2013 FALL Semester Midterm Examination
For General Chemistry II (CH103)
Date: October 24 (Thu),
Time Limit: 14:30 ~ 17:00
Write down your information neatly in the space provided below; print your Student ID in the upper
right corner of every page.
Professor
Name
Class
Problem
points
1
2
3
4
5
6
Student I.D. Number
Problem
/13
/6
/14
/7
/13
Name
TOTAL pts
points
7
8
9
10
11
/12
/9
/6
/9
/100
/5
/6
** This paper consists of 13 sheets with 11 problems (page 11: fundamental constants, page 12: claim form,
page 13: periodic table). Please check all page numbers before taking the exam. Write down your work and
answers in the sheet.
Please write down the unit of your answer when applicable. You will get 30% deduction for a missing unit.
NOTICE: SCHEDULES on RETURN and CLAIM of the MARKED EXAM PAPER.
(채점답안지 분배 및 이의신청 일정)
1. Period, Location, and Procedure
1) Return and Claim Period: October 28 (Mon, 6:30 ~ 7:30 p.m.)
2) Location: Room for quiz session
3)
Procedure:
Rule 1: Students cannot bring their own writing tools into the room. (Use a pen only provided by TA)
Rule 2: With or without claim, you must submit the paper back to TA. (Do not go out of the room with it)
If you have any claims on it, you can submit the claim paper with your opinion. After writing your opinions on the
claim form, attach it to your mid-term paper with a stapler. Give them to TA.
2. Final Confirmation
1) Period: October 31 (Thu) – November 1 (Fri)
2) Procedure: During this period, you can check the final score of the examination on the website.
** For further information, please visit General Chemistry website at www.gencheminkaist.pe.kr.
1
<The Answers>
Problem
1
2
3
4
5
6
1. (Total 13 pts)
(a) (6 pts)
points
6+3+4/13
3+3/6
4+4+6/14
1+3+3/7
3+3+7/13
3+3/6
Problem
7
8
9
10
11
TOTAL pts
points
6+6/12
3+3+3/9
3+3/6
2+4+3/9
5/5
/100
(b) (3 pts)
(c) (4 pts)
2. (Total 6 pts)
(a) (3 pts) At phase transition, temperature (T) does not change (constant), and the total
heat needed to change the phase is q. Then, ∆So vap = ∆Ho vap / T b because of ∆S = q/T.
CH 3 CH 3 ,
CH 3 F
CH 3 OH
∆Ho vap (kcal/mol)
3.51
3.99
8.42
T b (K)
184.5
194.8
337.7
∆So vap (kcal/K mol)
0.0190
0.0205
0.0249
(b) (3 pts) The liquid state with the most order will have the largest entropy change when
its order is disturbed. Thus methanol has the most ordered liquid state, and ethane has
the least ordered liquid state. Methanol (CH 3 OH) has the highest ∆So vap and T b due to
the hydrogen bonding of the alcohol group, and ethane (CH 3 CH 3 ) has only dispersion
forces.
3. (Total 14 pts)
(a) (4 pts) n(biphenyl) = 5.50 g/154.2g mol-1 = 0.0357 mol
m biphenyl) = 0.0357 mol / 0.1000kg = 0.357 mol kg-1
Kb = ΔTb/m = 0.903 K / 0.357 mol kg-1 = 2.53 K kg mol-1 for benzene
(b) (4 pts) m = ΔTb/Kb = 0.597 K / 2.53 K kg mol-1 = 0.236 mol kg-1
n = 0.236 mol kg-1 X 0.1500 kg = 0.0354 mol
M (molar mass of solute) = 6.30 g / 0.0354 mol = 178 g mol-1
(c) (6 pts)
4. (Total 7 pts)
(a) (1 pt) Q = 0, Q < K, so the forward reaction favored.
(b) (3 pts) NH 4 Cl(s)  NH 3 (g) + HCl(g)
Initial
0
0
Change
+x
+x
Equilibrium
+x
+x
-2
So, K = P NH3 P HCl = 1.04 X 10
x2 = 1.04 X 10-2; x = 0.102
Answer: P NH3 = P HCl = 0.102 atm.
