Practice Final Exam
CH 201
Name: ______________________________________
(please print)
Student I D : __________________
Instructions - Read Carefully
1.
Please show your work, and put your final answers in the spaces provided.
2.
Point values for each question are given in parentheses at the beginning of each question
pKa’s HCN : 9.40
HClO : 7.46
1-
H2S : 7.00
HS : 12.89
HClO4 : << 0 H3BO3 : 9.27
1A
1
H
HBrO : 8.70
NH41+ : 9.25
H2C2O4 : 1.23
HC2O41-
H3C2O2H: 4.7
HF: 3.2
HNO3 : -1.4
HNO2 : 3.40
: 4.19 HCl : -6.1
HN3 : 4.72
8A
2
He
3
Li
2A
4
Be
3A
5
B
4A
6
C
5A
7
N
6A
8
O
7A
9
F
6.941
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
Na
12
Mg
5B
6B
7B
15
P
16
S
17
Cl
18
Ar
24.31
4B
14
Si
22.99
3B
13
Al
19
K
20
Ca
21
Sc
22
Ti
23
V
24
Cr
25
Mn
26
Fe
27
Co
39.10
40.08
44.96
47.88
50.94
52.00
54.94
55.85
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
43
Tc
85.47
87.62
88.91
91.22
92.91
95.94
55
Cs
56
Ba
57
La
72
Hf
73
Ta
74
W
132.9
137.3
138.9
178.5
181.0
183.8
87
Fr
88
Ra
89
Ac
104 105 106
(223)
226.0
227.0
Rf
Bh
Hs
1.008
(261)
Db
(262)
Sg
(263)
10
Ne
1B
2B
26.98
28.09
30.97
32.07
35.45
39.95
28
Ni
29
Cu
30
Zn
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
58.93
58.69
63.55
65.39
69.72
72.61
74.92
78.96
79.90
83.80
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
(98)
101.1
102.9
106.4
107.9
112.4
114.8
118.7
121.8
127.6
126.9
131.3
75
Re
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
186.2
190.2
192.2
195.1
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
107
108 109 110 111
(264)
8B
4.003
Mt
(265)
(268)
Ds
Rg
(272)
(269)
Pledge: I have neither given nor received help during this exam.
Signature: ______________________________
Name (print): ___________________________
1
Please solve all problems
1. A. Calculate Eo for the cell
Given
Solution:
2 𝐶𝐶𝐶𝐶 + 3 𝑆𝑆𝑆𝑆4+ ↔ 3 𝑆𝑆𝑆𝑆2+ + 2 𝐶𝐶𝐶𝐶 3+
𝑆𝑆𝑆𝑆4+ + 2𝑒𝑒 − → 𝑆𝑆𝑆𝑆2+
𝐶𝐶𝐶𝐶 3+ + 3𝑒𝑒 − → 𝐶𝐶𝐶𝐶
𝐸𝐸 0 = 0.15 𝑉𝑉
𝐸𝐸 0 = −0.74 𝑉𝑉
𝑜𝑜
𝑜𝑜
𝑜𝑜
= 𝐸𝐸𝑟𝑟𝑟𝑟𝑟𝑟
− 𝐸𝐸𝑜𝑜𝑜𝑜
= 0.15 + 0.74 = 0.89 𝑉𝑉
𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑜𝑜
𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
=___________________________.
B. Calculate ∆Go for the reaction as well.
Solution:
𝑜𝑜
= −(6) �96,485
∆𝐺𝐺 𝑜𝑜 = −𝑛𝑛𝑛𝑛𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐶𝐶
� (0.89𝑉𝑉) = −515.7 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚
∆𝐺𝐺 𝑜𝑜 =___________________________.
𝑜𝑜
2. At 25 oC calculate the voltage of the cell Ag(s)|Ag+(aq, 0.1M)|| Cu2+(aq,0.01 M)|Cu if 𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
=
0.003 V, and given
𝐴𝐴𝐴𝐴+ + 𝑒𝑒 − → 𝐴𝐴𝐴𝐴
𝐸𝐸 0 = +0.337 𝑉𝑉
𝐸𝐸 0 = +0.340 𝑉𝑉
𝐶𝐶𝐶𝐶2+ + 2𝑒𝑒 − → 𝐶𝐶𝐶𝐶
Solution: We need the balanced equation for the cell.
