Name CHM 1045 Spring 2018 April 30 EXAMINATION FOUR SOLUTIONS VI VII VIII IX X

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1 Name CHM 1045 Spring 2018 April 30 EXAMINATION FOUR SOLUTIONS I II III IV V VI VII VIII IX X Total This exam consists of several problems. Rough point values are given. The total will be scaled to 175 points after the exams are marked. Glance over the entire exam, and then attempt the problems in the order of your choice. For calculations, give your answer to the correct number of significant figures, and be sure to include the correct units for your answer. You must show your work to receive any credit for a calculated answer. Additional information is provided in a separate information handout; you can use the back for scratch work. I. (5 points) Did you miss taking any of the hour exams (Exam 1, Exam 2, or Exam 3)? If so, which exam(s), and why? II. (20 points) Trifluoromethyl peroxynitrate (CF 3 O 2 NO 2 ) has been studied as a potential intermediate in the atmospheric degradation of fully halogenated hydrocarbons (e.g., CF 4, CF 3 Cl, CF 3 Br, CF 3 I). A. The dot structure for CF 3 O 2 NO 2 is shown on the right. Assign all nonzero formal charges. B. Now assign oxidation numbers for the atoms identified in the structure. 7 8 = (-1) 6 7 = (-1) 6 8 = (-2) 6 8 = (-2) 4 0 = (+4) 6 7 = (-1) 5 0 = (+5) C. What is the difference between formal charge and oxidation number? Do not simply explain how they are calculated differently. What is the conceptual difference between formal charge and oxidation number? Formal Charge assumes 100% perfect covalent bonding. The shared electrons are evenly shared, so you split them equally between the two atoms connected by that covalent bond. Oxidation Number assumes 100% ionic bonding, so the shared electrons in chemical bonds are counted as belonging to the most electronegative atom.

2 CHM 1045, Exam Four, Spring 2018 page 2 III. (36 points) A novel method for using ammonia (NH 3 ) as a fuel was recently reported. A balanced chemical equation for the combustion of ammonia is shown below. 2NH 3 (g) + 3 / 2 O 2 (g) N 2 (g) + 3H 2 O(g) One advantage of using ammonia as a fuel is that it does not produce carbon dioxide. Instead, nitrogen gas is produced ( A. The reaction shown above is expected to be (circle one): endothermic exothermic Explain your reasoning. combustion of a fuel is exothermic B. Now use bond energies to determine H for this reaction. 2NH 3 (g) + 3 / 2 O 2 (g) N 2 (g) + 3H 2 O(g) H rxn = 6 H BE (N-H) + 3 / 2 H BE (O=O) [ H BE (N N) + 6 H BE (O-H)] = 6(390 kj/mol) + 3 / 2 (498 kj/mol) [946 kj/mol + 6(464 kj/mol)] = -643 kj/mol C. Is the H you calculated in Part B consistent with your answer to Part A? D. Assign oxidation numbers for the reactants and products in this reaction. (-3) (+1) (0) (0) (+1) (-2) \ / \ / 2NH 3 (g) + 3 / 2 O 2 (g) N 2 (g) + 3H 2 O(g) E. Which reactant was reduced (circle one)? NH 3 (g) O 2 (g) F. The enthalpy of combustion for methane (CH 4 ) is given below. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) How much energy is released by combusting 3.0 kg of methane? H = kj/mol 3.0 kg x (1000 g/kg) x (1 mol/16 g) x ( kj/mol) = x 10 5 kj G. How much methane (in kg) would be required to release the amount of energy you calculated in Part F x 10 5 kj x(2 mol NH 3 /643 kj) x (17 g/mol) x (1 kg/1000 g) = 8.7 kg H. Another advantage of ammonia is that it can be compressed into a liquid at a relatively low pressure compared to other fuel gases such as hydrogen. Why is that property an advantage compared to hydrogen? Like liquid propane, ammonia does not require a heavy tank to transport.