(c) (3 pts) n NH3 = n HCl = 1.02 X 10-1 atm X 1.000L/0.08206 L atm mol-1 K-1 X 548.2K
= 2.268 X 10-3 mol
The mass of NH 4 Cl consumed = 2.268 X 10-3 mol X 53.49 g mol-1 = 0.121 g
So, the remaining mass = 0.859 g
Answer: NH4 Cl = 0.859 g
5. (Total 14 pts)
K2
∆H  1 1 
K
K1
 −  → ∆H 0 =
ln 2 = −
K1
R  T2 T1 
1 1
 − 
 T1 T2 
K
T1 = 298 K , T2 = 308 K , 2 = k
K1
R ln
0
∆H 0 =
8.3145 JK −1 ln k
= 76 kJ mol −1 × ln k
1 
 1
−


 298 K 308 K 
Therefore,
(a) (3 pts) k = 2, ΔH0 = 76 kJ mol-1 × ln2 = 53 kJ mol-1
(b) (3 pts) k = 1/2, ΔH0 = 76 kJ mol-1 × ln(1/2) = -53 kJ mol-1
(c) (7 pts) 2NO 2
N2O4
P NO2 = P – x and P N2O4 = x
K = x/(P- x)2 0≤ x ≤1
Using
ln
ln
K 2 = 6.96
x 2 – 2.1435 x + 1.00 = 0, x = 0.686
P N2O4 = 0.686 atm, P NO2 = 0.314 atm
6. (Total 6 pts)
Conjugate base is stabilized by inductive effect of electronegative element.
(a) (3 pts) The –CCl 3 group that is bonded to the carboxyl group, -COOH, in trichloroacetic acid, is
more electron withdrawing than the CH 3 group in acetic acid. Thus, trichloroacetic acid is the
stronger acid. CH3 COOH <
ClCH 2 COOH < Cl3 CCOOH
(b) (3 pts) More electrons are withdrawn from the XOH as the electronegativity of the halogen
increase.
ClOH > BrOH > IOH
7. (Total 12 pts)
(a) (6 pts)
H 2 A sc (aq) ⇄ H+(aq) + HA sc -(aq)
K a1 = 7.9×10-5
HA sc -(aq) ⇄ H+(aq) + A sc 2-(aq)
Starting
K a2 = 1.6×10-12
H 2 A sc (aq) ⇄ H+(aq) + HA sc -(aq)
Change
0.1
0
0
-x
+x
+x
x
x
Equilibrium 0.1-x
K a1 = [H+(aq)][HA sc -(aq)]/[H 2 A sc (aq)] = 7.9×10-5 = x2/(0.1 – x)
Assuming x to be much smaller than 0.1
x2/0.1 = 7.9×10-5
x = 0.0028
pH = -log[H+] = -log[0.0028] = 2.55
(b) (6 pts)
Starting
HA sc -(aq) ⇄ H+(aq) + A sc 2-(aq)
0.0028
0.0028
0
-y
+y
+y
Change
Equilibrium 0.0028-x 0.0028+x
y
K a2 = [H+(aq)][ A sc 2-(aq)]/[HA sc -(aq)] = 1.6×10-12
= (0.0028 + y)y/(0.0028 – y) ~ 0.0028y/0.0028 = y
[Asc 2-(aq)] = 1.6×10-12
8. (Total 9 pts)
For first ionization,
H 2 CO 3 (aq) + H 2 O(l)  HCO 3 -(aq) + H 3 O+(aq)
Initial
Change
0.034
-x
Equilibrium 0.034-x
0
+x
+x
x
x
So, Ka1 = x2/(0.034-x) = 4.3 X 10-7,
X = [H 3 O+] = [HCO 3 -] = 1.2 X 10-4 M
[H 2 CO 3 ] = 0.034 M
For second ionization,
~0
K a2 = 1.2 X 10-4 X [CO 3 2-]/1.2 X 10-4 = 4.8 X 10-11
[CO 3 2-] = 4.8 X 10-11 M
Answer: H2 CO 3 = 0.034 M, HCO 3 - = H 3 O+ = 1.2 X 10-4 M, CO 3 2- = 4.8 X 10-11 M
9. (Total 6 pts)
(a) (3 pts) K sp = [Zn2+][OH-]2 4.5 x 10-17 = (S)(2S)2
4S3 = 4.5 x 10-17, [Zn2+] = 2.2 x 10-6 M, 2.2 x 10-6 x 99.4 g/mol x 0.5 L = 1.09 x 10-4 g
(b) (3 pts) At pH = 6, [OH-] = 1 x 10-8 M
[Zn2+] = K sp /[OH-]2 = (4.5 X 10-17)/ (1 x 10-8) 2 = 0.45 M
At pH= 6.00, Zn(OH) 2 is far more soluble.
