Solution Manual for Physical Chemistry 2nd Editionby David W. Ball

Chapter 1

Gases and the Zeroth Law of Thermodynamics 1.2.

1.4.

A system is any part of the universe under observation. The “surroundings” includes everything else in the universe. Consider a solution calorimeter in which two aqueous solutions are mixed and temperature changes are recorded. In this case the solutes (reactants and products) would be considered the “system”. The water, calorimeter, and rest of the lab and world would be the “surroundings”.

1000 mL 1 cm 3   12,560 cm 3  1.256  10 4 cm 3 (a) 12.56 L  1L 1 mL (b) 45ºC + 273.15 = 318 K (c) 1.055 atm 

1.01325 bar 100,000 Pa   1.069  10 5 Pa 1 atm 1 bar

(d) 1233 mmHg 

1 torr 1 atm 1.01325 bar    1.644 bar 1 mmHg 760 torr 1 atm

1 cm 3  125 cm 3 (e) 125 mL  1 mL (f) 4.2 K – 273.15 = –269.0°C (g) 25750 Pa  1.6.

1 bar  0.2575 bar 100,000 Pa

patm = pmouth + hg, where hg correspond to the pressure exerted by the liquid patm - pmounth = hg = (1.0x103 kg/m3)(0.23m)(9.80m/s2) = 2254 N/m2 = 2.3 x103 Pa

1.8.

In terms of the zeroth law of thermodynamics, heat will flow from the (hot) burner or flame on the stove into the (cold) water, which gets hotter. Then heat will move from the hot water into the (colder) egg.

1.10.

For this sample of gas under these conditions, F(T) = 2.97 L  0.0553 atm = 0.164 Latm.

If the pressure were increased to 1.00 atm: 0.164 Latm = (1.00 atm)  V; therefore V = 0.164 L.

1.12.

V  R

pV nT , which rearranges to R  . p nT

Therefore: R 

(2.66 bar)(27.5 L) L  bar  0.0830 (1.887 mol)(466.9 K) mol  K 1

Instructor‟s Manual

1.14.

V1=67 L, p1=1.04 atm. x atm 1 atm  ; x = 6.34 atm. Therefore p2 = 1.04 atm + 6.34 atm = 7.38 atm.

64.0 m 10.1 m p1V1 = p2V2; so V2 =

1.16.

1.82  1013 g SO 2 

nRT p  V

(1.04 atm)(67 L)  9 .4 L (7.38 atm)

1 mole SO 2  2.84  1011 moles SO 2 64.07 g L  atm )(273.15  17.0K ) mol  K  7  10 10 atm 21 8  10 L

(2.84  1011 moles SO 2 )(0.0821

L  atm 5 / 9 K L  atm    0.0456 mol  K 1 R mol  R

1.18.

0.0821

1.20.

Calculations using STP and SATP use different numerical values of R because the sets of conditions are defined using different units. It‟s still the same R, but it‟s expressed in different units of pressure, atm for STP and bar for SATP.

1.22.

The partial pressure of N2 = 0.80 

14.7 lb lb  11.8 2 2 in. in.

The partial pressure of O2 = 0.20 

14.7 lb lb  2.9 2 2 in. in.

1.24.

4.50torr (44.01g / mole)  0.01706 g / L  1.71 10 5 g / cm 3 m pM m pV   RT;   L  torr V RT M (62.36 )(273.15  87K ) mol  K 1.26.

Using the ideal gas law, the number of moles of CO2 = .965 atm 1.56 0.0621 moles atm 0.0821 mol 273.15 22.0 1 mole C6 H12 O6 180.16 g 0.0622 moles CO2 5.60 g C6 H12 O6 2 moles CO2 1 mole C6 H12 O6

2

Chapter 1

1.28.

Following the normal rules of derivation: (a) 3 y 2 

3w 2 z 3 xy 2 z 3 y2z3  (b) (c) w 32 y w2

3y 2 z 2 3y 2 z 2 w 3 z 3 2 xyz 3 6 xy   (d) Using the answer from part a, we get  (e)  (f) w w w 32 y 2 Using the answer from part e, we get

1.30

3y 2 z 2 . w2

p  n  (a)     V  T, P R  T

 T  V (b)     p  V ,n n  R

- pV  n  (c)    2  T  P, V RT RT  p  (d)    V  n  T ,V 1.32.

(a) V(T, p) 

nRT p

 V   V   dp (b) dV    dT    T  p  p  T (c)    Latm   Latm     1 mole  0.0821  1mole  0.0821  350K     nRT   nR  Kmole   Kmole    0.10atm  dp   dT    dV   10.0K    2 2     1.08 atm 1.08atm   p   p          = -1.70 L

1.34.

 V   nRT   2 V  2nRT   (a)  ;  2   2 p p3  p  n ,T  p  n ,T

nR   2 p   p   0 (b)    ; V  T 2  n ,V  T  n ,V

1.36.

(a) Z=1 for an ideal gas. (b) If the gas truly follows the ideal gas law, Z will always be 1 regardless of the pressure, temperature, or molar volume.

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Instructor‟s Manual

1.38.

Using equation 1.23, TB 

a , and using data from Table 1.6, we have: bR

L2 atm mol 2 for CO2: TB   1026 K L atm   0.04267 L/mol  0.08205  mol K   3.592

L2 atm mol 2 for O2: TB   521 K L atm   0.03183 L/mol  0.08205  mol K   1.360

L2 atm 1.390 mol 2 for N2: TB   433 K L atm   0.03913 L/mol  0.08205  mol K  

C . In order for the term to be unitless, C should have units of V2 (volume)2/(moles)2, or L2/mol2. The C‟ term is C‟p2, and in order for this term to have the L·mol same units as p ̅ (which would be Latm/mol , C‟ would need units of . (The unit atm bar could also be substituted for atm if bar units are used for pressure.)

1.40.

The C term is

1.42.

Gases that have lower Boyle temperatures will be most ideal (at least at high temperatures). Therefore, they should be ordered as He, H2, Ne, N2, O2, Ar, CH4, and CO2.

1.44.

(a) He: b=

0.0237L 1 m 3 m3   2.37  10 5 mole 1000 L mole

3 m3 1mole  29 m 2.37 10   3.94 10 mole 6.02 10 23 atoms atom 5

V

4

4 3 m3 r  3.94  10  29 ; r  2.11  10 10 m or 2.11Å 3 atom

(b) H2O: b=

0.03049L ; using a similar method to part a, r = 2.30 Å mole

(c) C2H6: b=

0.0638L ; using a similar method to part a, r = 2.94 Å mole

Chapter 1

1.46.

Let us assume standard conditions of temperature and pressure, so T = 273.15 K and p = 1.00 atm. Also, let us assume a molar volume of 22.412 L = 2.2412  104 cm3. Second virial coefficient terms can be calculated using values in Table 1.6.

Therefore, we have for hydrogen:

pV B 15.7cm3 / mol =1+ =1+ =1.00070 , which is a 0.070% increase in the RT V 2.2412 ´10 4 cm3 / mol compressibility.

For H2O, we have:

pV B -213.3cm3 / mol =1+ =1+ = 0.9905 , which is a 0.95% decrease in the RT V 2.2412 ´10 4 cm3 / mol compressibility with respect to an ideal gas.

1.48.

By comparing the two expressions from the text 2

B C a 1 b  Z  1 b        and Z  1   2   RT  V  V  V V  it seems straightforward to suggest that, at the first approximation, C = b2. Additional terms involving V 2 may occur in later terms of the first expression, necessitating additional corrections to this approximation for C. 1.50.

The Redlich-Kwong equation of state: p 

RT a  Vb T V ( V  b)

 RT a a  p       2 2 2 T V ( V  b)   V  T , n ( V  b) T V ( V  b)

an 2 1.52. The van der Waals equation of state: (p  2 )( V  nb)  nRT ; As V approaches ∞, the V an2/V2 term goes to 0 and the nb term becomes negligible. The equation then reduces to pV=nRT. 1.54.

The Redlich-Kwong equation of state: p 

RT a  Vb T V ( V  b)

As T approaches infinity, the second term on the right side goes to 0. The molar volume of a gas at high temperature will generally be high so the correction factor, b, is negligible. The equation reduces to pV  RT

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Instructor‟s Manual

RT 1.56. Using the ideal gas law, p  ,p  V

(0.0821

L  atm )(273.15K ) K  mol  1.00atm 22.41L

Using the Dieterici equation of state:

p  (0.0821

L  atm e )(273.15K ) K  mol

Latm    (10.91atm L2 ) /  ( 22.41L )( 0.0821 )( 273.15 K )  K mol  

(22.41L  0.0401L)

 0.981atm ; It varies

by ~2%.

1.58.

In terms of p, V, and T, we can also write the following two expressions using the cyclic rule:  T   p      p V  V   V   T V  and  . There are other constructions possible that       p  p  T  T   T  p       V  T  V  p would be reciprocals of these relationships or the one given in Figure 1.11.

1.60.

Since the expansion coefficient is defined as

1  V    ,  will have units of V  T  p

1 volume 1   , so it will have units of K-1. Similarly, the volume temperatur e temperatur e 1  V   , so  will have units of isothermal compressibility is defined as   V  p  T 1 volume 1   , or atm-1 or bar-1. volume pressure pressure 1.62.



 1  V  1 1  1   nRT  1  nRT  1  1        V      For an ideal gas,   2 V  P  T V  p  V  p  p  V p p

This is equal to 1 bar-1 at STP and SATP 1.64.

For an ideal gas,    expression becomes

6

nRT 1  V  1   nRT  1 nRT  V , this last        . Since 2 p V  p  T V p  p  V p

1V 1 T  for an ideal gas. The expression  is evaluated as V p p p

Chapter 1

T T 1  V  T   nRT  T nR    . For an ideal gas, the ideal gas law can be     p p V  T  p pV T  p  pV p

rearranged to give

nR V  , so we substitute to get that this last expression is p T

T V 1 , which  . Thus, the two sides of the equation ultimately yield the same pV T p expression and so are equal.

1.66.

RT . Therefore, the expression for density becomes, substituting p M pM  for the molar volume, d  . The derivative of this expression with respect RT / p RT pM  d  to temperature is  . Using the definition of V , this can be rewritten as   RT 2  T  p.n For an ideal gas, V 

M  d  .    VT  T  p.n

1.68.

1.70.

 g  m    m  Mgh  mol  s 2  pe  ; If we convert g to kg and recognize that a J= J RT    K   K  mole  2  kg  m      2 kg  m 2 Mgh  mol  s   , . All of the units cancel and the exponent is unitless.

RT s2  J   K   K  mol  

Mgh RT



Mgh

If we assume that the average molecular weight of air is 28.967 g/mole, p  e RT =e-X

kg  m J    X X   0.028967  9.8 2  432m    8.314 273.15  39.0K   0.0473; e  1.05atm mole  K  mole  s   

4 3  5   10 7 7  7.14 1.72. N=7 score    4 3    7 7 1.74.

The probability that the particle is in the higher state = e  E / RT . (a) At 200K, probability = e

  J  1000 J /   8.314   200 K   Kmole   

=0.548 7

Instructor‟s Manual

(b) At 500K, probability = e

  J  1000 J /   8.314  500 K   Kmole   

(c) At 1000K, probability = e

=0.786

  J  1000 J /   8.314  100 K   Kmole   

=0.887

probability . For the three temperatures 1  probability these ratios equal: 1.21, 3.67, and 7.85; as the temperature goes up.

To calculate ratios we can use the equation:

1.76.

3 RT; Erot = RT 2 3 3 (b) H2O is a non-linear molecule. Etrans = RT; Erot = RT 2 2 3 (c) Kr is an atom. Etrans = RT; Erot = 0 2 3 3 (d) C6H6 is a non-linear molecule. Etrans = RT; Erot = RT 2 2 (a) (CN)2 is a linear molecule. Etrans =

Chapter 2 The First Law of Thermodynamics 2.2.

2.4.

Work is defined as negative pV because if a system does work on the surroundings, the system loses energy.

1000 mL 1 cm 3 12.56 L    12,560 cm 3  1.256 10 4 cm 3 1 L 1 mL

605 J of work are done ON the system.

2.6.

(a) The work would be less because the external pressure is less. (b) The work would be greater because the external pressure is greater. (c) No work would be performed because the external pressure is (effectively) zero.

2.8.

First, we need to find the final volume of the CO2. Using P1V1=P2V2, we find that V2=105mL

1233 mmHg  2.10.

8

1 torr 1 atm 1.01325 bar    1.644 bar 1 mmHg 760 torr 1 atm

First, determine the T: 330 K – 298 K = 32 K.

Chapter 1

125 mL 

1 cm 3  125 cm 3 1 mL

Substituting: 25750 Pa 

1 bar  0.2575 bar .

100,000 Pa

2.12. First, we need to calculate the number of moles of phosphorous:

V

 R 

nT p

2.14. The kinetic energy of the hailstone, KE, is equal to: (2.66 bar)(27.5 L) L  bar R  0.0830 (1.887 mol)(466.9 K) mol  K Since all of the kinetic energy is converted into thermal energy, we can say that:

2.16. First, calculate the energy needed to warm the water by 1.00ºC: q = mcT = (1.00105 g)(4.18 J/gK)(1.00 K) = 4.18105 J Now, determine how many drops of a 20.0 kg weight falling 2.00 meters in gravity will yield that much energy. The amount of energy in one drop is mgh = (20.0 kg)(9.81 m/s2)(2.00 m) = 392.4 J. Therefore, .

2.18. True. As the gas expands, the average distance between molecules increases. For a real gas, work must be done to increase this distance due to intermolecular attractive forces between the gas molecules.

2.20. The major inaccuracy is the omission of the phrase “for an isolated system”, since for nonisolated systems energy can move in or out, giving the impression that that energy is created and/or destroyed. Can you find other inaccuracies?

2.22. q = -124.0 J

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Instructor‟s Manual

w=

= +53.3 J

Since U = q + w, U = -124.0 J + 53.3 J 2.24.

Reversibly: 0.0821

U = -70.7 J

5/9K L  atm L  atm   0 .

0456 mol  K 1 R mol   R

Irreversibly: 0.80 

14.7 lb lb  11.8 2 in. in.2

The reversible work is much larger than the irreversible work. This is one numerical example of the concept that the maximum amount of work is obtained by a reversible process.

10

Chapter 1

2.26. If any change in a system is isothermal, then the change in U must be zero. It doesn‟t matter if the process is adiabatic or not!

2.28. Of the two distances mentioned, the 9-mile distance between the two cities is analogous to a state function, because that distance is independent of how a trip is actually traveled between the two cities.

2.30. The keys to this problem are the stated conditions. If the processes are adiabatic, then q = 0. If the initial and final temperatures are the same, then U = 0. By the first law of thermodynamics, if U and q are 0, then w = 0 as well. While these values fit the conditions of the problem, do you think that a piston can even work under such conditions? Probably not.

2.32. First, we should determine the number of moles of gas in the cylinder. Assuming the ideal gas law holds:

14.7 lb lb 0.20   2.9 2 in. in.2  m  m pM 4.50tor ( 4.4 01g/mole) 5 3 (a) The final pressure can be determined using Charles‟ law: pVMRT; V  RT  Ltor 0.01706g/L1.7110 g/cm ( 2.6 36 )(273.1587 )K molK

3y 2



y 2 z 3 w

(b) w = 0 since the volume of the tank does not change. q = n ̅ T = (572 mol)(21.0 J/molK)(140.0ºC – 20.0ºC) = 1.44106 J 



U = q + w = 1.44106 J + 0 = 1.44106 J.

