BCIT Fall 2013 Chem 3615 Exam #2 Name: Attempt all questions in this exam. Read each question carefully and give a complete answer in the space provided. Part marks given for wrong answers with partially correct reasoning/calculations. A Periodic Table, Table of Electronegativity and formula sheet are attached at the back. Total points = 50 Good Luck. 1
Section I: (18 points total, 1 point each) Choose the BEST answer to the following questions. 1. Which of the following combinations of quantum numbers is not allowed? n m m s a) 1 1 0 ½ b) 3 0 0 ½ c) 2 1 1 ½ d) 4 3 2 ½ e) 4 2 0 ½ 2. Which quantum number determines the orientation of a hydrogen atomic orbital? a) n b) c) m d) m s e) more information is needed 3. Which one of the following electron configurations depicts an excited carbon atom? a) 1s 2 2s 1 2p 1 b) 1s 2 2s 2 2p 3 c) 1s 2 2s 2 2p 1 4s 1 d) 1s 2 2s 2 2p 1 2d 1 e) more than one of the above 4. Which of the following statements is false? a) Ionic bonding results from the transfer of electrons from one atom to another. b) Dipole moments result from the unequal distribution of electrons in a molecule. c) The electrons in a polar bond are found nearer to the more electronegative element. d) A molecule with very polar bonds can be nonpolar. e) Linear molecules cannot have a net dipole moment. 5. The first ionization energy of Mg is 735 kj/mol. The second ionization energy is a) 735 kj/mol b) less than 735 kj/mol c) greater than 735 kj/mol d) more information is needed to answer this question. e) none of the above 6. Of the following elements, which is the largest? a) Se b) Ba c) At d) Cl e) Mg 2
7. In the gaseous phase, which of the following diatomic molecules would be the most polar? a) CsF b) CsCl c) NaCl d) NaF e) LiF 8. Atoms which are sp 2 hybridized can form pi bond(s). a) 0 b) 1 c) 2 d) 3 e) 4 9. As the bond order of a bond decreases, the bond energy and the bond length. a) increases, increases b) decreases, decreases c) increases, decreases d) decreases, increases e) More information is needed to answer this question. 10. Which of the following statements is true? a) Electrons are never found in an antibonding MO. b) Antibonding MOs are higher in energy than the atomic orbitals of which they are composed. c) Antibonding MOs have electron density mainly outside the space between the two nuclei. d) None of a, b and c are true. e) Two of a, b and c are true. 11. The configuration ( 2s ) 2 ( 2s *) 2 ( 2py ) 1 ( 2px ) 1 is the molecular orbital description for the ground state of a) Li 2 + b) Be 2 c) B 2 d) B 2 2 e) C 2 + 3
12. Which of the following should have the lowest boiling point? a) Na 2 O b) HF c) NH 3 d) N 2 e) H 2 O 13. Which substance can be described as cations bonded together by mobile electrons? a) Ag(s) b) S 8 (s) c) Kr( ) d) KCl(s) e) HCl( ) 14. Which of the following substance will have the largest band gap? a) C (diamond) b) Si c) Ge d) GaAs e) GaN 15. Doping Ge with As would produce a(n) semiconductor with conductivity compared to pure Si. a) n type, increased b) n type, decreased c) p type, increased d) p type, decreased e) more information is needed 16. The specific heat of gold is 0.13 J g 1 K 1, and that of copper is 0.39 J g 1 K 1. Suppose that both a 25 g sample of gold and a 25 g sample of copper are heated to 80 C and then drop each into identical thermally isolated beakers containing 100 ml of water at 10 C. When the beakers reach thermal equilibrium which of the following will be true? a) both beakers will be the same temperature b) the beaker with the gold will be at a higher temperature c) the beaker with the copper will be at higher temperature d) more information is needed 4
