Chem 152 Final. You will have 1 hour and 50 minutes. Do not begin the exam until you are instructed to start. Best of luck.

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1 Chem 152 Final Section: Name: You will have 1 hour and 50 minutes. Do not begin the exam until you are instructed to start. Best of luck. Question 1 /80 Question 2 /20 Question 3 /20 Question 4 /20 Question 5 /20 Question 6 /20 Question 7 /20 Total /200 Page 1

2 1. Multiple Choice (4 pts. each): a. Which of the following statements about quantum theory is incorrect: A: The momentum and position of an electron can be determined simultaneously. B: Lower energy orbitals are filled with electrons before higher energy orbitals. C: Hund s rule dictates that electrons will occupy unfilled orbitals before occupied orbitals. D: No two electrons can have the same four quantum numbers. b. Of energy, enthalpy, heat, entropy, and the Gibbs energy how many are state functions? A: 2 B: 3 C: 4 D: 5 c. Using the following 1/2 cell reactions: 2H + + 2e - 0 H 2 E 12 = (you should know this) H 2 O 2 + 2H + + 2e - 0 2H 2 O E 12 = 1.78 V What is G for the following reaction (in kj): 2H 2 O H 2 O 2 + H 2 A: -343 B: 343 C: 687 D: -687 Page 2

3 d. For the following reaction under standard thermodynamic conditions: H 2 Ol ( ) H 2 O( g) H is more positive than E by 2.5 kj mol -1. This difference can be assigned to: A. The heat flow required to maintain a constant temperature B. The work done by the system C. The difference in the H-O bond strength between liquid and gaseous water. D. That s just how H is calculated. e. Consider the following reaction: CH 4 ()+ g 4Cl 2 () g CCl 4 ( g)+ 4HCl( g) H = 434 kj If 19.2 g of CH 4 undergoes this reaction at constant pressure, what amount of heat is transferred between system and surroundings? A: 520 kj of heat are released to the surroundings B: 520 kj of heat are absorbed by the system C: 360 kj of heat are released to the surroundings D: 360 kj of heat are absorbed by the system f. For the following reaction occurring in basic solution, in the balanced equation how many electrons are transferred? Fe(s) + NO 3 (aq) Fe + (aq) + NH 3 ( g) A: 3 B: 4 C: 5 D: 8 g. Which of the following electron configurations is correct? Page 3

4 A. Ga: [Kr]4s 2 3d 10 4p 1 B. Bi: [Xe]6s 2 4f 14 5d 10 6p 3 C. Ca: [Ar]4s 1 3d 10 D. Mo: [Kr]5s 2 4d 5 Consider the following half reactions for questions h, j, k, and l: Cr e Cr Br 2 + 2e 2Br E 0 = 0.73 V E 0 =+1.09 V h. What is E cell for a galvanic cell based on these 1/2 cells? A: V B: V C: 0.36 V D: V j. What is ln(k) for the galvanic cell? A. 425 B. 425 C. 142 D. 213 k. If [Cr +3 ] = 0.2 M, [Br - ] = 0.1 M, and [Br 2 ] = 0.5 M, what is the value of E cell at 298 K? A: 2.21 V B: 1.76 V C: 2.12 V D: 1.88 V l. Which of the following statements is true about this cell? A. Br - is reduced B. Cr +3 is reduced C. Cr is reduced D. Br 2 is reduced m. Which of the following compounds has the bond with the most ionic character? Page 4

5 A: LiCl B: KF C: KCl D: NaCl n. Which of the following molecules has no dipole moment? A: NO 2 - B: O 3 C: PCl 5 D: SO 2 o. Which of the following molecules does not contain a double or triple bond? A: N 2 B: H 2 CO C: C 2 H 6 D: SCN - p. Which of the following molecules has a Lewis structure most like CO 3 2-? A: CO 2 B: SO 3 2- C: NO 3 - D: O 3 q. Consider the following reactions: 2ClF + O 2 Cl 2 O + F 2 O H = kj mol -1 2ClF 3 + 2O 2 Cl 2 O + 3F 2 O H = kj mol -1 2F 2 + O 2 2F 2 O H = 43.4 kj mol -1 At the same temperature, what is H for the following reaction? ClF + F 2 ClF 3 A: kj mol -1 B: kj mol -1 C: kj mol -1 D: kj mol -1 r. For a certain reaction, H = 40 kj and S = 50 J K -1. This reaction will be: Page 5

6 A: spontaneous at temperatures greater than 800 K B: spontaneous at temperatures less than 10 K C: spontaneous at all temperatures D: no temperature at which the reaction is spontaneous s. Based on Lewis dot structures, which of the following would you expect to not be a stable molecule? A: NH 3 B: N 2 H 2 C: N 2 H 4 D: N 2 H 6 t. In nonpolar KrCl 4, the Cl-Kr-Cl bond angle is: A: 120 B: 90 C: 180 D: 109 u. For which of the following reactions is S expected to be most positive? A: O 2 ()+ g 2H 2 () g 2H 2 Og ( ) B: H 2 Ol () H 2 O() g C: 2NH 4 NO 3 (s) 2N 2 ( g)+ O 2 ( g)+ 4H 2 Og D: N 2 O 4 () g 2NO 2 ( g) ( ) /80 Page 6

