Chemistry II
Midterm Exam
22 April, 2011
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. You have a 10.40-g mixture of table sugar (C12H22O11) and table salt (NaCl). When
this mixture is dissolved in 150. g of water, the freezing point is found to be
-2.24°C. Kf of water is 1.89 K·kg/mol, Kb of water is 0.51 K·kg/mol. Assume
that the density of the solution is 1.143 g/mL.
(a). What is the molality of all solutes in this solution? (2%)
(b). What is the boiling point of the solution? (2%)
(c). Calculate the percent by mass of sugar in the original mixture. (4%)
(d). What is the molarity of all solutes in this solution? (4%)
(e). What is the osmotic pressure of the solution at 25℃? (4%)
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2. The vapor pressure of H2O doubles when the temperature is raised from 50℃ to
64.26℃. Assume Δ Hovap and Δ Sovap of water are constant.
(a). Please calculateΔ Hovap of water. (4%)
(b). The vapor pressure of water at 25℃ is 23.76 torr. Please calculateΔ Sovap
of water. (4%)
(c). The pressure of a pressure cooker is 2.5 bar at cooking. Please find the
boiling temperature of pure H2O in the pressure cooker. (4%)
3. A solution contains two volatile liquids A and B. Please complete the following
table, in which the symbol
indicates attractive intermolecular forces
Relationship of Attractive Forces
Deviation from Raoult’s Law Δ Hsol
(none, negative or positive)
(>0, <0, = 0)
(A A, B B ) >> A B
a (1%)
b (1%)
(A A, B B ) << A B
c (1%)
d (1%)
(A A, B B ) = A B
e (1%)
f (1%)
4. The equilibrium constant of reaction I2(g)
2I(g) is 6.8 at 1200K. The partial
pressure of I2(g) and I(g) in a reactor contains are 0.33 bar and 1.2 bar
respectively at 1200K.
(a). What is the reaction quotient? (2%)
(b). What is the spontaneous direction of the reaction? (3%)
(c). What is the standard reaction Gibbs free energy at 1200K? (4%)
(d). What is the reaction Gibbs free energy at 1200K? (4%)
(e). What is the total pressure at equilibrium? (4%)
5. An important reaction step used in the production of sulfuric acid is the oxidation
of sulfur dioxide to sulfur trioxide: 2SO2(g) + O2(g)
2SO3(g).
(a). The equilibrium constant K of reaction is
at 500K and
at 298 K. Is this a exothermic reaction or endothermic
reaction? (2%) Please explain it. (2%)
(b). If the reaction reaches equilibrium, what happens to the partial pressure of
SO3 when the partial pressure of SO2 is decreased? (2%)
(c). If the reaction reaches equilibrium, what happens to the partial pressure of
O2 when the partial pressure of SO3 is increased? (2%)
6. A reaction used in the production of gaseous fuels from coal, which is mainly
carbon, is C(s) + H2O(g)
CO(g) + H2(g).
(a) Evaluate K and KC at 900K, given that the standard Gibbs free energies of
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formation of H2O(g) and CO(g) at 900K are -198.08 kJ/mol and -191.28
kJ/mol, respectively. (4%)
(b) A 5.2 kg sample of graphite and 125 g of water were placed into a 10L
container and heated to 900K. What are the equilibrium concentrations?
(4%)
7. (8%) When iodide ions (I) react with iodate ions (IO3) in basic aqueous triiodide
ions (I3) are formed. Write the oxidation, reduction, and net ionic equations for
the reaction. (Hint: the same product is obtained in each half-reaction.)
8. (8%) Acidified aqueous permanganate (MnO4) solutions and acidified aqueous
dichromate (Cr2O7) solutions are powerful oxidizing agents. Suppose solutions of
the two reagents are prepared and serve the two half-cells in a galvanic cell with
platinum electrodes that generates a current in an external circuit. (a) Determine
the standard potential of the cell constructed from these half-cells. (b) Write the
net ionic equation for the cell having a positive standard potential. (c) Write the
cell diagram of the reaction.
9. (7%) A tin electrode in 0.015M Sn(NO3)2(aq) is connected to a hydrogen electrode
in which the pressure of H2 is 1.0 bar. (a) Write down the cell diagram of the
reaction and determine its standard cell potential. (b) If the potential is 0.061V at
25 C, what is the pH of the electrolyte at the hydrogen electrode?
10. (10%) A fuel cell generates electricity directly from a chemical reaction, as in a
battery, but uses reactants that are supplied continuously, as in an engine. (a)
Write the oxidation (anodic) and reduction (cathodic) half-reactions of a H2/O2
fuel cell under basic condition (electrolyte: KOH(aq)) and determine its standard
cell potential. (b) What is the maximum electrical work that it can perform at the
standard state condition by oxidation of 1 mol H2?
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Ans 1:
(a) The molality of all solutes = 2.24/1.89 =1.185 (2%)
(b) The boiling point of the solution = 100 + 1.185*0.51 = 100.6 ℃ (2%)
(c) Assume that there is y g of sucrose
y=5.7 g
the percent by mass of sugar in the original mixture = 5.7/10.4*100% = 54.8%
(4%)
(d) The molarity of all solutes =
(4%)
(e) the osmotic pressure of the solution = 1.267*0.0821*298 = 30.99 atm (or =
1.297*0.08314 = 31.38 bar) (4%)
Ans 2:
(a) Using Clausius-Clapeyron Equation, we can get
Δ Hovap = 44023 J/mol ≈ 44 kJ/mol
(b) 8.314 J  K 1  mol1  ln
(4%)
23.76 Torr
44023