17. For a process in which one mole of a gas is expanded from state A to state B, which statement is true? a) When the gas expands from state A to state B, the surroundings are doing work on the system. b) The amount of work done in the process must be the same, regardless of the path. c) It is not possible to have more than one path for a change of state. d) The final volume of the gas will depend on the path taken. e) The amount of heat released in the process will depend on the path taken. 18. What is H for the following reaction? CBr 4 (g) + CCl 4 (g) CBr 3 Cl(g) + CBrCl 3 (g) Consider the Lewis structures of the molecules. a) H < 0 b) H = 0 c) H is close to 0 d) H > 0 e) H will depend on the reaction path taken Section II: Short answer calculations do not need to be shown (17 points total). 19. What are all possible values for the four quantum numbers of an electron in a hydrogen 4s orbital? (2 points) N = 4, = 0, m = 0, m s = +½ N = 4, = 0, m = 0, m s = ½ 20. The n=7 set of d orbitals consists of 5 orbitals. (1 point) 21. Write the electron configuration for the following and circle if they are diamagnetic or paramagnetic: (4 points) a) a zirconium atom (Zr) Diamagnetic Paramagnetic 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 2 or [Kr]5s 2 4d 2 b) an O 2 molecule Diamagnetic Paramagnetic 1s 2 1s 2 2s 2 2s 2 2p 2 2p 4 2p 2 5
22. a. Circle the dominant intermolecular force in each of the following molecules: (2 points) CH 3 OCH 3 Dispersion Dipole Dipole H Bond CH 3 CH 2 OH Dispersion Dipole Dipole H Bond b. Circle the molecule with the higher boiling point: (1 point) CH 3 OCH 3 CH 3 CH 2 OH 23. For each of the following molecules and ions draw the best Lewis structure, Name the ideal geometry (from the VSEPR model), give the ideal bond angles, hybridization of underlined atom and indicate if the molecule is polar or nonpolar. The central atom is underlined. (7 points) Molecule or Ion Best Lewis Structure (show all the electrons and resonance structures) Molecular Geometry Ideal Bond Angle Hybridization of underlined atom Polar or Nonpolar S SO 3 S Trigonal Planar 120 sp² nonpolar S ClF 2 + F Cl F Bent 109.5 sp³ polar 6
Section III: Calculations and reasoning must be shown (15 points total) 24. How much heat is required to raise the temperature of a 4.48 g sample of iron from 25.0 C to 79.8 C? The specific heat of iron is 0.450 J g 1 K 1. (2 points) 4.48 0.450 79.8 25.0 110 25. What is H for the following reaction? 2 H 2 C=CH 2 (g) H 3 C CH=CH CH 3 (g) Bond Energies: D C C = 347 kj/mol, D C=C = 611 kj/mol, D C H = 414 kj/mol. (2 points) H = (Energy of bonds broken) (Energy of bonds formed) ΔH =2( D c=c + 4D C H ) (2D c c + D C=C + 8D C H ) ΔH = D c=c 2D c c = 611 kj/mol 2(347 kj/mol) = 83 kj/mol 26. Given the following heats of reaction: H (kj/mol) 2O 3 (g) 3O 2 (g) 428 O 2 (g) 2O(g) 496 NO(g) + O 3 (g) NO 2 (g) + O 2 (g) 199 Calculate H for the reaction: NO(g) + O(g) NO 2 (g). (3 points) H (kj/mol) 3 / 2 O 2 (g) O 3 (g) 214 O(g) ½O 2 (g) 248 NO(g) + O 3 (g) NO 2 (g) + O 2 (g) 199 NO(g) + O(g) NO 2 (g) 233 kj/mol 7
27. During strenuous exercise the body can generate 5500 kj of energy in an hour. How many liters of water would have to evaporate through sweating to rid the body of the heat generated in 2.0 hours of exercise? Assume the density of sweat is 1.00 g/ml and ΔH vap of water is 44.0 kj/mol. (2 points) 5500 2 44.0 18.0 1.00 4.5 1000 28. Consider the following reaction: 2 Al(s) + Fe 2 O 3 (s) Al 2 O 3 (s) + 2 Fe(s) a. Calculate the H for this reaction. H f of Al 2 O 3 (s) is 1676 kj/mol and H f of Fe 2 O 3 (s) is 825.5 kj/mol. (2 points) H = H f [Al 2 O 3 (s)] + 2 H f [Fe(s)] 2 H f [Al(s)] H f [Fe 2 O 3 (s)] H = 1676 kj/mol + 2(0) 2(0) ( 825.5 kj/mol) = 850.5 kj/mol b. The specific heats of Al 2 O 3 and Fe are 0.79 J g 1 C 1 and 0.45 J g 1 C 1, respectively. The molar mass of Al 2 O 3 is 102 g/mol The melting point of iron is 1535 C and the H fus for iron is 15.1 kj/mol. Calculate the energy needed to raise 1 mole of reaction products from 25 C to the melting point of iron and to melt all the iron. (3 points) 102 0.79 2 55.85 0.45 1535 25 2 15.1 / 198 30.2 228 c. Will the reaction produce molten iron? Explain your answer. (1 point) Since the energy needed to heat up and melt all the iron, 228 kj, is much less than the energy released in the reaction, 850.5 kj, all the iron could melt 8
Equations and Formulas R = 0.0820575 L atm mol 1 K 1 = 8.314 J mol 1 K 1 k = 1.38x10 23 J/K c = 3.00x10 8 m/s m e = 9.11x10 31 kg h = 6.63x10 34 J s = 1.06x10 34 J s N A = 6.02x10 23 h mv c = f PV = nrt T(K) = T( C) + 273.15 q = nc m T = mc T = C T Heat capacity of water = 4.184 J g 1 C 1 w = P V for PV work E = q + w H = E + PV H = (Energy of bonds broken) (Energy of bonds formed) H p Products n H o f r Reactants n H o f 9
Electronegativity Values 10
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