𝐶𝐶𝐶𝐶2+ + 2 𝐴𝐴𝐴𝐴 → 𝐶𝐶𝐶𝐶 + 2 𝐴𝐴𝐴𝐴+
This essentially a concentration cell since the cell potential is nearly zero. Both Ag and Cu are
noble metals and their reduction potentials are nearly equal.
𝑅𝑅𝑅𝑅
0.0257
[𝐴𝐴𝐴𝐴+]2
𝑜𝑜
𝐸𝐸 = 𝐸𝐸𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 +
𝑙𝑙𝑙𝑙𝑙𝑙 = 0.003 +
𝑙𝑙𝑙𝑙 �
�
𝑛𝑛𝑛𝑛
𝑛𝑛
[𝐶𝐶𝐶𝐶2+]
0.0257
0.12
𝑙𝑙𝑙𝑙 �
� = 0.003 𝑉𝑉
2
0.01
It just happens that Q = 1 so the cell potential is the potential, and therefore also equal to the
voltage of the cell.
𝑉𝑉 =___________________________.
𝐸𝐸 = 0.003 +
2
3. Balance the equation
𝐶𝐶𝐶𝐶𝑂𝑂3− + 𝑁𝑁𝑂𝑂2− ↔ 𝐶𝐶𝐶𝐶 − + 𝑁𝑁𝑂𝑂3−
Solution: Step 1. Write the equation with coefficients for each species
𝑎𝑎 𝐶𝐶𝐶𝐶𝑂𝑂3− + 𝑏𝑏 𝑁𝑁𝑂𝑂2− ↔ 𝑥𝑥 𝐶𝐶𝐶𝐶 − + 𝑦𝑦 𝑁𝑁𝑂𝑂3−
Step 2. Determine the element equations based on number of each element in each species.
Cl: a = x
N: b = y
O: 3a+2b=3y
3x + 2y = 3y
3x = y
Step 3. Make one choice for a value that will fix all of the stoichiometric coefficients.
If x = 1, then y = 3, a = 1 and b = 3
𝐶𝐶𝐶𝐶𝑂𝑂3− + 3 𝑁𝑁𝑂𝑂2− ↔ 𝐶𝐶𝐶𝐶 − + 3 𝑁𝑁𝑂𝑂3−
Note: charge balance must also be satisfied, and this procedure guarantees it.
Balanced equation ______________________________________________________________.
4. A. How much heat is required to boil 1 liter of water starting from room temperature of 25 oC?
The specific heat of water is 4.18 J/g- oC.
Solution: For mass, m, and specific heat, s, the required heat, q, is:
4.18𝐽𝐽
𝑞𝑞 = 𝑚𝑚𝑚𝑚∆𝑇𝑇 = (1000 𝑔𝑔) �
� (75 o C) = 313.5 kJ
𝑔𝑔
q = _________________________.
B. If the water is heated on a propane stove, then the heat is transferred from the combustion of
propane. How many liters of propane are required if the propane is delivered in a pressurized
tank with a pressure of 10 atm, assuming that you may treat it as an ideal gas at 298 K. The
combustion of propane is given below.
𝐶𝐶3 𝐻𝐻8 + 5 𝑂𝑂2 → 3 𝐶𝐶𝑂𝑂2 + 4 𝐻𝐻2 𝑂𝑂 ∆𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐻𝐻 𝑜𝑜 = −2220 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
NOTE: Do not worry about the sign of the heat. Obviously, the heat output from the propane
must balance the heat input to the water.
Solution: Step 1. Calculate the number of moles of propane required.
𝑞𝑞 = 𝑛𝑛∆𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐻𝐻 𝑜𝑜
So
𝑞𝑞
313.5
𝑛𝑛 =
=
= 0.141
∆𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐻𝐻𝑜𝑜 2221
Step 2. Use the ideal gas law to solve for the volume:
𝑉𝑉 =
𝑛𝑛𝑛𝑛𝑛𝑛 (0.141)(0.008206)(298)
=
= 0.344 𝐿𝐿
𝑃𝑃
10
Volume = ____________________________.
3
5. Given the following thermochemical equation, calculate the heat involved in decomposing
12.5 g of water to the elements.
∆H° = -571.6 kJ
2H2 (g) + O2 (g)  2H2O (l)
Solution: First, we must determine the number of moles of H2O.
