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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 Class Student I.D. Number Name Name Problem points Problem points TOTAL pts 1 /13 7 /12 2 /6 8 /9 3 /14 9 /6 4 /7 10 /9 5 /13 11 /5 6 /6 /100 ** 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. ( 채점답안지분배및이의신청일정 ) 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 1

2 <The Answers> Problem points Problem points TOTAL pts / / / / / / / / / /5 / /6 1. (Total 13 pts) (a) (6 pts)

3 (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, S o vap = H o vap / T b because of S = q/t. CH 3 CH 3, CH 3 F CH 3 OH H o vap (kcal/mol) T b (K) S o vap (kcal/k mol) (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 S o vap and T b due to the hydrogen bonding of the alcohol group, and ethane (CH 3 CH 3 ) has only dispersion forces.

4 3. (Total 14 pts) (a) (4 pts) n(biphenyl) = 5.50 g/154.2g mol -1 = mol m biphenyl) = mol / kg = mol kg -1 Kb = ΔTb/m = K / mol kg -1 = 2.53 K kg mol -1 for benzene (b) (4 pts) m = ΔTb/Kb = K / 2.53 K kg mol -1 = mol kg -1 n = mol kg -1 X kg = mol M (molar mass of solute) = 6.30 g / 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 So, K = P NH3 P HCl = 1.04 X 10-2 x 2 = 1.04 X 10-2 ; x = Answer: P NH3 = P HCl = atm. (c) (3 pts) n NH3 = n HCl = 1.02 X 10-1 atm X 1.000L/ L atm mol -1 K -1 X 548.2K = X 10-3 mol The mass of NH 4 Cl consumed = X 10-3 mol X g mol -1 = g So, the remaining mass = g Answer: NH 4 Cl = g

5 5. (Total 14 pts) K ln K T = 298K, T 1 H H = R T2 K = 308K, K JK ln k = K 308K Therefore, 1 H T = k 0 = 76kJ mol K2 R ln K1 = 1 1 T1 T 2 1 ln k (a) (3 pts) k = 2, ΔH 0 = 76 kj mol -1 ln2 = 53 kj mol -1 (b) (3 pts) k = 1/2, ΔH 0 = 76 kj mol -1 ln(1/2) = -53 kj mol -1 (c) (7 pts) 2NO 2 N 2 O 4 P NO2 = P x and P N2O4 = x K = x/(p- x) 2 0 x 1 Using ln ln K 2 = 6.96 x x = 0, x = P N2O4 = atm, P NO2 = 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. CH 3 COOH < ClCH 2 COOH < Cl 3 CCOOH (b) (3 pts) More electrons are withdrawn from the XOH as the electronegativity of the halogen increase. ClOH > BrOH > IOH

6 7. (Total 12 pts) (a) (6 pts) H 2 A sc (aq) H + (aq) + HA - sc (aq) K a1 = HA - sc (aq) H + (aq) + A 2- sc (aq) K a2 = H 2 A sc (aq) H + (aq) + HA - sc (aq) Starting Change -x +x +x Equilibrium 0.1-x x x K a1 = [H + (aq)][ha - sc (aq)]/[h 2 A sc (aq)] = = x 2 /(0.1 x) Assuming x to be much smaller than 0.1 x 2 /0.1 = x = ph = -log[h + ] = -log[0.0028] = 2.55 (b) (6 pts) HA - sc (aq) H + (aq) + A 2- sc (aq) Starting Change -y +y +y Equilibrium x x y K a2 = [H + (aq)][ A 2- sc (aq)]/[ha - sc (aq)] = = ( y)y/( y) ~ y/ = y [A 2- sc (aq)] = (Total 9 pts) For first ionization, H 2 CO 3 (aq) + H 2 O(l) HCO 3 -(aq) + H 3 O+(aq) Initial ~0 Change -x +x +x Equilibrium x x x So, Ka1 = x 2 /(0.034-x) = 4.3 X 10-7, X = [H 3 O+] = [HCO 3 -] = 1.2 X 10-4 M [H 2 CO 3 ] = M For second ionization,

