O O O O F Si F H. Structure A Structure B Structure C (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 (A) 2 (B) 4 (C) 5 (D) 6 (E) 8

Transcription

1 For Questions 1 7 consider the Lewis structures below and assume that each one is drawn correctly. (Any additional resonance forms are not shown here.) H O O O O F Si F H C C C H Structure A Structure B Structure C O (3 pts) 1. How many of these structures would also have one or more additional resonance forms? (A) 0 (B) 1 (C) 2 (D) 3 (3 pts) 2. What type of hybridization is required for the silicon atom in Structure B? (A) sp 4 (B) sp 3 (C) sp 2 (D) sp (3 pts) 3. How many p bonds are present in Structure C? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 (3 pts) 4. How many s bonds are present in Structure C? (A) 2 (B) 4 (C) 5 (D) 6 (E) 8 (2 pts) 5. True or false: The two carbon oxygen bonds in Structure C are the same length. (A) True (B) False (4 pts) 6. One of these structures is an ion. Which choice correctly identifies the ion and its charge? (A) Structure A, 1+ (B) Structure B, 1 (C) Structure C, 1 (D) Structure A, 1 (E) Structure C, 1+ (5 pts) 7. What are the correct molecular shapes for Structures A and B? (A) Structure A is linear, Structure B is tetrahedral (B) Structure A is trigonal planar, Structure B is square planar (C) Structure A is linear, Structure B is see-saw shaped (D) Structure A is trigonal planar, Structure B is square planar (E) Structure A is bent, Structure B is tetrahedral A2

2 (4 pts) 8. How many of the following statements are true? NAME: (i) (ii) (iii) (iv) The smaller the band gap in a material, the better its conductivity will be. Doping silicon with aluminum produces a p-type semiconductor. Metals have large band gaps. The conduction band is always at lower energies than the valence band. (A) 0 (B) 1 (C) 2 (D) 3 (E) 4 (4 pts) 9. Which of the following band diagrams would be expected for a material made by doping germanium (Ge) with gallium (Ga)? (A) (B) (C) (D) For Questions10 14, determine whether each of the following should be positive, negative, or zero. (2 points each) 10. q for converting 100 g of liquid water at 0 C being converted to ice at 0 C 11. G for the melting of 100 g of ice at 1 atm and 25 C 12. S for the separation of CO2 into atoms: CO2(g) C(g) + 2 O(g) 13. H for the separation of CO2 into atoms: CO2(g) C(g) + 2 O(g) 14. S for any element in its standard state A3

3 Use the following thermodynamic data as needed to answer Questions C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(g) DH = kj Compound Hf (kj/mol) S (J/mol/K) C2H2(g)? O2(g) CO2(g) H2O(g) (5 pts) 15. What is Hf for C2H2(g)? (A) kj/mol (B) kj/mol (C) kj/mol (D) kj/mol (E) kj/mol DH = 4 DHf (CO2) + 2 DHf (H2O) 2 DHf (C2H2) 5 DHf (O2) = kj 2 DHf (C2H2) =4 DHf (CO2) + 2 DHf (H2O) 5 DHf (O2) kj = 4 mol ( kj/mol) + 2 mol( kj/mol) 5(0) kj = kj DHf (C2H2) = kj/2 mol = kj/mol (5 pts) 16. What is S for the combustion reaction? (A) J K 1 (B) J K 1 (C) J K 1 (D) J K 1 (E) J K 1 DS = 4 S (CO2) + 2 S (H2O) 2 S (C2H2) 5 S (O2) =4 mol (213.7 J/mol/K) + 2 mol(188.7 J/mol/K) 2 mol( J/mol/K) 5 mol(205.0 J/mol/K) = J/K A4

