Chemistry II
Midterm Exam (Total Score: 105 points)
19 April, 2013
Periodic Table of Elements
1
2
H
He
1.0
4.0
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.9
9.0
10.8
12.0
14.0
16.0
19.0
20.2
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
23.0
24.3
27.0
28.1
31.0
32.1
35.5
40.0
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.1
40.1
45.0
47.9
50.9
52.0
54.9
55.8
58.9
58.7
63.5
65.4
69.7
72.6
74.9
79.0
79.9
83.8
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.5
87.6
88.9
91.2
92.9
95.9
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
(98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
132.9 137.3 138.9 178.5 181.0 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222)
Constants
R = 8.314 J/mol·K = 0.08314 L·bar/K·mol = 0.0821 L·atm/K·mol = 8.314 L·kPa/K·mol
1 bar = 750.06 torr = 0.9869 atm
F = 9.6485×104 C/mol
1.
Consider an apparatus shown below, in which A and B are two 1.00-L spherical containers
joined by a stopcock C.
The volume of the stopcock is negligible. Initially, A and B are evacuated, the stopcock C is
1
closed, and 1.50 g of diethyl ether, C2H5OC2H5, is introduced into container A. The vapor
pressure of diethyl ether is 57 Torr at –45°C and 534 Torr at 25°C. Assume ideal behavior.
(a) (4 points) If the stopcock is left closed and the container A is brought to equilibrium at 25°C,
what will be the pressure of diethyl ether in container A?
(b) (4 points) If the temperature of the assembly is cooled to –45°C and the stopcock C is
opened, what will be the pressure of diethyl ether in the apparatus?
(c) (6 points) Please calculate the normal boiling point of diethyl ether.
2.
A 1.00 points by mass MgSO4(aq) solution has a freezing point of -0.192℃.
𝑘𝑓 of water is 1.86 K ∙ kg/mol.
(a) (3 points) Estimate the van’t Hoff i factor from the data.
(b) (3 points) Determine the total molality of all solute species including undissociated and
dissociated solutes.
(c) (5 points) Calculate the percentage dissociation of MgSO4(aq) in this solution.
3.
The temperature-composition diagram of the mixture of chloroform, CHCl3, and acetone,
(CH3)2CO, is shown below. Please answer
(a)
(2 points) Apparently, the mixture of chloroform and acetone is not an ideal solution. Does
this mixture show a positive deviation or a negative deviation from Raoult’s law?
(b)
(2 points) Identify the kinds of possible intermolecular forces in the liquid of chloroform.
(c)
(2 points) Identify the kinds of possible intermolecular forces in the liquid of acetone.
(d)
(5 points) Identify the kinds of possible intermolecular forces between chloroform and
acetone. And explain the deviation of the mixture from the ideal solution.
2
4.
Consider a simple reaction, A  B , with an equilibrium constant K =
aB
to answer the
aA
following questions.
(a)
(5 points) If the molar Gibbs free energy of a substance J (= A or B) can be expressed a
Gm (J)  Gmo (J)  RT ln aJ
where the activity aJ is defined as in the text book, derive an equation to express the relation
between Gro ( Gmo (B)  Gmo (A)) and K in terms of RT.
(b)
(6 points) Assuming that the equilibrium constants at T1 and T2 are K1 and K2, respectively,
derive the van’t Hoff equation based on Gr,o i  H r,o i  Ti Sr,o i with i = 1 or 2 . (assuming
also that H ro and Sro are both independent of temperature over the range of temperatures
of interest)
(c)
(4 points) Use van’t Hoff equation derived above to explain Le Chatelier’s principle for the
effect of temperature on an equilibrium in two cases: endothermic and exothermic processes.
(d)
(5 points) By combining the relation for KC in terms of K with the van’t Hoff equation, find
