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2018 FALL Semester Midterm Examination For General Chemistry II (CH103)

Date: October 17 (Wed), Time Limit: 19:00 ~ 21:00 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 /10 7 /8

/100

2 /10 8 /9

3 /9 9 /8

4 /10 10 /7

5 /10 11 /9

6 /10 ** This paper consists of 15 sheets with 11 problems (pages 13: fundamental constants, page 14: periodic table, page 15: claim form). 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 22 (Mon, 7:00 ~ 8:00 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 25 (Thu) – October 26 (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.

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1. (total 10 points) Which one is stronger acid?

(a) HF and HCl

(Answer)

(b) HClO4 and HClO3

(Answer)

(c) CH3COOH and HCOOH

(Answer)

(d) C6H5COOH and C6H5OH

(Answer)

(e) HClO3 and HBrO3

(Answer)

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2. (total 10 points) 1.144 grams of CH3COONa (MW = 82.03 g) are added to 0.5 L of 0.1 M solution of

CH3COOH (Ka = 1.76ⅹ10-5) at 25 oC.

(a) What is the pH of the resulting buffer solution? Use the Henderson-Hasselbalch equation.

(Answer)

(b) What is the pH after the addition of 0.0005 mole of HCI to the buffer?

(Answer)

(c) What is the pH after the addition of 0.0005 mole of NaOH to the original solution?

(Answer)

(d) What is the pH of a 0.2 M solution of CH3COONa at 25 °C?

(Answer)

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3. (total 9 points) A propionic acid (CH3CH2COOH) is a weak monoprotic acid, and its Ka value is

1.26×10-5.

(a) Calculate the pH of the solution which contains 0.1 M of propionic acid and 0.05 M of sodium

propionate (CH3CH2COONa).

(Answer)

(b) Calculate the pH after 50 mL of 0.08 M NaOH is added to 350 mL of the solution mentioned in (a).

(Answer)

(c) Calculate the pH after 50 mL of 0.08 M HCl is added to 350 mL of the solution mentioned in (a).

(Answer)

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4. (10 points) We can predict an unknown metal hydroxide solution using titration with standard HCl

solution. The procedure is as follows:

1) Prepare 10 mL of saturated unknown metal hydroxide solution.

2) Fill 50 mL of 0.020 M HCl in the burette and drop it to the unknown solution slowly.

3) Draw a graph and find the equivalence point.

4) Analyze hydroxide ion concentration by measuring the volume of HCl solution required.

5) Find the appropriate unknown metal hydroxide.

Metal Hydroxide, 𝑴𝑴(𝑶𝑶𝑶𝑶)𝒏𝒏 𝑲𝑲𝒔𝒔𝒔𝒔 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 [𝐴𝐴𝐴𝐴+][𝑂𝑂𝑂𝑂−] = 1.5 × 10−8

𝐅𝐅𝐅𝐅(𝑶𝑶𝑶𝑶)𝟑𝟑 [𝐹𝐹𝐹𝐹3+][𝑂𝑂𝑂𝑂−]3 = 1.1 × 10−36 𝐙𝐙𝐙𝐙(𝑶𝑶𝑶𝑶)𝟐𝟐 [𝑍𝑍𝑍𝑍2+][𝑂𝑂𝑂𝑂−]2 = 4.5 × 10−17 𝐂𝐂𝐂𝐂(𝑶𝑶𝑶𝑶)𝟐𝟐 [𝐶𝐶𝐶𝐶2+][𝑂𝑂𝑂𝑂−]2 = 5.5 × 10−6

Find the appropriate unknown metal hydroxide using experimental data and explain your reasoning.

(Answer)

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5. (total 10 points) Ksp for Pb(OH)2 is 4.2×10-15, and Kf for Pb(OH)3- is 4×1014.

(a) Suppose a solution whose initial concentration of Pb2+(aq) is 0.050 M is bought to pH 13.0 by the

addition of solid NaOH. Will solid Pb(OH)2 precipitate, or will the lead be dissolved as Pb(OH)3-(aq)?

What will be [Pb2+] and [Pb(OH)3-] at equilibrium?

(Answer)

(b) Ka of H2S is 9.1×10-8. Now, suppose that the solution is saturated with H2S at a concentration of

[H2S] = 0.10 M, and consider the below chemical equilibrium of lead sulfide. Will solid Pb(OH)2 and/or

PbS precipitate, or will the lead be dissolved as Pb(OH)3-(aq)? What will be [Pb2+] and [Pb(OH)3

-] at

equilibrium?

PbS (s) + H2O (l) ⇄ Pb2+ (aq) + OH- (aq) + HS- (aq), K† = 3×10-28

(Answer)

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6. (total 10 points)

(a) Calculate the equilibrium constant at 25 °C for the reaction.

Zn(s) + Cu2+(aq) ⇄ Zn2+(aq) + Cu(s)

The corresponding standard cell potential is 1.10 V.

