CHEM 10113, Exam 4. All equations must be balanced and show phases for full credit. Significant figures count, and box your answers!

Size: px

Start display at page:

Download "CHEM 10113, Exam 4. All equations must be balanced and show phases for full credit. Significant figures count, and box your answers!"

Kristina Mitchell
6 years ago
Views:

1 CHEM 10113, Exam 4 November 30, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, and box your answers! 1. (10 points) Consider the phase diagram for iodine shown here and answer each of the following questions. a. What is the normal boiling point for iodine? C b. What is mp for iodine at 1 atm? C c. What state is present at rt and normal atmospheric pressure? solid d. What state is present at 186 C and 1.0 atm? gas e. If a dynamic equilibrium is established between the solid and liquid phases, name a change you can make to increase the amount of liquid. Increase the temperature 2. (16 points) SHOW ALL WORK. Which of the following molecules are polar? IF 5 Polar BH 3 Nonpolar 1

2 PF 5 Nonpolar SBr 2 Polar ClO 4 Nonpolar O 2 2 Nonpolar IF 4 Nonpolar TeF 4 Polar 2

3 3. (18 points) SHOW ALL WORK. Benzene has a heat of vaporization of kj/mol and a normal bp of 80.1 C. At what temperature does benzene boil when the external pressure is 445 torr? Clausius-Claperon: ln P 2 = - H vap (1/T 2 1/T 1 ) P 1 R Let; P 2 =760 torr, T 2 =353.25K, P 1 =445 torr, T 1 =x ln 760 torr = J/mol 445 torr J/mol K (1/ /x) ln(1.7078) = ( )(1/ /x) = ( ) (1/ /x) = 1/ /x = /x = 1/x x = T 2 = K = C 4. (12 points) For each of the following pairs, decide which has the greater bond angle, or if they are equal, and defend your answer: [note; no credit without an explanation] a) NO 2 or NO 3 angle <120 due to lone pair 3

4 angle = 120 due to resonance all are equivalent => larger bond angle b) NO 2 or NO 2 + angle ~120 angle = 180 =>larger bond angle 5. (5 points) When a thin glass tube is put into water, the water rises 1.4 cm. When the same tube is put into hexane, the hexane rises only 0.4 cm. Explain the difference. Water can generate strong adhesive interactions with the glass (due to the dipoles at the surface of the glass), but hexane is nonpolar and cannot interact strongly with the glass surface. 6. (12 points) For the following organic molecule, give the valence bond description of the bonding. For the sketch, show only one example of each type of bond. (e.g. in H 2 O only draw one O-H bond) Formula EG at central atom Hybridization at central atom Draw the orbital interaction(s) NF 3 4 sp 3 Is bond or? ClOF 4 6 sp 3 d 2 4

7. (12 points) a) Label the strongest type of intermolecular attraction for each material, then b) explain the melting point data in terms of molecular attractive forces.

5 7. (12 points) a) Label the strongest type of intermolecular attraction for each material, then b) explain the melting point data in terms of molecular attractive forces. a) mp Strongest intermolecular attractive force F C London Dispersion Force PH C Dipole-dipole HBr -87 C Dipole-dipole CH 3 OH 98 C H-bonding b) Explain: As the strength of the intermolecular force increases, the melting point increases. Also, between HBr and PH 3, HBr has a larger dipole moment and so its dipoledipole interactions are stronger. MP increases with stronger forces because it takes more energy to break apart strong forces. 8. (14 points) SHOW ALL WORK. According to MO theory, which has the highest bond order, bond energy, and shortest bond length? O 2, O 2, or O 2 +? Include an MO diagram in your answer. 5