2.34. q = -2690 J (given) 

U = q + w = -2690 J + 2689 J = -1 J



H = U + (pV) Since the process occurs at constant temperature, Boyle‟s law applies and (pV) = 0. Therefore, H = -1 J.

2.36. Since we‟re at the normal boiling point, the vaporization is a constant-pressure process (1 atm at the normal boiling point). Therefore, H = qp and H = +2260 J/g.

For work, we need change in volume. Assuming the ideal gas law holds, the volume of 1 gram of steam at 100ºC (373.15 K) is:

11

Instructor‟s Manual

6 xy 

w3 z 3 2 xyz 3  w 32 y 2



3 y 2 z w

2

The volume of a gram of water, 1.00 mL, is negligible compared to this. Therefore, let us use V = 1.720 L. Therefore: 

3 y 2 z 2 w

Since U = q + w, we have U = +2260 J – 172 J = 2088 J. 2.38.

In terms of pressure and volume:

3y 2 z 2 w2

.

p  n   For enthalpy:  .  R T  V  T, P 2.40.

 T       p  V , n



V n  R

; There is no change in volume. Assuming the heat capacity of an ideal gas:

Using the ideal gas law:

2.42.

Since it is a constant pressure process (assumed 1 atm), H=-6.13x104J. Now, we need to calculate the volume change of the coffee:

12

Chapter 1

2.44. Start with

and

: Since p

. Substituting for

= RT for an ideal gas, we also have

(equation 2.32) and

(equation 2.27):

C 2 V

R is a constant, so it can be removed from the  term:

Now all terms can be divided by T to get the desired relationship: 0.0237L 1m 3 m3   2.37  10 5 mole 1000 L mole

.

2.46. isobaric = constant pressure isochoric = constant volume isenthalpic = constant enthalpy isothermal = constant temperature. A gaseous system that has all these conditions simultaneously probably isn‟t undergoing any physical change! Can you conceive of a process in which all of these conditions are satisfied at once?

2.48. Actually, the ideal gas law can be used to determine the Joule-Thomson coefficient for an ideal gas, but it will turn out that the Joule-Thomson coefficient for an ideal gas is zero! m3 1mole 29 m3  .3 9410 : mole .6 021023 atoms atom

2.50. Using the approximate version .2 37105  

p = 0.95 atm – 200.00 atm = -199.05 atm Since JT = 0.150 K/atm: V

4 m3 r 3  3.94  10  29 ; r  2.11  10 10 m or 2.11 3 atom

If the initial temperature is 19.0ºC and the temperature drops by 29.9 degrees, the final temperature should be about 19.0 – 29.9 = -10.9ºC. 2.52. µJT for Argon at 0°C and 1atm is 0.4307K/atm. From Eqn 2.35 we know that:

æ ¶H ö æ J öæ K ö J ; ç ÷ = -C P × m JT = - ç 20.8 ÷ç 0.4307 ÷ = -8.96 è ¶P øT è K × mol øè atm ø mol × atm equal to

for a monatomic ideal gas is

.

2.54. Although U and H have similar behavior for isothermal processes of ideal gases, they won‟t necessarily for real gases. Therefore, a Joule-Thomson coefficient defined in terms of U can be defined but will not have the same numerical value as one defined in terms of H. In addition, the Joule-Thomson experiment is originally defined (as is JT) for an isenthalpic process, not one for which U is constant. Therefore, that definition of JT would probably not be proper.

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Instructor‟s Manual

2.56.

2.58.

Since

:

for a monatomic ideal gas is equal to

. Using the equation

: -75 J =



(0.122 mol)(12.47 J/molK )T and solve for T: T = -49 K. Since the initial temperature was 235ºC = 508 K, the final temperature is 508 – 49 = 459 K.

2.60.

 RT a a   2 . For a monatomic ideal gas, (V  )b 2 TV(V  )b 2 TV (V  )b

2.62.

an 2 (p  2 )(V  nb)  nRT V

(a) CO2 is a linear polyatomic molecule where N=3 atoms. In the low temperature limit it only has contributions from translational and rotational motion so

p V  RT . In the high temperature limit, vibrational L  atm

motion contributes as well so: p  RT , p  (0.0821 K  mol )(273.15K )  1.00atm 22.41L

V

(b) H2O is a non-linear polyatomic molecule where N=3 atoms. In the low temperature limit it only has contributions from translational and rotational motion so L  atm e p  (0.0821 )(273.15K) K  mol

Latm   (10.91atm L2 ) /  ( 22.41L )( 0.0821 )( 273.15 K )  Kmol  

(22.41L  0.0401L)

 0.981atm . In the high temperature limit, vibrational

V  motion contributes as well so:   T

14

  

 p



 p   T p    V

  V   T

Chapter 1

2.64. Using Eqn. 2.47:

 T    p    V   V      p   T    T    V  p

1  V    The final temperature is 63% of the initial temperature.

V  T  p 2.66. For an adiabatic change,

1 volume 1 1   . Therefore,  V volume temperature temperature

 V   p 

   T

Therefore, the absolute temperature will drop to 55% of its initial temperature.

2.68. As vibrational energies begin to have more significance, decreases. As decreases, the exponent in the equation in Example 2.14 become smaller. This makes the final temperature in the second example higher than the first as is demonstrated.

2.70. (a) N2 and CO have nearly identical molar masses, M. What is different between the two gases is over a temperature range. By measuring the speed of sound at several different temperatures of the gases, one should be able to differentiate between the two. (b At 100K:

1 volume 1   volume pressure pressure At 500 K:

 1  V      V  P  T 2.72. The volumes of both water and ice at 0ºC are: water: Vl =

18.02 g 18.02 g =18.02 mL ; ice: Vs = =19.66 mL .

0.99984 g/mL 0.9168 g/mL

Therefore, V = 19.66 mL – 18.92 mL = 0.74 mL, so that the work is

æ 1atm ö æ 1L öæ 101.32J ö w = -pext DV = - (1bar ) ç ÷ ´ ( 0.74mL ) ç ÷ç ÷ = -0.074J . è 1.01325bar ø è 1000mL øè 1L·atm ø

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Instructor‟s Manual

2.74.

According to Table 2.3, 1 gram of H2O gives up 2260 J when it condenses. Since each gram of ice requires 333.5 J to melt, we get 1 V

T  p



T 1  V  p V  T

nR p 2.80.



T p

2.76.

2.78.

V p

   p



 1 p

T pV

 T

 nRT   p 

    

T pV

nR p

V T



The reactions are: 2[NaHCO3 (s)  Na (s) + ½ H2 (g) + C (s) + 3/2 O2 (g)] 2 -fH = +1901.62 kJ fH = -1130.77 kJ

2 Na (s) + C (s) + 3/2 O2 (g)  Na2CO3 (s)

fH = -393.51 kJ

C (s) + O2 (g)  CO2 (g)

fH = -285.83 kJ

H2 (g) + ½ O2 (g)  H2O (l)

This yields the overall reaction (you can verify that), and the overall rxnH is the sum of the values on the right: rxnH = 91.51 kJ. 2.82.

The reaction is 2 Al (s) + Fe2O3 (s)  Al2O3 (s) + 2 Fe (s). The rxnH is (using data from the appendix): (2 mol)(0 kJ/mol) + (1 mol)(-1675.7 kJ/mol) – (1 mol)(-825.5 kJ/mol) – (2 mol)(0 kJmol) = -850.2 kJ.

2.84.

In this case, since the combustion occurs in open atmosphere in (assumed) constant pressure, this time qp = H = -31,723 J. If the expansion of gases is done against a constant atmospheric pressure of 1 atm, to determine work we need to know the net volume of gas produced. From exercise 2.55, we found that there was a net change of –0.00491 mol of gas produced (ignoring the volume of the benzoic acid itself). Thus,

T pV

V T

,

which



1 p

Using this change in volume to calculate work: RT V  p Finally, to calculate U:

d

16



M RT /

p



pM RT

Chapter 1

2.86. This problem is very similar to Example 2.19, so we will follow that example, taking data from Table 2.1.

The heat needed to bring the reactants from 500ºC (or 773 K) to 298 K is: 

H1 = q = (2 mol)(2.02 g/mol)(14.304 J/gK)(-475 K) + (1 mol)(32.00 g/mol) (0.918 J/gK)(-475 K) = -41403 J The heat of reaction is 2(fH[H2O(g)]) = 2 mol  -241.8 kJ/mol = -483.6 kJ = H2 The heat needed to bring the products from 298 K to 500ºC (or 773 K) is:



H3 = q = (2 mol)(18.02 g/mol)(1.864 J/gK)(475 K) = +31,910 J The overall rxnH is the sum of these three parts. Converting all energy values to kJ:



rxnH = -41.403 kJ – 483.6 kJ + 31.910 kJ = -493.1 kJ

Chapter 3 The Second and Third Laws of Thermodynamics 3.2.

The dissolving of several simple salts is an endothermic process, as mentioned in the text. There are commercially-available cold packs, some of which use an endothermic chemical reaction (as opposed to simply storing them in a freezer). Melting and boiling are both endothermic processes that are spontaneous at certain temperatures.

3.4.

The individual steps must be carried out under the proper conditions: reversible & adiabatic (steps 2&4) or reversible and isothermal (steps 1&3). If they are, then we can calculate the efficiency as

e=-

3.6.

wcycle (-445 - 99 + 360 + 99) J -85 === 0.191 =19.1% q1 445 J 445

efficiency  1  0.440  1 

3.8.

Tlow Thigh

150  273.15 ; T

(Thigh  273.15)

high

 482  C

For a heat engine, the high-temperature reservoir must indeed be at a higher temperature than the low-temperature reservoir. An engine converts thermal energy to work. It is impossible to convert all of the thermal energy to work so the “leftover” heat is dumped into the lowtemperature reservoir. It is possible to reverse the cycle and move thermal energy from a lower temperature to a higher temperature sink. This is not called an “engine”, but a heat pump. Energy must be added to the process to make this work. A familiar example is a household air conditioner.

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Instructor‟s Manual

3.10.

For the highest efficiency, an engine should have the highest TH possible and the lowest TL. In practicality, TL is usually more limited than TH by the ambient temperature of the surroundings (air, water, etc.). To increase efficiency of an engine, it is most practical to make the high temperature reservoir as hot as possible.

3.12.

In Figure 3.2, steps 1&3 are isotherms, steps 2&4 are adiabats.

3.14.

All definitions of efficiency are applicable to real gases as well as ideal gases. Efficiency‟s definition is independent of the type of material involved in a process.

3.16.

S 

3.18.

A better statement of the second law of thermodynamics includes the conditions under which the second law is strictly applicable: for an isolated system, a spontaneous change is always accompanied by an increase in entropy.

3.20.

The units are standard, so let us simply substitute into the proper expression

q rev (3.87 mol)(10,480 J/mol) 40,557.6 J    74.5 J/K T (271.3  273.15 K) 544.45 K

 25.69  7.32  10 4 T  4.58  10 6 T 2   dT   T  295 K 

1273 K

S  2.5 

1273 K

 2 .5 

 25.69

  T

295 K

  7.32  10  4  4.58  10 6 T dT 

1273 K 1273 K   Tf 1273 K 1 4 6 2   2.5  25.69 ln  7.32  10 T  (4.58  10 T ) 295 K Ti 295 K 2  295 K    2.5  (37.56  0.7158  3.512)  100.9 J/K

3.22. Since this is an isothermal process, U=0 and q= -w. w   p ext V.

æ 101.32J ö w = -pext DV = - (1.75atm) ( 5.50L - 2.00L ) ç ÷ = -620.6J . So, q=+620.6J. Since this heat è 1L·atm ø  620.6J J came from the surroundings, we can calculate Ssurr   2.08 .

298.15K K 

Ssys must be calculated using a reversible q. Although this process is not reversible, we can use a reversible process to calculate it since S is a state function.

w = -nRT ln

Vf æ 5.50L öæ 101.32J ö 5.50L = - ( P1V1 ) ln = - (8.00atm) ( 2.00L ) ç ln ÷ç ÷ = -1640J. è 2.00L øè 1L·atm ø Vi 2.00L

So qsys=+1640J. DSsys =

18

+1640J J J = +5.50 and DSuniv = DSsys + DSsurr = +3.42 .

298.15K K K

Chapter 1

Since this is an irreversible process, the overall entropy of the universe has increased, as expected.  nRT an 2  3.24. Since this is an isothermal process, U=0 and q= -w. w  pdV    2 dV.  V  nb V  0.0750 L 0.0750 L æ é dq nRT an 2 ö 1 an 2 ù DSsys = ò rev = ò ç - 2 ÷ dV = ênR ln (V - nb) + ú T V øT TV û 0.0250 L ë 0.0250 L è V - nb

æ æ L öö 0.0750L - (0.00100mol) ç 0.05636 ÷÷ ç æ J ö ç è mol ø ÷ = 0.00100mol ç8.314 + ÷ ln è Kmol ø ç 0.0250L - (0.00100mol) æ 0.05636 L ö ÷ ç ÷÷ ç è mol ø ø è æ atmL2 ö (0.00100mol)2 ç 6.714 2 ÷ æ 101.32J öæ ö mol 1 1 è ø -3 J ç ÷ç ÷ = 9.10 ´10 è 1Latm øè 0.0750L 0.0250L ø 298.15K K Since the gas expands very, very, slowly, we can assume this is a reversible process so Ssurr= J 9.10 103 and Suniv=0.

K 3.26. Since this is an isothermal process, U=0 and q= -w.

   1atm   46.0psi 15.6L   p   101.32J  14.7 J  P V  14.7  14.696psi  S air  nR ln f   1 1  ln    19.12   ln  pi 295.15K K  T  15.6   1Latm  46.0     3.28. A calculation is not needed for this one. If you recognize that the initial and final conditions are exactly the same, then by using the concept of state function (which S is), then the overall change in S for the complete process is exactly zero.

3.30. 5/2 R is the heat capacity of an ideal gas under conditions of constant pressure. The overall process in Example 3.5 is not a constant-pressure process, but it is broken down into two steps. Since entropy is a state function the sum of the entropy changes from these two steps equals the entropy change for the overall process. Step 2 (where the 5/2R is used) is a constant pressure process.

3.32. The chemical processes can be represented as: Ar (4.00 L, 298 K, 1.50 atm) + He (2.50 L, 298 K, 1.50 atm)  Ar, He (6.50 L, 298 K, 1.50 atm) Ar, He (6.50 L, 298 K, 1.50 atm)  Ar, He (20.0 L, 298 K, 0.488 atm)

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Instructor‟s Manual

The new pressure was determined by simply applying Boyle‟s law. Since entropy is a state function, the change in entropy for the overall process can be determined by calculating the entropy changes for each process (a mixing process and an expansion process), then adding the two values. To determine the Smix, the number of moles and the mole fractions of Ar and He are needed:

For Ar : n 

pV (1.50 atm)(4.00 L)   0.245 mol RT (0.08205 L  atm/mol  K)(298 K)

For He : n 

pV (1.50 atm)(2.50 L)   0.153 mol RT (0.08205 L  atm/mol  K)(298 K)

The total number of moles of gas is 0.245 + 0.153 = 0.398 mol, so the mole fractions of 0.245 Ar and He are xAr   0.616 , so xHe = 1 – 0.616 = 0.384.

0.398 The entropy change of the mixing step is therefore:

S mix  (8.314 J/mol  K)(0.245 mol)ln(0.61)  (0.153 mol)(ln(0.384)  2.20 J/K The entropy change for the expansion step is:

S exp  nR ln

Vf Vi

 (0.398 mol)(8.314 J/mol  K) ln

20.0 L  3.72 J/K 6.50 L

Therefore, the total entropy of the process is S = +2.20 + 3.72 = +5.92 J/K. 3.34.