(Answer)

(b) Which is the potential at 25 °C of the following cell?

Zn(s) l Zn2+(0.001M) ∥ Cu2+(10.0 M) l Cu(s)

(Answer)

(c) Which happens when we dip a strip of zinc into a solution of CuSO4?

(Answer)

(d) What happens when we dip a strip of copper into a solution of ZnSO4?

(Answer)

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7. (total 8 points) To induce a current flow, a lithium air battery is based on the chemistry that uses

oxidation of lithium metal, Li (s), at the anode and reduction of oxygen, O2 (g), at the cathode to form

lithium peroxide, Li2O2 (s).

(a) Write the half-cell and the overall cell reactions during discharge of the battery, and calculate the

standard potential for the cell at 298 K. The standard Gibbs free energy of formation of the lithium

peroxide is ∆G°f = -578.9 kJ/mol.

(Answer)

(b) Assume that 30 gram of O2 (g) is consumed as the cell operates for 600 seconds. Calculate the

number of moles of electrons that pass through the cell. Also, calculate the average current, in

amperes, that passes through the cell. (atomic weight of O = 16 g/mol)

(Answer)

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8. (total 9 points) A galvanic cell is constructed by linking a Co2+|Co(s) half-cell to an Ag+|Ag(s) half-

cell through a salt bridge and then connecting the cobalt and silver electrodes through an external

circuit. When the circuit is closed, the cell potential is measured to be 1.08 V, and silver is seen to

plate out while cobalt dissolves.

(a) Write the half-cell reactions that occur at the anode and at the cathode and overall balanced

equation, including the state of each chemical species.

(Answer)

(b) The cobalt electrode is weighed after 150 minutes of operation and is found to have decreased in

mass by 0.360 g. By what amount has the silver electrode increased in mass? (MW of Co = 58.93; MW

of Ag = 107.87)

(Answer)

(c) What is the average current drawn from the cell during this period?

(Answer)

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9. (total 8 points) If Tris[4-(dimethylamino)phenyl]methylium chloride (crystal violet) reacts with

hydroxide ion, it loses its purple color and becomes transparent. During the reaction, we plot a graph,

using time (t) as the x-axis and the natural log of absorbance as the y-axis at different temperatures.

The graph is shown below.

Use these relationships.

1) The reaction is first order with respect to crystal violet concentration. Rate = k[crystal violet]

2) Absorbance ∝ Concentration

(a) Calculate the rate constant (k) and the half life time (𝑡𝑡1/2) of the reaction at each temperature.

(Answer) (b) Calculate the activation energy (kJ mol-1) of the reaction.

(Answer)

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10. (total 7 points) The following mechanism has been proposed for the gas phase reaction of

chloroform and chlorine.

2Cl(g)

Cl(g) + CHCl3(g) HCl(g) + CCl3(g)

Cl(g) + CCl3(g) CCl4(g)

Cl2(g) (fast)

(slow)

(fast)

step 1

step 2

step 3

k1

k-1k2

k3

(a) What is the overall reaction?

(Answer)

(b) What are the intermediates in the mechanism?

(Answer)

(c) What is the rate law and overall order of the reaction predicted by this mechanism?

(Answer)

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11. (total 9 points) Consider following enzymatic reaction in steady state:

(a) Applying steady-state approximation for ES, express Michaelis-Menten constant in terms of [E],

[S], and [ES].

(Answer)

(b) What is the concentration of substrate when the reaction rate is half of Vmax?

(Answer)

(c) Plot graph of 1/[S] vs. 1/v0. Express 1/v0 and 1/[S] intercepts in terms of KM and Vmax.

(Answer)

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Physical Constants

Avogadro’s number NA = 6.02214179 ⅹ 1023 mol-1

Bohr radius a0 = 0.52917720859 Å = 5.2917720859ⅹ10-11 m

Boltzmann’s constant KB = 1.3806504 ⅹ 10-23 J K-1

Electronic charge e = 1.602176487 ⅹ 10-19 C

Faraday constant F = 96485.3399 C mol-1

Masses of fundamental particles:

Electron me = 9.10938215 ⅹ 10-31 kg

Proton mP = 1.672621637 ⅹ 10-27 kg

Neutron mn= 1.674927211 ⅹ 10-27 kg

Permittivity of vacuum εo = 8.854187817 ⅹ 10-12 C-2 J-1 m-1

Planck’s constant h = 6.62606896 ⅹ 10-34 J s

Ratio of proton mass to electron mass mP / me = 1836.15267247

Speed of light in a vacuum c = 2.99792458 ⅹ 108 m s-1 (exactly)

Standard acceleration of terrestrial gravity g = 9.80665 m s-2 (exactly)

Universal gas constant R = 8.314472 J mol-1 K-1 = 0.0820574 L atm mol-1 K-1

Values are taken from the 2006 CODATA recommended values, as listed by the National Institute of Standards and Technology. Conversion factors