6 Bond Order = ½ [# bonded e - - # antibonded e - ] O 2 = ½ [8-4] = 2 O 2 - = ½ [8-5] = 1.5 O 2 + = ½ [8-3] = 2.5 6

7 Bond Length: O 2 - > O 2 > O 2 + Bond Energy: O 2 + > O 2 > O 2-9. (6 points) In each blank write >, < or = as appropriate. Strength of intermolecular forces in H 2 O < Strength of intramolecular forces in CH 3 OH Viscosity of CH 4 < Viscosity of CH 3 Cl bp of CO 2 < bp of NO (12 points) Briefly define the following terms. axial (in VSEPR theory) Location of atoms that are above & below the equatorial plane in a trigonal bipyramidal structure (labeled as a in pic) a e a e e bond head-to-head overlap along the bond axis with zero nodes along the bonding axis triple point T and P where all 3 phases coexist in equilibrium 11. (18 points) SHOW ALL WORK. The vapor pressure of CCl 3 F at 300 K is 856 torr. If 12.6 g of CCl 3 F is enclosed in a 1.0-L container, will any liquid be present? If so, what mass of liquid? 856 torr x 1 atm = atm 760 torr n = PV = atm x 1.0 L = mol RT L atm x 300 K K mol mol x g = g 1 mol 12.6 g = 6.3 g 7

8 yes liquid will be present 12. (15 points) SHOW ALL WORK. Given g of hot tea at 80.0 C, what mass of ice at 0 C must be added to obtain iced tea at 10.0 C? For tea, C s = J/g C and H fus for ice is 6.01 kj/mol. [note; to simplify, ignore the contribution of the melted ice warming to 10 C.] q=m C s T q=(400.0 g)(4.184 J/g C)(70.0 C) = J needed J x 1 mol = mol ice x g/mol = 351 g 6010 J 8

9 CHEM 10113, Exam 4 November 30, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, and box your answers! 1. (10 points) Consider the phase diagram for substance A shown here and answer each of the following questions. a. What is the normal boiling point for A? 72 C b. What is mp for A at 1 atm? 28 C c. What state is present at rt and normal atmospheric pressure? solid d. What state is present at 75 C and 1.0 atm? gas e. If a dynamic equilibrium is established between the solid and liquid phases, name a change you can make to increase the amount of solid. Decrease the temperature 72 C 28 C 2. (16 points) SHOW ALL WORK. Which of the following molecules are polar? ClOF 4 Polar CO 2 Nonpolar 9

10 PCl 3 F 2 Nonpolar NF 3 Polar NO 3 Nonpolar NO Polar AsCl 4 + Nonpolar ICl 3 Polar 10

11 3. (18 points) SHOW ALL WORK. Nitrogen has a heat of vaporization of 5.92 kj/mol and a normal bp of -196 C. At what temperature does nitrogen boil when the external pressure is 1028 torr? Clausius-Claperon: ln P 2 = - H vap (1/T 2 1/T 1 ) P 1 R Let; P 2 =1028 torr, T 2 =x, P 1 =760 torr, T 1 =77 K ln 1028 torr = J/mol (1/x 1/77) = ln(1.3526) = ( )(1/x 1/77) 760 torr J/mol K = - (712.05) (1/x 1/77) = 1/x 1/77 = 1/x = 1/x x = T 2 = 80 K = -193 C 4. (12 points) For each of the following pairs, decide which has the greater bond angle, or if they are equal, and defend your answer: [note; no credit without an explanation] a) ClO 4 or ClO 3 angle = => Larger Bond Angle 11

12 angle = <109.5 b) NO 2 or NO 2 angle ~120 => Larger Bond Angle; larger angle because the single e - repels less than a pair of e - angle < (5 points) When a thin glass tube is put into water, the water rises 1.4 cm. When the same tube is put into hexane, the hexane rises only 0.4 cm. Explain the difference. Water can generate strong adhesive interactions with the glass (due to the dipoles at the surface of the glass), but hexane is nonpolar and cannot interact strongly with the glass surface. 6. (12 points) For the following organic molecule, give the valence bond description of the bonding. For the sketch, show only one example of each type of bond. (e.g. in H 2 O only draw one O-H bond) Formula EG at central atom Hybridization at central atom Draw the orbital interaction(s) CCl 4 4 sp 3 Is bond or? 12