10. (Total 9 pts)
(a) (2 pts)
(b) (4 pts)
(c) (3 pts)
11. (5 pts)
1. (Total 13 pts) The normal boiling point of iodomethane, CH 3 I, is 42.43 oC, and its vapor pressure
at 0.00 oC is 140 Torr.
(a) (6 pts) Calculate the standard enthalpy of vaporization of iodomethane.
(Answer)
(b) (3 pts) Calculate the standard entropy of vaporization of iodomethane.
(Answer)
(c) (4 pts) Calculate the vapor pressure of iodomethane at 25.0 oC.
(Answer)
2
2. (Total 6 pts)
(a) (3 pts) Compare ∆So vap for CH 3 CH 3 , CH 3 F, and CH 3 OH as calculated from ∆Ho vap and T b
(boiling temperature) of the compounds.
∆H
o
-1
vap
(kcal mol )
T b (K)
CH 3 CH 3 ,
CH 3 F
CH 3 OH
3.51
3.99
8.42
184.5
194.8
337.7
(Answer)
(b) (3 pts) Which of these compounds (CH 3 CH 3 , CH 3 F, and CH 3 OH) shows the most ordered liquid
state and the least ordered liquid state? Explain why.
(Answer)
3
3. (Total 14 pts)
(a) (4 pts) When 5.50 g of biphenyl (C 12 H 10 ) is dissolved in 100.0 g of benzene, the boiling point
increases by 0.903 oC. Calculate K b for benzene.
(Answer)
(b) (4 pts) When 6.30 g of an unknown hydrocarbon is dissolved in 150.0 g of benzene, the boiling
point of the solution increases by 0.597 oC. What is the molar mass of the unknown substance?
(Answer)
(c) (6 pts) A 0.40 g sample of a polypeptide dissolved in 1.0 L of an aqueous solution at 27 oC gave
rise to an osmotic pressure of 3.74 Torr. What is the molar mass of the polypeptide?
(Answer)
4
4. (Total 7 pts) Solid ammonium chloride is in equilibrium with ammonia and hydrogen chloride
gases:
NH 4 Cl(s)
NH 3 (g) + HCl(g)
The equilibrium constant at 275 oC is 1.04 × 10-2. We place 0.980 g of solid NH 4 Cl into a closed
vessel with volume 1.000 L and heat to 275 oC.
(a) (1 pt) In what direction does the reaction proceed?
(Answer)
(b) (3 pts) What is the partial pressure of each gas at equilibrium?
(Answer)
(c) (3 pts) What is the mass of solid NH4 Cl at equilibrium?
(Answer)
5
5. (Total 13 pts)
(a) (3 pts) What is the standard enthalpy of a reaction for which the equilibrium constant is doubled,
when the temperature is increased by 10 K at 298 K?
(Answer)
(b) (3 pts) What is the standard enthalpy of a reaction for which the equilibrium constant is halved,
when the temperature is increased by 10 K at 298 K?
(Answer)
(c) (7 pts) Suppose that nitrogen dioxide (NO 2 ) gas was allowed to dimerize into N 2 O 4 gas until the
reaction reached equilibrium 1.00 atm at 25oC. What are the partial pressure of NO2 and N 2 O 4 ? The
equilibrium constant K is 8.06 x 105 at -75oC and the standard enthalpy change for this reaction (∆Ho)
is -57.2 kJmol-1.
(Answer)
6
6. (Total 6 pts) Arrange the followings in order of acid strength and briefly explain your answer.
(a) (3 pts) CH 3 COOH, ClCH 2 COOH, Cl 3 CCOOH
(Answer)
(b) (3 pts) ClOH, BrOH, IOH
(Answer)
7. (Total 12 pts) Ascorbic acid (Vitamin C) is a diprotic acid, H 2 C 6 H 6 O 6 . The acid ionization
constants are K a1 = 7.9×10-5 and K a2 = 1.6×10-12.
