Final Exam. CHEM 181: Introduction to Chemical Principles December 14, 2015 Answer Key N C S C

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1 Final Exam EM 181: Introduction to hemical Principles December 14, 2015 Answer Key 1. The compound 4 5 NO has the connectivity shown below O N On the next page: Draw all reasonable resonance structures and label them as major or minor. Label all non-zero formal charges. Rank the resonance structures: 1=best, 2=next best, etc. If two resonant structures are equivalent or very close to equivalent, you can assign them the same number in the ranking. If you prefer, you can skip writing the carbons and the hydrogens attached to N the carbons, like so: O 1

2 There is only one structure with zero formal charges; everything else is minor. Ranking of minor structures is based on electronegativity of the atom getting the positive charge. O N 1, major O N 3 O N 2, minor O N 3 O N 2 O N 4 2

3 2. A compound containing only,, and O with a molecular weight under 110 amu has the following 13 NMR spectrum (and note that peak heights in this 13 NMR spectrum are not meaningful): IR spectrum: 3

4 Dl 3 QE and 1 NMR spectrum: expanded to show detail: Draw a Lewis structure for the compound. Also mark on each spectrum: (a) which protons correspond to which peaks in the 1 NMR spectrum, and (b) labels for peaks in the IR spectrum that can be assigned unambiguously. 4

5 3. The electron configurations for the ground state and first three excited states of N 2 are shown. Use the following information about energies and bond lengths to figure out the electron configurations for the ground and excited states of N. (All answers that are logical and self-consistent will be marked as correct. You can use the MO diagram for N 2 on the next page the MO diagram for N will have states with different energies, but the order and names will be the same.) N 2 energy (aj) bond length (nm) σ 2s σ 2s π 2px, π 2py σ 2pz π 2p x, π 2p y σ 2p z N energy (aj) bond length (nm) σ 2s σ 2s π 2px, π 2py σ 2pz π 2p x, π 2p y σ 2p z

6 There are a few ways to answer this in a self-consistent manner. ere is one line of reasoning: The 0.18 aj state must come from a transition not available to N 2. Promoting a π 2px (or y) to a σ 2pz meets this requirement. Both are bonding, so the overall bond order is the same, and the bond length changes only slightly. The 1.08 aj and 1.17 aj states involve significant increases in bond length, and should involve moving a bonding electron to an antibonding orbital. Match the 1.08 aj state for N to the 1.00 aj state for N 2, and the 1.17 aj to the 1.18 aj state. The 0.51 aj also has no clear counterpart in N 2. ince bond length decreases slightly, this means either bond order must stay the same or increase. For that to be the case, an electron must move out of the σ2s orbital, and the relatively low energy of this state would place it into the σ 2pz. The 2s orbital for will be significantly higher in energy than the N 2s, which will bring the σ2s much closer in energy to the states derived from the 2p orbitals. 6

7 4. Use the following enthalpies of formation f 2 6 (g) 83.7 N 2 4 (g) and the following bond enthalpies bond N N N N N alculate the enthalpy of formation of 3 N 2 (g). N The reaction 2 6 (g) + N 2 4 (g) 2 3 N 2 (g) involves breaking a and an N N bond, and making two N bonds; viz., 2 6 (g) + N 2 4 (g) 3 (g) + 3 (g) + N 2 (g) + N 2 (g) 2 3 N 2 (g) o rxn = bond(reactants) bond(products) = bond( ) + bond(n N) 2 bond( N) = = 7 kj mol 1 But we can also express rxn in terms of formation enthalpies: rxn = f (products) f (reactants) 7 kj mol 1 = 2 f ( 3 N 2 (g)) [ f ( 2 6 (g)) + f (N 2 4 (g)) ] 7 kj mol 1 = 2 f ( 3 N 2 (g)) [ 83.7 kj mol kj mol 1] 2 f ( 3 N 2 (g)) = 7 kj mol kj mol kj mol 1 = 26 kj mol 1 f ( 3 N 2 (g)) = 13 kj mol 1 7

8 5. Molecular iodine (I 2 ) is slightly soluble in water and somewhat more soluble in cyclohexane ( 6 12 ): K 20 K 50 I 2 (s) 2O(l) I 2 (aq) M M I 2 (s) 6 12 (l) I 2 ( 6 12 sol.) M M Water and cyclohexane do not mix, and iodine will distribute between the two solvents according to the reaction: I 2 (aq) I 2 ( 6 12 sol.) For this reaction, what is K at 20, and what are and? (Assume that and are constant over this temperature range.) Our two reactions have very simple equilibrium constant expressions: K sp (aq) = [I 2 (aq)] and K sp ( 6 12 ) = [I 2 ( 6 12 sol.)] For we have I 2 (aq) I 2 ( 6 12 sol.) K = [I 2( 6 12 sol.)] [I 2 (aq)] = = 7.76 This is at 20 ; at 50, K = We can use the 20 value to find G : G = RT ln K = (8.314 J mol 1 K 1 ) (293 K) ln 7.76 = 4.99 kj mol 1 The van t off equation gives us using the values of K at the two different 8