12.5 𝑔𝑔
= 0.694 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
18 𝑔𝑔
Note from the stoichiometry that the indicated enthalpy is for 2 moles of H2O. Also, the sign is
for formation of water and problem is asking for the reverse process
1
𝐻𝐻2 𝑂𝑂 → 𝐻𝐻2 + 𝑂𝑂2
∆𝐻𝐻 𝑜𝑜 = +285.8 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
2
The heat is thus
𝑛𝑛𝐻𝐻2𝑂𝑂 =
𝑞𝑞 = 𝑛𝑛∆𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝐻𝐻 𝑜𝑜 = (0.694) �285.8
𝑘𝑘𝑘𝑘
� = 198 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚
q = _________________________.
6. Will CaSO4 precipitate if 50 mL of 0.0010 M CaCl2 is added to 50 mL of 0.010 M Na2SO4?
(Ksp for CaSO4 = 2.4 x 10-5)
Solution: You may ignore the Na+ and Cl- ions (they are spectators). The solubility equilibrium
is
𝐶𝐶𝐶𝐶𝐶𝐶𝑂𝑂4 ↔ 𝐶𝐶𝐶𝐶2+ + 𝑆𝑆𝑂𝑂42−
Given the dilution factor of two in this problem the concentrations are:
[𝐶𝐶𝐶𝐶2+] = 0.001 𝑀𝑀
[𝑆𝑆𝑂𝑂42−] = 0.01 𝑀𝑀
And
𝑄𝑄 = [𝐶𝐶𝐶𝐶2+][𝑆𝑆𝑂𝑂42−] = 10−5
There are two ways to think about this. First, you may write
∆𝐺𝐺 = ∆𝐺𝐺 𝑜𝑜 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅
In this case you must calculate ∆𝐺𝐺 𝑜𝑜 from the known Ksp.
∆𝐺𝐺 𝑜𝑜 = −𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝐾𝐾𝑠𝑠𝑠𝑠 = −(8.31)(298)𝑙𝑙𝑙𝑙(2.4 × 10−5 ) = +26.3 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
Then calculate ∆𝐺𝐺
𝑘𝑘𝑘𝑘
+ (8.31)(298)𝑙𝑙𝑙𝑙(10−5 ) = −2160 𝐽𝐽/𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚
Since the solution is spontaneous the will be no precipitation in this case.
A second way is to calculate Q and determine whether Q < Ksp. If Q < Ksp then there will
be no precipitation (as was proven the hard way by calculating the free energy. When in doubt,
calculate the free energy.
Answer = _________________________.
∆𝐺𝐺 = −26.3
4
7. Given the thermochemical equation
2NO(g) + O2(g)  2NO2(g)
∆H° = -114.0 kJ,
calculate ∆H° for the following reaction:
NO2(g)  NO(g) + 1/2 O2(g)
Solution: The second equation is just the back reaction of the first (so the sign will be positive)
and the number of moles is half as great.
∆H° = +57 kJ
∆H° = _________________________.
8. A 15.0-g block of aluminum at an initial temperature of 27.5°C absorbs 0.678 kJ of heat. What
is the final temperature of the block? (The specific heat of Al is 0.902 J g-1 °C-1.)
Solution: We use the equation
And solve for
𝑞𝑞 = 𝑚𝑚𝑚𝑚∆𝑇𝑇
𝑞𝑞
678 𝐽𝐽
=
= 50 𝑜𝑜 𝐶𝐶
𝐽𝐽
𝑚𝑚𝑚𝑚 (15𝑔𝑔) �0.902
�
𝑔𝑔 − 𝑜𝑜 𝐶𝐶
The final temperature is 77.5 oC.
∆𝑇𝑇 =
T = ________________________.
9. Use the following thermochemical equations
C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(g) ∆H° = -1300 kJ
C2H6(g) + 7/2 O2(g)  2CO2(g) + 3H2O(g) ∆H° = -1560 kJ
H2(g) + 1/2 O2(g)  H2O(g) ∆H° = -286 kJ
to calculate ∆H° for the following reaction:
C2H2(g) + 2H2(g)  C2H6(g)
Solution: Use the additivity of enthalpies (Hess’s Law) in two steps.
Step 1. Generate an equation that has ethane as reactant and ethane as product.
Eqn 1. [C2H2(g) + 5/2 O2(g)  2CO2(g) + H2O(g)]
∆H° = -1300 kJ
-Eqn. 2 [C2H6(g) + 7/2 O2(g)  2CO2(g) + 3H2O(g)]
- ∆H° = -1560 kJ
- - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - -----------∆H° = +260 kJ
C2H2(g) + 2 H2O(g) +  C2H6(g) + O2(g)
5
Step 2. Eliminate water.