4 NAME: (6 pts) 17. How much heat would be released by the combustion of 358 g of C2H2? (You should still be using information given on the previous page here!) (A) 7.01 kj (B) 17,300 kj (C) 34,500 kj (D) 3120 kj (E) 2510 kj 358 g 1 mol 2511 kj = 17, 300 kj g 2 mol For Questions 18 & 19, consider a particular chemical reaction for which DH = 96.6 kj and DS = 382 J/K. (5 pts) 18. What is the approximate value of DG for this reaction at T = 100 C? (Assume that both DH and DS are independent of temperature.) (A) 58.4 kj (B) kj (C) kj (D) 45.9 kj (E) 239 kj G = H T S = kj ( K) > kj? = kj K (5 pts) 19. At what temperatures would this reaction be predicted to be spontaneous? (A) T > 253 K (B) T > 298 K (C) T = 298 K (D) T < 298 K (E) T < 253 K G = H T S = 0, so H = T S and T = H S kj = kj = 253 K K That s the T at which DG changes sign. Both DH and DS are negative, so the reaction is spontaneous only at temperatures BELOW 253 K. A5

5 (10 pts) 20. Draw Lewis structures for each of the species listed below. Show all important resonance structures where appropriate. Please draw your final structures neatly in the boxes! (a). XeF5 + Graded as 4 points for the top one and 6 for the bottom, because of the resonance. (b). NO2F (The 3 other atoms are all bound to the nitrogen.) A6

6 21. NAME: (5 pts) (a). The thermodynamic standard states of titanium and iodine are Ti(s) and I2(s). Write a chemical equation for the formation reaction of TiI3(s). Be sure to include the correct physical states in your equation. Ti(s) + 3 I2(s) TiI3(s) 2 (8 pts) (b). The heat of formation (DHf ) for TiI3(s) is 328 kj/mol, and DH = 839 kj for the following reaction: 2 Ti(s) + 3 I2(g) 2 TiI3(s) Find the heat of sublimation for I2. (Recall that sublimation is a phase change from solid to gas, like the dry ice demo that we saw in class last week. So the heat of sublimation would be DH for the process I2(s) I2(g).) There are a couple of ways to do this. One is to use the formation reaction from (a) and the reaction above and do Hess s law. The sublimation equation is: 2 3 {Ti(s) I2(s) TiI3(s)} 1 3 {2 Ti(s) + 3 I2(g) 2 TiI3(s)} So DH = 2 3 ( 328 kj) 1 3 ( 839) = 61.0 kj/mol Another way is to realize that the sublimation reaction is the formation reaction for I2(g). So you can do: DH = 839 kj = 2DHf (TiI3) 2DHf (Ti) 3DHf (I2(g)) DHf (I2(g)) = 1 3 { 2DHf (TiI3) 2DHf (Ti) } = 1 3 { 2 ( 328) 2 (0) } = 61.0 kj/mol A7

7 (10 pts) 22. A particular ring contains both 18 K gold and sterling silver, with the two metals in an intricate design resembling braided ropes. The total mass of the ring is 15.0 g, and a jeweler wants to determine the masses of gold and of silver present. The clever jeweler places the ring into some hot water, heating it to a temperature of 87.6 C, and then drops the hot ring into 18.0 g of water at 21.8 C. When the ring and the water reach thermal equilibrium, the final temperature is 24.0 C. What mass of gold is present in the ring? Specific heats: 18 K gold: J g 1 C 1 Sterling silver: J g 1 C 1 Water: J g 1 C 1 I think the simplest way to do this is to first find the specific heat of the ring, and then use that to find mass of each metal. qw = qring (mcdt)w = (CDT)ring (18.0 g)(4.184 J g 1 C 1 )(2.2 C) = Cring( 63.6 C) J = 63.6 g C * Cring Cring = J C 1 Here that Cring is like a calorimeter constant it is the heat capacity of the object. We can relate it to the masses and to the specific heats of the individual metals. Cring = J C 1 = maucau + magcag We also know that mau + mag = 15, so we can write: J C 1 = maucau + (15 mau) CAg = 0.132mAu + 15(0.245) 0.245mAu = 0.113mAu mau = 9.47 g Alternatively, if we don t start with specific heat of the ring, we could merge the 2 steps. qw = qring (mcdt)w = {(mcdt)au + (mcdt)ag)} (18.0 g)(4.184 J g 1 C 1 )(2.2 C) = {(mcdt)au + (mcdt)ag)} J = {(mcdt)au + (mcdt)ag)} We still know that mau + mag = 15, so we can write: = {(mau CAu) + (mag CAg))} DT = {(0.132mAu) + (15 mau) 0.245} ( 63.6) = 0.132mAu mAu = 0.113mAu mau = 9.47 g A8