13 SOF 4 5 sp 3 d 1 7. (12 points) a) Label the strongest type of intermolecular attraction for each material, then b) explain the melting point data in terms of molecular attractive forces. a) mp Strongest intermolecular attractive force He -272 C London Dispersion Force Kr -157 C London Dispersion Force H 2 Te -49 C Dipole-dipole MgF C Ion-ion (ionic bonds) b) Explain: As the strength of the intermolecular forces increase, the mp increases. Also, the LDF of Kr is greater than that of He due to its larger size, which is why Kr has the higher melting point. mp increases with stronger forces because it takes more energy to break apart strong forces. 8. (14 points) SHOW ALL WORK. According to MO theory, which has the highest bond order, bond energy, and shortest bond length? C 2, C 2 +, C 2? Include an MO diagram in your answer. 13

14 14

15 Bond Order = ½ [# bonded e - - # antibonded e - ] C 2 = ½ [6-2] = 2 C 2 + = ½ [5-2] = 1.5 C 2 - = ½ [7-2] = 2.5 Bond Length: C 2 + > C 2 > C 2 - Bond Energy: C 2 - > C 2 > C (6 points) In each blank write >, < or = as appropriate. Strength of intermolecular forces in MgCl 2 = Strength of intramolecular forces in MgCl 2 Viscosity of CH 3 OH > Viscosity of H 2 CO bp of CS 2 > bp of CO (12 points) Briefly define the following terms. equatorial (in VSEPR theory) In the tbp shape, the three terminal positions that are coplanar and at 120 angles from each other. (labeled e in the pic) 15

16 e a a e e bond side-to-side overlap of p orbitals deposition phase change from gas to solid 11. (18 points) SHOW ALL WORK. The vapor pressure of CCl 3 F at 300 K is 856 torr. If 11.5 g of CCl 3 F is enclosed in a 1.0-L container, will any liquid be present? If so, what mass of liquid? 856 torr x 1 atm = atm 760 torr n = PV = atm x 1.0 L = mol RT L atm x 300 K K mol mol x g = g; so 11.5 g g= 5.2 g 1 mol yes; liquid will be present 12. (15 points) SHOW ALL WORK. Given g of hot tea at 78.0 C, what mass of ice at 0 C must be added to obtain iced tea at 15.0 C? For tea, C s = J/g C and H fus for ice is 6.01 kj/mol. [note; to simplify, ignore the contribution of the melted ice warming to 15 C.] q=m C s T q=(425.0 g)(4.184 J/g C)(63.0 C) = J needed J x mol = mol ice x g/mol = 336 g 6010 J 16

17 CHEM 10113, Exam 4 November 30, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, and box your answers! 1. (10 points) Consider the phase diagram for substance X shown here and answer each of the following questions. a. What is the normal boiling point for X? 211 C b. What is mp for X at 1 atm? 121 C c. What state is present at rt and normal atmospheric pressure? solid d. What state is present at 213 C and 1.0 atm? gas e. If a dynamic equilibrium is established between the solid and liquid phases, name a change you can make to increase the amount of liquid. Increase the temperature ( C) 2. (16 points) SHOW ALL WORK. Which of the following molecules are polar? SOF 4 polar XeO 3 Polar 17

18 CCl 4 Nonpolar ClO 3 Polar BrO Polar PCl 4 + Nonpolar SO 3 2 Polar SO 4 2 Nonpolar 18

19 3. (18 points) SHOW ALL WORK. Ammonia has a heat of vaporization of 24.7 kj/mol and a normal bp of C. At what temperature does ammonia boil when the external pressure is 456 torr? Clausius-Clapeyron: ln P 1 = - H vap (1/T 1 1/T 2 ) P 2 R Let; P 1 =760 torr, T 1 = K, P 2 = 456 torr, T 2 = x K ln 760 torr = J/mol 456 torr J/mol K (1/ /x) ln(1.6667) = ( )(1/ /x) = ( ) (1/ /x) = 1/ /x = /x = 1/x T 2 = K = C 4. (12 points) For each of the following pairs, decide which has the greater bond angle, or if they are equal, and defend your answer: [note; no credit without an explanation] a) NO 2 or NO 2 + Angle <120 angle = 180 => Larger bond angle b) NO 3 or ClO 3 19