Ångström 1 Å= 10-10 m

Atomic mass unit 1 u = 1.660538782 ⅹ 10-27 kg 1 u = 1.492417830 ⅹ 10-10 J = 931.494028 MeV (energy equivalent form E =

mc2) Calorie 1 cal = 4.184 J (exactly)

Electron volt 1 eV = 1.602177 ⅹ 10-19 J = 96.485335 kJ mol-1 Foot 1 ft = 12 in = 0.3048 m (exactly)

Gallon (U. S.) 1 gallon = 4 quarts = 3.785412 L (exactly)

Liter 1 L = 10-3 m-3 = 103 cm3 (exactly)

Liter-atmosphere 1 L atm = 101.325 J (exactly)

Metric ton 1 t = 1000 kg (exactly)

Pound 1 lb = 16 oz = 0.4539237 kg (exactly)

Rydberg 1 Ry = 2.17987197 x 10-18J = 1312.7136 kJ mol-1 = 13.60569193 eV

Standard atmosphere 1 atm = 1.01325 x 105 Pa = 1.01325 x 105 kg m-1 s-2 (exactly)

Torr 1 torr = 133.3224 Pa

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Claim Form for General Chemistry Examination Page ( / )

Class: , Professor Name: , I.D.# : , Name: If you have any claims on the marked paper, please write down them on this form and submit this with your paper in the assigned place. (And this form should be attached on the top of the marked paper with a stapler.) Please, copy this sheet if you need more before use.

By Student By TA

Question # Claims Accepted? Yes(√) or No(√)

Yes: □ No: □ Pts (+/-) Reasons

<The Answers>

Problem points Problem points TOTAL pts

1 2x5/10 7 4+4/8

/100

2 3+2+2+3/10 8 3x3/9

3 3x3/9 9 4+4/8

4 2x5/10 10 2+2+3/7

5 5+5/10 11 2+2+5/9

6 3+3+2+2/10

전체 기준: 전개과정은 맞으나 답이나 unit 이 틀리면 -1

식을 전혀 쓰지 않고 (혹은 흔적이 전혀 없고) 답만 맞았을 때 -1 (3 pts), -2 (4 pts 이상)

1. (total 10 points) 2 pts for each (a) HCl (b) HClO4 (c) HCOOH (d) C6H5COOH (e) HClO3 2. (total 10 points) (a) (3 pts)

pKa of acetic acid pKa=-log(1.76ⅹ10−5)=4.754

1.44g CH3COONa ⅹ 1𝑚𝑚𝑚𝑚𝑚𝑚82.03𝑔𝑔

= 0.01395 mol

[CH3COO−] = 0.01395 𝑚𝑚𝑚𝑚𝑚𝑚0.5𝐿𝐿

= 0.0279 M

pH= pKa + log[𝐶𝐶𝐶𝐶3𝐶𝐶𝐶𝐶𝐶𝐶−][𝐶𝐶𝐶𝐶3𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶]

= 4.754 + log 0.02790.1

= 4.2 (b) (2 pts)

[HCI]= 0.0005 mol0.5L

= 0.001M [CH3COO−] = 0.0279M – 0.001M = 0.0269M [CH3COOH] = 0.1M + 0.0001M= 0.101M pH = pKa+ log[𝐶𝐶𝐶𝐶3𝐶𝐶𝐶𝐶𝐶𝐶−]

[𝐶𝐶𝐶𝐶3𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶] = 4.754 + log 0.0269

0.101 = 4.179

(c) (2 pts)

pH =pKa + log[𝐶𝐶𝐶𝐶3𝐶𝐶𝐶𝐶𝐶𝐶−][𝐶𝐶𝐶𝐶3𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶]

= 4.754 + log0.02890.099

= 4.219

(d) (3 pts) CH3COO−(aq) + H2O(ℓ) ⇄ CH3COOH(aq) + OH− (aq)

Kb = KwKa

= 1.0 ⅹ 10−14 1.76ⅹ10−5

= 5.7 ⅹ 10−10

[CH3COOH] [OH−] [CH3COO−]

= 𝒳𝒳2

0.2−𝒳𝒳 = 5.7ⅹ10−10

𝒳𝒳2

0.2 = 5.7 ⅹ 10−10

𝒳𝒳 = 1.07 ⅹ 10−5= [OH−]

[H3O+] = Kw [OH −]

= 1.0 ⅹ10−14

1.07 ⅹ10−5 = 9.4 ⅹ 10−10

pH = -log[H3O+] = 9.03 3. (total 9 points)

(a) (3 pts)

pKa of the propionic acid = -log10 (1.26×10-5) = 4.90

Using Henderson-Hasselbalch Equation,

pH = pKa - log10 [CH3CH2COOH]0/[CH3CH2COO-]0 = 4.90 - log10 0.1/0.05 = 4.90 - 0.30 = 4.60