7 K a2 = 1.2 X 10-4 X [CO 3 2- ]/1.2 X 10-4 = 4.8 X [CO 3 2- ] = 4.8 X M Answer: H 2 CO 3 = M, HCO 3 - = H 3 O + = 1.2 X 10-4 M, CO 3 2- = 4.8 X M 9. (Total 6 pts) (a) (3 pts) K sp = [Zn 2+ ][OH - ] x = (S)(2S) 2 4S 3 = 4.5 x 10-17, [Zn 2+ ] = 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 [Zn 2+ ] = K sp /[OH - ] 2 = (4.5 X )/ (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)

8 11. (5 pts)

9 1. (Total 13 pts) The normal boiling point of iodomethane, CH 3 I, is o C, and its vapor pressure at 0.00 o C is 140 Torr. (a) (6 pts) Calculate the standard enthalpy of vaporization of iodomethane. (b) (3 pts) Calculate the standard entropy of vaporization of iodomethane. (c) (4 pts) Calculate the vapor pressure of iodomethane at 25.0 o C. 2

10 2. (Total 6 pts) (a) (3 pts) Compare S o vap for CH 3 CH 3, CH 3 F, and CH 3 OH as calculated from H o vap and T b (boiling temperature) of the compounds. CH 3 CH 3, CH 3 F CH 3 OH H o vap (kcal mol -1 ) T b (K) (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. 3

11 3. (Total 14 pts) (a) (4 pts) When 5.50 g of biphenyl (C 12 H 10 ) is dissolved in g of benzene, the boiling point increases by o C. Calculate K b for benzene. (b) (4 pts) When 6.30 g of an unknown hydrocarbon is dissolved in g of benzene, the boiling point of the solution increases by o C. What is the molar mass of the unknown substance? (c) (6 pts) A 0.40 g sample of a polypeptide dissolved in 1.0 L of an aqueous solution at 27 o C gave rise to an osmotic pressure of 3.74 Torr. What is the molar mass of the polypeptide? 4

12 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 o C is We place g of solid NH 4 Cl into a closed vessel with volume L and heat to 275 o C. (a) (1 pt) In what direction does the reaction proceed? (b) (3 pts) What is the partial pressure of each gas at equilibrium? (c) (3 pts) What is the mass of solid NH 4 Cl at equilibrium? 5

13 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? (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? (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 25 o C. What are the partial pressure of NO 2 and N 2 O 4? The equilibrium constant K is 8.06 x 10 5 at -75 o C and the standard enthalpy change for this reaction ( H o ) is kjmol -1. 6

14 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 (b) (3 pts) ClOH, BrOH, IOH 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 = and K a2 = (a) (6 pts) What is the ph of a 0.1 M solution? (b) (6 pts) What is the concentration of ascorbate ion, C 6 H 6 O 2-6? 7

15 8. (Total 9 pts) Calculate the equilibrium concentrations of H 2 CO 3, HCO - 3, CO 2-3, and H 3 O + in a saturated aqueous solution of CO 2, in which the original concentration of H 2 CO 3 is M. K a1 and K a2 are 4.3 X 10-7 and 4.8 X 10-11, respectively. 9. (Total 6 pts) Zinc hydroxide (Zn(OH) 2 ) is sparingly soluble base (K sp = 4.5 X ). (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. (b) (3 pts) Compare the solubility in pure water with that in a solution buffered at ph

16 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 K a for the acid? (b) (4 pts) What is the initial concentration of the acid? (c) (3 pts) What is the concentration of base in the titrant? 9

17 11. (5 pts) Describe briefly the names and their achievements of the Nobel Laureates in Chemistry

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19 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 2 /8 7 /8 3 /10 8 /11 4 /10 9 /11 5 /10 10 /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) 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 1

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 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

20 2013 FALL Semester Final Examination for General Chemistry II (CH103) Problem points Problem points TOTAL pts / / / / / / / /11 / / /8 1. (12 points) 2 points 4 points 2 points (d) At ph = 2, hydronium ion concentration is M, and n = v /10 x log ([Mn 2+ ] 2 [Zn 2+ ] 5 )/([MnO 4 - ] 2 [H 3 O + ] 16 ) = 2.25v v/10 x log (5.3 x ) = 2.14 v 4 points 2

21 2. (8 points) AgCl(s) + e - Ag(s) + Cl- (aq) cathode Cu(s) Cu 2+ (aq) + 2 e - anode Q = It Chemical amount of electrons (0.500 C s -1 )(6.06 X 10 3 s)/(96,485 C mol -1 ) = (3.14 X 10-2 mol e - ) (3.14 X 10-2 mol e - ) X ( 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) = g Cu dissolved 4 points 3. (10 points) 1 point per each (e) (total 6 points) 3 points 3 points 3

