1

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 Student I.D. Number Name

Problem points Problem points TOTAL pts

1 /13 7 /12

/100

2 /6 8 /9

3 /14 9 /6

4 /7 10 /9

5 /13 11 /5

6 /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.

<The Answers>

Problem points Problem points TOTAL pts

1 6+3+4/13 7 6+6/12

/100

2 3+3/6 8 3+3+3/9

3 4+4+6/14 9 3+3/6

4 1+3+3/7 10 2+4+3/9

5 3+3+7/13 11 5/5

6 3+3/6

1. (Total 13 pts)

(a) (6 pts)

(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, ∆Sovap = ∆Ho

vap / Tb because of ∆S = q/T.

CH3CH3, CH3F CH3OH

∆Hovap (kcal/mol) 3.51 3.99 8.42

Tb (K) 184.5 194.8 337.7

∆Sovap (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 (CH3OH) has the highest ∆Sovap and Tb due to

the hydrogen bonding of the alcohol group, and ethane (CH3CH3) 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) NH4Cl(s) NH3(g) + HCl(g)

Initial 0 0

Change +x +x

Equilibrium +x +x

So, K = PNH3 PHCl = 1.04 X 10-2

x2 = 1.04 X 10-2; x = 0.102

Answer: PNH3 = PHCl = 0.102 atm.

(c) (3 pts) nNH3 = nHCl = 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 NH4Cl consumed = 2.268 X 10-3 mol X 53.49 g mol-1 = 0.121 g

So, the remaining mass = 0.859 g

Answer: NH4Cl = 0.859 g

5. (Total 14 pts)

kmolkJ

KK

kJKH

kKKKTKT

TT

KKR

HTTR

HKK

ln76

3081

2981

ln3145.8

,308,298

11

ln11ln

11

0

1

221

21

1

2

0

12

0

1

2

×=

−

=∆

===

−

=∆→

−

∆−=

−−

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) 2NO2 N2O4

PNO2 = P – x and PN2O4 = x

K = x/(P- x)2 0≤ x ≤1

Using ln

ln

K2 = 6.96

x 2 – 2.1435 x + 1.00 = 0, x = 0.686

PN2O4 = 0.686 atm, PNO2 = 0.314 atm

6. (Total 6 pts)

Conjugate base is stabilized by inductive effect of electronegative element.

(a) (3 pts) The –CCl3 group that is bonded to the carboxyl group, -COOH, in trichloroacetic acid, is

more electron withdrawing than the CH3 group in acetic acid. Thus, trichloroacetic acid is the

stronger acid. CH3COOH < ClCH2COOH < Cl3CCOOH

(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)

H2Asc(aq) ⇄ H+(aq) + HAsc-(aq) Ka1 = 7.9×10-5

HAsc-(aq) ⇄ H+(aq) + Asc

2-(aq) Ka2 = 1.6×10-12

H2Asc(aq) ⇄ H+(aq) + HAsc-(aq)

Starting 0.1 0 0

Change -x +x +x

Equilibrium 0.1-x x x

Ka1 = [H+(aq)][HAsc-(aq)]/[H2Asc(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)

HAsc-(aq) ⇄ H+(aq) + Asc

2-(aq)

Starting 0.0028 0.0028 0

Change -y +y +y

Equilibrium 0.0028-x 0.0028+x y

Ka2 = [H+(aq)][ Asc2-(aq)]/[HAsc

-(aq)] = 1.6×10-12

= (0.0028 + y)y/(0.0028 – y) ~ 0.0028y/0.0028 = y

[Asc2-(aq)] = 1.6×10-12

8. (Total 9 pts)

For first ionization,

H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq)