Eqn. 1 [C2H2(g) + 2 H2O(g) +  C2H6(g) + O2(g)]
∆H° = +260 kJ
+ 2 xEqn. 2 [H2(g) + 1/2 O2(g)  H2O(g)]
+ 2 x ∆H° = -286 kJ
- - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - -----------C2H2(g) + 2H2(g)  C2H6(g)
∆H° = -312 kJ
∆H° = ____________________________.
10. A volume of 50.0 mL of 0.400 M HBr at 24.35°C is added to 50.0 mL of 0.400 NaOH, also
at 24.35°C. The final temperature is 27.06°C. Calculate the enthalpy change, ∆H, in kJ for the
following reaction:
HBr(aq) + NaOH(aq) --> NaBr(aq) + H2O (l)
The heat capacity of the system is 0.418 J/°C.
Solution: In this problem we know that the increase in temperature is due to the enthalpy of
reaction. We calculate the heat (which is a constant pressure) and then set this equal to ∆H. The
heat is
𝐽𝐽
𝑞𝑞 = 𝐶𝐶∆𝑇𝑇 = �0.418 𝑜𝑜 � (27.07 − 24.35) = 13 𝐽𝐽
𝐶𝐶
The total volume is 100 mL, which permits us to calculate the number of moles of each reactant
to be n = 0.4 M x 0.1 L = 0.04 moles. Therefore, the molar enthalpy change is
∆𝐻𝐻 =
𝑞𝑞
13 𝐽𝐽
=
= 325 𝐽𝐽/𝑚𝑚𝑚𝑚𝑚𝑚
𝑛𝑛 0.04 𝑚𝑚𝑚𝑚𝑚𝑚
∆H = ____________________________.
11. The molar solubility of PbCl2 in 0.10 M NaCl is 1.7 x 10-3 moles in a liter
(that is 1.7 x 10-3 moles of PbCl2 will dissolve in 1 liter of the solution).
What is the Ksp of PbCl2?
Solution: The solubility equilibrium is
And
𝑃𝑃𝑃𝑃𝐶𝐶𝐶𝐶2 ↔ 𝑃𝑃𝑃𝑃 2+ + 2𝐶𝐶𝐶𝐶 −
𝐾𝐾𝑠𝑠𝑠𝑠 = [𝑃𝑃𝑃𝑃 2+][𝐶𝐶𝐶𝐶 −]2 = (1.7 × 10−3 )(2 × 1.7 × 10−3 )2 = 1.9 × 10−8
Ksp = ____________________________.
6
12. Determine the equilibrium constant for the reaction,
𝐻𝐻𝐻𝐻 + 𝐶𝐶𝐶𝐶 −1 ↔ 𝐻𝐻𝐻𝐻𝐻𝐻 + 𝐹𝐹 −1
𝐾𝐾𝑎𝑎 (𝐻𝐻𝐻𝐻) 10−3.2
𝐾𝐾 =
=
= 10−9.3 = 5 × 10−10
6.1
𝐾𝐾𝑎𝑎 (𝐻𝐻𝐻𝐻𝐻𝐻)
10
K = ____________________________.
Calculate ∆G at 298 K.
o
∆𝐺𝐺 𝑜𝑜 = −𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅(𝐾𝐾) = −(8.31)(298) ln(5 × 10−10 ) = 53.0 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
∆Go = ____________________________.
13. Acid rain is caused by industrial emissions of SO3, which combine with water to form H2SO4.
If the SO3 pressure is 6 x 10-7 atm and the equilibrium constant for hydration,
Is
𝑆𝑆𝑂𝑂3 (𝑔𝑔) ↔ 𝑆𝑆𝑂𝑂3 (𝑎𝑎𝑎𝑎)
𝐾𝐾 =
[𝑆𝑆𝑂𝑂3 ]
= 5 𝑀𝑀/𝑎𝑎𝑎𝑎𝑎𝑎
𝑃𝑃𝑆𝑆𝑆𝑆3
NOTE: We write the equilibrium constant this way with units, but we know that these are
relative to the standard state of 1 atm of pressure and 1 molar of solution. K actually has no units,
but the concentration [𝑆𝑆𝑂𝑂3 ] is in units of molarity and 𝑃𝑃𝑆𝑆𝑆𝑆3 is in units of atm. Assume that the
equilibrium lies far to the right so all of the dissolved SO3 forms H2SO4.