(b) (3 pts)

Chemical reaction upon the addition of strong base:

CH3CH2COOH (aq) + OH- (aq) → CH3CH2COO- (aq) + H2O

Total solution volume = 50 mL + 350 mL = 400 mL

Thus, the concentrations are

[CH3CH2COOH]0 = (0.1×350 - 0.08×50)/400 = 0.0775 M

[CH3CH2COO-]0 = (0.05×350 + 0.08×50)/400 = 0.05375 M

Using Henderson-Hasselbalch Equation,

pH = pKa - log10 [CH3CH2COOH]/[CH3CH2COO-]

= 4.90 - log10 0.0775/0.05375 = 4.90 - 0.16 = 4.74

(c) (3 pts)

Chemical reaction upon the addition of strong base:

CH3CH2COO- (aq) + H3O+ (aq) → CH3CH2COOH (aq) + H2O

Total solution volume = 50 mL + 350 mL = 400 mL

Thus, the concentrations are

[CH3CH2COOH]0 = (0.1×350 + 0.08×50)/400 = 0.0975 M

[CH3CH2COO-]0 = (0.05×350 - 0.08×50)/400 = 0.03375 M

Using Henderson-Hasselbalch Equation,

pH = pKa - log10 [CH3CH2COOH]/[CH3CH2COO-]

= 4.90 - log10 0.0975/0.03375 = 4.90 - 0.46 = 4.44

4. (total 10 points) for each item, 2 pts Volume of HCl solution in equivalence point = 11mL 11mL × 0.02M = 10mL × [𝑂𝑂𝑂𝑂−] [𝑂𝑂𝑂𝑂−] = 2.2 × 10−2𝑀𝑀 Calculate each of hydroxide ion concentrations in saturated metal hydroxide solutions.

① AgOH 𝐾𝐾𝑠𝑠𝑠𝑠 = [𝐴𝐴𝐴𝐴+][𝑂𝑂𝑂𝑂−] = 1.5 × 10−8 = 𝑆𝑆2

𝑆𝑆 = [𝑂𝑂𝑂𝑂−] = 1.2 × 10−4M ② Fe(𝑂𝑂𝑂𝑂)3

𝐾𝐾𝑠𝑠𝑠𝑠 = [𝐹𝐹𝐹𝐹3+][𝑂𝑂𝑂𝑂−]3 = 1.1 × 10−36 = 27𝑆𝑆4 3𝑆𝑆 = [𝑂𝑂𝑂𝑂−] = 1.3 × 10−9M

③ Zn(𝑂𝑂𝑂𝑂)2 𝐾𝐾𝑠𝑠𝑠𝑠 = [𝑍𝑍𝑍𝑍2+][𝑂𝑂𝑂𝑂−]2 = 4.5 × 10−17 = 4𝑆𝑆3

2𝑆𝑆 = [𝑂𝑂𝑂𝑂−] = 4.5 × 10−6M ④ Ca(𝑂𝑂𝑂𝑂)2

𝐾𝐾𝑠𝑠𝑠𝑠 = [𝐶𝐶𝐶𝐶2+][𝑂𝑂𝑂𝑂−]2 = 5.5 × 10−6 = 4𝑆𝑆3 2𝑆𝑆 = [𝑂𝑂𝑂𝑂−] = 2.2 × 10−2M

∴ 𝐂𝐂𝐂𝐂(𝑶𝑶𝑶𝑶)𝟐𝟐 5. (total 10 points) (a) (5 pts)

Since the initial [Pb2+]0 of 0.050 M is less than 0.17 M, no precipitate of Pb(OH)2 (s) can form. The

Ksp equilibrium is not in effect. Essentially all of the Pb2+ ion is tied up in the complex, making the

concentration of Pb(OH)3- equal 0.050 M. Put this value into the Kf mass-action expression

(b) (5 pts)

K† is much smaller than Ksp of Pb(OH)2, implying that PbS is much more insoluble than Pb(OH)2.

Suppose PbS does precipitate.

At pH 13.0, where [OH-] = 0.10 M, [HS-] is determined from the acid-base equilibrium, i.e.,

Ka = 9.1×10-8 = [HS-][H3O+]/[H2S]

Substitute [H3O+]= 10-13 M and [H2S] = 0.10 M, and then [HS-] is given as 9.1×104. Then,

K† = 3×10-28 = [Pb2+][OH-][HS-] = [Pb2+](0.10)(9.1×104)

This means [Pb2+] is locked at 3.3×10-32 M if solid PbS is present. By substituting [Pb2+] = 3.3×10-32 M

and [OH-] = 0.10 M into the mass-action expression for the complexation equilibrium,

Kf = 4×1014 = [Pb(OH)3-]/[Pb2+][OH-]3

we obtain [Pb(OH)3-] = 1.3×10-20 M. Since the initial [Pb2+]0 of 0.050 M far exceeds 1.3×10-20, PbS

(s) precipitates. At equilibrium, [Pb2+] = 3.3×10-32 M and [Pb(OH)3-] = 1.3×10-20 M.