9 temperatures: ln K 2 K 1 = R ( 1 T 2 1 T 1 ) ln = J mol 1 K (8.314 J mol 1 K 1 ) = ( K 1 ) = 1.48 kj mol 1 ( K 1 ) 293 K ince the equilibrium constant increases (albeit slightly) with temperature, Lehâtelier s principle says should be positive (albeit slightly): add heat to raise the temperature, and equilibrium shifts towards products in order to take up some of that heat. Then, G = T 4.99 kj mol 1 = 1.48 kj mol 1 (293 K) = 6.47 kj mol K = 22.1 J mol 1 K 1 9

10 6. ere are the chemical reactions to consider in the reaction of baking soda (NaO 3 ) with vinegar ( 3 OO(aq)): O 2 (g) O 2 (aq) K = M atm 1 O 2 (aq) + 2 O(l) 2 O 3 (aq) K = O 3 (aq) O 3 (aq) + + (aq) K a1 = M O 3 (aq) O 2 3 (aq) + + (aq) K a2 = M 3 OO(aq) 3 OO (aq) + + (aq) K a = M 2 O(l) + (aq) + O (aq) K w = M 2 (a) 1 L of a 0.8 M aqueous solution of 3 OO has NaO 3 added to it until it reaches p 7, at which point it is at equilibrium with respect to all of the above reactions, and with atmospheric O 2 (g) with P O2 = atm. ow many moles of NaO 3 were added to reach this equilibrium? (int: one way to do this is to figure out [Na + ] at equilibrium.) Equilibrium constant expressions for all of the above: [O 2 (aq)] P O2 (g) = M atm 1 [ 2 O 3 (aq)] [O 2 (aq)] = [ + ][O 3 (aq)] [ 2 O 3 (aq)] = M [ + ][O 2 3 (aq)] [O 3 (aq)] = M [ + ][ 3 OO (aq)] [ 3 OO(aq)] = M [ + (aq)][o (aq)] = M 2 We have P O2 in: fixed at an equilibrium value of atm. Plugging [O 2 (aq)] P O2 (g) = M atm 1 [O 2 (aq)] atm = M atm 1 [O 2 (aq)] = M [ 2 O 3 (aq)] = M 10

11 At p 7, [ + ] must be 10 7 at equilibrium, so [ + ][O 3 (aq)] [ 2 O 3 (aq)] = M 10 7 [O 3 (aq)] = M [O 3 (aq)] = M [O 2 3 (aq)] = M On the other hand, at p 7 almost all of the acetic acid is in its conjugate base form: [ + ][ 3 OO (aq)] [ 3 OO(aq)] 10 7 [ 3 OO (aq)] [ 3 OO(aq)] [ 3 OO (aq)] [ 3 OO(aq)] = M = M = 180 o at equilibrium, the negative ions and their concentrations in solution are just under 0.8 M of 3 OO, M of O 3, 10 7 M of O, and M of O 2 3. There is only 10 7 M of +, so a neutral solution requires 0.8 moles of NaO 3 were added. [Na + (aq)] 0.8 M 11

12 (b) 1 L of a 0.8 M aqueous solution of 3 OO has NaO 3 added to it until it reaches p 7, at which point it is at equilibrium with respect to all of the above reactions. In this case, the reaction takes place in a sealed 4 L container. ompared to part (a), how much NaO 3 was added: more, slightly more, exactly the same amount, slightly less, or less? Explain your answer. In a sealed container, P O2 is going to be much higher, resulting in significant concentrations of all of the carbonate species, and in particular O 3 (aq). The extra negative charge must be balanced by extra sodium, so we will need more NaO 3. Alternately, keeping higher pressures of O 2 (g) will act to acidify the solution, so you must add more base (NaO 3 ) to counteract this. (c) alculate how many moles of NaO 3 were added in part (b). (Use T = 298 K. int to simplify computations: the 3 L space left in the container will hold P /atm moles of gas at this temperature.) With a O 2 (g) pressure of P at equilibrium, the following come from our equilibrium constant expressions: [O 2 (aq)] = (P/atm) M [ 2 O 3 (aq)] = (P/atm) M [O 3 (aq)] = (P/atm) M [O 2 3 (aq)] = (P/atm) M Each O 3 that is protonated ends up either as an 2 O 3 (aq), a O 2 (aq), or a O 2 (g). ince [ 3 OO (aq)] 0.8 M, there are 0.8 moles of protons to be accounted for. The number of protons making 2 O 3 (aq) is more or less cancelled out by the number released forming O 2 3 (aq), so 0.8 mol n O2 (aq) + n O2 (g) = (P/atm) M 1 L (P/atm) mol = (P/atm) mol P = 5.1 atm The O 3 concentration must be [O 3 (aq)] = (P/atm) M = 0.45 M In contrast to the situation in part (a), our negative-ion species in solution are 12

13 just under 0.8 M of 3 OO, 0.45 M of O 3, negligible amounts of O and O 2 3. We must have added 1.25 M of NaO 3. 13