𝑆𝑆𝑂𝑂3 (𝑎𝑎𝑎𝑎) + 𝐻𝐻2 𝑂𝑂 ↔ 𝐻𝐻2 𝑆𝑆𝑂𝑂4 (𝑎𝑎𝑎𝑎)
Solution: Step1. Calculate the concentration of [𝑆𝑆𝑂𝑂3 ]
𝑀𝑀
� (6 × 10−7 ) = 3 × 10−6 𝑀𝑀
[𝑆𝑆𝑂𝑂3 ] = 𝐾𝐾𝑃𝑃𝑆𝑆𝑆𝑆3 = �5
𝑎𝑎𝑎𝑎𝑎𝑎
Since H2SO4 is a strong acid we can assume that the first dissociation is complete and the [H+] is
equal to the concentration of the acid.
[𝑆𝑆𝑂𝑂3 ] = [𝐻𝐻2 𝑆𝑆𝑂𝑂4 ] = [𝐻𝐻 +] = 3 × 10−6 𝑀𝑀
𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 [𝐻𝐻 +] = −𝑙𝑙𝑙𝑙𝑙𝑙10 (3 × 10−6 ) = 5.5
pH = ____________________________.
7
14. Acetic acid buffers are used to buffer pH values less than 7.
A. What is the optimum buffer pH for acetic acid?
B. What is the range of the acetic acid buffer system (i.e. what are the limits in terms of low and
high pH)?
C. Starting with a total of 0.1 M acetic acid, determine the concentration of [acetate] at three pH
values, the low end, optimum and high end of the buffer region.
Solution:
A. The optimum buffer value is when pH = pKa. For acetic acid this is 4.7.
B. The range of a buffer is ±1 pH unit, thus from 3.7 to 5.7.
C. The ratio of acetic acid to acetate is 50:50 when pH = pKa.
To go to the low extreme the ratio is,
And at the high end
[𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎]
=1
[𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎]
[𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎]
= 0.1
[𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎]
[𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎]
= 10
[𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎]
Starting with concentration, c, and ratio, a, we can solve the general equation
𝑥𝑥
= 𝑎𝑎
𝑐𝑐 − 𝑥𝑥
For x,
𝑥𝑥 = 𝑎𝑎(𝑐𝑐 − 𝑥𝑥)
We can make a table
Ratio
0.1
1
10
𝑥𝑥 =
𝑎𝑎𝑎𝑎
1 + 𝑎𝑎
[acetate]
0.009
0.05
0.09
A. Optimum pH = _____________________________.
B. Low and high pH = __________________________.
C. Acetate concentration (M) under three pH conditions
Low _________ Optimum _________ High _________
8
15. Determine the pH when 60 mL of a 0.1 M solution of NaCN is mixed with 10 mL of a
solution of 0.4 M NaOH. The pKa of HCN is 9.4.
Solution: This is an weak acid equilibrium addition of a strong base. NaOH is completely
dissociated. Thus, [OH-] = 0.4 M prior to dilution. The equilibrium we need to consider is
𝐻𝐻𝐻𝐻𝐻𝐻 + 𝑂𝑂𝑂𝑂 − ↔ 𝐻𝐻2 𝑂𝑂 + 𝐶𝐶𝐶𝐶 −
Since this is the conjugate base equilibrium we need to calculate Kb from Ka. Recall that
𝑝𝑝𝐾𝐾𝑏𝑏 + 𝑝𝑝𝐾𝐾𝑎𝑎 = 𝑝𝑝𝐾𝐾𝑤𝑤
𝑝𝑝𝐾𝐾𝑏𝑏 = 𝑝𝑝𝐾𝐾𝑤𝑤 − 𝑝𝑝𝐾𝐾𝑎𝑎 = 14 − 9.4 = 4.6
Therefore,
𝐾𝐾𝑏𝑏 = 10−𝑝𝑝𝐾𝐾𝑏𝑏 = 10−4.6 = 2.5 × 10−5
To obtain the initial concentration for a reaction table we need to first determine the dilution
factors
0.06
[𝐻𝐻𝐻𝐻𝐻𝐻] = �
� (0.1 𝑀𝑀) = 0.0857 𝑀𝑀
0.07
0.01
[𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁] = �
� (0.4 𝑀𝑀) = 0.0571 𝑀𝑀
0.07
Conc.