Regarding on the equilibrium of Pb(OH)2 dissolution,

Q = [Pb2+][OH-]2 = (3.3×10-32)(0.10)2 = 3.3×10-34 < Ksp (=4.2×10-5)

thus no precipitate of Pb(OH)2 can form.

6. (total 10 points) (a) (3 pts) ΔG˚ = -nFE˚cell = -2.303RT log K -2 x 9.65 x 104C/mol x 1.10V = -2.303 x 8.31J/(mol·K) x 298K x log K log K = 37.2 K = 𝟏𝟏.𝟔𝟔𝟔𝟔𝟏𝟏𝟔𝟔𝟑𝟑𝟑𝟑

(b) (3 pts)

(c) (2 pts) A spontaneous redox reaction occurs, driven forward by a fall in free energy. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) ΔG˚ = -212kJ E˚ = 1.10V (d) (2 pts) Nothing. Cu(s) + Zn2+(aq) → Cu2+(aq) + Zn(s) ΔG˚ = 212kJ E˚ = -1.10V is not spontaneous. Zn2+ does not oxidize copper.

7. (total 8 points) (a) (4 pts) Anode: Li (s) → Li+ + e- Cathode: 2Li+ + O2 (g) + 2e- → Li2O2 (s) Overall: 2Li (s) + O2 → Li2O2 (s) Using ∆G° = -nFE°, E°cell = -∆G°f /nF = -(-578.9×1000)/(2)(96,485) = 3.00 V (b) (4 pts) Equivalent mass of O2 in the cathode reaction is Weq = (16×2)/2 = 16. Thus, during 600 seconds, n = 30/16 = 1.875 moles of electrons passed through the cell. Also, the average current i = nF/t = (1.875×96,485)/600 = 301.5 A passed through the cell. 8. (total 9 points) (a) (3 pts) Anode: Co(s) Co2+(aq) + 2e- …….(1 pt) Cathode: Ag+(aq) + e- Ag(s) …….(1 pt) Overall: 2Ag+(aq) + Co(s) 2Ag(s) + Co2+(aq) …….(1 pt) (b) (3 pts)

Total amount of electron: Amount of Ag(s) = 12.2 mmol = 12.2 mmol * 107.868 (mg/mmol) = 1.32 g Ans) 1.32 g. Correct answer = 3 pts. Total amount of electron only = 1 pt. (c) (3 pts) Total charge of electron = 12.2 mmol * 96.485 C/mmol= 1180 C. Average current = 1180 C / 9000 sec = 0.131 A Ans) 0.131 A. Correct answer = 3 pts. Total charge of electron only = 1 pt. 9. (total 8 points) (a) (4 pts) The reaction is first order to crystal violet concentration. The equation, therefore, C = 𝐶𝐶0𝐹𝐹−𝑘𝑘𝑘𝑘 (C: crystal violet concentration, k: rate constant) is satisfied. Using second relationship, we can rearrange the equation. A = 𝐴𝐴0𝐹𝐹−𝑘𝑘𝑘𝑘(A: Absorbance) If we take natural log both sides, the equation is satisfied. lnA = −kt + ln𝐴𝐴0 We can calculate rate constant (k) using slopes of the graph. ∴ 𝟐𝟐.𝟒𝟒𝟔𝟔 × 𝟏𝟏𝟔𝟔−𝟑𝟑𝒔𝒔−𝟏𝟏 (𝟓𝟓℃), 𝟒𝟒.𝟑𝟑𝟔𝟔 × 𝟏𝟏𝟔𝟔−𝟑𝟑𝒔𝒔−𝟏𝟏 (𝟏𝟏𝟓𝟓℃) Half life time

𝑡𝑡1/2 =𝑙𝑙𝑍𝑍2𝑘𝑘

∴ 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 (𝟓𝟓℃), 𝟏𝟏𝟔𝟔𝟏𝟏𝒔𝒔 (𝟏𝟏𝟓𝟓℃) (b) (4 pts) k = A𝐹𝐹−

𝐸𝐸𝑎𝑎𝑅𝑅𝑅𝑅 ⇒ ln𝑘𝑘1 = 𝑙𝑙𝑍𝑍𝐴𝐴 − 𝐸𝐸𝑎𝑎

𝑅𝑅𝑇𝑇1, ln𝑘𝑘2 = 𝑙𝑙𝑍𝑍𝐴𝐴 − 𝐸𝐸𝑎𝑎

𝑅𝑅𝑇𝑇2

⇒ ln(𝑘𝑘1𝑘𝑘2

) = −𝐸𝐸𝑎𝑎𝑅𝑅

( 1𝑇𝑇1− 1

𝑇𝑇2)

ln(2.46 × 10−3

4.30 × 10−3) = −

𝐸𝐸𝑎𝑎8.314𝐽𝐽𝑚𝑚𝑚𝑚𝑙𝑙−1𝐾𝐾−1 (

1278.15𝐾𝐾

−1

288.15𝐾𝐾)

