Name: Student code: The average kinetic energy of an argon atom is the same as that of a water droplet.

Transcription

1 T-11 ame 5 pts Student code The mass of a water droplet m = V ρ = [(4/3) π r 3 ] ρ = (4/3) π (0.5x10-6 m) 3 (1.0 g/cm 3 ) = 5.2x10-16 kg 10 Average kinetic energy at 27 o C KE = mv 2 /2 = (5.2x10-16 kg) (0.51x10-2 m/s) 2 /2 = 6.9x10-21 kg m 2 /s 2 = 6.9 x10-21 J The average kinetic energy of an argon atom is the same as that of a water droplet. KE becomes zero at 273 o C. From the linear relationship in the figure, KE = at (absolute temperature) where a is the increase in kinetic energy of an argon atom per degree. a = KE/T = 6.9x10-21 J/(27+273K) = 2.3x10-23 J/K 2 S specific heat of argon number of atoms in 1g of argon S = 0.31 J/g K = a x = S/a = (0.31 J/g K) / (2.3x10-23 J/K) = 1.4x Avogadro s number ( A ) umber of argon atoms in 40 g of argon A = (40)(1.4x10 22 ) = 5.6 x

2 T-21 ame Student code 5 pts mass of a typical star = (4/3)(3.1)(7x10 8 m) 3 (1.4 g/10-6 m 3 ) = g mass of protons of a typical star = ( g)(3/4 + 1/8) = g number of protons of a typical star = ( g)( /g) = number of stellar protons in the universe = ( )(10 23 ) = Partial credits on principles Volume = (4/3)(3.14)radius 3 density; 4 1 mole = ; 4 Total number of protons in the universe = number of protons in a star ; 2 Mass fraction of protons from hydrogen = (3/4)(1/1); Mass fraction of protons from helium = (1/4)(2/4); E(2 3) = C(1/4-1/9) = C λ(2 3) = nm E(1 2) = C(1/1-1/4) = 0.75 C λ(1 2) = (656.3)(0.1389/0.75) = nm o penalty for using Rydberg constant from memory. 1 penalty if answered in a different unit (z, etc.) 2-3. T = ( m K)/ m = K λ = m/ = 0.21 m T = ( m K)/0.21 m = K e ( 17 ) , acceptable

3 T-31 ame Student code 5 pts k des = A exp(-e des /RT) = (1x10 12 s -1 )(5x10-32 ) = 5x10-20 s -1 at T = 20 K 10 surface residence time, τ residence = 1 / k des = 2x10 19 s = 6x10 11 yr 20 (full credit for τ half-life = ln2 / k des = 1x10 19 s = 4x10 11 yr) residence time = 2x10 19 s The distance to be traveled by a molecule x = πr = 300 nm. k mig = A exp(-e mig /RT) = (1x10 12 s -1 )(2x10-16 ) = 2x10-4 s -1 at T = 20 K average time between migratory jumps, τ = 1 / k mig = 5x10 3 s the time needed to move 300 nm = (300 nm/0.3 nm) jumps x (5x10 3 s/jump) = 5x10 6 s = 50 days 1 (Full credit for the calculation using a random-walk model. In this case t = τ (x/d) 2 = 5 x 10 9 s = 160 yr. The answer is still (b).) (a) (b) (c) (d) (e) 10 k(20 K) / k(300 K) = exp[(e/r) (1/T 1-1/T 2 )] = e -112 = ~ for the given reaction ).) 1 The rate of formaldehyde production at 20 K = ~ molecule/site/s = ~ molecule/site/ yr 10 (The reaction will not occur at all during the age of the universe (1x10 10 yr).) rate = molecules/site/yr 3-4. circle one (a) (b) (c) (a, b) (a, c) (b, c) (a, b, c) (1, all or nothing) 3

4 T-41 ame Student code 5 pts P umber of atoms ( 11.3 ) 1 10 Theoretical wt % ( 3.43 ) 10 guanine cytosine each, 20 for three (10 on each) adenine thymine cytosine thymine guanine adenine cytosine cytosine thymine thymine guanine thymine cytosine adenine thymine thymine guanine guanine adenine adenine for each bracket 2 adenine guanine 2 Uracil 2 cytosine C ( 5 ) ( 5 ) ( 4 ) ( 4 ) 2 ( 0 ) ( 1 ) ( 2 ) ( 1 ) 4