n
 c o RT 
the analog of the van’t Hoff equation for KC. (Hint: K   o  K C )
 P 
(e)
(5 points) Think of the vaporization of a liquid as an equilibrium reaction, e.g., for substance
A, A(𝑙) ↔ A(𝑔). This reaction has an equilibrium constant expression K  PA because PA
is the pressure of the gas above a 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. How does the vapor pressure of a liquid vary with
temperature?
5.
(4 points) At 400 C, the equilibrium constant K for the reaction 2SO2 (𝑔) + O2 (𝑔) →
2SO3 (𝑔) is 3.1  104, determine the value for KC at this temperature.
6.
(4 points) If Q = 1.0  1040 for the reaction C(s) + O2 (g) → CO2 (g) at 25 ℃, will the reaction
have a tendancy to form products or reactants or will it be at equilibrium?
(Hint: ∆𝐺𝑓° (CO2 , g) = −394.36 kJ ∙ mol−1 )
3
7.
A student constructed an electrochemical cell from a piece of metallic Zn, a piece of metallic
Cu, and a lemon which contained 4.7 g of citric acid in 100 mL of its juice. When the Zn metal
was replaced by a piece of Mg, the measured voltage increased. On the other hand, the
measured voltage did not vary when the Cu metal was changed in to a piece of graphite. Answer
the following questions.
(a)
(6 points) Identify the redox couples, the anode, and the cathode for this cell.
(b)
(2 points) Draw a symbolic cell diagram for the device using IUPAC convention.
(c)
(6 points) Write down the reduction half-reactions and the standard potentials for the anode
and the cathode. What is the standard cell potential Eocell?
(d)
(8 points) By assuming that each citric acid molecule releases one proton, the concentration
of Zn2+(aq) is 1.0 x 10-9 M, and the concentrations of Cu ions are negligible initially,
calculate the cell potential E. Assume that the pressure of H2 is 1 bar. (Show your step-by
step calculations)
(e)
(4 points) Do you expect the cell potential E will decrease or increase after the cell is
operated for a period of time? Explain why.
References:
Citric acid:
8.
C6H8O7, M.W. 192.124 g/mol
(10 points) By using the standard potential table (see page 5), please find the data of both
Eo (Au3+/ Au) and Eo (Au+/ Au). Then, write down all related half-reactions and calculate the
standard potential Eo for Au3+/ Au+.
4
Standard Potential Table (at 25℃)
5
Answers 1 (14 points total)
(a)
534
mass of diethyl ether
(760 ) (1) = (
74.12
) × 0.082 × 298
Mass of diethyl ether = 2.13 g > 1.50 g. (2 points)
In this case, there is not enough diethyl ether to achieve equilibrium so all of the ether will
vaporize. The pressure will be
1.5
) × 0.082 × 298
74.12
P = 0.495 atm (or 376 Torr) (2 points) (4 points total)
P×1=(
(b)
(
57
760
mass of diethyl ether
) (2) = (
74.12
) × 0.082 × 228
Mass of diethyl ether = 0.594 g < 1.50 g. (2 points)
Because there are 1.50 g of diethyl ether, the vapor pressure will be achieved, the pressure will
(c)
be 57 Torr. (2 points) (4 points total)
First, the enthalpy of vaporization could be obtained by using the Clausius-Clapeyron
Equation
°
−∆𝐻𝑉𝑎𝑝
57
1
1
ln (
)=
(
−
)
543
8.314 273 − 45 273 + 25
57
1
1
°
∆𝐻𝑉𝑎𝑝
= −8.314 × ln (
)÷(
−
)
543
228 298
J
°
∆𝐻𝑉𝑎𝑝
= −18190 mol (3 points)
The normal boiling point of diethyl ether is the temperature at which the vapor pressure of
diethyl ether is 760 torr.
760
−18190 1
1
ln (
)=
(
−
)
543
8.314 𝑇𝑛𝑏 298
1
1
8.314
760
=
−
ln (
) = 0.003202
𝑇𝑛𝑏 298 18190
543
𝑇𝑛𝑏 = 312.3 𝐾 = 39.3 ℃ (3 points) (6 points total)
6
Answers 2 (11 points total)
(a) A 1.00 points aqueous solution of MgSO4 will contain 1.00 g of MgSO4 for 99.0 g of water. To
use the freezing point depression equation, we need the molality of the solution.