∴ 𝑬𝑬𝒂𝒂 = 𝟑𝟑𝟑𝟑.𝟐𝟐 𝒌𝒌𝒌𝒌𝒎𝒎𝒎𝒎𝒎𝒎−𝟏𝟏

10. (total 7 points) (a) (2 pts)

Cl2(g) + CHCl3(g) HCl(g) + CCl4(g) (b) (2 pts)

Cl(g), CCl3(g)

(c) (3 pts) Rate = k2[CHCl3][Cl] K1[Cl2] = k-1[Cl]2, [Cl] = {k1/k-1 [Cl2]}1/2

Rate = k2(k1/k-1)1/2 [CHCl3][Cl2]1/2 Overall order is 1.5 11. (total 9 points) (a) (2 pts)

(No partial points) (b) (2 pts)

(No partial points) (c) (5 pts)

1/v0 intercept= 1/[S] intercept= Graph shape (if slope is +, 1/v0 intercept is +, and 1/[S] intercept is -) (1 pt) / each intercept (2 pts).

1

2018 FALL Semester Final Examination For General Chemistry II (CH103)

Date: December 12 (Wed), Time Limit: 19:00 ~ 21:00 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 /7 7 /6

/100

2 /10 8 /10

3 /8 9 /8

4 /12 10 /10

5 /13 11 /7

6 /9 ** This paper consists of 13 sheets with 11 problems (page 11: fundamental constants, page 12: periodic table, page 13: claim form). 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.

(채점 답안지 분배 및 이의신청 일정)

1. Period, Location, and Procedure 1) Return and Claim Period: December 14 (Fri, 12:00 ~ 14:00 p.m.) 2) Location: Room in Creative Learning Bldg. (E11)

Class Room Class Room Class Room Class Room

A 202 B 207 C 203 D 208

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) (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 will be uploaded on December 14 on the web.)

2. Final Confirmation 1) Period: December 15 (Sat) – December 16 (Sun) 2) Procedure: During this period, you can check the final score of the examination on the website.

2

1. (Total 7 pts) Following is a derivative of aspartame.

NH

O

C

NH2

COOH

OCH3

O

OH

(a) Name all the functional groups in this compound.

(Answer)

(b) How many chiral centers does the compound have? Draw * on each chiral center of the above

structure.

(Answer)

3

2. (Total 10 pts) The molecule shown in skeletal representation below is menthol, a crystalline organic

compound having local anesthetic and counterirritant properties.

HO

(a) What is the formula of menthol?

(Answer)

(b) How many chiral centers (asymmetric carbons) does it have? Indicate them on the structure.

(Answer)

(c) How many stereoisomers are possible (do not count conformational or geometrical isomers)?

(Answer)

4

3. (Total 8 pts) Styrene monomer undergoes addition polymerization with peroxide radical initiators.

H2C CH R O O R

Styrene Peroxide Initiator

(a) Write the initiation reactions of this polymerization.

(Answer)

(b) Write the propagation reactions of this polymerization.

(Answer)

(c) Write the termination reactions of this polymerization.

(Answer)

5

4. (Total 12 pts) How many lines appear in the proton NMR spectrum? Include signals arising from

both the chemical shift and J couplings.

(a) C2H4 (Answer)

(b) (Answer)

C C

Br

H

H

CCl3

(c) CH3CH2COCH2CH3 (Answer)

(d) CH3CHCl2 (Answer)

(e) (Answer)

OCH3

CH

OCH3

H3C

(f) (Answer)

H2C

H2C

CH2

O

6

5. (Total 13 pts)

(a) Tropomyosin is a two-stranded ɑ-helical coiled coil 70 kd muscle protein. Estimate an approximate

length of the tropomyosin molecule (molecular weight of 1 amino acid is 110 dalton and the rise per

residue of an ɑ-helix is 1.5 angstrom).

(Answer)

(b) Tautomer form of thymine(T) nucleotide is shown below. How this tautomer is base paired during

the replication of DNA?

N

N

OH

CH3

O

(Answer)

(c) Explain why human cannot eat cellulose in terms of structural unit.

(Answer)

(d) Write the nucleotide sequence of mRNA produced from the DNA template below.

5'-ATCGTACCGTTA-3'

(Answer)

7

6. (Total 9 pts) Below two molecules are Rubrene (≡ 5,6,11,12-tetraphenyltetracene) and Ir(mppy)3 (≡

Tris[2-(p-tolyl)pyridine]iridium(III)), which are widely used for organic light-emitting diode (OLED)

applications.