20 angle = 120 => Larger bond angle angle = < (5 points) When a thin glass tube is put into water, the water rises 1.4 cm. When the same tube is put into hexane, the hexane rises only 0.4 cm. Explain the difference. Water can generate strong adhesive interactions with the glass (due to the dipoles at the surface of the glass), but hexane is nonpolar and cannot interact strongly with the glass surface. 6. (12 points) For the following organic molecule, give the valence bond description of the bonding. For the sketch, show only one example of each type of bond. (e.g. in H 2 O only draw one O-H bond) Formula EG at central atom Hybridization at central atom Draw the orbital interaction(s) SBr 2 4 sp 3 Is bond or? IF 5 6 sp 3 d 2 7. (12 points) a) Label the strongest type of intermolecular attraction for each material, then b) explain the melting point data in terms of molecular attractive forces. a) 20

21 mp Strongest intermolecular attractive force SiH C London Dispersion Force Kr -157 C London Dispersion Force HBr -87 C Dipole-dipole NCl 3-40 C Dipole-dipole b) Explain: dipole-dipole > London Dispersion Force The dipole-dipole in NCl 3 is greater than that of HBr due to NCl 3 being larger than HBr, which is why it has the higher mp. Also, the LDF in Kr is greater than that of SiH 4 due to Kr being larger than SiH 4, which is why it has the higher mp. mp increases with stronger forces because it takes more energy to break apart strong forces. 8. (14 points) SHOW ALL WORK. According to MO theory, which has the highest bond order, bond energy, and shortest bond length? N 2, N 2, or N 2 +? Include an MO diagram in your answer. 21

22 Bond Order = ½ [# bonded e - - # antibonded e - ] N 2 = ½ [8-2] = 3 N 2 + = ½ [8-3] =

23 N 2 - = ½ [7-2] = 2.5 Bond Length: N 2 + = N 2 - > N 2 Bond Energy: N 2 > N 2 + = N 2-9. (6 points) In each blank write >, < or = as appropriate. Strength of intermolecular forces in CH 4 < Strength of intramolecular forces in CH 3 OH Viscosity of > Viscosity of CH 3 OH CH 3 CH 2 CH 2 OH bp of NH 3 > bp of CH (12 points) Briefly define the following terms. hybridization In valence bond theory, combination of 2 or more atomic orbitals on the same atom to form a new set of Hybrid Atomic Orbitals used in bonding. antibonding orbital In molecular orbital theory, an orbital overlap resulting in destructive interference where energy increases as atoms are brought closer together, indicating a net repulsion rather than a net attraction. critical point T and P upper limits on the liquid-gas curve of a phase diagram 11. (18 points) SHOW ALL WORK. The vapor pressure of CCl 3 F at 300 K is 856 torr. If 12.2 g of CCl 3 F is enclosed in a 1.0-L container, will any liquid be present? If so, what mass of liquid? 856 torr x 1 atm = atm 760 torr n = PV = atm x 1.0 L = mol RT L atm x 300 K K mol mol x g = g; so = 5.9 g 1 mol yes; liquid will be present 23

24 12. (15 points) SHOW ALL WORK. Given g of hot tea at 75.0 C, what mass of ice at 0 C must be added to obtain iced tea at 15.0 C? For tea, C s = J/g C and H fus for ice is 6.01 kj/mol. [note; to simplify, ignore the contribution of the melted ice warming to 15 C.] q=m C s T q=(450.0 g)(4.184 J/g C)(60.0 C) = J needed J x mol = mol ice x g/mol = 339 g 6010 J 24

PLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET. c) SeF 4

Chem 130 EXAM 4 Fall 99 PLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET QUESTIONS 1-5 MAY HAVE MORE THAN ONE POSSIBLE ANSWER CIRCLE ALL CORRECT RESPONSES TO EACH QUESTION