(a) (6 pts) What is the pH of a 0.1 M solution?
(Answer)
(b) (6 pts) What is the concentration of ascorbate ion, C 6 H 6 O 6 2-?
(Answer)
7
8. (Total 9 pts) Calculate the equilibrium concentrations of H 2 CO 3 , HCO 3 -, CO 3 2-, and H 3 O+ in a
saturated aqueous solution of CO 2 , in which the original concentration of H 2 CO 3 is 0.034 M. K a1
and K a2 are 4.3 X 10-7 and 4.8 X 10-11, respectively.
(Answer)
9. (Total 6 pts) Zinc hydroxide (Zn(OH) 2 ) is sparingly soluble base (K sp = 4.5 X 10-17).
(a) (3 pts) A 1.0 g sample of solid Zn(OH) 2 is shaken with 0.5 L of water. Calculate the mass (grams)
of Zn(OH) 2 that dissolves.
(Answer)
(b) (3 pts) Compare the solubility in pure water with that in a solution buffered at pH 6.00.
(Answer)
8
10. (Total 9 pts) Following is the titration curve for the neutralization of 25 mL of a monoprotic acid
with a strong base. Answer the following questions about the reaction and explain your reasoning in
each case.
(a) (2 pts) What is Ka for the acid?
(Answer)
(b) (4 pts) What is the initial concentration of the acid?
(Answer)
(c) (3 pts) What is the concentration of base in the titrant?
(Answer)
9
11. (5 pts) Describe briefly the names and their achievements of the Nobel Laureates in Chemistry
2013.
(Answer)
10
11
2013 FALL Semester Final Examination
for General Chemistry II (CH103)
Date: December 19 (Thu), Time Limit: 2:30 ~ 4:30 p.m.
Write down your information neatly in the space provided below; print your Student ID in the upper
right corner of every page.
Professor
Name
Class
Problem
points
Student I.D. Number
Problem
1
2
/12
Name
TOTAL pts
points
/12
/8
6
7
3
/10
8
/11
4
/10
9
/11
5
/10
10
/8
/8
/100
** This paper consists of 12 sheets with 10 problems (page 10,11: constants & periodic table, page 12: claim
form). Please check all page numbers before taking the exam. Write down your work and answers in the space.
Please write down the unit of your answer when applicable. You will get 30% deduction for a missing unit.
NOTICE: SCHEDULES on RETURN and CLAIM of the MARKED EXAM PAPER.
1. Period, Location and Procedure
1)
Return and Claim Period: Dec 20 (Friday, 12:00-14:00)
2)
Location: Creative Learning Bldg.(E11)
3)
Class
Room
Class
Room
A
406
D
410
B
407
E
411
C
409
Claim Procedure:
Rule 1: Students cannot bring their own writing tools into the room. (Use a pen only provided by TA)
Rule 2: With or without claim, you must submit the paper back to TA. (Do not go out of the room with it)
(During the period, you can check the marked exam paper from your TA and should hand in the paper with a FORM for claims if
you have any claims on it. The claim is permitted only on the period. Keep that in mind! A solution file with answers for the
examination will be uploaded on 12/20 on the web.)
2. Final Confirmation
1) Period: Dec 21(Sat) – 22(Sun)
2) Procedure: During this period, you can check the final score of the examination on the website again.
To get more information, visit the website at www.gencheminkaist.pe.kr.
1
2013 FALL Semester Final Examination
for General Chemistry II (CH103)
Problem
points
1
2
2+4+2+4/12
Problem
points
2+6+4/12
4+4/8
6
7
3
1+1+1+1+6/10
8
3+4+4/11
4
4+4+2/10
9
2+3+6/11
5
6+2+2/10
10
TOTAL pts
4+4/8
/100
2+2+2+2/8
1. (12 points)
2 points
4 points
2 points
(d) At pH = 2, hydronium ion concentration is 0.010 M, and n = 10.