Rubrene Ir(mppy)3

(a) Predict which molecule will show faster light emission.

(Answer)

(b) Explain why that molecule in (a) shows faster light emission.

(Answer)

7. (Total 6 points) The percentage transmittance of light at 250 nm through a certain aqueous solution

is 20.0% at 25 °C. The experimental cell length is 1.0 cm, and the concentration of the solution is 5 ×

10-4 mol L-1. Calculate the absorbance. Calculate the molar absorption coefficient.

(Answer)

8

8. (Total 10 pts) Graphite has a layered structure, and each layer, also known as a graphene, consists

of the sp2 carbon atoms arranged in a honeycomb lattice. In such graphitic systems, there exists a

characteristic in-plane vibrational modes which is usually labeled as “G peak”.

(a) Decide which spectroscopic technique in between the IR absorption measurement and Raman

scattering should be employed to detect the “G peak”, and explain why.

(Answer)

(b) The vibrational frequency of the “G peak” is 4.75 × 1013 s-1. Calculate the wavenumber of the “G

peak” of the graphene. Also, predict the wavenumber of the “G peak” of the graphene synthesized

using 13C instead of 12C.

(Answer)

9

9. (Total 8 pts) Iron(III) and Platinum(IV) form octahedral complexes. Sketch the structures of all the

distinct isomers of following compounds, and indicate which pairs of structures are mirror images of

each other.

(a) [Fe(en)2Cl2]+ (en = ethylenediamine)

(Answer)

(b) Pt(NH3)2Cl2F2

(Answer)

10

10. (Total 10 pts) The octahedral complex ion [MnCl6]3- has more unpaired spins than the octahedral

complex ion [Mn(CN)6]3-.

(a) Which compound is paramagnetic?

(Answer)

(b) How many unpaired electrons are present in each species?

(Answer)

(c) In each case, express the crystal field stabilization energy in terms of Δ0.

(Answer)

11. (Total 7 pts)

(a) An aqueous solution of Fe(NO3)3 has only a pale color, but an aqueous solution of K3[Fe(CN)6] is

bright red. Do you expect a solution of K3[FeF6] to be brightly colored or pale? Explain your reasoning.

(Answer)

(b) Would you predict a solution of K2[HgI4] to be colored or colorless? Explain.

(Answer)

11

Physical Constants

Avogadro’s number NA = 6.02214179 ⅹ 1023 mol-1

Bohr radius a0 = 0.52917720859 Å = 5.2917720859ⅹ10-11 m

Boltzmann’s constant KB = 1.3806504 ⅹ 10-23 J K-1

Electronic charge e = 1.602176487 ⅹ 10-19 C

Faraday constant F = 96485.3399 C mol-1

Masses of fundamental particles:

Electron me = 9.10938215 ⅹ 10-31 kg

Proton mP = 1.672621637 ⅹ 10-27 kg

Neutron mn= 1.674927211 ⅹ 10-27 kg

Permittivity of vacuum εo = 8.854187817 ⅹ 10-12 C-2 J-1 m-1

Planck’s constant h = 6.62606896 ⅹ 10-34 J s

Ratio of proton mass to electron mass mP / me = 1836.15267247

Speed of light in a vacuum c = 2.99792458 ⅹ 108 m s-1 (exactly)

Standard acceleration of terrestrial gravity g = 9.80665 m s-2 (exactly)

Universal gas constant R = 8.314472 J mol-1 K-1 = 0.0820574 L atm mol-1 K-1

Values are taken from the 2006 CODATA recommended values, as listed by the National Institute of Standards and Technology. Conversion factors

Ångström 1 Å= 10-10 m

Atomic mass unit 1 u = 1.660538782 ⅹ 10-27 kg 1 u = 1.492417830 ⅹ 10-10 J = 931.494028 MeV (energy equivalent form E =

mc2) Calorie 1 cal = 4.184 J (exactly)

Electron volt 1 eV = 1.602177 ⅹ 10-19 J = 96.485335 kJ mol-1 Foot 1 ft = 12 in = 0.3048 m (exactly)

Gallon (U. S.) 1 gallon = 4 quarts = 3.785412 L (exactly)

Liter 1 L = 10-3 m-3 = 103 cm3 (exactly)

Liter-atmosphere 1 L atm = 101.325 J (exactly)

Metric ton 1 t = 1000 kg (exactly)

Pound 1 lb = 16 oz = 0.4539237 kg (exactly)

Rydberg 1 Ry = 2.17987197 x 10-18J = 1312.7136 kJ mol-1 = 13.60569193 eV

Standard atmosphere 1 atm = 1.01325 x 105 Pa = 1.01325 x 105 kg m-1 s-2 (exactly)

Torr 1 torr = 133.3224 Pa

12

13

Claim Form for General Chemistry Examination Page ( / )

Class: , Professor Name: , I.D.# : , Name: If you have any claims on the marked paper, please write down them on this form and submit this with your paper in the assigned place. (And this form should be attached on the top of the marked paper with a stapler.) Please, copy this sheet if you need more before use.