1.00 g

1 
120.37 g  mol 
molality  
 0.0839 mol  kg 1
0.0990 kg
∆𝑇𝑓 = 𝑖𝑘𝑓 𝑚 = 𝑖(1.86)(0.0839) = 0.192
i = 1.23 (3 points)
(b) molality of all solute species (undissociated MgSO4(aq) plus
Mg2+(aq)  SO4 2 (aq))  1.23  0.0839 mol  kg 1  0.103 mol  kg 1 (3 points)
(c) If all the MgSO4 had dissociated, the total molality in solution would have been
0.168 mol  kg 1 , giving an i value equal to 2. If no dissociation had taken place, the molality in
solution would have equaled 0.0839 mol  kg 1.
MgSO4 (𝑎𝑞)
↔
0.0839 mol  kg 1  x
Mg 2+ (𝑎𝑞) +
𝑆𝑂42− (𝑎𝑞)
x
x
Total molality = (0.0839-x)+x+x = 0.103 (2 points)
0.0839 + x = 0.103

x=0.019
Percentage of dissociation = 0.019/0.0839 * 100% = 23% (3 points) (5 points total)
Answers 3 (11 points total)
(a) A negative deviation from Raoult’s law. (2 points)
(b) London force, dipole-dipole (2 points)
(c) London force, dipole-dipole (2 points)
(d) 以下兩組答案皆對
London force, dipole-dipole, hydrogen bonding (3 points)
或是 London force, strong dipole-dipole(3 points)
Because the intermolecular forces between chloroform and acetone are stronger than either pure
chloroform or pure acetone, the mixtures of chloroform and acetone would show a negative
deviation from Raoult’s law. (2 points)
7
Answers 4 (25 points total)
(a) (5 points)
Gm (A)  Gmo (A)  RT ln aA
Gm (B)  Gmo (B)  RT ln aB
Gr  Gm (B)  Gm (A)  Gmo (B)  RT ln aB   Gmo (A)  RT ln aA 
 Gmo (B)  Gmo (A)   RT ln
aB
a
 Gro  RT ln B
aA
aA
At equilibrium when Gr  0 and K =
aB
, then Gro   RT ln K .
aA
(b) (6 points)
 RT1 ln K1  Gr,o 1  H r,o 1  T1Sr,o 1
 RT2 ln K 2  Gr,o 2  H r,o 2  T2 S r,o 2
o
o
o
o
o
o
1  Gr, 1 Gr, 2 
1  H r, 1  T1S r, 1 H r, 2  T2 S r, 2 
ln K1  ln K 2   


 

R  T1
T2 
R 
T1
T2


ln
o
o

H ro  1 1 
1  H r, 1 H r, 2 
o
o



S


S


  r, 1

  
r, 2  
R  T1
T2 
R  T1 T2 

K1
H ro

K2
R
1 1 
K 2 H ro  1 1 

or
ln



  
T
T
K
R
 1
2
1
 T1 T2 
(c) (4 points) For endothermic condition, H ro > 0, if T2 > T1, then ln
K1
> 1 or K2 > K1, this implies
K2
that an increase in temperature favors the formation of product for an endothermic reaction (亦
K
即升溫有助於吸熱反應). For exothermic condition, H ro < 0, if T2 < T1, then ln 1 > 1 or K2 >
K2
K1, this implies that a decrese in temperature favors the formation of product for an exothermic
reaction (亦即降溫有助於放熱反應).
(d) (5 points)
K 2 H r   1 1 
ln

  
K1
R  T1 T2 
  RT2 n K c 2  H   1 1 
r
ln 

  
  RT n K 
R
 T1 T2 
1
c1 

8
 K c 2  H r   1 1 
 T2 
n ln    ln 

  
R  T1 T2 
 T1 
 K c1 
 K  H r   1 1 
 T2 
ln  c 2  
    n ln  
R  T1 T2 
 T1 
 K c1 
(e) (5 points)
K 2 H r  1 1 