-Initial
Change
Equilibrium
OH+
0.0571
-x
0.0571-x
HCN
0.0857
-x
0.0857-x
𝐾𝐾𝑏𝑏 =
CN0
x
x
[𝐶𝐶𝐶𝐶 −]
𝑥𝑥
=
−
[𝐻𝐻𝐻𝐻𝐻𝐻][𝑂𝑂𝑂𝑂 ] (0.0571 − 𝑥𝑥)(0.0857 − 𝑥𝑥)
0.00489 − 1.1428𝑥𝑥 + 𝑥𝑥 2 = 0
1.1428 ± �1.14282 − 4(0.00489)
= 0.00429
𝑥𝑥 =
2
[𝑂𝑂𝑂𝑂 −] = 0.0571 − 0.00489 = 0.05221 𝑀𝑀
𝑝𝑝𝑝𝑝𝑝𝑝 = −𝑙𝑙𝑙𝑙𝑙𝑙10 (0.05221) = 1.28
𝑝𝑝𝑝𝑝 = 14 − 𝑝𝑝𝑝𝑝𝑝𝑝 = 14 − 1.28 = 12.72
pH = ____________________________.
9
16. 5 grams of an unknown solute were dissolved in 40.0 g of tetrahydrofuran (C4H8O).
The boiling point of the solution was measured to be 67.0 oC. What is the molar mass of the
unknown? The physical data are given below.
Boiling point Tvap = 66 oC
Enthalpy of vaporization ∆vapHo = 32 kJ/mol.
Kb = 0.413 K/molal
Solution: Using the boiling point elevation formula we have
𝒎𝒎𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 =
The number of moles of solute is
∆𝑇𝑇 = 𝐾𝐾𝑏𝑏 𝑚𝑚𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
(1 𝑜𝑜 𝐶𝐶 )
∆𝑇𝑇
=
= 2.42 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝐾𝐾𝑏𝑏 �0.413 𝐾𝐾 �
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑛𝑛𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚) = 𝑚𝑚 𝑇𝑇𝑇𝑇𝑇𝑇 (𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔)𝒎𝒎𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 �
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
2.42 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔� = (40 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔) �
�
1000
1000 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
𝑛𝑛𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚) = 0.096 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
The molar mass of the unknown is
𝑀𝑀𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
5 𝑔𝑔
𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔
= 52
0.096 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
𝑀𝑀𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =______________________.
17. An non-ionic solution of has a rise in height of 0.1 cm in the molar mass osmometer.
A. Calculate the concentration of the solute.
Solution:
kg
m
−3
Π
ρgh �1000 m3 � �9.8 s2 � (10 m)
𝑐𝑐 =
=
=
= 3.9 × 10−3 mol/L
J
RT RT
� (298 K)
�8.31
mol − K
Concentration = _____________________________.
B. The technician dissolved 5 mg./mL to obtain this result. What is the molar mass of the ionic
solute.
Solution:
5 𝑔𝑔/𝐿𝐿
𝑀𝑀𝑚𝑚 =
= 1,266 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚
3.9 × 10−3 mol/L
𝑀𝑀𝑚𝑚 =__________________________.
10
18. A solution is prepared by dissolving 40.5 g of LiF in 225 g of water. If the vapor pressure of
pure water is 21.8 torr at this temperature, what is the vapor pressure of the solution?
Solution: We need to calculate the mole fraction of LiF in order to obtain the vapor
pressure lowering by Raoult’s law.
40.5 𝑔𝑔
𝑛𝑛𝐿𝐿𝐿𝐿𝐿𝐿 =
= 1.56 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
25.94 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚
However, we note that the actual number of moles of the ionic solution formed by Li+
and F- is twice as great as 𝑛𝑛𝐿𝐿𝐿𝐿𝐿𝐿 so 𝑛𝑛𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝐿𝐿𝐿𝐿𝐿𝐿 = 3.12 moles
𝑛𝑛𝐻𝐻2𝑂𝑂 =
The mole fraction is
𝑥𝑥𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝐿𝐿𝐿𝐿𝐿𝐿
=
225 𝑔𝑔
= 12.5 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
18 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚
𝑛𝑛𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝐿𝐿𝐿𝐿𝐿𝐿
=
3.12
= 0.2
12.5 + 3.12
𝑛𝑛𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝐿𝐿𝐿𝐿𝐿𝐿 + 𝑛𝑛𝐻𝐻2𝑂𝑂
The vapor pressure of the water solvent is
∗
= 0.8(21.8 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇) = 17.4 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝑃𝑃𝐻𝐻2𝑂𝑂 = 𝑥𝑥𝐻𝐻2𝑂𝑂 𝑃𝑃𝐻𝐻2𝑂𝑂