(a) After the piece of copper is dropped into the calorimeter, a state will be reached, where the is no more energy exchange between the water and copper. According to the zeroth law, in this state the copper and the water will have the same temperature. According to the first law, the energy of the isolated system remains constant. Therefore, any energy lost by one part of the system will be gained by another part of the system. In this case, the hot copper will lose energy and the cooler water will gain energy. In terms of the second law, this spontaneous change will occur only if the total entropy of the system increases. (b) Heat lost = heat gained. Note that heat lost is negative, while heat gained is positive.

Heat lost =

 m  c  T  m  c  (T f  Ti )  (5.33 g)(0.385 Heat gained = 3

J )(T f  372.85 K)  2.052T f  765.107 gK

y 2 z 2 w

2

Equate the two quantities and solve for Tf: 2.052Tf  765.107 = 416.03T f  123,042.470 123,807.577 = 418.082Tf Tf = 296.13 K = 23.0ºC. Tf 296.13 K  (5.33 g) (0.385 J/g  K) ln  0.473 J/K  entropy loss of Cu (c) S  mc ln Ti 372.85 K

20

Chapter 1

(d) S  mc ln

Tf Ti

 (99.53 g) (4.18 J/g  K) ln

296.13 K  0.534 J/K  entropy gain of water 295.75 K

(e) The total entropy change for the system is S = -0.473 + 0.534 = 0.061 J/K. (f) Because the overall entropy change of the isolated system is positive, so we would expect that the process – the equalization of temperatures – would be spontaneous.

3.36. Parts c, d, and e are probably fairly accurate despite the fact that neither Cu (s) nor H2O (l) are ideal gases. The derivation of the equation used to calculate S did not assume that the system being described was limited to an ideal gas.

3.38.

T   0.455kg  1000g  0.45J  923  5   ln S  n C ln final   800lbs    1.88  10 J / K  1lb  1kg  gK  293   Tinitial 

3.40. At 37ºC = 310 K, the number of moles of gas is pV (1 atm)(1 L) n   0.0393 moles . The change in entropy is RT (0.08205 L  atm/mol  K)(310K) then 590 mmHg S  (0.0393 mol)(8.314 J/mol  K) ln  0.0827 J/K 760 mmHg 3.42. (a) In Example 3.2, the entropy change for the vaporization of benzene (C6H6) was calculated as 1.12 J/K for 1 gram. Since there are 78.0 grams in one mole of benzene, the molar entropy change is 1.12  78.0 = 87.4 J/molK, which is fairly close to 85. (b) Determine the molar entropy change for 1 mole of water: q 40,700 J/mol   109 J/mol  K , which is off quite a bit and so does not T 373.15 K support Trouton‟s rule. The reason is that the hydrogen bonding between water molecules makes water deviate from ideality much more than in benzene, in which no hydrogen bonding exists. (c) According to Trouton‟s rule: S 

85 J/mol  K 

q 30,100 J/mol . Solving for T:  T T

30,100 J/mol  354 K  81C , which is very close to the measured boiling point 85 J/mol  K of 80.7ºC. T

3.44. If S = k ln , where  is the number of possible combinations, the number of combinations can never be less than 1 for any real system. The logarithm of 1 is zero, so it may be possible that the entropy of a system is zero. However, numbers greater than 1 have positive logarithms, and if k is a positive constant (which it is), then the absolute amount of entropy S can never be negative. (This does not preclude that changes in S might have negative values.)

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Instructor‟s Manual

3.46.

(a) the dirty kitchen (b) the blackboard with writing on it (c) 10 gram of ice (d) If perfectly crystalline, both have the same entropy (zero) (e) 10 grams of ethanol at 22ºC. 1000

3.48. S1000 K  S 298.15 K 

dT

J



J



1000

J

 C T  126.04 K  mol   20.78 K  mol  ln 298.15  151.2 K  mol p

298.15

This is assuming Cp for He is equal to 5/2R=20.78

J .

K  mol

3.50.

If helium is indeed a liquid at absolute zero, it would not have zero entropy. That‟s because according to the third law of thermodynamics, only a perfect crystal (that is, a solid) can have zero entropy at absolute zero.

3.52.

S  S 2  S1  k ln  2  k ln 1  1.00

k ln ln

J Kmol

2 J  1.00 1 K  mol

2 1   7.25  10 22  23 1 1.38  10

22 2  e 7.2510 1

This is a HUGE number. Even for a small increase in molar entropy, the increase in the number of available states beyond our ability to easily calculate.

3.54.

For the first case: S  nR ln

Vf J J   2.00L  (1mole) 8.314  5.76  ln Vi K  mole  1.00L K 

For the second case: S  nR ln

Vf J J   11.00L  (1mole) 8.314  0.792  ln Vi K  mole  10.00L K 

For any process going from state 1 to state 2. S  S 2  S1  k ln  2  k ln 1 S  k ln

2  and 2  e S / k 1 1 5.76

J

/k

0.792

J

/k

For case 1:  2  e K 1 and for case 2:  2  e K 1 . Case 1 has a greater change in the availability of states. This makes sense because you are doubling the available volume in case 1 and increasing it by a smaller percentage in case 2.

3.56.

22

First, we need the formation reaction for each compound, then we can do a „productsminus-reactants‟ action using values of S from the appendix.

Chapter 1

(a) The reaction is H2 (g) + ½ O2 (g)  H2O (l) 



S = 69.91 – 130.68 – ½ (205.14) = -163.34 J/K

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Instructor‟s Manual

(b) The reaction is H2 (g) + ½ O2 (g)  H2O (g) 



S = 188.83 – 130.68 – ½ (205.14) = -44.42 J/K

(c) The reaction is 2 Fe (s) + 3 S (s) + 6 O2 (g)  Fe2(SO4)3 (s) 



S = 307.46 – 2(27.3) – 3(32.054) – 6(205.14) = -1074.1 J/K

(d) The reaction is 2 Al (s) + 3/2 O2 (g)  Al2O3 (s) 



S = 50.92 – 2(28.30) – 3/2 (205.14) = -313.39 J/K

3.58.

The balanced chemical reaction is 2 Al (s) + Cr2O3 (s)  Al2O3 (s) + 2 Cr (s)



S = [50.92 + 2(23.62)] – [2(28.30) + 80.65] = -39.09 J/K

3.60.

A phase transition can be approximated as a constant-pressure reversible process for which: J  H mol  0.31 J  trans S  trans  Ttrans 653.15K K  mol  200

3.62.

S  S (C, diamond )  S (C, graphite)  2.377

J J J  5.69  3.31 K  mol K  mol K  mol

The sign on this value is negative which corresponds to a decrease in entropy or an increase in order. This makes sense because diamond is a more ordered structure than graphite. Diamond has covalent bonding in three dimensions. Graphite is covalently bound in two dimensional sheets.

3.64.

First, we need a balanced equation: 6CO 2 (g)  6H 2 O(l)   C 6 H12 O 6 (s)  6O 2 (g)





 rxn S  6  S (O 2 (g))  S (C 6 H12 O 6 (s))  6  S (CO 2 (g))  6  S (H 2 O(l))  6mol  205.14  262.14

J J J J    1mol  209.19  6mol  213.785  6mol  69.91 K  mol K  mol  K  mol K  mol 

J K

Chapter 4 Gibbs Energy and Chemical Potential 4.2.

24

Processes occur with change in energy as well as changes in entropy. Therefore, spontaneity conditions usually have to be determined with respect to both. However, if

Chapter 1

entropy changes are to be used as the sole, strict spontaneity condition, an isolated system is required, which prohibits changes in energy. Therefore both U and H must be zero.

4.4.

The total change in U or H is required to be negative for these spontaneity conditions, not any one component (as given by a partial derivative).

4.6.

Starting with the expression dU + pdV – TdS  0, use the definition dA = dU – TdS – SdT that we get by derivating eq. 4.5, solve for dU and substitute: dU = dA + TdS + SdT; therefore, dA + TdS + SdT + pdV – TdS  0, which simplifies to dA + SdT + pdV  0. Under conditions of constant T and V, the second and third terms are zero, so this spontaneity condition simplifies to (dA)T,V  0.

4.8.

For internal energy, dU = 0 under conditions of constant volume and entropy.

For enthalpy, dH = 0 under conditions of constant pressure and entropy.

For Helmholtz energy, dA = 0 under conditions of constant volume and temperature.

4.10.

The reaction can do up to 237.13 kJ of work for every mole of H2O formed.

4.12.

U = H = 0 (for an isothermal process). To determine the pressure at the bottom of the ocean, if the water pressure increases by 1 atm for every 10.55 meters of depth and the 1 atm depth is 10,430 meters, then the pressure increase is 10,430 m   989 atm . If 10.55 m this is the pressure increase, and assuming that the pressure on the surface is 1atm, then the pressure at that depth must be 989 + 1 = 990 atm. Therefore: w  nRT ln

pi 990 atm  (1 mol)(8.314 J/mol  K)(273 K) ln  15,700 J . pf 1 atm

Because q = -w, q= 15,700 J Since this is a reversible process, this equals A as well. Finally, for S: S  nR ln

pi 990 atm  (1 mol)(8.314 J/mol  K) ln  57.3 J/K . pf 1 atm

4.14.

For NaHCO3 (s)  Na+ (aq) + HCO3− (aq),



H = [−240.12 + (−691.99)] – (−950.81) = +18.70 kJ.



S = [59.1 + (91.2)] – (101.7) = +48.6 J/K.



G = [−261.88 + (−586.85)] – (−851.0) = +2.3 kJ. For Na2CO3 (s)  2 Na+ (aq) + CO32− (aq),



H = [2(−240.12) + (− 676.3)] – (−1130.77) = − 25.8 kJ.

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Instructor‟s Manual



S = [2(59.1) + (−53.1)] – (138.79) = − 73.7 J/K.



G = [2(−261.88) + (−528.1)] – (−1048.01) = − 3.9 kJ.

4.16. The first method to calculate G uses the Gibbs free energies of formation, fG:

 rxn G   n   f G(products)   n   f G(reac tan ts)

 12   f G  CO 2 (g)   6   f G  H 2O(l)     2   f G  C6 H 6 (l)   15   f G  O 2 (g)    (12mole  394.35kJ / mole  6  237.14kJ / mole)  (2 124.4kJ / mole  0)  6404kJ The second method uses the equation: G  H  TS . First, we need to calculate H and S.

 rxn H   n   f H(products)   n   f H(reac tan ts)  12   f HCO 2 (g)   6   f HH 2 O(l)   2   f HC 6 H 6 (l)   15   f HO 2 (g)   (12mole  393.51kJ / mole  6  285.83kJ / mole)  (2  48.95kJ / mole  0)  6535kJ

 rxn S   n  So (products)   n  So (reac tan ts)



 



 12  S o CO 2 (g )   6  S o H 2 O(l)   2  S o C 6 H 6 (l)   15  S o O 2 (g ) 

 (12mole  213.785kJ / mole  6  69.91kJ / mole)  (2  173.26kJ / mole  15  205.14kJ / mole)  438.7 J G  (6535000J )  (298K )(438.7 J )  6404kJ

The values are identical.

4.18.

26

G for the reaction is zero, because under the conditions given the phase change between liquid and solid water is an equilibrium. Data given in the appendix are given for 25ºC, not 0ºC. Solid H2O is not thermodynamically stable at 25ºC, so a lot of data is not given for that phase under that condition. However, since we know that 0ºC is the normal melting point of H2O under standard pressure, we expect that the value of G would be exactly 0.

Chapter 1

4.20. First we need to calculate rxnG:

 rxn G   n   f G (products)   n   f G (reac tan ts)

 6   f G CO 2 (g)   6   f G H 2 O(l)   1   f G C 6 H 12 O 6 (s)   6   f G O 2 (g )   (6  394.35kJ / mole  6  237.14kJ / mole)  (910.4kJ / mole  0)  2879kJ / mole For each mole of glucose utilized, 2878kJ of energy are available to do work.

 1hr   1mole glu cos e  2878kJ   120g glu cos e   0.918hrs 180g   1mole glu cos e  2090kJ  4.22. No. If G=0, the system is at equilibrium.

4.24. Since G=0 for a phase change, S 

H 12610J / mole J   9.43 1064  273.15K  T K  mol

4.26. A=0 for a constant volume phase change. This could be a good approximation for a solid-liquid change, but not when one of the phases is a gas.

4.28. The two heat capacities can be defined, but not very easily. The heat capacity is a change in U or H with change in temperature. However, the natural variable equations for U and H do not have temperature as a variable; rather, they both have temperature as a constant.

Changes in U and H with respect to temperature can be written analytically, but you wouldn‟t get a simple, useable expression for CV or Cp. 4.30.

First, we write a „natural variable‟ relationship for dS in terms of V and T:  S   S  dS    dV    dT . This gives us the two differentials in the correct places;  V  T  T V now we need only derive the coefficients on the differentials. Let us consider the first  S   p  coefficient,   . According to Maxwell‟s relationships, this equals   . Using the  V  T  T V V / T  p chain rule of partial differentiation, we can show that this equals  . V / p T 1 V / T  p V Multiplying both the numerator and denominator by 1/V, this becomes  , 1 V / p T V which is /. This gives us the first coefficient. For the second coefficient, we go backwards and use two different Maxwell relationships to substitute for the numerator S / p V  V / T S  and denominator: . This last expression is the chain-rule T / p V V / ST

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Instructor‟s Manual

 S  expansion for   . Thus, we‟ve shown that on a term-by-term basis,  T V S / p V   S   S  dS   dT .  dV    dT  dV  T / p V   V  T  T  V 4.32.

4.34.

4.36.

 G    V . Volume is always a positive quantity. As p increases (p Equation 4.25 is:   p  T positive), G must also be positive to make the volume positive.

 G   H    V , where the bars have   V and equation 4.25 is:  Equation 4.21 is:   p  S  p  T been added to represent molar quantities. Since the molar volume of a gas is so much greater than that of a solid or a liquid, H or G must be much greater for a gas for the same p.

We will check each of these using the cross-derivative equality requirement of exact differentials:

  F     F   1  F  1  F  (a)    ;    . If        then the expression will be y  x  y  x  y  x  y  x  y x  y  x y x considered an exact differential. In this case,

  F     F          =0. So this y  x  y  x  y  x  y x

expression is an exact differential.

 1   F   1 1  F  1   F    F  (b)    ;    ;     2 ;     2  x  y y  y  x x y  x  y  x y x  y  x  y x This expression is not an exact differential.  F    F     F    F  2 2 3 (c)    2 x 2 y 2 ;    3x 3 y 3 ;     4x y;     9x y  x  y  y  x  x  y  y  x   x  y   y  x

This expression is not an exact differential.

 F    F    F    F  2 2  2 2 (d)    2 x 2 y 3 ;    2 x 3 y 2 ;     6x y ;     6x y y  x  y  x  y  x  y  x  y  y  x x This expression is an exact differential.  F    F     F    F  (e)    x n ;    y n ;     0;     0 y  x  y  x  y  x  y  x  y  y  x x

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Chapter 1

This expression is an exact differential.  F   F  (f)    x 3 cos y;    x 3 sin y ;  x  y  y  x

  F     F   3 2      x sin y;     3x sin y y  x  y  x  y  x  y x

This expression is not an exact differential.