Initial 0.034 0 ~0

Change -x +x +x

Equilibrium 0.034-x x x

So, Ka1 = x2/(0.034-x) = 4.3 X 10-7,

X = [H3O+] = [HCO3-] = 1.2 X 10-4 M

[H2CO3] = 0.034 M

For second ionization,

Ka2 = 1.2 X 10-4 X [CO32-]/1.2 X 10-4 = 4.8 X 10-11

[CO32-] = 4.8 X 10-11 M

Answer: H2CO3 = 0.034 M, HCO3- = H3O+ = 1.2 X 10-4 M, CO3

2- = 4.8 X 10-11 M

9. (Total 6 pts)

(a) (3 pts) Ksp = [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+] = Ksp/[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)

2

1. (Total 13 pts) The normal boiling point of iodomethane, CH3I, 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)

3

2. (Total 6 pts)

(a) (3 pts) Compare ∆Sovap for CH3CH3, CH3F, and CH3OH as calculated from ∆Ho

vap and Tb

(boiling temperature) of the compounds.

CH3CH3, CH3F CH3OH

∆Hovap (kcal mol-1) 3.51 3.99 8.42

Tb (K) 184.5 194.8 337.7

(Answer)

(b) (3 pts) Which of these compounds (CH3CH3, CH3F, and CH3OH) shows the most ordered liquid

state and the least ordered liquid state? Explain why.

(Answer)

4

3. (Total 14 pts)

(a) (4 pts) When 5.50 g of biphenyl (C12H10) is dissolved in 100.0 g of benzene, the boiling point

increases by 0.903 oC. Calculate Kb 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)

5

4. (Total 7 pts) Solid ammonium chloride is in equilibrium with ammonia and hydrogen chloride

gases:

NH4Cl(s) NH3(g) + HCl(g)

The equilibrium constant at 275 oC is 1.04 × 10-2. We place 0.980 g of solid NH4Cl 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 NH4Cl at equilibrium?

(Answer)

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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 (NO2) gas was allowed to dimerize into N2O4 gas until the

reaction reached equilibrium 1.00 atm at 25oC. What are the partial pressure of NO2 and N2O4? 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)

7

6. (Total 6 pts) Arrange the followings in order of acid strength and briefly explain your answer.

(a) (3 pts) CH3COOH, ClCH2COOH, Cl3CCOOH

(Answer)

(b) (3 pts) ClOH, BrOH, IOH

(Answer)

7. (Total 12 pts) Ascorbic acid (Vitamin C) is a diprotic acid, H2C6H6O6. The acid ionization

constants are Ka1 = 7.9×10-5 and Ka2 = 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, C6H6O62-?

(Answer)

8

8. (Total 9 pts) Calculate the equilibrium concentrations of H2CO3, HCO3-, CO3

2-, and H3O+ in a

saturated aqueous solution of CO2, in which the original concentration of H2CO3 is 0.034 M. Ka1

and Ka2 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 (Ksp = 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)

9

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)

10

11. (5 pts) Describe briefly the names and their achievements of the Nobel Laureates in Chemistry

2013.

(Answer)

11

1

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 Student I.D. Number Name

Problem points Problem points TOTAL pts

1 /12 6 /12

/100

2 /8 7 /8

3 /10 8 /11

4 /10 9 /11

5 /10 10 /8 ** 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)

Class Room Class Room

A 406 D 410

B 407 E 411

C 409 3) 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.

2

2013 FALL Semester Final Examination for General Chemistry II (CH103)

Problem points Problem points TOTAL pts

1 2+4+2+4/12 6 2+6+4/12

/100

2 4+4/8 7 4+4/8

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 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 )/([MnO4-]2 [H3O+]16 )

= 2.25v – 0.0592v/10 x log (5.3 x 1018) = 2.14 v 4 points

3

2. (8 points) AgCl(s) + e- Ag(s) + Cl- (aq) cathode Cu(s) Cu2+(aq) + 2 e- anode 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

4

4. (10 points)