7 Section II: Long-Answer/Numerical Questions 2 (20 pts.) Lewis Dot Structures. Write the Lewis dot structures for the following compounds. Include all structural isomers, and resonance structures if appropriate. Underline the one you believe is the most-reasonable structure. Finally, describe the 3D geometry of your most-reasonable structure. An example is provided below: Ex: Cl 2 O Cl O Cl Cl Cl O Cl O Cl Cl Cl O a) IO 2 - geometry _bent. b) CCl 4 geometry. geometry. Page 7

8 Problem 2 continued. c) ONF d) N 2 O geometry. geometry. /20 Page 8

9 3 (20 pts.) One mol of an ideal monatomic gas at standard temperature and pressure undergoes reversible isochoric heating until a final pressure of 5 atm is reached. Next, the gas undergoes reversible isothermal expansion until the original pressure is reached. Calculate E, q, w, H, and S for this thermodynamic process. Use C v (Ne) = R if you do not remember the constantvolume heat capacity for an ideal monatomic gas. Step 1: Isochoric Heating ( ) 3 2 R E = nc V T = 1 mol H = nc P T = 1 mol q = E w = 0 T i ( ) 5 2 R S = nc V ln T f = 1 mol ( ) 3 2 R Step 2: Isothermal Expansion ( 1490 K 298 K)= 14.9 kj ( 1490 K 298 K)= 24.8 kj ln 5 ()= 20.1 J K -1 E = H = 0 q = w = nrt ln V f = ( 1 mol) ( R)1490 ( K)ln ()= kj S = nrln V f = ( 1 mol) ( R)ln 5 V i V i ()= 13.4 J K-1 q: 34.8 kj w: kj E: 14.9 kj H: 24.8 kj S:33.5 J K -1 (more room on next page) /20 Page 9

10 4) (more room for problem 3) Page 10

11 4. (20 pts) Consider the following reaction: 2NO 2 (g) N 2 O 4 (g) One initially begins a reaction with only N 2 O 4 present at a pressure of 10 atm. What is the pressure of NO 2 at equilibrium? The following thermodynamic data will prove useful, and you can employ the standard assumptions of equilibrium chemistry. H f 0 (kj mol -1 ) S 0 (J mol -1 K -1 ) NO 2 (g) N 2 O 4 (g) H 0 = H f 0 (N 2 O 4 ) 2 H f 0 (NO 2 ) = 58 kj S 0 = S 0 (N 2 O 4 ) 2S 0 (NO 2 ) = 176 J K -1 ( ) 1 kj G 0 = H 0 T S 0 = 58 kj -( 298 K) 176 J K J = 6 kj K = e G0 RT = e +2.4 = 11.3 K = 11.3 = P N 2 O 4 P = 10 x 2 NO2 2x 4x 2 = x = 0.47 = 2x = 0.94 atm P NO2 ( ) 10 ( ) 2 ( 2x) 2 /20 Page 11

12 5. (20 pts.) Using the bond-enthalpy method, determine H rxn for the following: C 2 H 4 + 3O 2 2CO 2 + 2H 2 O Bond enthalpies (in kj mol -1 ) C C 347 C C 614 C C 839 O O 146 O O 495 C O 799 C O 1072 O H 467 C H 413 H = ( CC double )+ 4( CH )+ 3(OO double ) 4( CO double ) 4( OH ) = ( 614 kj mol -1 )+ 4( 413 kj mol -1 )+ 3( 495 kj mol -1 ) 4( 799 kj mol -1 ) kj mol -1 = 1313 kj mol -1 ( ) /20 Page 12

13 6. (20 pts) For a number of years it was unclear whether mercury cation existed in solution as Hg + or Hg A electrochemical cell was constructed using mercury electrodes at both the anode and cathode, A solution of g of mercury nitrate (HgNO 3 ) in 100 ml of water was placed at the anode, and a corresponding solution of g of mercury nitrate in 50 ml of water was placed at the cathode. A voltage of V was measured for this cell. How does mercury cation exist in solution, as Hg + or Hg 2 +2? You must show all work to receive any credit on this problem. 0 E cell = E cell V log( Q) n V = V log( Q) n Q = [] anode [] cathode = g ( 0.1 L) MW g 0.05 L ( ) ( )( MW ) V = V log( 0.167)= V ( 0.777)= n n V= V n n = 2 = (note: MW cancels out in Q; therefore, it is irrelevant) Need cation that corresonds to a two electron reduction: Hg e Hg /20 Page 13

14 7. (20 pts.) a) Consider the following reaction: N 2 (g) + O 2 (g) 2NO(g) 0 Given that G f ( NO(g) )= 86.7 kj mol -1, determine the equilibrium constant for this reaction. 0 G rxn 0 = 2 G f ( NO)= kj K = e G rxn 0 RT = Page 14

15 b. (12 pts) Given that S 0 ( N 2 (g))= 192 J mol -1 K -1, S 0 ( O 2 (g))= 205 J mol -1 K -1, and S 0 ( NO(g) )= 211 J mol -1 K -1 0, determine H f ( NO(g) ). S 0 = 2S 0 (NO) S 0 (O 2 ) S 0 (N 2 ) = 25 J K -1 G 0 = H 0 T S 0 ( ) 1 kj H 0 = G 0 + T S 0 = kj + ( 298 K) 25 J K J = kj H 0 = 2 H f 0 (NO) kj = 2 H f 0 (NO) 90.4 kj mol -1 = H f 0 (NO) /20 Page 15