Final Exam. CHEM 181: Introduction to Chemical Principles Answer Key. 1. The major resonance structure of 1,4-diazepine is shown below:

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1 Final Exam EM 181: Introduction to hemical Principles Answer Key 1. The major resonance structure of 1,4-diazepine is shown below: or Show all reasonable resonance structures; label them as major (if equivalent to the one above), minor, or very minor. Label all non-zero formal charges. (If you prefer, you can skip writing the carbons and their attached hydrogens, as shown above on the right.) There are seven possible arrangements that rotate the three double bonds around the ring. There is one minor structure: which places positive and negative formal charges on the two nitrogens. The following five structures are all very minor (positive nitrogen and negative carbon), and roughly though not exactly equivalent: 1

2 2. onsider the following resonance structures of cytosine: There are other (more minor) resonance structures that are not important for this problem. (next page) 2

3 When an acid protonates cytosine, the + is added only at one particular nitrogen. Draw the Lewis structure for protonated cytosine, including resonance structures and formal charges. Explain the reason that this site is protonated while the others are not. A simple approach is to look at the resonance structures for the unprotonated molecule, and note that for both the top and the bottom nitrogens, the lone pair is not present in all resonance structures that is, the lone pair is actually involved in molecular orbitals associated with bonding. This is not true for the center nitrogen. Protonating the center nitrogen leaves the major and minor resonance structures alone: 3

4 Protonating either of the others prevents the formation of either one or two of the minor structures: 4

5 3. itric acid ( ) is a triprotic acid. Abbreviating citrate as it = , the K a values are 3 it(aq) 2 it (aq) + + (aq) 2 it (aq) it 2 (aq) + + (aq) it 2 (aq) it 3 (aq) + + (aq) K a1 = K a2 = K a3 = Given orange juice with a p of 3.5, what are the relative concentrations (expressed as percentages) of 3 it, 2 it, it 2, and it 3? The equilibrium constant expressions are [ + ][ 2 it ] [ 3 it] [ + ][it ] [ 2 it ] [ + ][it 3 ] [it 2 ] = = = and the [ + ] concentration is = Since this is the + concentration at equilibrium, we can plug it in to the above to get [ 2 it ] [it ] = 2.66, [ 3 it] [ 2 it ] = 0.057, and [it3 ] [it 2 ] = If we set [ 3 it] = 1, that gives us [ 2 it ], which gives us [it 2 ], which gives us [it 3 ]: [ 3 it] = 1 [ 2 it ] = 2.66 [it 2 ] = [it 3 ] = Whatever the actual overall concentration, equilibrium concentrations of each species will always have these same proportions. Divide by the sum (3.81) and multiply by 100 to get percentages: 26.2% 3 it, 69.7% 2 it, 4.0% it 2, and 0.05% it 3 5

6 4. Use the following reactions and equilibrium constants: Ag 3 As 4 (s) 3Ag + (aq) + As 3 4 (aq) K sp = AgBr(s) Ag + (aq) + Br (aq) K sp = What are the equilibrium concentrations of Ag + (aq), Br (aq) and As 3 4 (aq) in a solution where there is an excess of both solids? (An approximate answer within 10% of the correct value will receive full credit if you also state whether your approximation is above or below the exact value.) Equilibrium expressions: [Ag + ] 3 [As 3 4 ] = [Ag + ][Br ] = Stoichiometry says that for every three silvers, there must be three Br or one As 4 : [Ag + ] = [Br ] + 3[As 3 4 ] (Arithmetic check: if we just had Ag 3 As 4, the silver concentration would be three times the As 4 concentration.) ow, divide the top equilibrium expression by the cube of the second: [Ag + ] 3 [As 3 4 ] ([Ag + ][Br ]) 3 = ( ) 3 [As 3 4 ] ([Br ]) 3 = [As 3 4 ] = ([Br ]) 3 Plug in to the stoichiometry: [Ag + ] = [Br ] ([Br ]) 3 and plug this into the AgBr equilibrium: [Ag + ][Br ] = ( [Br ] [Br ] 3) [Br ] = [Br ] 4 + [Br ] = 0 6

7 This is a quadratic in [Br ] 2 (or you can solve it with a calculator): Plug and chug [Br ] 2 = and [Br ] = [Br ] = M [Ag + ] = M [As 3 4 ] = M heck the stoichiometry: [Ag + ] = [Br ] + 3[As 3 4 ] = = You can also take an approximate approach. It s important to figure out which reaction to start with first, though; if you try the AgBr reaction, you get [Ag + ][Br ] = [Ag + ] = [Br ] [Ag + ] 2 = [Ag + ] = owever, if you plug this value for [Ag + ] into the other reaction, things do not work out: [Ag + ] 3 [As 3 4 ] = ( ) [As 3 4 ] = [As 3 4 ] = Since there must be at least three silver ions for every As 3 4, this cannot be 7