4.38. Starting with dH = TdS + Vdp, divide both sides by dp and hold T constant:

æ ¶H ö æ ¶S ö 1 dS T ¶S . Using ) . All we need to do is show that T =  ç ÷ = T ç ÷ +V = V (1+ V dp V ¶p è ¶p øT è ¶p øT the definition of :



1  V  1  S   , where we have used a Maxwell relation as a substitution.      V  T  p V  p  T

Multiplying by T: T 

T  S    , which is the desired relationship. Therefore, V  p  T

 H     V (1  T ) .  p  T

4.40. Simply put, by multiplying through by the denominator in the partial derivative, we get that d(U) = -(p)dV, which is a form of the definition for work. Recall that changes in U manifest in only two forms: as heat and/or as work. This expression is consistent with the work part of U. 4.42. To demonstrate the cyclic rule of partial derivatives, let us evaluate the derivatives listed in Figure 1.11 using the ideal gas law:

  nRT  nR   nRT  nR  p   V             p  T  p T  p   T V T  V  V  V    nRT  nRT        2 and substitute: p  p  T p  p 

V / T  p nR (nR / p ) nR p  p   which reduces to   or pV = nRT,    2 (V / p ) T V  (nRT / p ) V T  T V verifying the cyclic rule.

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Instructor‟s Manual

4.44. Substituting for the definitions of  and :  V  (1 / V )   T  p  V    V       . Using a Maxwell relationship, substitute for the final   S  T  V   S  T   (1 / V )  p  T partial derivative, and rewrite the denominator as its reciprocal in the numerator (note that the 1/Vs cancel):  V   p   T   . Inspection of the three partial derivatives shows that the three       T  p  V  T  p V variables are organized in such a way as to give the cyclic rule of partial differentiation, so that the product of the three derivatives is –1: -(-1) = 1.

4.46.

 U   p  From Example 4.11:    T   p  V  T  T  V 2  nRT  an  For a van der Waals gas, p    2  V  nb  V 2 2  U   nR   nRT  an  an    T     2   2  V  T  V  nb   V  nb  V  V

an 2 For an isothermal process in an ideal gas, U=0. 2 is the correction factor for the V pressure and represents the magnitude of the interaction between the gas particles. This is not present for the ideal gas.

4.48.

 p   T  According to a Maxwell relationship,      . For an ideal gas,  S  T  V  p   pV  p  T   . For a van der Waals gas:      V  nR  p nR  V  p  an 2   p  2 V  nb  V   1   2an 2  an 2   T      V  nb   p  2  .

T Therefore : -     nR nR  V 3  V   V  p 

Here, we have had to take the derivative of the two V-containing terms using the chain rule.

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Chapter 1

4.50. The derivation of a Gibbs-Helmholtz expression for A is exactly the same as that given in section 4.7 in the text, except that A is substituted for G and U is substituted for H. Rather   A U than redo the entire derivation, here we provide the final answer:     2 , or T  T  p T   A  U    2 . T  T  p T

4.52.

pf

Vi 25.0 L  (0.988 mol)(8.314 J/mol  K)(350 K) ln  967 J . pi Vf 35.0 L The substitution for volumes instead of pressures was made using Boyle‟s law. G  nRT ln

 nRT ln

4.54. The easiest way to verify the connection between equations 4.41 and 4.42 is to start with equation 4.42, perform the derivation in the expression, and show that we get equation 4.41: 

 G H     2  Taking the derivative of the fraction, we must apply the chain rule of T  T  T differentiation, since both G and T depend on T:

 H  1 1   H  1  1 G   2 . Now, we  G  G    2 and simplifying: G    2      T T  T  T T  T  T  T T  G G H can multiply through by –T:   , which is equation 4.41.

T T T

4.56.

To show that diamond is unstable with respect to graphite, we need to show that G is positive at all temperatures for the following reaction: C(graphite)  C(diamond ) . We know that at 25°C, H=1.897kJ and G=2.90kJ. If we use Eqn 4.44 to calculate G at another temperature, T, we get: G 2.90kJ  T 298.15K  1.897kJ . Or: G  1.00kJ  1.897kJ .

1  T 298.15K T 1     T 298.15 

Since T is always positive, G must be positive for all temperatures.

 A  4.58. We start with    p . Integrating, we get:  V  T Vf

V nRT dV  nRT ln f V Vi Vi

A   

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Instructor‟s Manual

4.60.

We need to use the equation G  Vp .

 1000J  1L  atm   1.00kJ   G 1kJ  101.32J     548atm (a) p  V  18.02cm 3   1mL  1L   1 mole   3   1cm  1000mL   mol  (b) G  nRT ln

ln

pf p J    (1.00mol ) 8.314 298K  ln f  1.00kJ pi K  mol  pi 

pf pf  0.404;  1.50 pi pi

The final pressure is 1.5 times the initial pressure. Therefore ∆p

0.5pi

 G    V , we can see that in order to obtain a small molar (c) Using the equation:   p  T volume for a condensed phase, p has to be large for a fixed G. An ideal gas has a much larger molar volume so p can be smaller.

4.62.

(a) Using Eqn. 3.25: S mix  R

G mix  RT

no .of gases

 n ln x , Gmix=Hmix-TSmix. Since Hmix=0, i 1

i

i

no .of gases

 n ln x i 1

i

i

(b) R and T are both always positive. Since the xi‟s are mole fractions, they are always less than one. The natural log of anything less than one is negative so the sum (and therefore Gmix) is always less negative.

J  1 2 3   (c) G mix   8.314 308K 1.0 ln  2.0 ln  3.0 ln   15539J / mol K  mol  6 6 6  

4.64.

 G   G  We know that    S . We also know that    . We can take the partial  T  p  n  T ,P

derivative of both sides with respect to T and we get:

  G           . Since  T  n  T ,P  p  T  p

we also know that:

 2G  2G   G      S   , we can substitute in and we find that:         Tn nT  T  p n  T  p  n  T ,P

32

Chapter 1

4.66.  is intensive because it is an inherently molar amount. Partial molar volume and partial molar entropy are also intensive. All of these quantities may change depending on the proportions of different components in a system, but for a given system they are amountindependent.

4.68. A gas that expands by a factor of 10 at constant temperature decreases its pressure by a factor of 10; that is, pf = 1/10 pi. p 1 (a) m final - minitial = RT ln f = (8.314 J/mol× K)(100 K)ln = -1914 J/mol . pi 10 p 1 (b) m final - minitial = RT ln f = (8.314 J/mol× K)(300 K)ln = -5743 J/mol . pi 10 4.70. Equation 4.62 should not be used to determine  for an ideal gas. It is unnecessary –  for an ideal gas is 1.

4.72. Oxygen should deviate from ideality more than helium at any pressure, although as the pressure decreases the behavior of both gases should approach that of an ideal gas.

Chapter 5 Introduction to Chemical Equilibrium 5.2.

A static equilibrium is an equilibrium in which forward and reverse processes have stopped. For example, a balanced seesaw is a static equilibrium. A dynamic equilibrium is an equilibrium in which two opposing processes are still occurring but cancel each other out so that no net change occurs. An example would be a chemical reaction at equilibrium, or a bucket with a hole in it that is dripping water and being filled with water, but the water level stays constant. The similarity in the two situations is that there is no net change in the state of the system.

5.4.

The reactions are Ca(C2H3O2)2 (aq, supersat)  Ca(C2H3O2)2 (s) + Ca(C2H3O2)2 (aq, sat) followed by the equilibrium Ca(C2H3O2)2 (s)

5.6.

(a) 12.56 L 

Ca(C2H3O2)2 (aq, sat).

1000 mL 1 cm 3   12,560 cm 3  1.256 104 cm 3 (b)  cannot equal 3 because we would need 18 moles of 1 L 1 mL

H2 to react, and we only have 10.0 moles. In this case, the maximum value of  is 1.67.

5.8.

The heme + CO reaction lies farther towards products. In fact, one can show that the equilibrium constant for the reaction hemeO2 + CO hemeCO + O2 is 2.31023/9.21018 = 2.5104, indicating a strong preference by heme for CO – explaining its potential toxicity

5.10. False. pº is the standard pressure for gaseous substances in a system, defined as either 1 atm or 1 bar.

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Instructor‟s Manual

5.12.

It could be true, but not in general. Assume we have a reaction1.055 atm  1atm  1bar  1.06910 Pa . 1233 mmHg  1.01325 bar 100,000 Pa

5

1 tor 1atm .1 01325 bar    .1 644 bar .

1mmHg 760 tor 1atm

3

If all of the pressures drop by ½, 125 mL  11cm  125 cm mL

3

5.14. rxnGº is for the reaction under standard conditions of pressure. Using data from the appendix: rxnGº = (2 mol)(−394.35 kJ/mol) – [(2 mol)(−137.16 kJ/mol) + 0] = −514.38 kJ. Using this value and the instantaneous conditions given, we can calculate rxnG:



rxnG = −539.3 kJ

5.16.

(a) The equilibrium constant for this expression is V  R 

nT . SO doesn‟t appear in the 3 p

expression because it‟s in a condensed phase. (b) From the data in the appendix: G° = −368 – [−300.13 + ½(0)] = −68 kJ. (c) G° = −RT ln K. –68,000 J = −(8.314 J/mol·K)(298.15 K) ln K

pV ln K = 27.432 K = 8.21011. (d) The reaction would move nT toward the direction of more products. ln K = R 

5.18.

First we need to calculate Go=120.0−75.0 =45kJ. (a) G° = −RT ln K. 45000 J = −(8.314 J/mol·K)(298.15 K) ln K ln K = R  (b)

x atm 1 atm (2.66 bar)(27.5 L) L  bar ln K = −18.15 K = 64.0 m  10.1m .  0.0830 (1.887 mol)(466.9 K) mol  K

(1.04 atm )( 67 L )  9 .4 L (7.38 atm )

1moleSO2 1 5.20. For the reaction aA+bB cC+dD, the true definition of 1.82 10 g SO   2.8410 molesSO2 where p is the 2 64.07g 13

o

standard pressure (1 bar). Since each partial pressure in the reaction is divided by the standard pressure, Q (or K) is unitless.

5.22.

False. For the reaction A+B

1

V

equal to 1.

34

Latm

C+D, p  nRT  (2.8410 molesSO2)(0.0821molK)(273.1517.0 )K  71010atm. If pC=pA=1 atm and pD=pB=2 atm, K is still 81021L

Chapter 1

5.24. The reaction will reverse when G equals 0. Let p be the pressure at which G = 0:

0.0821

L  atm 5 / 9 K L  atm .7 lb lb  11.8    0.0456  . Simplify and solve for p: 0.80  14in. in. mol  K 1 R mol  R 2

0.20 

14.7 lb lb  2.9 2 in. in.2

2

for the

reaction to reverse.

5.26. Addition of an inert gas to the equilibrium should not affect the position of the equilibrium because the gas does not participate in the reaction and the partial pressures of the gases involved in the reaction do not change.

5.28. rxnGº = [−41.9 – 73.94] – [2(51.30) – 228.61] = 10.2 kJ. m pV   RT; M

m pM   V RT

4.50torr (44.01g / mole) (62.36

L  torr )( 273.15  87K ) mol  K

 0.01706 g / L  1.71  10 5 g / cm 3

ln K = −4.1148… K = 1.6310−2.

5.30. It is not appropriate. mixG takes into account Gibbs energy changes due to the mixing of the chemical species and does not account for chemical reactions. rxnG takes the chemical potentials of the chemical species into account.

5.32. (a) fGo[AgCl(s)]= −109.80kJ/mol; rxnGo=−RTln K= 3 y





2



y

2

z w

3

rxnGo=fGo[Ag+(aq)+Cl−(aq)]−fGo[AgCl(s)] 55620J/mol=fGo[Ag+(aq)+Cl−(aq)]−(−109800J/mol); fGo[Ag+(aq)+Cl−(aq)]= −54180J/mol

(b) fGo[Ag+(aq)+Cl−(aq)]= fGo[Ag+(aq)]−131300J= −54180J  5.34.



fGo[Ag+(aq)]=77120J/mol 3w 2 z 3 32 y



xy 2 z 3 w2

ln K = −1.16991… K = 0.310.

On the basis of this result, graphite is the stable phase of carbon at atmospheric pressure.

5.36. The balanced reaction is 60 C (s) 6 xy



ln K = −9.67396…

C60 (s). The equilibrium constant for this reaction is

w3 z 3 32 y 2



2 xyz 3 w

K = 6.310−5.

35

Instructor‟s Manual

2

3y z w

2 2 3 y z ;  w

2

5.38.



5.40.

We know that 3 yw z

5.42.

To answer this question, plot ln K versus 1/T and get the slope of the line. That slope equals –H/R (from equation 5.19). The graph is left to the student, but a value of –77 kJ would be a good estimate for H.

5.44.

We need eqn. 5.2: Tp  nVR 

; p=940 atm.

p  n  so    . Calculations are left to the student. If the solution is very  V  T, P R  T dilute,  approaches 1. As the molality increases,  decreases. Since  is a coefficient related to the deviation from ideality, this makes sense. The solution behaves less−ideally if the ionic concentration is high.

2

2

2



V ,n

- pV  n    ; RT 2  T  P, V

 p     n  T , V



RT V

So K2=0.338K1. Or, the equilibrium constant decreases by approximately 66% with the rise in  temperature.Tp  nVR V ,n

5.46.

Use equation 5.20 and let K2 = 2K1:

V (T , p ) 

nRT  V   V   dp Solve for dV    dT    T  p p  p  T

T2:    Latm   Latm     1mole 0.0821    1mole 0.0821 350K    nR   nRT   Kmole  10.0K     Kmole  0.10atm dV   dT   2 dp      1.08 atm 1.08atm2  p   p         

T2 = 293 K.

To determine a temperature for the equilibrium constant is 10 times the original value, let K2 = 10K1 and perform the same calculations:  V   nRT   2 V  2nRT  p  nR   2 p        ;   ;  0 Solve   2   p 2  V  p2 p3  T  n , V  p  n ,T   n ,T  T  n , V for T2:

TB

36



a T2 = 282 K. bR

Chapter 1

For a reaction whose H = −20 kJ, repeat both sets of calculations using this value of H. Letting K2 = 2K1: L2 atm L2 atm 1.360 2 mol mol 2 TB   1026 K TB   521 K L atm    0.04267 L/mol  0.08205 0.03183 L/mol  0.08205 L atm   mol K  mol K    3.592

L2 atm mol 2 TB   433 K  0.03913 L/mol  0.08205 L atm  mol K   1.390

Solve for T2:

T2 = 274 K.

37

Instructor‟s Manual

To determine a temperature for the equilibrium constant is 10 times the original value, let K2 = 10K1 and perform the same calculations:

C 2 V

Solve for

T2: 0.0237L 1 m 3 m3 T2 = 232 K.   2.37  10 5 mole 1000 L mole

5.48.

We need eqn. 5.2:

 T  V     p V,n n  R

3 m3 1mole 29 m 2.37 10   3.94 10 mole 6.02 1023 atoms atom 5

3 4 3  29 m V  r  3.94  10 ; r  2.11 10 10 m or 2.11 3 atom

5.50.

First we need to find K.

0.03049L mole

After the volume change, we need to make up a table to find the new equilibrium pressures: Pressure(bar) Initial Equilibrium

Br2(g) 0.334 0.334−x

2Br(g) 0.113 0.113+2x

0.0638L ; x=0.0208bar. So the equilibrium pressure of Br2 is 0.3132bar and that mole

of Br is 0.155bar. This agrees with e Chatelier‟s principle because when the pressure decreased, the equilibrium shifted to the side with more gas molecules.