By Student By TA

Question # Claims Accepted? Yes(√) or No(√)

Yes: □ No: □ Pts (+/-) Reasons

<The Answers>

Problem points Problem points TOTAL pts

1 5+2/7 7 3+3/6

/100

2 1+6+3/10 8 4+6/10

3 2+2+4/8 9 3+5/8

4 2x6/12 10 2+4+4/10

5 3+4+3+3/13 11 4+3/7

6 4+5/9

전체 기준: 전개과정은 맞으나 답이나 unit 이 틀리면 -1

1. (Total 7 pts) (a) (5 pts) 1 pt for each

Carboxylic acid, amine, amide, ester, phenol (alcohol)

(b) (2 pts)

2 chiral centers

NH

O

C

NH2

COOH

OCH3

O

OH

**

2. (Total 10 pts) (a) (1 pts) C10H20O

(b) (6 pts) 3 chiral centers (3 pts).

HO* *

*

(3 pts)

(c) (3 pts) 8 isomers.

3. (Total 8 pts) (a) (2 pts)

R O + H2C CH C COR

H

H H

(b) (2 pts)

C COR

H

H H+ H2C CH C COR

H

H H

C C

H

H

H

(c) (4 pts) Coupling (2 pts)

C COR

H

H H

C C

H

H

H

C COR

H

H H

C C

H

H

H

+ C COR

H

H H

C C

H

H

H

C C C

H H

H

C

H

O

H

H

R

n

mn

m

Disproportionation (2 pts)

C COR

H

H H

C C

H

H

H

C COR

H

H H

C CH

H

H

H

+ C COR

H

H H

C C

H

H

H

C C C

H H

H

C

H

O

H

H

R

+n

m

4. (Total 12 pts) 2 pts for each (a) 1

(b) 4

(c) 7

(d) 6

(e) 7

(f) 8

5. (Total 13 pts) (a) (3 pts)

(b) (4 pts)

Hydrogen bonded to guanine(G) instead of adenine(A).

(c) (3 pts)

Cellulose is a linear polymer of β-glycoside unit. Therefore, human who has an α-glycosidase cannot

digest β-glycoside unit.

(d) (3 pts)

5'-UAACGGUACGAU-3'

6. (Total 9 pts) (a) (4 pts)

The Rubrene will show faster light emission than the Ir(mppy)3

(b) (5 pts)

Due to the presence of the heavy metal element of Ir, Ir(mppy)3 has much stronger spin-orbit coupling,

which facilitates the intersystem crossing. (2 pts)

This enables the Ir(mppy)3 to exhibit a phosphorescence, while the Rubrene consisting of light

elements of C and H exhibits a fluorescence. (3 pts) Thus, the Rubrene will show faster light emission

than the Ir(mppy)3.

7. (Total 6 pts) absorbance 3 pts, molar absorption coefficient 3 pts

More accurate calculation yields ε = 1.40 × 103 L mol-1 cm-1.

8. (Total 10 pts) (a) (4 pts)

The vibration mode of the “G peak” involves no change in the dipole moment, but change in the

polarizability. Thus Raman scattering is the proper method to detect the “G peak”.

(b) (6 pts) wavelength 3 pts, 13C 3 pts

𝜈𝜈 =𝑐𝑐𝜆𝜆

= 𝑐𝑐𝜈𝜈�

Consequently,

𝜈𝜈� =𝜈𝜈𝑐𝑐

=4.75 × 1013 s−1

3 × 1010 cm s−1 = 1,583 cm−1

Using the relation of 𝜈𝜈 = 12𝜋𝜋�

𝑘𝑘𝜇𝜇,

𝜈𝜈�′ = �𝜇𝜇𝜇𝜇′𝜈𝜈� = �12

13× 1,583 = 1,521 cm−1

9. (Total 8 pts) (a) (3 pts) 1 pt for each isomer

(b) (5 pts) 5 pts for 6 isomers, 4 pts for 5 isomers … 1 pts for 1 or 2 isomers

Pt

H3N Cl

H3N F

Cl

Fmirror

10. (Total 10 pts) (a) (2 pts)

Both [MnCl6]3- and [Mn(CN)6]3-.

(b) (4 pts) 2 pts for each

[MnCl6]3-

4 unpaired electrons. [Mn(CN)6]3-

2 unpaired electrons. (c)(4 pts) 2 pts for each

[MnCl6]3- : t2g3eg

1.: -3/5 Δ0

[Mn(CN)6]3- : t2g4. : -8/5 Δ0

11. (Total 7 pts) (a) (4 pts) Answer 1 pt, Reasoning 3 pts

(b) (3 pts) Answer 1 pt, Reasoning 3 pts