3 CHM 1045, Exam Four, Spring 2018 page 3 IV. (24 points) A recent study looked at hydrogen bonding in nonmetal hydrides such as GeH 4, AsH 3, and H 2 S. Hydrogen bonding is covered in General Chemistry II, but an understanding of hydrogen bonding requires dot structures and molecular geometry. Write the dot structure and draw the molecular geometry for these molecules. Include a descriptive name for the molecular shape: bent, T-shaped, see-saw, trigonal planar, trigonal pyramid, trigonal bipyramid, square planar, square pyramid, tetrahedral, or octahedral. GeH 4 tetrahedral dot structure molecular geometry descriptive name AsH 3 H 2 S trigonal pyramidal dot structure molecular geometry descriptive name bent dot structure molecular geometry descriptive name Did you notice they wrote the hydrogen atoms in front for H 2 S, but not for GeH 4 and AsH 3? Why did they do that? Ionizable hydrogen atoms are written at the front of the formula, so H 2 S must be acidic. (You will begin to explore why certain molecules are acidic or basic next semester, which is a lot better than memorizing all the acids and bases, and their strengths.) V. (15 points) Write the best dot structure you can for the following molecular species, exceeding the octet rule, where allowed, to minimize formal charge. A. OClF 3 (Cl is the central atom) B. BrO 2 C. NNN

4 CHM 1045, Exam Four, Spring 2018 page 4 VI. (18 points) For each of the following hypothetical dot structures sketch the molecules, using wedges and dashed lines where necessary, and include a descriptive name for the molecular shape: bent, T-shaped, see-saw, trigonal planar, trigonal pyramid, trigonal bipyramid, square planar, square pyramid, tetrahedral, or octahedral. trigonal bipyramid see-saw T-shaped Now give an example of a real molecule that would have the same dot structure. For example... PF 5 SF 4 ClF 3 VII. (30 points) The Super Trans-Iron Galactic Element Recorder (SuperTIGER) is an instrument designed to study rare heavy nuclei, with the goal of understanding how cosmic rays attain speeds approaching the speed of light. The most common cosmic ray particles are protons traveling at speeds between 45% and 99.6% of the speed of light (masterclass.icecube.wisc.edu/en/icetop/measuring-cosmic-rays). A. What would be the wavelength expected for a proton traveling at 75% of the speed of light? λ = h/(mv) = x J s / [ x g x (1 kg/1000 g)] x (0.75 x m/s) = 1.76 x m B. The SuperTIGER instrument described in the last problem is lifted to the necessary altitude by a helium balloon. Before launching the balloon is inflated to a volume of roughly L (assuming 2 sig figs). The balloon is launched from McMurdo Station, Antarctica, and the current temperature there is 26 F, corresponding to 3 C (you re welcome). The balloon is launched to reach 39,000 m above sea level, where the temperature is typically 23 C and the absolute pressure is 0.18 atm. ( Determine the volume of the balloon when it reaches an altitude of 39,000 m above sea level. PV=nRT n and R are constant => P 1 V 1 /T 1 = P 2 V 2 /T 2 => [(1 atm)(5.0 x 10 7 L)/( K)] = [(0.18 atm)(v 2 )/( K)] => V 2 = 3.0 x 10 8 L How many moles of helium does the balloon contain? PV=nRT => n = PV/RT = (1 atm)(5.0 x 10 7 L)/( L atm/k mol)( K) = 2.3 x 10 6 mol

5 VIII. A technique was recently reported for increasing the power of simple LED lasers by matching the wavelength to groves on the semiconductor surface ( Assuming the spacing in the groves, 27 μm, matches the wavelength of the laser... A. what is the frequency? CHM 1045, Exam Four, Spring 2018 page 5 c = => = c/ = (3.0 x 10 8 m/s) (27 μm x (1 m/10 6 μm) = 1.1 x s -1 B. and what is the energy of one photon? E = h = x J s x 1.1 x s -1 = 7.4 x J IX. ODDS AND ENDS A. For each of the following orbitals, give n, l, and the number of orbitals: 1. 6p n = ; 6 l = ; 1 # of orbitals d n = ; 4 l = ; 2 # of orbitals 5 B. Why are there no 3f orbitals? n = 3, so l can be 0, 1, and 2, corresponding to s, p, and d orbitals, respectively. C. Write balanced molecular equations (ME) and net ionic equations (NIE) for the following reactions. Be sure to include the correct states in your final equations and charges for ions in the NIE. HNO 3 (aq) + NH 3 (aq) ME: HNO 3 (aq) + NH 3 (aq) NH 4 NO 3 (aq) NIE: H + + NO NH 3 NH NO 3 - NIE: H + (aq) + NH 3 (aq) NH 4 + (aq) KCl(aq) + Pb(NO 3 ) 2 (aq) ME: 2KCl(aq) + Pb(NO 3 ) 2 (aq) PbCl 2 (s) + 2KNO 3 (aq) NIE: 2K + + 2Cl - + Pb NO 3 - PbCl 2 (s) + 2K + + 2NO 3 - NIE: Pb 2+ (aq) + 2Cl - (aq) PbCl 2 (s)

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