5.52.

(a First we need to use the ideal gas law to calculate the initial pressure of Br2 in the flask.

We need to make up a table to find the equilibrium pressures.

Pressure(atm) Initial Equilibrium

Br2(g) 0.9852 0.9852−x

2Br(g) 0 2x

So the equilibrium pressure of Br2 is 0.352 atm and that of Br is 1.266 atm.

38

Chapter 1

(b) When the volume of the flask doubles, the pressures are cut in half. We need to make up a table to find the new equilibrium pressures.

Pressure(atm)

Br2(g)

2Br(g)

Initial

0.352/2

1.266/2

Equilibrium

.176−x

0.633+2x

a  1   b  Z  1 b     RT  V  V 

2



So the equilibrium pressure of Br2 is 0.121 atm and that of Br is 0.743 atm.

5.54. (a) First we need to set up a table to calculate our equilibrium constant: Pressure(atm) Initial Equilibrium

Z  1

O2(g) 1.00 1−3x

O3(g) 0 2x

B C  2  V V

The easiest way to solve this cubic equation is by guessing an x and successively approximating the correct value. Using this method, we find that V 2 The equilibrium pressure of O2=0.634atm and that of O3 is 0.244atm. (b) Since Kr is an inert gas, it does not participate in the equilibrium. Therefore there is no change in the equilibrium partial pressures.

5.56. Since the isoelectric point lies about halfway between the two pKas, we need to determine the averages of all the pKas of the amino acids and see which one lies closest to 7. The average for proline, 6.295, lies closest to 7.

5.58. The equilibrium constant we need to find is:

p 

RT  Vb

a T V ( V  b)

From Table 5.1 we know that for glycine the pKa1=2.34 and the pKa2=9.60. These correspond to the following equations:

  

p V

   T ,n



We find that the overall

39

Instructor‟s Manual

 RT a a   2 2 2 ( V  b) T V ( V  b) T V ( V  b) (p 

an 2 )( V  nb)  nRT V2

The zwitterion is the most prevalent form in solution.

5.60.

p V  RT Assuming that G=Go, p  (0.0821

L  atm e )(273.15K ) K  mol

p

RT ,p  V

(0.0821

L  atm )(273.15K ) K  mol  1.00atm 22.41L

Latm    (10.91atm L2 ) /  ( 22.41L )( 0.0821 )( 273.15 K )  K mol  

(22.41L  0.0401L)

 0.981atm

This large value suggests that disulfide bond formation is very favorable in proteins.

Chapter 6 Equilibria for Single Component Systems 6.2.

When freeze-drying a complex mixture like coffee, not just water but other volatile components can be removed from the mixture. The net result is that when water is added back to the dried coffee (but none of the other volatile components), the coffee probably will not taste the same.

6.4.

Solid and liquid phases of a substance can exist in the same closed, adiabatic system if the conditions of the system (in particular, the temperature and the pressure) are such that those conditions correspond to the melting or freezing point of the substance. At the melting point (or freezing point), both the solid and liquid phases of the substance can coexist in an equilibrium state.

6.6.

(a) The equilibrium shifts to the liquid phase. (b) The equilibrium shifts to the gas phase. (c) The equilibrium shifts to the solid phase. (d) The system remains in the solid phase (unless there is a stable allotrope or crystal form of the substance that is more stable at lower temperature. Metallic tin is an example where this occurs.)

40

Chapter 1

6.8.

The algebraically equivalent way is to simply move the negative sign to the other side of the equation: -dnliquid = dnsolid. The point is that either side of the equation can have the negative sign.

6.10. Assume we have a chemical compound, A. Sublimation of A can be represented by the equation: A(s)  A(g) with an enthalpy of sublimation subH. We can also represent the fusion and vaporization of A by the following equations:

A(s)  A(l)

 fus H

We will add these two equations together using Hess‟s aw.

A(l)  A(g)  vap H A(s)  A(g)  fus H   vap H

Therefore subH=fusH+vapH

6.12. The heat of fusion given up by the freezing water can be transferred (at least in part) to the citrus fruit, keeping them warmer and (hopefully) keeping the fruit itself from freezing.

41

Instructor‟s Manual

6.14.

We will assume that 1g of the vapor is removed. The heat lost by the liquid is:

kJ   1mole     2.44kJ q  H vap  n   43.99 1g  mol   18.01g   Assuming that the temperature of the flask stays constant the only source of this energy is from the conversion of water to ice (an exothermic process):

 1mole   . x=7.32 g of ice formed.

2.44kJ  H fus x g   18.01g 

qrev +30, 700 J/mol = = +86.9 J/mol·K , fairly close to what would be T (80.1+ 273.15) K predicted by Trouton‟s rule.

6.16. DS =

6.18.

Using the definition of entropy for this isothermal change: S 

124.7 J/mol  K 

510,400 J/mol T

q rev  fus H  and substituting: T T

T  4093 K or 3820C.

6.20.

Imagine taking water from ice at −5°C to liquid water at 5°C. We would need to work this in 3 steps: ice changing temp from −5°C to 0°C, the phase transition, and the liquid changing temperature from 0°C to 5°C. For both the first and third step we would use the equation q=mcT, but the c used in step one is different from the c used in step 3! We know that  H  Cp    . If we plotted the molar enthalpy of a substance vs. temperature, the slope of that  T  P graph would be Cp. At phase transitions, the plot would be a vertical line, the slope of which is equal to infinity. Therefore the heat capacity is infinite for a phase transition.

6.22.

The expression “dphase 1 = dphase 2” in the derivation of the Clapeyron equation does not necessarily imply a closed system because we‟re considering chemical potential, which is an intensive variable.

6.24.

We should recognize that thermodynamic values measured at standard conditions, like S and H and V, may not be very applicable to extreme conditions of temperature and pressure, since most common thermodynamic values are measured at normal pressure (1 atm) and normal temperatures (like 298 K). However, increasing the pressure by about 6 atm is scarcely a very extreme change in conditions, especially for a solid substance.

Therefore, our answer should be fairly accurate.

6.26.

(a) Yes, because a gas phase is involved. (b) Yes, because a gas phase is involved. (c) No, because no gas phase is involved. (d) No, because no gas phase is involved.

42

Chapter 1

(e) No, because the process isn‟t an equilibrium phase change. (f) No, because no gas phase is involved. (g) No, because no gas phase is involved. (h) Yes, because a gas phase is involved.

6.28. First we need to calculate the molar volumes of Ga(s) and Ga(l):

æ 69.723g öæ 1cm3 öæ 1mL öæ 1L ö 1mole Ga(s) : ç ÷ç ÷ = 0.0118L / mol ÷ç 3 ÷ç è 1mole Ga øè 5.91g øè 1cm øè 1000mL ø æ 69.723g öæ 1cm3 öæ 1mL öæ 1L ö 1mole Ga(l) : ç ÷ç ÷ = 0.0114L / mol ÷ç 3 ÷ç è 1mole Ga øè 6.09g øè 1cm øè 1000mL ø

Using

So, DV = (0.0114 - 0.0118)L / mol = -3.51´10-4 L / mol and DT = 298.1- 302.9K = -4.8K. J 18.45 æ 1L × atm ö DS K·mol DT = Eqn.6.10, Dp = ÷ = 2490atm (-4.8K ) çè -4 DV -3.51´10 L / mol 101.32J ø Approximately 2490 atm of pressure is required to only slightly change the melting point of the solid.

6.30. Using the fact that V was –0.0070 L/mol for the phase change in exercise 6.23, and the fact that the monoclinic-to-rhombic phase change must have a H of –0.368 kJ/mol:

æ DH ö Tf æ -368 J/mol öæ (100 + 273.15) K öæ 1 L × atm ö Dp = ç ÷ç ln ÷ p = 6.3 atm, so ÷ç ÷ ln = ç è DV ø Ti è -0.0070 L/mol øè (95.5 + 273.15) K øè 101.32 J ø that increasing the pressure by 6.3 atm to 7.3 atm should make the rhombic phase more stable.

6.32. No, the behavior of chemical hot packs cannot be described using the Clapeyron equation or the Clausius-Clapeyron equation because supersaturated solutions are not equilibrium systems, nor does the process involve a phase change (it involves a solubility change).

6.34. We want to calculate the boiling point of benzene at 1 atm. For this we need to use Eqn. 6.14: ln

p2 H  1 1     p1 R  T2 T1 

1atm ln  0.241atm

J  1 mol  1  ; T2  351.6K  78.4  C  J  T2 40  273.15  8.314 Kmol

33.9  10 3

The measured normal boiling point of benzene is around 80°C.

43

Instructor‟s Manual

dp H  p as the form of the Clausius-Clapeyron equation, we can  dT RT 2 substitute directly the values given in the problem: dp (71,400 J/mol)  (7.9  10 -5 bar)   7.8  10 6 bar/K. dT (8.314 J/mol  K)(22.0  273.15 K) 2

6.36.

If we use

6.38.

æ 46.07g öæ 1mL öæ 1L ö The molar volume of ethanol can be found: ç ÷ç ÷ = 0.058L / mol ÷ç è 1mole øè 0.789g øè 1000mL ø 0.058

p  pe *

6.40.

V ( l ) p / RT





Latm    0.0821   293.15 K  Kmol  



 5.95  10 Pa e 3



 9.86910 6 atm  L  8.5310 6 Pa    mol 1Pa  

 9.869  10 6 atm    0.072atm  7298Pa 1Pa  

Since applied pressure is not a vapor pressure, we will use eqn. 6.17.

p * V (l)P ln  ; P  p RT

p*  Latm   2.00   0.0821 298.15K  ln  p Kmol  1.47      63.9atm L V ( l) 0.118 mol

RT ln

Where the molar volume has been calculated using:

æ 147.0g öæ 1mL öæ 1L ö 1, 4 - dichlorobenzene : ç ÷ç ÷ = 0.118L / mol ÷ç è 1mole øè 1.25g øè 1000mL ø 6.42.

ln

 p1 40,660 J/mol  1 1     1 atm 8.314 J/mol  K  (300  273.15 K 373.15 K 

p1  96.9 atm .

A pressure of 96.9 atm (which is an increase of 95.9 from normal atmospheric pressure) corresponds to a depth of (95.9)(10 m) = 959 meters of depth, or over 3100 feet. The deepest ocean point is found in the Mariana Trench (36,000 ft). Underwater volcanoes this deep can still boil water because haven‟t exceeded the critical pressure of water.

6.44.

We want to boil methanol at a temperature of 298.15K. The equation we need is: ln

p2 H  1 1     p1 R  T2 T1 

J p 1 1  mol  ln 2     ; p 2  0.096atm J 1atm 298.15K 351K   8.314 Kmol 38.56  10 3

The pressure has to be reduced to approximately 10% of normal atmospheric pressure for methanol to boil at room temperature.

44

Chapter 1

6.46.

(a) In order for the term dA to have units of energy, and dA has units of m2,  must have units of J/m2. (b) The derivative of A, in terms of r, is dA = 8r dr and the derivative of V in terms of r is dV = 4r2 dr. Substituting into the right side of the equation:

2dV 2(4r 2 dr)   8rdr  dA . (c) By simply rearranging the expression, we can r r dA  r 2dV get dV  . (d) According to the expression dA  , droplets with smaller 2 r radii will contribute to a larger dA, which will in turn contribute to a larger value of dG. Since this expression relates the change in area to the change in volume – which relates directly to how fast the droplet is evaporating, we use this expression for dA, not the expression in part b. (e) According to our analysis in part d, smaller droplets should evaporate faster than large droplets. (f) See part e.

6.48. Large bodies of ice can move because large amounts of pressure at the bottom of an ice body can favor the liquid phase over the solid phase. While it‟s doubtful that liquid water exists at the bottom of a glacier, the „solid‟ ice may not be as solid as we may think!

dp  S  , we can dT V see that the derivative must be negative since entropy increases for the solid to liquid phase transition. This derivative is just the slope of the solid/liquid equilibrium line in the phase diagram of Figure 6.5.

6.50. For H2O, Vsolid  Vlqd so V  Vlqd  Vsolid is negative. Using eqn 6.9,

6.52. In Figure 6.16 the regions: solid, liquid, and gas, have 2 degrees of freedom since there is only one phase present. The lines have one degree of freedom since there are two phases present, and the triple point has 0 degrees of freedom.

6.54. This situation is similar to that of a gas adsorbed onto a surface (a 2-D gas). The equation of state for a 2-D gas is A  nRT where  is the surface pressure and A is the surface area. The Gibbs phase rule does still apply since there are three independent variables: , T, and A.

   6.56. For equation 6.20,     S , the units of  are J/mol and the unit on T is K, so the  T  p ,n overall unit on the left side is J/molK, which are the units for molar entropy. For    equation 6.21,    V , the units of  are J/mol and the unit on p is atm or bar,  p  T ,n while the unit on the molar volume is L/mol. However, if we remember that the unit J can be written in terms of Latm, then the units of  can be written as Latm/mol, which when divided by the unit atm equal units of L/mol, which are the units of molar volume.

6.58. Figure 6.6 shows 12 phase boundaries.

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Instructor‟s Manual

6.60.

If there were such a thing as a single-axis phase diagram, equation 6.17 would probably be rewritten as „degrees of freedom 2 – P‟, because there would be one less variable needed to determine the exact state of the system.

6.62.

Higher pressures would be necessary to keep a vapor in a liquid state. This is evident from equations 6.21 and 6.24, since vapor phases always have much higher molar volumes than the liquid phases. This is not always the case for liquid-solid phase changes (see exercise 6.58, and remember the phase diagram of water).

6.64.

Yes. At these intersections three phases of the compound are in equilibrium with each other. That is the definition of a triple point.

6.66.

The answer will depend on the Figure and the line chosen. All of them are

T p

derivatives, but the phases involved vary with the line.

6.68.

As sulfur goes from rhombic to monoclinic, the entropy should be increasing, just like it would if it were melting or vaporizing as the temperature is increased.

Chapter 7 Equilibria for Multiple Component Systems 7.2.

With only a single component and phase, if only Fe2(SO4)3 were present in our system, we would have C – P + 2 = 1 – 1 + 2 = 2 degrees of freedom.

7.4.

For a single-component system, there will never be negative degrees of freedom. The maximum number of phases that can exist in equilibrium is three, so the Gibbs phase rule would yield 1 – 3 + 2 = 0 degrees of freedom (like might occur at the triple point). For a two-component system, the maximum number of phases that can exist in equilibrium is four. There will never be negative degrees of freedom.

7.6.

In this reaction, there are only three independent components (the amount of the fourth can be determined from the equilibrium constant) and 3 phases, two solids and a gas mixture. Therefore, there are 3 – 3 + 2 = 2 degrees of freedom.

7.8.

For water: n 

pV (23.76 torr)(5.00 L) 1 atm    6.39  10 3 mol of RT (0.08205 L  atm/mol  K)(298.15 K) 760 torr water is needed to ensure that there is a liquid and gas phase. This equals 0.115 grams.

For methanol:

46

Chapter 1

pV (125.0 torr)(5.00 L) 1 atm    3.36  10  2 mol of methanol is RT (0.08205 L  atm/mol  K)(298.15 K) 760 torr needed to ensure that there is a liquid and gas phase. This equals 1.08 grams. n

7.10. Since, according to equation 7.11, ai  ai 

pi pi

*

, and since pi* = 760 torr, we can calculate ai:

748.2 torr  0.984 760 torr

7.12. Equation 7.19 is derived from equation 7.18 in the text. The key rearrangement starts * x1 p1 with the expression y1  , where we distribute the p2* in the * * x1 p1  (1  x1 ) p 2 denominator: y1 

x1 p1

*

x1 p1  p 2  x1 p 2 *

*

*

. Next, we factor the x1 from the first and third terms in the *

x1 p1 . * * * p2  ( p1  p2 ) x1 7.14. If it‟s a 1:1 molar ratio of hexane to cyclohexane, then the mole fraction of each is 0.50.