2.25v - 0.0592/10 x log ([Mn2+]2 [Zn2+]5 )/([MnO 4 -]2 [H 3 O+]16 )
4 points
= 2.25v – 0.0592v/10 x log (5.3 x 1018) = 2.14 v
2
2. (8 points)
AgCl(s) + e-  Ag(s) + Cl- (aq) cathode
Cu(s)  Cu2+(aq) + 2 eanode
Q = It
Chemical amount of electrons (0.500 C s-1)(6.06 X 103 s)/(96,485 C mol-1) = (3.14 X 10-2 mol e-)
(3.14 X 10-2 mol e-) X (107.87 g mol-1) = 3.39 g Ag deposited 4 points
(3.14 X 10-2 mol e-) X (63.55 g mol-1) X (1/2) = 0.998 g Cu dissolved 4 points
3. (10 points)
1 point per each
(e) (total 6 points)
3 points
3 points
3
4. (10 points)
kA
k1
k2
A →
I1 →
I 2 →
P
d [A]
= −k A [A], [A] = [A]0 e −k At , [A]0 = initial concentration of A
dt
d [I1 ]
= k A [A] − k1[I1 ]
dt
d [I 2 ]
= k1[I1 ] − k 2 [I 2 ]
dt
d [P]
= k2 [I 2 ]
dt
3 points
By using the steady-state approximation,
d [I1 ]
k
k
= k A [A] − k1[I1 ] = 0 → [I1 ] = A [A] = A [A]0 e −k At
dt
k1
k1
d [I 2 ]
k
k
= k1[I1 ] − k 2 [I 2 ] = 0 → [I 2 ] = 1 [I1 ] = A [A]0 e −k At
dt
k2
k2
4 points
d [ P]
= k 2 [I 2 ] = k A [A]0 e −k At
dt
∴ [P] = [A]0 (1 − e −k At )
When the concentration of P reaches to one-half of the initial concentration of A,
[P] = [A]0 (1 − e
−k At
)=
ln 2
= 35 s
t =
0.020 s −1
[A]0
ln 2
→t =
2
kA
3 points
5. (10 points)
4
Derivation: 4 points
2 points
2 points
2 points
6. (12 points)
(a) [Co(en) 3 ]3+ : optical isomers 2 points
(b) [Co(en) 2 Br 2 ]+ : cis with optical isomers, and trans without optical isomer
Br
Br
Br
Br
6 points
Br
Br
5
(c) [Co(NH 3 ) 3 Br 3 ] : mer- and facBr
Br
NH3
Br
H3N
NH3
Br
Br
4 points
NH3
NH3
NH3
Br
7. (8 points)
each configuration 2 points, total 4 points
On the other hand, Cl- is a weak filed ligand, and [FeCl 4 ]- and [FeCl 4 ]2- ions form tetrahedral
structures; here we expect the d electrons to be distributed in both the e and t 2 orbitals. Upon
reduction, one more electron is added, resulting in only one electron being paired in the e level,
leading to a complex that is still paramagnetic:
t2
t2
e
e
each configuration 2 points, total 4 points
6
8. (11 points)
(a) 3 points
+1
+2
N
H
N
+
H
A
(b) 4 points
+1
B
+1
H
N
-
-
O N O
+1
H
N
H
N
O
A
O
N
H
O
B
N
O
+1
Structure of NO 2 +: +1, Electron pushing of arene: +1, aromatic regeneration: +1, Base-acid: +1
No credit for wrong direction of pushing arrow
(c) 3 points
NH3
NH2
NH3
+1
+1
Electron deficient at ortho-, para-position or electron rich at meta-position: +1
7
9. (11 points)
(a) 2 points Addition
(b) 3 points
chiral carbon
H2
C
C
H3C
n
H
(no reduction for missing n)
(c) 6 points
isotactic
syndiotactic
atactic
10. (8 points)
methyl ester unit
L-aspartic acid unit
L-phenylalanine unit
amide bonding for a dimer
2 points
2 points
2 points
2 points
8
1. (12 points)
An aqueous solution of potassium permanganate (KMnO 4 ) appears deep purple. In aqueous acidic
solution, the permanganate ion can be reduced to the pale-pink manganese(II) ion (Mn2+). Under
standard conditions, the reduction potential of [MnO 4 - | Mn2+] half-cell is ε0 = 1.49 V. Suppose this
half-cell is combined with [Zn2+ | Zn] half-cell (ε0 = -0.76 V) in a galvanic cell, with [Zn2+] = [MnO 4 –] =
[Mn2+] = [H 3 O+] = 1 M.