  
K1
R  T1 T2 
P H vap  1 1 
 ln 2 
  
P1
R  T1 T2 
ln
where P1 and P2 are vapor pressures of A in T1 and T2 , respectively, and H vap is the standard
enthalpy of vaporization.
Answers 5 (4 points)
n
 c o RT 
K   o  K C with c o  1 mol L1 and P o  1 bar
 P 
Po
1

 12.03
o
c R 1 0.083145
n
 T 
 K 
 KC
 12.03 
n
1
 12.03 
 12.03 
4
6
 KC  
 K 
  3.1 10  1.73  10
 T 
 673 
Answers 6 (4 points)
G   RT ln K
G
ln K  
RT
394.36  103 J  mol1
ln K  
 159.17
(8.314 J  K 1  mol1 )(298 K)
K ~ 1  1069
Because Q < K, the reaction will tend to proceed to produce products.
Answers 7 (26 points total)
In the lemon cell, Zn oxidized while H+(aq) reduced. Cu did not participate in the reaction because
there were no Cun+ ions in the lemon. This was also shown by the fact that the measured voltage
did not change when Cu was changed in to graphite. If a student uses Cu 2+/Cu instead of H+/H2 for
the calculations, the maximum points he/she can get is 10.
9
(a) Redox couples: Zn2+/Zn, H+/H2 (2 points each, 4 points total)
Anode: Zn, cathode: Cu (1 point each, 2 points total)
(b) 以下兩組答案皆對
Zn(s) | Zn2+(aq) || H+(aq) | H2(g) | Cu(s) (2 point)
或是
Zn(s) | Zn2+(aq) | H+(aq) | H2(g) | Cu(s) (2 point)
(c) Anode
Eo = -0.76V
Cathode
Eo = 0 V (1 point each, 4 points total)
Eocell = 0.76 V (2 points)
(d) Use Nernst equations: E = Eo – (0.025693 V /n) ln Q
E = E o – (0.05916 V /n) log Q (1 point)
or
[Zn2+(aq)] = 1.0 x 10-9 M (1 point)
[H+(aq)] = (4.7 g / 100 mL) x (1 mol / 192.124 g) x ( 1000 mL / 1 L)
= 0.2446 mol/L = 2.446 x 10-1 M (2 points)
Q = [Zn2+(aq)] / [H+(aq)]2 = 1.0 x 10-9 / (2.446 x 10-1)2 (1 point)
n = 2 (1 point)
E
= 0.76 V - (0.025693 V /2) ln 1.0 x 10-9 / (2.446 x 10-1)2
= 0.76 V - (-0.23 V) = 0.99 V (2 points) (8 points total)
(e) E will decrease because Q will be increased by the following reasons. (2 points) After a period of
time, the concentration of Zn2+(aq) will increase (1 point) while the concentration of H+(aq) will
decrease. (1 point) (4 points total)
Answers 8 (10 points total)
To calculate the standard potential Eo, it has to be converted into Go , Go = -nFEo
For (Au3+/ Au),
Au3+(aq) + 3 e- → Au(s) EoA = 1.4 V (1 point)
For (Au+/ Au),
Au+(aq) +
e- → Au(s)
EoB = 1.69 V
For (Au3+/ Au+),
Au3+(aq) +
For (Au3+/ Au),
For (Au+/ Au),
For (Au3+/ Au+),
Au3+(aq) + 3 e- → Au(s) the free energy change is GoA (1 point)
Au+(aq) + e- → Au(s) the free energy change is GoB (1 point)
Au3+(aq) + 2 e- → Au+(aq) the free energy change is GoC (1 point)
2 e- → Au+(aq)
(1 point)
EoC = ? (1 point)
GoA = GoB + GoC (2 points)
-nAF EoA = (-nBF EoB) + (-nCF EoC)
-3 EoA = (-1 EoB) + (-2 EoC)
-3 x 1.4 V = (-1 x 1.69 V) + (-2 x EoC)
10
EoC
= (4.2 V + 1.69 V ) / 2 = 1.255 V
(2 points)
11