Therefore, the total pressure is p = (0.50)(151.4) + (0.50)(97.6) torr = 124.5 torr. denominator to get equation 7.19: y1 

7.16. p(ethanol) = (0.0006)(115.5 torr) = 0.0693 torr = 0.07 torr (1 sig fig) 7.18. (a) 1 3 p A   45.9 torr   11.5torr; p B   99.2 torr   74.4 torr; p tot  11.5  74.4  85.9 torr 4 4 (b) 3 1 p A   45.9 torr   34.4 torr; p B   99.2 torr   24.8torr; p tot  34.4  24.8  59.2 torr 4 4

7.20. p = (xmeth)(p*meth) + (xeth)(p*eth) = (xmeth)(p*meth) + (1 – xmeth)(p*eth) 350.0 mmHg = (xmeth)(413.5) + (1 – xmeth)(221.6) mmHg 350.0 = 413.5x – 221.6x + 221.6 128.4 = 191.9x xmeth = 0.669 Therefore, xeth = 0.331.

7.22. Equation 7.19 is y1  denominator: y1 

x 1 p1

*

p 2  ( p1  p 2 ) x 1 *

*

x1 p1

*

*

p 2  x1 p1  x1 p 2 *

*

. First, let us multiply through the x1 in the

*

. Now, multiply the denominator over to the other

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Instructor‟s Manual

side of the equation: y1 ( p 2 *  x1 p1*  x1 p 2 * )  x1 p1* , or * * * * y1 p 2  y1 x1 p1  y1 x1 p 2  x1 p1 . Collect all terms having x1 in it on one side of the equation: y1 p 2 *  x1 p1*  y1 x1 p1*  y1 x1 p 2 * and factor the x1 out of all three terms on the right side: y1 p 2 *  x1 ( p1*  y1 p1*  y1 p 2 * ) . Now, divide the parenthetical terms over to the other side of the equation (which we will flip): x1 

y1 p 2

*

p1  y1 p1  y1 p 2 *

*

*

. Now factor

out the y1s from the last two terms in the denominator: x1 

y1 p 2

*

p1  ( p 2  p1 ) y1 *

*

*

, which is our ultimate expression.

*

7.24.

Using Equation 7.19, y1 

y1 

x1p1 . If #1 is hexane, * * * p 2  (p1  p 2 ) x1

2  151.4torr 3 2 97.6torr  (151.4  97.6 torr )  3

 0.76 y 2  1  0.76  0.24

*

7.26.

Consider the equation p tot 

p 2 p1

*

p1  ( p 2  p1 ) y1 *

*

*

. If y1 goes to zero, then the entire *

second term in the denominator is zero, and the expression becomes

p 2 p1

*

, which * p1 reduces to p2*, as it should when the system contains only component #2. If y1 goes to 1, * * * * p 2 p1 p 2 p1 * then the expression for ptot becomes * , which becomes  p1 , as it * * * p2 p1  p 2  p1 should when the system contains only component #1.

7.28.

7.30.

Equation 7.24 can‟t be used immediately because the equation uses the mole fraction of the vapor phase, while the example gives mole fractions of the liquid phase. Before using equation 7.24, one must first calculate the mole fractions in the vapor phase.

 1mole    0.347moleC 5 H12 25.0gC 5 H 12   72.15g   1mole    0.522moleC 6 H14 45.0gC 6 H14   86.18g   1mole    0.654moleC 6 H 12 ; Total moles = 1.523 moles 55.0gC 6 H 12   84.16g 

48

Chapter 1

J    0.347 0.347 0.522 0.522 0.654 0.654   mix G  RT  i ln  i   8.314 ln  ln  ln 310.15K  Kmol    1.523 1.523 1.523 1.523 1.523 1.523  J  2751  1.523moles  4190J mol  G  4190J J  mix S  mix   13.5 T  310.15K K

7.32. If an initial composition of x1 = 0.1 were used, the tie lines that you would draw would eventually lead you to the minimum-boiling azeotrope, near the middle of the phase diagram. Thus, the azeotrope will be your ultimate product.

7.34. An azeotrope can be distinguished from a pure component by determining its temperature profile as it freezes. If it is a pure component, it will have a distinct freezing point for the entire sample. However, if it is two (or more) components, then each component should freeze at distinctly different temperatures.

7.36. Diagrams are left to the student. CCl4/HCOOH will produce a minimum boiling azeotrope. HCOOH/pyridine will produce a maximum boiling azeotrope.

7.38. The mixture of water and ethylene glycol has both a higher boiling point and lower freezing point than pure water alone.  101325Pa  9 3 7.40. p i  K i  i ; 103.6atm    4.03  10 Pa  i ;  i  2.60  10 1 atm   7.42. Hydrogen chloride is a diatomic gas, whereas hydrochloric acid is HCl that has been dissolved in water. Since HCl is a strong acid and virtually 100% ionized in solution, it is doubtful that a solution of HCl will act ideally. HCl is very polar and unlikely to act ideally.

7.44. A mole fraction of 4.1710-5 implies that there are 4.1710-5 moles of CCl2F2 in 0.9999583 moles of water. Assuming no volume change when such a small amount of solute is added to water, the volume of 0.9999583 moles of water is 18.01 g 1.00 cm 3 0.9999583 mol    18.0 cm 3 , or 0.0180 L. Therefore the molarity of 1 mol 1.00 g

4.17  10 5 mol  0.00232 M . To determine the Henry‟s law constant, use 0.0180 L 1 atm = 101,325 Pa as the pressure: this solution is

101,325 Pa = Ki(4.1710-5) Ki = 2.43109 Pa. 7.46. (a) The mole fraction of nitrogen in water is about 90% of the mole fraction of air in water. Since nitrogen is 80% of air, it should not be surprising that the majority of the mole fraction of air in water is composed of nitrogen.

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Instructor‟s Manual

(b) If air is 20% oxygen and the mole fraction of air in water is 1.38810-5, let us assume that oxygen would be 20% of that mole fraction (assuming that the composition of the air dissolved in water is the same as the composition of gaseous air). Therefore, the mole fraction of oxygen in water would be about (0.20)(1.38810-5) = 2.7810-6. (c) If oxygen is 20% of 1 atm = 101,325 Pa, then the partial pressure of oxygen is (0.20)(101,325 Pa) = 20,265 Pa. Therefore: 20,265 Pa = Ki(2.7810-6) Ki = 7.29109 Pa. This is the same answer we got in exercise 7.30, and it should be – all we did was multiply both numerical values by 0.20, so the calculated value of Ki should be the same. However, Table 7.1 shows that the actual value of Ki is somewhat less than this (4.34109 Pa), indicating that nitrogen and oxygen do not dissolve in water to an extent proportional to their composition of air.

7.48.

We will assume that N2 is 78% of the air mixture that the diver is breathing. p N 2  0.78(4.0atm)  3.12atm The mole fraction of N2 in the diver‟s blood is calculated

3.12atm  101325Pa 

pi  1atm   3.69  10 5. If we assume that the molar volume  Ki 8.57  10 9 Pa of the blood is equal to the molar volume of water we can approximate the molarity of ci 3.69 ´10-5 the solution by: = = 2.05´10-3 M Vwater 0.01801L / mol

by:  i 

7.50.

If 87.0 grams of phenol can be dissolved in 100 mL of water, to calculate the molarity we need to know the moles of phenol and the total volume of the solution. The number of 1 mol  0.925 mol phenol . The volume of the phenol is moles of phenol are 87.0 g  94 g given by

1 mL  82.1 mL . If we assume that the volumes are additive, then the total 1.06 g volume of the solution is 100 + 82.1 mL = 182.1 mL = 0.1821 L. Therefore, the molarity 0.925 mol of the solution is  5.08 M .

0.1821 L 87.0 g 

7.52.

50

(a) The calculated mole fraction of naphthalene in toluene is 0.311, so there are 0.311 mol of naphthalene in 0.689 mol of toluene. The molecular weight of toluene (C6H5CH3) is 92.0 g/mol, and using the density of toluene as given, we can calculate the volume of toluene used: 92.0 g 1 mL 0.689 mol    73.2 mL of volume . We do the same for naphthalene 1 mol 0.866 g (C10H8):

Chapter 1

128.0 g 1 mL   38.8 mL of volume . Thus, the total volume is 73.2 + 38.8 1 mol 1.025 g 0.311 mol = 112.0 mL = 0.1120 L. Determining the molarity: M   2.78 M . (b) 0.112 L Because the ideal solubility is calculated using only properties of the solute, the calculated mole fraction solubility of naphthalene in n-decane is the same as in toluene: xsolute = 0.311. However, the concentrations expressed in other units will be different. To get the solubility in grams per 100 mL of solvent: 0.311 mol 

0.311 mol 128.0 g C10 H 8 1 mol 0.730 g 39.8 g naphthalene 100 / 134 29.7 g       0.689 mol mol 142.0 g decane 1 mL 134 mL decane 100 / 134 100 mL

In terms of molarity, again we need to determine the total volume of the two components:

0.689 mol 

142.0 g 1 mL   134.0 mL of volume . We do the same for naphthalene 1 mol 0.730 g

(C10H8):

128.0 g 1 mL   38.8 mL of volume . Thus, the total volume is 134.0 + 1 mol 1.025 g 0.311 mol 38.8 = 172.8 mL = 0.1728 L. Thus, the molarity is: M   1.80 M .

0.1728L 0.311 mol 

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Instructor‟s Manual

7.54.

7.56.

Of the four solutions listed, the C20H42 in cyclohexane is probably the closest to ideal.

The sodium chloride/water and sucrose/water solutions deviate because of polar interactions between solute and solvent, and the water/carbon tetrachloride combines a polar solute with a nonpolar solvent. Therefore, the calculated properties of the C20H42/cyclohexane solution will probably be closest to actual properties.

14,900 J/mol  1 1 1  1    4.828314...  1792    8.314 J/mol  K  298.15 K T   298.15 T  0.002694  (0.003354  1 / T ) 1/T 0.006600…. T = 1515 K = 1242ºC. ln(8.0  10 3 )  

7.58.

We can tell by looking at a phase diagram of the NaCl-H2O system and see if the temperatures and relative concentrations involved point to a eutectic or to the colligative property. But as mentioned in the chapter, the percent of NaCl in the salt/water eutectic is 23% salt (that is, about 1 part NaCl to 3 parts H2O). Personal experience suggests that salting roads doesn‟t use that much salt with respect to water, so the melting phenomenon is actually due to the colligative property of freezing-point depression.

7.60.

The drawing is left to the student.

7.62.

ln xsolute  

 2,600 J/mol  1 1   ln x = -0.30184…  8.314 J/mol  K  273.15 K (97.8  273.15) K 

x = 0.739.

7.64.

Vapor pressure depression and boiling point elevation are related. When a non-volatile solute is added to a solvent, the solute molecules make it more difficult for the solvent molecules to go into the vapor phase. For a fixed temperature, this lowers the vapor pressure of the solution when compared to the pure solvent.

Boiling occurs when the vapor pressure of a liquid is equal to the external pressure. Since we now have to add more energy to the solution (when compared to the pure solvent) to get it‟s vapor pressure to equal the external pressure, the boiling point of the solution is elevated.

7.66.

Nonvolatile solutes ALWAYS lower the vapor pressure of the solution when compared to the pure liquid. The solute prevents solvent molecules from escaping into the vapor phase. A volatile solute SOMETIMES lowers the vapor pressure of the solution. In some cases the mixture of the two volatile compounds shows a positive deviation from Raoult‟s aw. In other cases there is a negative deviation. See Figures 7.12 and 7.13.

 1mole anthracene   1mole CCl 4    0.140mole; 250.0g CCl 4    1.63mole 7.68. 25.0g anthracene 178.23g    153.81g  p solv   solv p *solv 

52

1.63 72.2torr   66.5torr 1.63  .140

Chapter 1

7.70. First we have to solve for the mole fraction of the solvent:  solv 

p solv 97.23   0.978 ; and the number of moles of MEK: p *solv 99.40

 1moleMEK    1.39molesMEK 100.0gMEK  72 .

11 g   So, solute=0.022=

n solute n solute  ; n solute  .03127moles n solute  n MEK n solute  1.39moles

The molecular weight can be calculated from the number of grams added and the 12.00g g determined number of moles of solute: M solute   384 0.03127moles mole 7.72. First, let us calculate Kf and Kb:

M RT (18.02 g/mol)(8.3 14 J/mol  K)(273.15 K) 2 K f  solv MP   1.86 K/molal 1000 fus H (1000 g/kg)(6009 J/mol) 2

M solv RTBP (18.02 g/mol)(8.3 14 J/mol  K)(373.15 K) 2 Kb    0.513 K/molal 1000 vap H (1000 g/kg)(40,6 60 J/mol) 2

For the freezing point depression: T = (2 particles)(1.86 K/molal)(1.08 molal) = 4.02 K. Therefore, the freezing point goes down by 4.02º: FP = -4.02ºC. For the boiling point elevation: T = (2 particles)(0.513 K/molal)(1.08 molal) = 1.11 K. Therefore, the boiling point goes up by 1.11º: BP = 101.11ºC. For the osmotic pressure, we need the mole fraction of the solute. If there is 1.08 moles of “NaCl” per liter of solution yielding 2.16 mol of particles per liter , and a liter of solution contains approximately 1000 mL of water (assuming negligible volume change from the solute , that‟s 55.5 moles of H2O. Therefore, the mole fraction of solute is 2.16  0.0375. The molar volume of the solution is approximately the same as the 57.65 molar volume of water, or 0.01802 L. Using the van‟t Hoff equation, equation 7.56:

V  xRT

(0.01802 L)  (0.0375)(0.08314 L  bar/mol  K)(298.15K)

Solve for : 51.5 bar.

7.74. Plant cell membranes are semipermeable and there are some dissolved species (salts, enzymes, etc.) inside the cell. When the celery becomes dehydrated it becomes limp. If you place the limp celery in the pure water, the water will be hypotonic (lower solute concentration) when compared to the cells. Water will flow into the cells and when the pressure inside the cells increases, the celery becomes crisper. If you place the limp celery in a salt solution (hypertonic compared to the inside of the cell), water will flow out of the celery cells and make the celery even limper.

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Instructor‟s Manual

g  J   2  60.05  8.314 289.15K  M RT K  kg mol  Kmol    3.60 7.76. (a) K f  solvent g  J  1000   fus H mole 1000 11700  kg  mol  2 MP

 1 mole I 2    0.108mole I 2 (b) 27.6g I 2   253.81g  K  kg   0.108mole   Tf  K f msolute   3.60   5.30K  mole   0.0733kg  

The freezing point of the solution is 16.0°C-5.34°C=10.70°C. 7.78.

If x

0.739, we can use the van‟t Hoff equation:

V  xRT

(0.0152 L)  (0.739)(0.08314 L  bar/mol  K)(273.15K)

Solve for :  = 1104 bar.

M solv RTMP (159.8 g/mol)(8.3 14 J/mol  K)( 273.15-7.2 K) 2 Kf    8.89 K/molal 1000 fus H (1000 g/kg)(10,5 70 J/mol) 2

7.80.

