2O4-2 is oxidize to CO2.
-2
5 C2O 4
→ 10 CO2 + 10 e−
_________________________________
2 MnO4
0.2482 g Na2 C2O 4 ∗
-
+ 16H+
mol Na2 C2O 4
134.00 g
∗
+
10e− → 2Mn+2 +
2 mol KMnO4
5 mol Na2 C2O 4
∗
8 H2O
L
0.0296 M KMnO 4
n = 10
= 0.0250L = 25.0 mL
€
14
(10 pts) Compare and contrast a voltaic cell and an electrolytic cell with respect to each of the following:
Voltaic Cell
Electrolytic Cell
i) Sign of the free energy change:
ΔG (-) spontaneous
ΔG (+) non spontaneous
( - ) or ( + )
ii) Type of half-reaction at the cathode:
Reduction Rxn
Reduction Rxn
Oxidation or Reduction
iii) Type of half-reaction at the anode:
Oxidation Rxn
Oxidation Rxn
Oxidation or Reduction
iv) Charge on the electrode labeled "anode" &
Anode (-) Cathode (+)
Anode (+) Cathode (-)
“cathode”.
(+) or (-)
v) Electrode from which electron enters the cell.
e- enters cathode (+) terminal
e- enters cathode (-) pulled by battery
enter or exit
e- exits anode (-) source of e-
e- exits anode (+) pushed by the battery
15
(18 pts) Consider the phase changes of water according to the diagram shown.
Complete the table by writing (+), (-) or (0) under the thermodynamic
parameter for the corresponding phase change as labeled in the
diagram at standard state conditions.
Process
ΔH°
ΔS°
ΔG°
16
(3 pts)
1
( + )
( + )
( + )
2
( - )
( - )
( - )
3
( - )
( - )
( + )
4
( + )
( + )
( + )
5
( + )
( + )
( - )
6
( - )
( - )
( + )
(15 pts) Solve the following:.
Mg(s) + Ag+(aq)  Mg2+(aq) + Ag(s)
i) Calculate E° cell:
Mg / Mg+2 (E° = + 2.375 V)
Ag+/ Ag (E° = + 0.7994 V)
E° cell = + 3.174 V
€
(4 pts)
ii) Calculate the E° for Ir3+/ Ir: Pd(s) | Pd2+(aq), || Ir3+(aq) | Ir(s)
E° cell = + 0.156 V
Pd+2 / Pd (E° = +1.00 V)
Ir+3/ Ir (E°=
? V)
E°cell = + 0.156 V, E° = ____ - 1.00V = + 0.156 V
Std Red potential : Ir+3 / Ir E° = 1.156 V
€
(8 pts)
iii) Calculate the pH: Ag(s)|Ag+(0.010 M)|| NiO2(s) | Ni2+(0.010 M)| Pt(s)
Ecell = 0.960 V
Net Rxn : 4H+(aq) + 2Ag(s) + NiO2(s) → 2 Ag+(aq) + Ni+2(aq) + 2H2O(l) E°cell = 0.9006 V
0.960 = 0.9006 -
0.0592
2
 [Ag + ]2 [Ni+2 ] 

log
[H+ ]4






.9006-.960 (2)
- 0.0594 (2)




 [Ag + ]2 [Ni+2 ]   [0.01]2 [0.01] 
-2.0068)

 = 
 = 10 .0592  = 10 .0592  = 10(
= 9.846 ∗10−3
+ 4
+ 4
[H ]
[H ]

 

 [0.01]2 [0.01] 
 = 1.0157 •10−4
[H+ ]4 = 
−3 
 9.846 •10 
€
[H+ ] = 0.100M
pH = 1.00
17
(16 pts) Arrange each of the following groups in order of increasing state function, X°. The order should be smallest or most negative
to largest or most positive. Place numerical value to each choice, 1 (smallest), 2 or 3 (largest) after placing them in order to justify
your answer.
i) Entropy: Ag (s), Cu (s), Al (s)
Al (s), S° = 28.32 J (1) <
Cu (s), S° = 33.30 J (2) <
ii) Standard reduction potential: Cr+3/Cr, Al+3/Al, Ga+3/Ga
Ag (s), S° = 42.55 J (3)
(1) Al+3/Al (-1.66) < (2) Cr+3/Cr (-0.74) < (3) Ga+3/Ga (-0.53)
iii) Free Energy: O (g), H2O (l), H (g)
(1) H2O (l) –237.13 < (2) H (g) 203.26 < (3) O (g) 230.1 kJ
iv) Enthalpy: (a) 17 g of NH3 (g),
(b) 0.10 mole of CO2 (g),
(1) NH3 : 17.0 g ∗
(c) 11.2 L of HCl(g) at 0°C, 1 atm)
1 mole
(2) HCl : 0.5 mole ∗
(3) CO2 : 0.1 mole ∗
17 g
-46.19 kJ
∗
mol
-92.30 kJ
= - 46.15 kJ
mol
-393.5 kJ
mol
= - 46.19 kJ
= - 39.35 kJ
€
18
(15 pts) Consider the following reaction:
H2O(g) + Cl2O(g) → 2HOCl(g)
Keq (298) = 0.090
a) Calculate ΔG° for the reaction using the equation ΔG° = -RT ln (keq)
b) Calculate ΔH° for the reaction.
c) Using the results from part a and b calculate ΔS° for the reaction.
d) Estimate the value of keq at 500K.
e) Estimate ΔG at 25°C when PH2O = 18 torr, PCl2O = 2.0 torr and PHOCl = 0.10 torr.
a) ΔG° = - RT lnK eq = - 8.314
J
mol•K
∗ 298 K ∗ ln (0.090)
b)
H-O -H
c) ΔG° = ΔH° - TΔS° , TΔS° = ΔH° - ΔG°,
ΔH° - ΔG°
T
=
0 - (+5970 J)
298 K
ΔS° =
= -20.02
ΔG = − 14.3 KJ
19
€
J
mol • K
ΔH° - ΔG°
→
2 H - O - Cl
d) ΔG500 = ΔH° - TΔS° = 0 - (500K) (-20.02
T
J
K
) = +10.0 kJ
 ΔG
 10000 J 
 = .0902
Ke q = exp -   = eexp - 
 RT 
 (8.314) (500)
J
K
e) ΔG = ΔG° + RT ln Q = + 5965.86 J + 8.314
ΔG = + 5965.86 J + 8.314
Cl - O - Cl
2 O- H
+ 2 Cl - O
2 H - O + 2 O - Cl
ΔH = Bond EnergiesProducts - Bond EnergiesReactant = 0 kJ
ΔG° = + 5965.86 J = +5.97 kJ
ΔS° =
+
J
mol • K
(298K) ln
0.102
(18) (2.0)
(298K) (-8.234) = + 5,965.86 J - 20,288.07 J
(10 pts) Find the element in bold with the lowest and highest oxidation states.
KOsCl4 (+3)
F2O (+2)
(NH4)2Ce(SO4)3 (+4)
IO4 (+7)
ClO3- (+5)