5 T-51 ame Student code 5 pts (20 ) 1st ionization is complete 2 S S 4 - [ 2 S 4 ] = 0 2nd ionization [ + ][S 4 2- ]/[S 4 - ] = K 2 = 1.2 x 10-2 (1) Mass balance [ 2 S 4 ] + [S 4 - ] + [S 4 2- ] = 1.0 x 10-7 (2) Charge balance [ + ] = [S 4 - ] + 2[S 4 2- ] + [ - ] (3) Degree of ionization is increased upon dilution. [ 2 S 4 ] = 0 Assume [ + ] 2S4 = 2 x 10-7 From (1), [S 2-4 ]/[S - 4 ] = 6 x 10 4 (2nd ionization is almost complete) [S - 4 ] = 0 From (2), [S 2-4 ] = 1.0 x 10-7 [] From (3), [ + ] = (2 x 10-7 ) /[ + ] [ + ] = 2.4 x 10-7 (p = 6.6) [8 ] [ - ] = /(2.4 x 10-7 ) = 4.1 x 10-8 [2 ] From (1), [S 4 - ] = [ + ][S 4 2- ]/K 2 = (2.4 x 10-7 )(1.0 x 10-7 )/(1.2 x 10-2 ) = 2.0 x [] Check charge balance 2.4 x 10-7 (2.0 x ) + 2(1.0 x 10-7 ) + (4.1 x 10-8 ) Check mass balance x x x 10-7 Species Concentration - S x S x x x

6 T-52 ame Student code 5-2. (20 ) mmol 3 P 4 = ml 1.69g/mL 1 mol/98.00 g 1000 = 51.0 [] The desired p is above pk 2. A 11 mixture of 2 P - 4 and P 2-4 would have p = pk 2 = If the p is to be 7.40, there must be more P 4 than 2 P - 4. We need to add a to convert 3 P 4 to 2 P - 4 and to convert to the right amount of 2 P - 4 to P P P P P The volume of 0.80 a needed to react with to to convert 3 P 4 to 2 P - 4 is 51.0 mmol / 0.80M = ml [] To get p of 7.40 we need 2 P P 4 Initial mmol 51.0 x 0 Final mmol 51.0-x 0 x p = pk 2 + log [P 4 2- ] / [ 2 P 4 - ] 7.40 = log {x / (51.0-x)}; x = mmol [] The volume of a needed to convert mmol is mmol / 0.80 M = ml The total volume of a = = ml, 103 ml [] Total volume of 0.80 M a (ml) 103 ml 6

7 T-53 ame Student code 5-3. (20 ) pk = 3.52 p = pk a + log ([A - ]/[A]) [A - ]/[A] = 10 (p-pka) [] In blood, p =7.40, [A - ]/[A] = 10 ( ) = 7586 Total ASA = = 7587 [] In stomach, p = 2.00, [A - ]/[A] = 10 ( ) = 3.02x10-2 Total ASA = x10-2 = 1.03 [] Ratio of total aspirin in blood to that in stomach = 7587/1.03 = 7400 [] Ratio of total aspirin in blood to that in stomach

8 T-61 ame 5 pts Student code () e (g) (or e - 2 (g) ) 6-2. () e - (or 2 1/ e - ) 6-3. () Cu Cu e (20 ) Reduction of sodium ion seldom takes place. It has a highly negative reduction potential of V. Reduction potential for water to hydrogen is negative (water is very stable). But, it is not as negative as that for sodium ion. It is V. Reduction of both copper ion and oxygen takes place readily and the reduction potentials for both are positive. In the present system, the reverse reaction (oxidation) takes place at the positive terminal. Copper is oxidized before water. Reduction potential for hydrogen ion is defined as V (1) p = = 9.16 [ - ] = 6.92 x K sp = [Cu 2+ ][ - ] 2 = x (6.92 x ) = 4.79 x

9 T-62 ame Student code 6-6. E = E o Cu2+/Cu + (0.0592/2) log [Cu 2+ ] = (0.0592/2) log [Cu 2+ ] = (0.0592/2) log (K sp / [ - ] 2 ) = (0.0592/2) log (K sp ) - (0.0592/2) log [ - ] 2 = (0.0592/2) log (K sp ) log [ - ], 3 By definition, the standard potential for Cu() 2 (s) + 2e - Cu(s) is the potential where [ - ] = E = E o Cu()2/Cu = (0.0592/2) log (K sp ) = (0.0592/2) log (4.79 x ) = = V ne may solve this problem as following. Eqn 1 Cu() 2 (s) + 2e - Cu E + o = E o Cu()2/Cu =? Eqn 2 Cu() 2 (s) Cu E o = ( /n) logk sp = ( /2) log( ) = V 3 Eqn 1 Eqn 2 Cu e - Cu E o o - = E + - E o = E o Cu2+/Cu = 0.34 V Therefore, E + o = E - o + E o = ( ) 2 = V V 6-7. Below p = 4.84, there is no effect of Cu() 2 because of no precipitation. Therefore, E = E Cu2+/Cu = (0.0592/2) log [Cu 2+ ] = (0.0592/2) log = = V V 9

10 T-63 ame Student code g graphite = mol carbon 6 mol carbon to 1 mol lithium; 1 g graphite can hold mol lithium To insert 1 mol lithium, coulombs are needed. Therefore, 1 g graphite can charge = 1340 coulombs coulombs / g = 1340 A sec / g = 1340 x 1000 ma (1 / 3600) h = 372 ma h / g 372 ma h / g 10