8 the As 3 4 concentration. Instead, you have to start with the other reaction: [Ag + ] 3 [As 3 4 ] = [Ag + ] = 3[As 3 4 ] ( 3[As 3 4 ] ) 3 [As 3 4 ] = [As 3 4 ] 4 = [As 3 4 ] = [Ag + ] = and then [Br ] = /[Ag + ] = Because this number is less than 10% of the As 3 4 concentration, it s reasonable to say that the amount of Ag + coming from this reaction is (at least) 10 times less than the amount coming from the As 3 4 reaction. By just plugging the Ag + concentration in, as we did above, we re pretending that somehow, producing Br doesn t produce more Ag +. This means that our approximate answer is going to predict less Ag + than there actually will be, and so must also be predicting more Br and As Definitions and constants: (a) The enthalpy of formation for a compound is the heat that goes into the system when pure elements in their most stable forms react to make the compound. The most stable forms for carbon, hydrogen, and oxygen are (s), 2 (g), and 2 (g). (b) The enthalpy of formation of 2 is 116 kj/mol. (c) The enthalpy of sublimation for carbon is the enthalpy change for the reaction: (s) (g) and is equal to 715 kj/mol. (d) The bond dissociation enthalpies for 2 and 2 are the heats that must go into the system to break the or bonds, forming gaseous atomic products. The exact values are bond ( 2 ) = 436 kj/mol and bond ( 2 ) = 498 kj/mol (e) The enthalpy of atomization is the heat that goes into a molecule to break every bond, forming only atomic gas-phase products. 8

9 alculate the enthalpy of atomization of 2. We want the reaction enthalpy for 2 (g) 2(g) + (g) + (g) The reactions we have to work with are 2 (g) + (s) (g) 2 (g) f = 116 kj mol 1 (s) (g) sub = 715 kj mol 1 2 (g) 2(g) bond = 436 kj mol 1 2 (g) 2(g) bond = 498 kj mol 1 What we want instead are 2 (g) 2 (g) + (s) (g) f = 116 kj mol 1 (s) (g) sub = 715 kj mol 1 2 (g) 2(g) bond = 436 kj mol (g) (g) bond = 249 kj mol 1 These add up to Alternately, ( ) kj mol 1 = 1516 kj mol 1 9

10 6. alcium hydroxide, a() 2, is somewhat soluble in water, but dissociates completely when it does dissolve: a() 2 (s) 2 a 2+ (aq) + 2 (aq) A saturated solution (with excess solid) of a() 2 at 25 is measured to have a p of The same solution heated to 50 has a p of What is the enthalpy of solution of a() 2? onvert p values to concentrations: [ ] = = K and [ ] = = K In both cases, [a 2+ ] = [ ]/2 by stoichiometry, giving us K 298 = and K 323 = Plug these values into the van t off equation: ln K 323 K 298 = R ( ) ln = R ( ) R = = R = ( J mol 1 K 1) = J mol 1 = 17.7 kj mol 1 Raising the temperature reduces the concentration, so it must be driving products back to reagents. Le hatelier s principle tells us that for this to be the case, must be negative. 10

11 7. The M diagram for l is formed from the 1s atomic orbital of and the 3s and 3p atomic orbitals of the l, and is shown below (vertical energy scale is not exact): l 4p l 4s l l 3d z 1s y x (or ) l 3p z s l 3s n the same energy scale as the molecular orbitals, draw in (horizontal lines) and label the 3s, 3p, 3d, 4s, and 4p energy levels for atomic chlorine and the 1s energy level for atomic hydrogen. The l 3s and 3p orbitals have to line up with the non-bonding l orbitals (which look exactly like the atomic orbitals). The 1s orbital must lie somewhere between the bonding and antibonding Ms. 11

12 The table below gives the wavelengths for the lowest-energy absorption features of l and l. Because some of the molecular orbitals of l look very much like some of the l atomic orbitals, some of the electronic transitions are very similar between l and l. Each transition corresponds to exciting a single electron from an initial to a final state. Example of how to describe transitions: the 3p 5p transition in chlorine causes an absorption at 105 nm, while the σ z l(5s) transition is at 100 nm. (These values will probably not be used.) λ (nm) l transition λ (nm) l transition 200 π x,y σ z 134 3p 4s 136 π x,y 4s 125 σ z σ z 120 3p 4p 120 π x,y 4p 116 3s 3p Fill in the above blanks with the starting and ending states for each transition. Things to consider when making these assignments: (a) Transitions must start with an orbital with at least one electron, and end in an orbital with at most one. (b) The chlorine 3s and 3p create non-bonding Ms, and for the same reasons so do the chlorine 4s and 4p. These will be at the same or nearly the same energy for l and l, so transitions that are matched in energy for l and l should both start and end from l non-bonding orbitals. (c) Since these are the lowest-energy absorption features, transitions to 4d, 5s, 5p, etc. orbitals should not be considered before transitions to 4s, 4p, etc. (Similarly, transitions from 1s, 2s, and 2p orbitals should not be considered.) 12

13 (d) Transitions that l has but l does not should either start at the σ z state (which is filled) or end at the σ z state (which is empty). (e) Transitions that end at the chlorine 3p orbital are the only ones that won t have a match with l this orbital is not completely filled for l, but the similar π x,y orbital is filled for l. 13