]I[]P[

]I[]I[]I[

]I[]A[]I[

A ofion concentrat initial]A[ ,]A[]A[],A[]A[PIIA

22

22112

111

00

2121

kdt

d

kkdt

d

kkdt

d

ekdt

d

A

tkA

kkk

A

A

=

−=

−=

==−=

→→→

−

3 points

By using the steady-state approximation,

)1(]A[]P[

]A[]I[]P[

]A[]I[]I[0]I[]I[]I[

]A[]A[]I[0]I[]A[]I[

0

022

02

12

122211

2

011

1111

tk

tkA

tkA

tkAAA

A

A

A

A

e

ekkdt

d

ekk

kkkk

dtd

ekk

kkkk

dtd

−

−

−

−

−=∴

==

==→=−=

==→=−=

4 points

When the concentration of P reaches to one-half of the initial concentration of A,

s35==

=→=−=

−

−

1

00

s020.0

2ln

2ln

2

]A[)1(]A[]P[

t

kte

A

tkA

3 points

5. (10 points)

5

Derivation: 4 points

2 points

2 points

2 points 6. (12 points) (a) [Co(en)3]3+ : optical isomers 2 points

(b) [Co(en)2Br2]+ : cis with optical isomers, and trans without optical isomer 6 points

Br

Br

Br

Br

Br

Br

6

(c) [Co(NH3)3Br3] : mer- and fac- 4 points

H3N NH3

Br NH3

Br

BrBr NH3

Br NH3

NH3

Br 7. (8 points)

each configuration 2 points, total 4 points On the other hand, Cl- is a weak filed ligand, and [FeCl4]- and [FeCl4]2- ions form tetrahedral structures; here we expect the d electrons to be distributed in both the e and t2 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:

e

t2

e

t2

each configuration 2 points, total 4 points

7

8. (11 points) (a) 3 points

NH+

A

NH

+1 +2

(b) 4 points

A

NH

N OO

NH

NO

-OH

B

NH

NO

-O

B

+1

+1

+1

+1

Structure of NO2

+: +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

8

9. (11 points) (a) 2 points Addition (b) 3 points

H2C

C

HH3C

chiral carbon

n

(no reduction for missing n) (c) 6 points

isotactic

syndiotactic

atactic

10. (8 points)

methyl ester unit 2 points L-aspartic acid unit 2 points L-phenylalanine unit 2 points amide bonding for a dimer 2 points

9

1. (12 points) An aqueous solution of potassium permanganate (KMnO4) 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 [MnO4

- | 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+] = [MnO4

–] = [Mn2+] = [H3O+] = 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 [MnO4

–] = 0.12 M, [Mn2+] = 0.0010 M, and [Zn2+] = 0.0015 M. Calculate the cell voltage ∆ε at 25 oC. (Answer)

10

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)

11

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 (Na2O2) is dissolved to form 200 mL of an aqueous solution, what would be the pH of the solution? For H2O2, Ka1 = 1.8×10-12 and Ka2 is negligible. (Answer)

12

4. (10 points) In the following sequential reaction, product (P) formation results from the formation and decay of two intermediate species, I1 and I2. Calculate the time required for the concentration of P to reach to one-half of the initial concentration of A, given that kA = 0.020 s-1 and k1 = k2 = 0.20 s-1.

PIIA 2121 →→→ kkk A

(Answer)

13

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 L-

1. 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)

14

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)2Br2]+

(Answer)

(c) [Co(NH3)3Br3] (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 [FeCl4]- ion is reduced to [FeCl4]2-, the ion remains paramagnetic. Explain these observations. (Answer)

15

8. (11 points) In general, the electron donating group activates the benzene ring to electrophilic attack while the withdrawing group deactivates.

N N

NO2

H+ 1) NO2+

A

2) Base

(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 NO2

+ is required). (Answer) (c) Briefly rationalize the experimental results for this reaction. (Hint: resonance of intermediate A) (Answer)

16

9. (11 points) Polymerization of propylene (CH2=CH-CH3) 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)

17

18

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