11 T-71 ame Student code 4 pts (10 ) n/v = P/RT = (80 x 10 6 / x 10 5 atm)/[(0.082 atm L/mol/K)(298K)] = 32 mol/l density = mass/volume = d = 32 x 2 g/l = 64 kg/m kg/m 3 2 (g) + 1/2 2 (g) 2 (l); rexn-1 = f [ 2 (l)] = -286 kj/mol = -143 kj/g 7 C(s) + 2 (g) C 2 (g); rexn-2 = f [C 2 (g)] = -394 kj/mol = -33 kj/g 7 (- rexn-1 ) / (- rexn-2 ) = 4.3 or (- rexn-2 ) / (- rexn-1 ) = or (g) + 1/2 2 (g) 2 (l) c = -286 kj/mol = -143 kj/g = -143 x 10 3 kj/kg G = T S S c = /2 = J/K/mol G c = -286 kj/mol + 298K x J/K/mol = -237 kj/mol = -1.2 x 10 5 kj/kg (a) electric motor W max = G c 1 kg = x 10 5 kj (b) heat engine W max = efficiency x c = (1 298/573) x (-143 x 10 3 kj) = -6.9 x 10 4 kj 119 x 10 3 kj = 1 W x t(sec) t = 1.2 x 10 8 sec = 3.3 x 10 4 hr = 1.4 x 10 3 days = 46 month = 3.8 yr G = -nfe n = # of electrons involved in the reaction F = 96.5 kc/mol 2 (g) + 1/2 2 (g) 2 (l) n = 2 E = - G/nF = 237 kj/mol / 2 / 96.5 kc/mol = 1.23 V I = W/E = 0.81 A (a) (-)1.2 x 10 5 kj, (b) (-)6.9 x 10 4 kj 1.2 x 10 8 sec or 3.3 x 10 4 hr or 1.4 x 10 3 days or 46 month or 3.8 yr I = 0.81 A 11

12 T-81 ame 5 pts Student code ( on each) 1 C 2 C 3 C Fe 2 3 (s) + 3C(g) 2Fe(s) + 3C 2 (g) 1 C(s) + 2 (g) C 2 (g) 1 = kj = f (C 2 (g)) 2 C 2 (g) + C(s) 2C(g) 2 = kj From 1 and 2, f (C(g)) = (1/2){ ( )} = kj f (Fe 2 3 ) = kj 3 = 3ⅹ f (C 2 (g)) - f (Fe 2 3 ) - 3ⅹ f (C(g)) = 3ⅹ( ) (-824.2) - 3ⅹ( ) = kj 7 S 3 =2ⅹ ⅹ ⅹ =15.36 J/K 3 G 3 = -T S =-24.8kJ-15.36J/Kⅹ1kJ/1000Jⅹ K= kj K = e (- G /RT) = e (47430J/(8.314J/Kⅹ K)) = (20 ) Balanced equation of 3 Fe 2 3 (s) + 3C(g) 2Fe(s) + 3C 2 (g) ne AB 2 4 unit has available 4 (= 1 + (1/4)ⅹ12) octahedral sites. K = (20 ) Since one face-centered cube in AB 2 4 represents one Fe 3 4 unit in this case, it has 8 available tetrahedral sites. In one Fe 3 4 unit, 1 tetrahedral site should be occupied by either one Fe 2+ (normal-spinel) or one Fe 3+ (inverse-spinel). Therefore, in both cases, the calculation gives (1/8) ⅹ100% = 12.5% occupancy in available tetrahedral sites (10 for d-orbital splitting, 10 for elec. distribution) 12.5% 12

13 T-91 ame 5 pts Student code answer for 8, two for 1 _ + _ + 3 C 3 C ( 10 ) 3 C C C 2 C 2 3 C C C 2 (10 ) (10 ) each + _

14 T-92 ame Student code ( 10 ) C C (40 ) 2 C C 3 C 3 C 2 3 C 3 C + C 3 _ B C D E (10 ) n F + 14

15 T-101 ame 9 pts Student code each M L C 2 C 2 C Me Me Me Me Me Me C 2 C C each for correct structures umber of possible structures CMe () () Me CMe Me Me () Me () 3 4 () () () e() 15

16 ame T-102 Student code each G I C 2 C 2 Me Me Me Me Me Me C 2 C umber of the correct structure for C from B C () 10 each () D J CMe CMe () (Me) Me Me Me () Me Me() Me Me 16

17 T-103 ame Student code C C C 17

18 T-111 ame 7pts Student code C C each a, c, d C C Transition State 18

19 T-112 ame Student code For the enzyme-catalyzed reaction, Arrehnius equation could be applied. k cat /k uncat = A exp (-E a, cat / RT) / A exp (-E a, uncat / RT) = exp [- E a, cat-uncat / RT] = exp [- E a, cat-uncat (J/mol) / (2,480 J/mol)] = 10 6 Therefore, - E a, cat-uncat = 34,300 J/mol 1 k uncat, T /k uncat, 298 = exp (- uncat / RT) / exp (- uncat / 298R) = exp [(- uncat /R)(1/T-1/298)] ln(k uncat, T /k uncat, 298 ) = 13.8 = [(-86900/8.32)(1/T-1/298)] Therefore, T = 491 K, or 218 o C 1 -E a, cat-uncat = 34,300 J/mol T = 491 K, or 218 o C 19