vapor pressure =
19. Determine the entropy and average energy of each of the systems shown in the figure below.
Give the energy in the units of the systems. We can just call them energy units. The entropy
can be calculated using
𝑆𝑆 = 𝑘𝑘𝐵𝐵 𝑙𝑙𝑙𝑙𝑙𝑙
Solution: There are 14 particles in total. The entropy is given by
𝑆𝑆 = 𝑘𝑘𝐵𝐵 𝑙𝑙𝑙𝑙𝑙𝑙
For system A we have
𝑁𝑁!
14!
𝑊𝑊 =
=
= 2 × 1011
𝑛𝑛0 ! 𝑛𝑛2 ! 𝑛𝑛4 ! 𝑛𝑛6 ! 5! 4! 3! 2!
11
For system B we have
𝑊𝑊 =
𝑁𝑁!
14!
=
= 2 × 1011
𝑛𝑛0 ! 𝑛𝑛1 ! 𝑛𝑛2 ! 𝑛𝑛3 ! … 5! 4! 3! 2!
The entropies are the same since the two distributions are identical. The fact that the energies are
different does not affect the entropy. For both systems
𝐽𝐽
𝐽𝐽
𝑆𝑆 = �1.38 × 10−23 � ln(2 × 1011 ) = 3,5 × 10−22
𝐾𝐾
𝐾𝐾
The averages are
𝐸𝐸𝐴𝐴 = 5(0) + 4(2) + 3(4) + 2(6) = 32
𝐸𝐸𝐵𝐵 = 5(0) + 4(1) + 3(2) + 2(3) = 16
S for system A = __________________________.
S for system B = __________________________.
E for system A = __________________________.
E for system B = __________________________.
20. Consider the following reaction:
2 HBr(g) + Cl2(g)  2 HCl(g) + Br2(g)
At the reaction temperature of 320 K, K = 250.
During the reaction, the reaction mixture was sampled and the following pressures were found:
P(HBr) = 0.15 atm
P(Cl2) = 0.020 atm P(HCl) = 0.48 atm
P(Br2) = 1.00 atm
a) What is the value of Q?
𝑄𝑄 =
𝑃𝑃𝐵𝐵𝑟𝑟2 𝑃𝑃𝐻𝐻𝐻𝐻𝐻𝐻 2
𝑃𝑃𝐶𝐶𝑙𝑙2 𝑃𝑃𝐻𝐻𝐻𝐻𝐻𝐻 2
(1)(0.48)2
𝑄𝑄 =
= 512
(0.02)(0.15)2
12
Q=_____
b) What is the value of ∆G at the reaction temperature?
∆𝐺𝐺 𝑜𝑜 = −𝑅𝑅𝑅𝑅 ln 𝐾𝐾
∆𝐺𝐺 𝑜𝑜 = −(8.31)(320) ln(250) = −14,630 𝐽𝐽/𝑚𝑚𝑚𝑚𝑚𝑚
∆𝐺𝐺 = ∆𝐺𝐺 𝑜𝑜 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑄𝑄 = −14,630 + (8.31)(320) ln(512) = 1,960 𝐽𝐽/𝑚𝑚𝑚𝑚𝑚𝑚
∆𝐺𝐺 =
21. Use the following data in the table to find the rate law for the reaction,
𝑑𝑑[𝐴𝐴]
= −𝑘𝑘[𝐴𝐴]𝑛𝑛
𝑑𝑑𝑑𝑑
Solution:
Time (sec)
0
10
20
30
40
50
60
70
kJ
[A](M)
0.522
0.366
0.258
0.181
0.128
0.090
0.063
0.045
ln([A])
-0.650088
-1.00512
-1.3548
-1.70926
-2.05573
-2.40795
-2.76462
-3.10109
1/[A]
1.91571
2.73224
3.87597
5.52486
7.8125
11.1111
15.873
22.2222
If we compare these two we can see that the ln([A]) plot is linear in time. This is shown below.
You do not necessarily have to graph these to tell that one is linear and one is not. Since
the time increments are regular (10 seconds for each data point), you can inspect the first and last
data points to see that the change is nearly the same for ln([A]), but quite different for 1/[A].