(a) Write equations for the reactions at the anode and the cathode.
(Answer)
(b) Write a balanced equation for the overall cell reaction.
(Answer)
(c) Calculate the standard cell potential difference, ∆ε0.
(Answer)
(d) Suppose the cell is operated at pH 2.00 with [MnO 4 –] = 0.12 M, [Mn2+] = 0.0010 M, and [Zn2+] =
0.0015 M. Calculate the cell voltage ∆ε at 25 oC.
(Answer)
9
2. (8 points)
An electrolytic cell is constructed in which the silver ions in silver chloride are reduced to silver at the
cathode and copper is oxidized to Cu2+(aq) at the anode. A current of 0.500 A is passed through the
cell for 101 min. Calculate the mass of copper dissolved and the mass of silver deposited.
(Answer)
10
3. (10 points)
State which atom of each of the following pairs is more electronegative:
(a) sulfur, phosphorus;
(Answer)
(b) selenium, tellurium;
(Answer)
(c) sodium, cesium;
(Answer)
(d) silicon, oxygen.
(Answer)
(e) If 2.00 g of sodium peroxide (Na 2 O 2 ) is dissolved to form 200 mL of an aqueous solution, what
would be the pH of the solution? For H 2 O 2 , K a1 = 1.8×10-12 and K a2 is negligible.
(Answer)
11
4. (10 points)
In the following sequential reaction, product (P) formation results from the formation and decay of
two intermediate species, I 1 and I 2 . Calculate the time required for the concentration of P to reach
to one-half of the initial concentration of A, given that k A = 0.020 s-1 and k 1 = k 2 = 0.20 s-1.
k1
k2
kA
P
I 2 →
A →
I1 →
(Answer)
12
5. (10 points)
The half-life for the second-order reaction of a substance A is 50.5 s when [A] 0 = 0.84 mol L1
. Calculate the time needed for the concentration of A to decrease to
(a) one-sixteenth;
(Answer)
(b) one-fourth;
(Answer)
(c) one-fifth of its original value.
(Answer)
13
6. (12 points)
Draw out all the isomers, geometric and optical, of the following compounds (en =
ethylenediamine).
(a) [Co(en) 3 ]3+
(Answer)
(b) [Co(en) 2 Br 2 ]+
(Answer)
(c) [Co(NH 3 ) 3 Br 3 ]
(Answer)
7. (8 points)
When the paramagnetic [Fe(CN) 6 ]3- ion is reduced to [Fe(CN) 6 ]4-, the ion becomes diamagnetic.
However, when the paramagnetic [FeCl 4 ]- ion is reduced to [FeCl 4 ]2-, the ion remains
paramagnetic. Explain these observations.
(Answer)
14
8. (11 points)
In general, the electron donating group activates the benzene ring to electrophilic attack while the
withdrawing group deactivates.
N
1) NO2+
H+
2) Base
N
NO2
A
(a) Please draw the missing intermediate A and provide an arrow pushing mechanism.
(Answer)
(b) Please propose an arrow pushing mechanism for the conversion of molecule A to the final
product (The structure of NO 2 + is required).
(Answer)
(c) Briefly rationalize the experimental results for this reaction. (Hint: resonance of intermediate A)
(Answer)
15
9. (11 points)
Polymerization of propylene (CH 2 =CH-CH 3 ) produces so-called polypropylene with molecular
weight of several hundred thousand.
(a) What kind of polymer it is, condensation or addition?
(Answer)
(b) Draw the repeating unit of polypropylene and identify the chiral carbon.
(Answer)
(c) Draw three isomeric structures of polypropylene including their name.
(Answer)
10. (8 points)
Aspartame is an artificial sweetener, which has been widely used as a sugar substitute in food
industry. It is a methyl ester of the dipeptide of L-aspartic acid and L-phenylalanine (α-carboxyl
group of phenylalanine has methyl ester attached to it through an ester linkage). Draw the
structure of aspartame.
(Answer)
16
17
18
19