M solv RTBP (159.8 g/mol)(8.3 14 J/mol  K)(273.15  58.78 K) 2   4.95 K/molal 1000 vap H (1000 g/kg)(29,5 60 J/mol) 2

Kb 

7.82.

If you‟re making an 0.0001000 molal solution with a polymer that has a molecular weight of 200,000 amu, you‟re going to measure out (0.0001000)(200,000) = 20.00 grams of the polymer and dissolve it in 1.000 kg of water. But, not all of the 20.00 grams will be 200,000-amu polymer molecules: 0.5% of the sample will be 100-amu molecules.

Therefore, the actual number of dissolved particles will not be as expected, and the osmotic pressure will be different than expected. First, let us determine the number of moles of particles expected for a pure sample:

1 mol  0.0001000 mol of particles . That number of particles will yield 200,000 g some osmotic pressure pure. Now let us calculate the number of particles in the impure sample: 20.00 g 

0.995´ 20.00 g ´

1 mol = 0.0000995 mol of particles 200,000 g

0.005  20.00 g 

1 mol  0.001000 mol of particles . Total number of particles: 100 g

0.0010995 mol.

This is 10.995 times higher than we would expect. If the impure sample has an osmotic pressure of impure, if there are 10.995 times as many particles as one would expect, then 54

Chapter 1

impure = 10.995pure. This means that your calculated molecular weight will be 1/10.995th of what it really is, or 200,000/10.995 or 18,200 – a substantial error! (Notice that we never did have to actually calculate the osmotic pressure itself.)

7.84. The derivation of equation 7.49 is parallel to the development of equation 7.47 in the text.

Rather than repeat the steps here, readers are referred to equation 7.41 through 7.49, with the substitution of TBP and vapH for TMP and fusH in the equations.

 1 mole   1 mole  7.86. (a) 10.0 g KBr    0.0840 mole KBr;100.0 g H 2O    5.55 mole H 2 O  119g   18.02g  The molar volume of water is 0.0180L and KBr separates into two species upon dissolution, so:

æ

Õ=

N c solute RT = V

( 2) çè

0.0840 öæ Latm ö ÷ç 0.0821 ÷ ( 293.15K ) 0.0840 + 5.55 øè Kmol ø = 39.9atm 0.0180L / mol

 1mole    0.1262moleSrCl2 . SrCl2 separates into three species upon (b) 20.0gSrCl 2   158.53g  dissolution, so:

æ

Õ=

N c solute RT = V

(3) çè

0.1262 öæ Latm ö ÷ç 0.0821 ÷ (353.15K ) 0.1262 + 5.55 øè Kmol ø =107atm 0.0180L / mol

(c) The 0.100molal solution contains 0.100mole of HCl in 1000g of water(55.5mol H2O) so the 0.100 mole fraction of the HCl is:  HCl   1.80  10 3 0.100  55.5

æ

Õ=

N c solute RT = V

Latm ö ÷ (310.15K ) Kmol ø = 5.09atm 0.0180L / mol

( 2) (1.80 ´10-3 ) çè 0.0821

Chapter 8 Electrochemistry and Ionic Solutions 8.2.

(a) 12.56 L 

1000 mL 1 cm 3   12,560 cm 3  1.256  10 4 cm 3 1L 1 mL

F = 3.541022 N (b) 3.54 ´10 22 N =

q2 4p (8.854 ´10-12 C2 /J × m)(1.494 ´1011 m)2

55

Instructor‟s Manual

Solve for q: q = 2.971017 C. At 1.60210−19 C per electron, this works out to 1.851036 electrons, or 3.081012 moles. The mass of that many electrons is 1 torr 1 atm 1.01325 bar 1233 mmHg     1.644 bar , or approximately 1700 metric tons − 1 mmHg 760 torr 1 atm 18 orders of magnitude less than the mass of the earth.

8.4.

Since 1 dyne equals 110−5 newton, the easiest way to determine how many statcoulombs in a coulomb is to determine what charge is needed to have a force of 110−5 N at a distance of 1 cm (0.01 m): 125 mL 

1 cm 3  125 cm 3 and solve for 1 mL

q: q = 3.33610−10 C. Therefore, if 1 statcoulomb = 3.33610−10 C, then there are 2.998109 statcoulombs in 1 coulomb.

8.6.

æ 1 ö C F = e × N A = 1.6021733´10-19 C ç 6.0221367 ´10 23 which is Faraday‟s ÷ = 96485 è mol ø mol constant.

(

)

8.8.

“Electromotive force” is not a force in that it is not a mass times an acceleration, or even the equivalent. Rather, “electromotive force” is a difference between two electric potentials, which have units of J/C and not newtons.

8.10.

(a) V



R 

R 

56

nT p

pV nT

Chapter 1 L  bar

( 2.66 bar)(27.5 L)

(b) R  (1.887 mol)(466.9 K)  0.0830 mol  K

x atm 1 atm  64.0 m 10.1 m (c)

(1.04 atm )( 67 L ) ( 7.38 atm )

 9 .4 L

1mole SO 2 1.82 10 g SO 2   2.84 1011 moles SO 2 64.07 g 13

L  atm

( 2.84  10 moles SO )(0.0821 )( 273.15  17.0K ) mol  K (d) p  nRT   7  10 atm V 8  10 L 11

2

10

21

L  atm 5 / 9 K L  atm 0.0821    0.0456  mol  K 1 R mol  R 

G° for the original reaction in (d) is one−half of the G° calculated since the reaction formed from the half−reactions is double the original.

8.12. The intensive and extensive variables are related through n, the number of electrons transferred in the process. It turns the intensive variable E into the extensive variable Gº.

8.14. In order for the electrochemical process to provide that much work, the G of the process must be greater than 5.00102 kJ. Therefore, we need to determine the E for each process and from that calculate G. (a) For this reaction, E = 1.1037 V and G = −213 kJ, so this reaction wouldn‟t provide enough energy to perform the process. (b) For this reaction, E = 2.868 V and G = −553 kJ, so this reaction would provide enough energy to perform the process. (c) For this reaction, E = 2.212 V and G = −427 kJ, so this reaction wouldn‟t provide enough energy to perform the process. (d) For this reaction, E = 1.096 V and G = −211 kJ, so this reaction wouldn‟t provide enough energy to perform the process.

8.16. For the standard hydrogen electrode, the spontaneous process would be 2 Li + 2 H+  2 Li+ + H2 The voltage of this process is 3.04 V. For the standard calomel electrode, the process is 2 Li + Hg2Cl2  2 Li+ + 2 Hg + 2 Cl−

57

Instructor‟s Manual

and the voltage is 3.31 V. Therefore, when the calomel half reaction is used as the reduction reaction, the voltage shifts up by 0.2682 V. However, consider the other half reaction with silver. With the hydrogen electrode, the spontaneous process is 2 Ag+ + H22 Ag + 2 H+ and the voltage is 0.7996 V. With the calomel electrode, the spontaneous reaction is 2 Ag+ + 2 Hg + 2 Cl− 2 Ag + Hg2Cl2 and the voltage is 0.5314 V, so the voltage shifts down by 0.2682 V. Therefore, the direction of the shift depends on how the electrode reaction is used.

8.18.

(a) The overall reaction can be broken down into two half reactions: Au3+ + 3 e−  Au

E = 1.498 V

Au  Au+ + e−

E = −1.692 V

In order to determine the voltage of the overall reaction, we need to convert these voltages into Gs and sum the Gs. For the first reaction: G = −(3 mol)(96,485 C/mol)(1.498 V) = −433.6 kJ. For the second reaction: G = −(1 mol)(96,485 C/mol)(−1.692 V) = +163.3 kJ. Therefore, the sum of the two reactions has a G of (−433.6 + 163.3) = −270.3 kJ. Converting this back into an E for the two−electron overall process: −270,300 J = −(2 mol)(96,485 C/mol)(E) E = 1.401 V. (b) The overall reaction can be broken down into two half reactions: Sn4+ + 2 e−  Sn2+

E = 0.151 V

−

2+

Sn + 2 e  Sn

E = −0.1375 V

In order to determine the voltage of the overall reaction, we need to convert these voltages into Gs and sum the Gs. For the first reaction: G = −(2 mol)(96,485 C/mol)(0.151 V) = −29.1 kJ. For the second reaction: G = −(2 mol)(96,485 C/mol)(−0.1375V) = +26.5 kJ. Therefore, the sum of the two reactions has a G of (−29.1 + 26.5) = −2.6 kJ. Converting this back into an E for the two−electron overall process: −2600 J = −(4 mol)(96,485 C/mol)(E) E = 0.0067 V. 8.20.

To determine if the process will work, we need to see if the following (unbalanced) reaction is 14.7 lb lb  11.8 spontaneous: 0.80  in. in.

2

2

Breaking it into its half−reactions:

0.20 

14.7 lb lb  2 .

9 in.2 in.2

E° for the overall reaction is −0.9002V. Since this is negative the process is non−spontaneous and the plan would not work.

58

Chapter 1

8.22. We need to separate the reaction into its half−reactions:  m  m pM 4.50torr (44.01g / mole) pV   RT;    0.01706 g / L  1.71 10 5 g / cm 3  M  V RT (62.36 L  torr )(273.15  87K) mol  K

E° for the overall reaction is −0.463V. Since this is negative the original reaction is non−spontaneous and gold is resistant to corrosion.

8.24. Because these half−reactions typically occur in an aqueous solvent, the interactions between the ionic species and the solvent molecules has an impact on the overall energy change (in terms of E and G) of the process. Although all alkali metal ions have a +1 charge, the smaller, higher−charge−density lithium ion interacts more strongly with water molecules, increasing the energy of the process.

8.26.

If we reverse the bottom reaction and add the two together, we get:

3y 2 8.28.

y 2 z3 w



3w 2 z 3 xy 2 z 3 Eº for this reaction is 1.1037 V, so we have  32 y w2 6 xy 



w3 z 3 32 y 2



2 xyz 3 w

3y 2 z 2 w

This rearranges to

. Unfortunately, we can‟t mathematically

determine specific concentrations without more information; we can only specify the ratio.

8.30. First we need to determine E° for standard conditions:



3y 2 z 2 w

Now, we can use the Nernst Equation to solve for E.

2

3 y z 2 w

2

59

Instructor‟s Manual

8.32.

p  n    . To determine E at 1700ºC,  V  T, P R  T

According to equations 8.27 and 8.28, 

we need to calculate the S for the reaction. Using data in the appendix: S = [50.92 + 2(27.3)] – [2(28.30) + 87.4] = −38.5 J/K. Six moles of electrons are transferred in the reaction. Substituting:  T   p 

    V ,n



V n  R

.

8.34.

- pV  n     RT 2  T  P, V

8.36.

(a) Since the hydrogen half reaction has a voltage of 0.000 V, the Eº for the reaction is RT  p    :  n V   T ,V

–0.044 V. Using the equation 

nRT . Solving for K: K = 3.2510−2. (b) p According to the equilibrium constant, D+ prefers to be in solution over H+.

V (T , p ) 

8.38.

 V   V  In order to use the equation dV    dT    dp , we will need the entropy change of the  T  p  p  T + reaction. The entropy of H (aq) is defined as zero, and we are assuming that S(D+, aq) is zero also. Therefore, the entropy change of the reaction is (using data from the appendix) 144.96 – 130.68 = 14.28 J/K. The number of electrons transferred is 2 so we have:    Latm   Latm     1 mole  0.0821  1mole  0.0821  350K     nRT   nR  Kmole   Kmole    0.10atm   dV   dT   2 dp   10.0K        1.08 atm 1.08atm2  p   p         

Solving for T: T = 595 K. Therefore, if

we raise the temperature from 298 K to (298 + 595) = ~893 K, the voltage of the reaction should be about 0.

60

Chapter 1

 V   nRT  2V  2nRT 8.40. The temperature coefficient is equal to:    2 ;  2   3 . We need to calculate S°:  p n,T p  p n,T p

nR   p   p  ;  2   0     T  n ,V V  T  n ,V 2

The 8 comes from an analysis of the half−reactions involved in this overall reaction.

8.42. Using equation 8.30: TB TB 

L2atm .3 592 2 mol  1026 K 0.04267L/mol .0 08205 Latm   mol K 



a , we can determine the temperature coefficient, bR

. First, we need to find E° for the reaction:

L2 atm mol 2 TB   521 K  0.03183 L/mol  0.08205 L atm  mol K   1.360

E° for the overall reaction is equal to 0.5355V.

L2 atm mol 2 TB   433 K L atm   0.03913 L/mol  0.08205  mol K   1.390

8.44. Since heat capacity (at constant pressure) is defined as

C , we can take the V 2

derivative of equation 8.30 with respect to temperature: , which simplifies to

0.0237L 1m 3 m3   2.37  10 5 . mole 1000 L mole

8.46. First we start with the two half−reactions: 3 3 m 1 mole m 2.37 105   3.94 1029 23 mole 6.02 10 atoms atom

61

Instructor‟s Manual

Using Eqn. 8.32 and assuming molar concentrations,

3 4 3 m V  r  3.94 1029 ; r  2.111010 m or 2.11 3 atom This value makes sense since it is the square of Kw=1.0x10−14. The sum of the two electrochemical half−reactions is double the equation used to find Kw. 8.48.

Following Example 8.7: the reaction can be written in terms of two half reactions: AgCl (s) + e−  Ag (s) + Cl− (aq) +

−

Ag (s)  Ag + e

E = 0.22233 V E = −0.7996 V

The overall voltage of the combination of the two reactions is –0.5773 V. Using equation 8.32:

0.03049L mole 8.50.

Solve for Ksp: Ksp = 1.7410−10.

Following Example 8.7: the reaction can be written in terms of two half reactions: Hg2Cl2 (s) + 2 e−  2 Hg (s) + 2 Cl− (aq) +2

−

2 Hg (s)  Hg2 + 2 e

E = 0.2682 V E = −0.7973 V

The overall voltage of the combination of the two reactions is –0.5291 V. Using equation 8.32:

0.0638L mole

Solve for Ksp: Ksp = 1.2910−18.

8.52.

Following Example 8.8,

8.54.

The Ksp for Hg2Cl2 was determined in exercise 8.28 and is 1.2910−18. If x moles per liter of Hg2Cl2 dissociates, one gets x M Hg22+ and 2x M Cl−. Therefore, we have: 1.2910−18 = (x)(2x)2 1.2910−18 = 4x3 x = 6.8610−7 Since the equilibrium concentration of Cl− is twice this, [Cl−] = 1.3810−6 M.

62

Chapter 1

2

8.56. For this we need equation: Z  1   b  a  1   b    . The two differences between chemical 

RT  V  V 

standard states and biochemical standard states are the temperature (37°C for biochem) and concentration of H+ (1x10−7M for biochem).

Z  1

B C   V V 2

2

V RT

a

8.58. Start with the definition that p  V  b  T V(V  b) , which is the definition of mean activity from equation 8.46 but without the standard molality in the denominator. First, distribute the  p   . Now we focus on the molality term. First, let us substitute for exponent:   V  T ,n

 RT

a





a

the mean ionic molality from equation 8.43: (V  b) T V ( V  b) T V (V  b) Since n = n+ + n−, the outermost exponent cancels with the 1/(n+ + n−) exponent inside the parentheses, )( V  nb)  nRT . For a salt having formula An Bn , for a certain molality m the leaving (p  an V + − molality of the positive ion, m+, is n+m and the molality of the negative ion, m−, is n−m.