Based on the model the process is first order (n = 1). Recall that a first-order process has
an exponential time decay,
[𝐴𝐴] = [𝐴𝐴]0 𝑒𝑒 −𝑘𝑘𝑘𝑘
While a second order process is hyperbolic
[𝐴𝐴]0
[𝐴𝐴] =
1 + [𝐴𝐴]0 𝑘𝑘𝑘𝑘
n = ______________________________.
13
22. For the reaction of propene with benzene to make cumene, the rate constant is observed to
increase from k = 10-4 M-1s-1 to k = 3 x 10-4 M-1s-1 at 25 oC and 45 oC, respectively. Determine
the Arrhenius parameters A and Ea for this process.
Solution: The temperatures in Kelvin are 298 K and 318 K. The activation energy is given by
𝑘𝑘2
�
(8.31)ln⁡
(3)
𝑘𝑘1
𝐸𝐸𝑎𝑎 = −
=−
= 43.3 𝑘𝑘𝑘𝑘/𝑚𝑚𝑚𝑚𝑚𝑚
1
1
1
1
�318 − 298�
�𝑇𝑇 − 𝑇𝑇 �
2
1
We can obtain A at either temperature. We will use 298 K.
𝑅𝑅𝑅𝑅𝑅𝑅 �
𝐴𝐴 =
𝑘𝑘
𝐸𝐸𝑎𝑎
𝑒𝑒𝑒𝑒𝑒𝑒 �− 𝑅𝑅𝑅𝑅
�
=
10−4
= 3900 𝑀𝑀−1 𝑠𝑠 −1
43300
𝑒𝑒𝑒𝑒𝑒𝑒 �−
�
(8.31)(298)
𝐸𝐸𝑎𝑎 =____________________________.
A = _____________________________.
23. The first-order decomposition of a colored chemical species, X, into colorless products is
monitered with a spectrophotometer by measuring changes in absorbance over time. Species X
has a molar absorptivity constant of 5 x 103 M-1cm–1 and the pathlength of the cuvette containing
the reaction mixture has a path length of 1.00 cm. The data from the experiment are given in the
table below.
[X](µM)
?
40.00
30.00
15.0
Absorbance
0.600
0.200
0.150
0.075
Time (min)
0.0
35.0
44.2
?
A.Calculate the initial concentration of the unknown species.
Solution: Use Beer’s law
𝐴𝐴
0.6
=
= 1.2 × 10−4 𝑀𝑀 = 120 𝜇𝜇𝜇𝜇
𝜀𝜀ℓ 5000(1)
You could also do this by noting that concentration depends linearly on absorbance. Since the
absorbance is three times as large at the initial time point as at 35 min. then the concentration
must also be three times as large.
𝑐𝑐 =
concentration = _________________________________.
14
B. Calculate the rate constant for the first order reaction using the values given for concentration
and time. Include units with your answers.
Solution: We use the first order rate law and solve for the rate constant, k
𝑘𝑘 = −
𝑙𝑙𝑙𝑙 �
[𝐴𝐴]
1
�
𝑙𝑙𝑙𝑙 �3�
[𝐴𝐴]0
=−
= 0.0313 𝑚𝑚𝑚𝑚𝑚𝑚−1
35 𝑚𝑚𝑚𝑚𝑚𝑚.
𝑡𝑡
k = _____________________________.
C. Calculate the number of minutes it takes for the absorbance to drop from 0.600 to 0.075.
Solution: Use the rate law and known rate constant to solve for the unknown time in this part.
[𝐴𝐴]
0.075
�
𝑙𝑙𝑙𝑙 �0.600�
[𝐴𝐴]0
=−
= 66.4 𝑚𝑚𝑚𝑚𝑚𝑚.
𝑡𝑡 = −
0.0313 𝑚𝑚𝑚𝑚𝑚𝑚−1
𝑘𝑘
t = ______________________________,
D. Calculate the half-life of the reaction. Include units with your answer.
Solution: Use the half-life formula
ln⁡
(2)
0.697
𝜏𝜏1/2 =
=
= 22.2 𝑚𝑚𝑚𝑚𝑚𝑚.
k
0.0313 𝑚𝑚𝑚𝑚𝑚𝑚−1
𝜏𝜏1/2 =________________________________.
𝑙𝑙𝑙𝑙 �
15