 S vap
750.06 Torr
298 K
Δ Sovap = 119.027 J/K/mol
(4%)
or
8.314 J  K 1  mol1  ln
23.76 Torr
44023

 S vap
760 Torr
298 K
Δ Sovap = 118.92 J/K/mol
(4%)
(c) Assume the boiling temperature of pure H2O in the pressure cooker is T2. Using
Clausius-Clapeyron Equation, we can get
T2. = 395.15K = 122.15 ℃
(4%)
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Or
T2. = 395.54K = 122.54 ℃
(4%)
Ans 3:
(a) Positive (1%)
(b) Δ Hsol>0 (1%)
(c) Negative (1%)
(e) none (1%)
(d) Δ Hsol<0 (1%)
(f) Δ Hsol=0 (1%)
Ans 4:
(c) Q=1.22/0.33 = 4.36 (2%,
如果將 Q 寫出單位以 0 分計算)
(d) Because Q(=4.36) < K(=6.8), forward (3%)
(e) ΔGor = -8.314*1200*ln6.8 = -19124.8 J/mol ≈ -19.12 kJ/mol (4%)
(f) ΔGr =ΔGor +8.314*1200*ln4.36 = -4434 J/mol ≈ -4.4 kJ/mol (4%)
(g)
I2(g)
 2I(g)
0.33
0.93
-x
1.2
0
+2x
0.93-x
+2x
2
K = (2x) /(0.93-x)=6.8
X = 0.668 or -2.3677 (meaningless)
Partial pressure of I2 is 0.93-0.668=0.262
Partial pressure of I is 2*0.668=1.336
Total pressure = 1.336+0.262 = 1.598 bar (4%)
Ans 5:
(a) Exothermic reaction (2%)
because K at 500 is smaller than K at 298K (2%)
(b) decrease (2%)
(c) increase (2%)
Ans 6:
(a) Gro =  RT lnK;
lnK = 
Gro
RT
Gro = (-191.28 kJ ·mol-1) – (-198.08 kJ ·mol-1) = 6.8 kJ ·mol-1
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
lnK =
K
6800 J  mol 1
= - 0.909;
(8.314 J  K 1  mol 1 )(900 K )
如果將 K 寫出單位以 0 分計算)
= 0.403 (2%,
KC
=
K(RT)-1 = 0.403/(0.008314*900) = 0.005386
(2%, 如果將 Kc 寫出
單位以 0 分計算)
(b) 5.20 103 g C 
1mol C
= 433. mol C
12.011g C
125 g H2O 
1mol H 2O
=
18.016g H 2O
6.94 mol H2O
H2O is limiting. Conc. of H2O = 6.94 mol/10.0 L = 0.694 mol·L-1
Concentration (mol  L1 )
initial
H2O (g)
CO(g)
0.694
H2(g)
0
0
change
x
x
x
final
0.694 - x
x
x
COH 2 
H 2O
KC =
=
( x)( x)
= 0.005386
(0.694  x)
x2
=
x2
+ 0.005386 x - 0.00374 = 0
0.00374 – 0.005386x;
x = + 0.0585 or  0.0639 (not meaningful)
[CO] = [H2] = 0.0585 M
[H2O] = 0.694 - 0.0585 = 0.636 M (4%)
Ans 7:
Let’s start with I + IO3I3 (unbalanced):
(1) Oxidation: 3I I3Iare oxidized to I2+ I3I I3 + 2e
(2) Reduction: 3IO3I3I are reduced to I2 + I
3IO3eI33IO3H2O + eI3OH
(3) Net equation = (1)x8 + (2)
24I + 3IO3H2O I3OH
The stoichiometric numbers are divided by 3 and put the phase notation
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for each species in the chemical equation.
8Iaq + IO3aqH2Ol I3aqOHaq
Ans 8:
Example 13.7
(a) MnO4aq + 8Haqe Mn2+aqH2OlE0 = +1.51 V
Cr2O7aq + 14Haqe Cr3+aqH2OlE0 = +1.33 V
E0(cell) = E0(R)- E0(L) = 1.51 - 1.33 = +0.18 V
(3%)
(b) 6MnO4aq + 48Haqe Mn2+aqH2Ol
Cr3+aqH2OlCr2O7aq + 70Haqe
6MnO4aq + 11 H2OlCr3+aqMn2+aqHaqCr2O7aq
 
(c) Pt(s)Cr2O7aqCr3+aqHaq∥HaqMnO4aqMn2+aqPt(s) (2%)
Ans 9:
Exercise 13.45
(a) Sn(s)Sn2+(aq,∥H(aq,pH =?H2 (g, 1.0 bar)Pt(s)
E0(cell) = E0(R)- E0(L) = 0 – (-0.14) = +0.14 V
0
2+
(3 %)
+ 2
(b) Nernst equation: E = E – 0.059/2log([Sn ]/[H ] )
log([Sn2+])  0.059pH
pH = 2.25 (4%)
Ans 10:
Box 13.1
(a) Anode: 2H2(g) + 4OH(aq) le
Cathode: O2(g) el4OH(aq) (2%)
Net reaction: 2H2(g) + O2(g) l
E0(cell) = E0(R)- E0(L) = 0.4 – (-0.83) = +1.23 V (2%直接寫出答案者不給分)
(b) we,max = G0 = nFE0 = 4964851.23 = 474.7 kJ
Note that this free energy is produced by oxidation of 2 moles of H2.
Let’s rewrite the chemical equation as H2(g) + 1/2O2(g) l, the
maximum work can be performed by the fuel cell is 237 kJ per mol of
hydrogen. (4%)
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