22 4. (10 points) ka k1 A I I 1 2 k2 P d[a] = k A[A], [A] = [A] 0e dt d[i1] = k A[A] k1[i1] dt d[i2] = k1[i1] k2[i2] dt d[p] = k2[i2] dt kat,[a] 0 = initial concentration of A 3 points By using the steady-state approximation, d[i1] = k dt A k [A] k1[i1] = 0 [I1] = k k [A] = k d[i2] k1 k = k1[i1] k2[i2] = 0 [I2] = [I1] = dt k k d[p] = k2[i2] = k A[A] 0e dt kat [P] = [A] (1 e ) 0 kat A 1 2 A 1 A 2 [A] [A] 0 0 e e kat kat 4 points When the concentration of P reaches to one-half of the initial concentration of A, [P] = [A] (1 e t = 0 ln s 1 k A t = 35 s [A] ) = 2 0 t = ln 2 k A 3 points 5. (10 points) 4

23 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 6 points Br Br Br Br Br Br 5

24 (c) [Co(NH 3 ) 3 Br 3 ] : mer- and fac- 4 points Br NH 3 Br NH 3 Br NH 3 H 3 N NH 3 Br Br NH 3 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: t 2 t 2 e e each configuration 2 points, total 4 points 6

25 8. (11 points) (a) 3 points N H + H N A (b) 4 points B H N H N H N O N O +1 A - O N O H B +1 - O N O 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 NH 3 NH 2 NH Electron deficient at ortho-, para-position or electron rich at meta-position: +1 7

26 9. (11 points) (a) 2 points Addition (b) 3 points chiral carbon H 2 C H 3 C C H n (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

27 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 (Mn 2+ ). Under standard conditions, the reduction potential of [MnO 4 - Mn 2+ ] half-cell is ε 0 = 1.49 V. Suppose this half-cell is combined with [Zn 2+ Zn] half-cell (ε 0 = V) in a galvanic cell, with [Zn 2+ ] = [MnO 4 ] = [Mn 2+ ] = [H 3 O + ] = 1 M. (a) Write equations for the reactions at the anode and the cathode. (b) Write a balanced equation for the overall cell reaction. (c) Calculate the standard cell potential difference, ε 0. (d) Suppose the cell is operated at ph 2.00 with [MnO 4 ] = 0.12 M, [Mn 2+ ] = M, and [Zn 2+ ] = M. Calculate the cell voltage ε at 25 o C. 9

28 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 Cu 2+ (aq) at the anode. A current of A is passed through the cell for 101 min. Calculate the mass of copper dissolved and the mass of silver deposited. 10

29 3. (10 points) State which atom of each of the following pairs is more electronegative: (a) sulfur, phosphorus; (b) selenium, tellurium; (c) sodium, cesium; (d) silicon, oxygen. (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 = and K a2 is negligible. 11

30 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 = s -1 and k 1 = k 2 = 0.20 s -1. k 1 A k k I1 I2 P A 2 12

31 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; (b) one-fourth; (c) one-fifth of its original value. 13

32 6. (12 points) Draw out all the isomers, geometric and optical, of the following compounds (en = ethylenediamine). (a) [Co(en) 3 ] 3+ (b) [Co(en) 2 Br 2 ] + (c) [Co(NH 3 ) 3 Br 3 ] 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. 14

33 8. (11 points) In general, the electron donating group activates the benzene ring to electrophilic attack while the withdrawing group deactivates. N H + 1) NO 2 + N 2) Base NO 2 A (a) Please draw the missing intermediate A and provide an arrow pushing mechanism. (b) Please propose an arrow pushing mechanism for the conversion of molecule A to the final product (The structure of NO 2 + is required). (c) Briefly rationalize the experimental results for this reaction. (Hint: resonance of intermediate A) 15

34 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? (b) Draw the repeating unit of polypropylene and identify the chiral carbon. (c) Draw three isomeric structures of polypropylene including their name. 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. 16

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1 /10 6 /10 2 /5 7 /5 3 /10 8 /10 4 /10 9 /10 5 /10 10 /20 2009 FALL Semester Midterm Examination For General Chemistry II (CHEM103) Date: October 21, 2009 (Wed), Time Limit: 7:00 ~ 9:00 p.m. Write down your information neatly in the space provided below; print