Substituting into the expression for a: p V  RT , which can be rearranged 2

2

2

2

2

by distributing the exponents: together:

L  atm e p  (0.0821 )(273.15K) K  mol

Latm   (10.91atm L2 ) /  ( 22.41L )( 0.0821 )( 273.15 K )  Kmol  

(22.41L  0.0401L)

p

RT ,p  V

 0.981atm

(0.0821

L  atm )(273.15K ) K  mol  1.00atm 22.41L

. Bringing the two molality terms

, and now realizing again that n = n+ + n−, we do one

final substitution for the exponent on m: we‟re looking for.

 V   T

   p





 p   T  p   V

  V   T

, which is the expression

8.60. Using the definition of ionic strength in equation 8.47: (a) (b) (c)

   

V p

1 V

   T

  





 T   p   T   V

V T

  

   V    p

p

1 volume 1   volume temperatur e temperatur e

(d) 

1 V

 V   p 

   T

8.62. We will assume that the concentrations are so low that we can assume that the molarity is equal to the molality. (a) 1 volume  volume pressure 



 1   V 



1 pressure

V P

   T

(b) Speculation is left to the student. Hint: osmoregulation

63

Instructor‟s Manual

8.64.

In the equation H2 (g) + I2 (s)  2 H+ (aq) + 2 I− (aq), the overall enthalpy of reaction is – 110.38 kJ. Using the concept of products−minus−reactants to determine H, we need the heats of formation of the products (one of which is the object of this calculation) and the heats of formation of the reactants. The two reactants are elements, so their fHs are zero.

By convention, the fH of H+ (aq) is also zero, so the only non−zero fH is that for I−.

Therefore, we have −110.38 kJ = (2 mol)(fH[I−]) fH [I−] = −55.19 kJ/mol.

8.66.

 1   nRT  1  nRT  1  1  V  1   1 2  p       V   p  V p p 

(a) The real equation of interest is:   V  p

.

1  V  1   nRT  1 nRT           2 V  p T V p  p  V p (b) The real equation of interest is:

nRT p

 V

1 V 1  V p p 8.68.

T p



T T 1  V  T   nRT  T nR       p p V  T  p pV T  p  pV p This value is very close to the value calculated in Example 8.11.

8.70.

64

As you go from F− to I− in column 7A, the ions become increasingly large. fHo‟s are obviously more negative for the smaller ions. This large negative number represents greater stability of the smaller ions. This makes sense because if the ion is smaller, water molecules can approach the ion more closely. From Coulomb‟s law, we know that this increases the binding energy between the ion and the water molecules and results in greater stability of the ion.

Chapter 1

8.72. The complete expression for A is worked out in the text. The student need simply verify that the numbers and units do reduce to 1.171 molal−1/2.

8.74.

0.9% NaCl implies 0.9 g NaCl in 99.1 g water. Assuming that the volume of the solution nR V is 100 mL = 0.100 L = 0.100 kg: . Therefore, the  p T ionic strength is

8.76.

T pV

V T

, which 

1 . p

From Coulomb‟s law, we know that the force of interaction for higher−charged ions RT (q=2,3..) is much greater than that of singly charged ions: V  In addition, the p ionic strength of a solution is much higher for higher−charged ions since the ionic strength is proportional to the square of the charge on the ion. Since the Debye−Huckel theory assumes solutions with very low ionic strength, solutions with many higher−charged ions can be problematic.

8.78. (a) Identity of the counterion is necessary to determine the ionic strengths of the solutions. (b) If the counterion were sulfate instead of nitrate, the ionic strengths would need to be recalculated. If sulfate were the counterion, the salt‟s formula is Fe2(SO4)3 and the sulfate concentration is 3/2 of the iron ion concentration: I (Fe3+ soln) = ½[(0.100)(+3)2 + (0.150)(−2)2] = 0.750 molal If sulfate were the counterion, the other salt‟s formula would be CuSO4 and the sulfate concentration would equal the copper ion concentration: I (Cu2+ soln) = ½[(0.050)(+2)2 + (0.050)(−2)2] = 0.200 molal Using equation 8.52 for each ion:

d



pM  d      RT 2  T  p .n

M RT /

p



pM RT

 (Fe3+) = 0.0388 Therefore, the activity of Fe3+ is (0.0388)(0.100

m) = 0.00388 m.

Similarly, for the Cu2+:

V M  d      VT  T  p . n

 (Cu2+) = 0.275 Therefore, the activity of Cu2+ is (0.275)(0.050 m)

= 0.0138 m.

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Instructor‟s Manual

Substituting these activities into the Nernst equation: p



e



Mgh RT

Mgh RT



g   m    m  2  mol  s  J    K   K  mole 

When you compare this to the voltage from the example (0.372 V), we see how the ionic strength of the solution, as influenced by the counterion, can have an influence on the voltage even though the counterions don‟t participate in the reaction.

8.80.

Equation 8.61 is

kg  m 2 s2

. Equation 8.5 shows that E has units of

N/C, e has units of C, z is unitless (it is simply the magnitude of the charge on the ion), the fraction (Ni/V) has units of 1/m3, A has units of m2, and in SI units, viscosity has units of kg/ms. The numbers 6 and  have no units. The radius r has units of m. Combining all of these units: Mgh RT

2   kg  m    2    mol  s   J    K   K  mol 

This can be rearranged to get

pe



Mgh RT . One of the

C units cancels, as do the three m units in the numerator and three of the four m units in  kg  m  J  the denominator. This results inX0. 28967 9.8 2432m8.314 273.1539.0K0. 473;eX 1.05atm. If we break down the newton unit into kgm/s2,  mole s   Kmole we have 4 3  5   10 7 7 score       7.14 4 3    7 7

. The kg and m units cancel, as do one of the second units in the numerator

and denominator. What‟s left is C/s, and one coulomb per second is an ampere, the unit of current.

8.82.

For a galvanic cell, oxidation (the loss of electrons) occurs at the anode and is considered the negative electrode. Therefore, I− is the current towards the cathode, and I+ must be the current towards the anode. In an electrolytic cell, oxidation (the loss of electrons) still occurs at the anode, so I− is still the current towards the cathode and I+ is the current towards the anode. For any given cell reaction, however, the identities of the cathode and anode are switched for galvanic and electrolytic cells.

Chapter 9 Pre-Quantum Mechanics 9.2.

66

There are two ways to determine the Hamiltonian for the system in exercise 9.1. The first is to use equation 9.11, the original definition of the Hamiltonian function, and define H in terms of the Lagrangian L. The second and easier way is to use equation 9.13 and

Chapter 1

recognize that H is the sum of the kinetic and potential energies. As mentioned in 1 exercise 9.1, the kinetic energy for a mass falling in the z direction is mz 2 and the 2 gravitational potential energy is mgz. Therefore, the Hamiltonian for this system is simply 1 H  mz 2  mgz .

2 (a) Newton‟s laws of motion are probably easiest to use to describe this system because Newton‟s laws involve forces more directly than other quantities.

9.4.

(b) Either agrange‟s or Hamilton‟s laws of motion would be better to use, especially if the kinetic and potential energies of the rocket were easily determined.

9.6.

Atomic line spectra: Every element absorbed and emitted discrete characteristic wavelengths of light.

Blackbody radiation: Classical theory predicted that as the wavelength of the emitted light approached 0, the intensity should go to infinity. Experimentally as the wavelength approached 0, the intensity approached 0.

The Photoelectric effect: Classical theory predicted that as the intensity of light shone on a metal surface increases, the kinetic energy of the ejected electrons should increase. It does not.

9.8.

(a) o

218 A´

1m

100 cm = 2.18´10-6 cm = l 10 1 m 1´10 A o

c = ln (b)

´

\

1

l

=

1 = 459, 000 cm -1 2.18´10-6 cm

2.9979 ´108 m/s = l ×8.077 ´1013s-1

2.9979 ´108 m/s l= = 3.712 ´10-6 m = 3.712 ´10-4 cm 13 -1 8.077 ´10 s



1





1  2694 cm -1 4 3.712  10 cm

1m 100 cm 1 1   3.31  10  4 cm    3020 cm -1 6 1m  3.31  10  4 cm 10 m If two spectra of two different compounds have some lines at exactly the same wavelength, then by irchhoff‟s and Bunsen‟s proposition, they must share one or more constituent element. (c)   3.31 m 

9.10.

9.12. To determine the series limit, assume that 1/n12 = 1/  2 = 0. For the Lyman series: 1  wavenumber  109,700 cm -1  2  0   109,700 cm -1 is the series limit. For the Brackett 1  series:

 1  wavenumber  109,700 cm -1  2  0   6856 cm -1 is the series limit. 4  67

Instructor‟s Manual

9.14.

The range of each series of hydrogen line spectra can be calculated using the Rydberg equation (9.17). For the Balmer series, n1=2, the line with the lowest energy would have n2=3. The high energy limit of the series is found by letting n2 ∞. The easiest way to solve this problem is using a spreadsheet. It can be found that the energy ranges for the series are: Energy (cm−1) series Lyman Balmer Paschen Brackett Pfund

n2

lowest 1 2 3 4 5

highest 82303 15241 5334 2469 1341

109737 27434 12193 6859 4389

Overlap begins with the Paschen and Brackett series.

9.16.

Using equation 9.34 with n=100 and a0: e 0 n2h2 r= = n 2 × ao = (100)2 × (5.29 ´10-11 m) = 5.29 ´10-7 m = 529nm 2 p mee

9.18.

(a) A single electron has a mass of 9.10910−31 kg. If we assume that a helium nucleus has the mass of two protons plus two neutrons (ignoring the mass defect that represents the energy stabilization of the nucleus), then the helium nucleus has a mass of (21.67210−27 + 21.67510−27) kg = 6.69410−27 kg. The ratio of these two masses will tell us how many beta particles it takes to equal the mass of 1 alpha 6.694  10 27  7349 beta particles to mass the same as one alpha particle. particle: 9.109  10 31 (b) A beta particle would be moving faster than an alpha particle of the same kinetic energy, since it has a much smaller mass. (c) This is consistent with the fact that beta particles are known to be the more penetrating particle.

68

Chapter 1

9.20.

For a flux of 1.00 W/m2:

1.00

W W  (5.6705  10 8 2 4 )(T 4 ) 2 m m K

Solving for T : T  65 K

For a flux of 10.00 W/m2:

10.00

W W  (5.6705  10 8 2 4 )(T 4 ) 2 m m K

Solving for T : T  115 K

For a flux of 100.00 W/m2:

100.00

W W  (5.6705  10 8 2 4 )(T 4 ) 2 m m K

Solving for T : T  205 K

9.22. As stars become hotter, their λmax shifts to lower wavelengths. Lower wavelengths correspond with higher energies and higher temperatures. Since blue light is higher energy than red light, the bluish star is hotter.

9.24. (a) power per unit area  (5.6705  10 -8 W/m 2 K 4 )(5800 K) 4  6.42  10 7 W/m 2 (b) 6.42107 W/m2  6.0871012 m2 = 3.911020 W (c) 3.911020 W = 3.911020 J/s 

365 d 24 hr 3600 s    1.23  10 28 J per year.

1 year 1 d 1 hr

9.26. (a) Using Wien‟s law: max5800 K = 2898 mK max = 0.4997 m = 4997Å. (b) 5000 Å = 0.5000 m, so using Wien‟s law: 0.5000 mT = 2898 mK T = 5796 K. (c) The two are very close, suggesting (but not proving!) that the eye may have evolved to take advantage of the brightest part of the sun‟s spectrum.

9.28.

Red light has a significantly lower energy than UV light.

9.30. (a) Using

ö ¶r 8p hc æ 1 = 5 ç hc/lkT ÷ : ¶l l èe -1 ø

 8hc  1   5  hc / kt    e 1 8 (6.626  10 34 J  s)(2.9979  10 8 m/s)  1   ( 6.6261034 Js)(2.9979108 m/s)/(5.00 10-7 m)(1.38110- 23 J/K)(1000 K)  -7 5 (5.00  10 m) 1 e  5.12  10 5 J/m 4

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Instructor‟s Manual

(b) 8 (6.626  10 34 J  s)(2.9979  10 8 m/s)  1   ( 6.6261034 Js)(2.9979108 m/s)/(5.00 10-7 m)(1.38110- 23 J/K)(2000 K)  -7 5 (5.00  10 m) 1 e  90.5 J/m 4

(c) 8 (6.626  10 34 J  s)(2.9979  10 8 m/s)  1   ( 6.6261034 Js)(2.9979108 m/s)/(5.00 10-6 m)(1.38110- 23 J/K)(2000 K)  -6 5 (5.00  10 m) 1 e  497.1 J/m 4

(d) 8 (6.626  10 34 J  s)(2.9979  10 8 m/s)  1   ( 6.6261034 Js)(2.9979108 m/s)/(1.00 10-5 m)(1.38110- 23 J/K)(2000 K)  -5 5 (1.00  10 m) 1 e  47.4 J/m 4

(e) Comparing to the answers to exercise 9.25, it seems clear that the agreement (which isn‟t great anyway is best for high temperatures and long wavelengths.

9.32.

Using dE  110−9 m:

2hc 2  1   hc / kt d , substituting for the variables and using d =  = 5  e 1

At 1000 K:

dE 

2(6.626  10 34 J  s)(2.9979  10 8 m/s) 2  1   ( 6.6261034 Js)(2.9979 108 m/s)/(3.50 510-7 m)(1.38110- 23 J/K)(1000 K)  1  10 9 m 7 5 (3.505  10 m) 1 e



dE = 1.04 ´10-10 W/m2.

At 3000 K: dE 

2(6.626  10 34 J  s)(2.9979  10 8 m/s) 2  1   ( 6.62610  34 Js)(2.9979 108 m/s)/(3.50 510- 7 m)(1.38110- 23 J/K)(3000 K)  1  10 9 m 7 5 (3.505  10 m)  1 e





dE = 81.1 W/m2.

At 10,000 K: dE 

2(6.626  10 34 J  s)(2.9979  10 8 m/s) 2  1   ( 6.62610  34 Js)(2.9979 108 m/s)/(3.50 510- 7 m)(1.381 10- 23 J/K)(10000 K)  1  10 9 m 7 5 (3.505  10 m)  1 e



dE = 1.19106 W/m2.

70





Chapter 1

9.34. All we need to do is convert the work function into J units, then use the Planck relationship and the definition of the speed of light to get an equivalent wavelength. For Li: 1.602  10 -19 J  4.65  10 19 J  (6.626  10 -34 J  s)   7.00  1014 s -1 1 eV 8 2.9979  10 m/s  (7.00  1014 s -1 )     4.27  10 -7 m  427 nm 2.90 eV 

1.602  10 -19 J  3.43  10 19 J  (6.626  10 -34 J  s)   5.17  1014 s -1 1 eV 2.9979  10 8 m/s  (5.17  1014 s -1 )     5.79  10 -7 m  579 nm

For Cs: 2.14 eV 

1.602  10 -19 J  8.01  10 19 J  (6.626  10 -34 J  s)   1.21  1015 s -1 1 eV 8 2.9979  10 m/s  (1.21  1015 s -1 )     2.48  10 -7 m  248 nm

For Ge: 5.00 eV 

9.36. Using the definition of the speed of light, Å

hc    KE electron  1J (6.626  10 34 J  s)(2.9979  10 8 m / s)    (2.90 eV )   KE electron 18 10  1  10 m   6.242  10 eV   (1850Å)  1Å 

h 

KEelectron = 6.09 ´10-19 J 9.38. (a) For this problem we assume that the kinetic energy (KE) of the ejected electron is 0 so: nh   , where n is the minimum number of electrons absorbed to overcome the work function.

nhc  

æ ö 1J -9 (4.67eV ) ç ÷ (776.5´10 m) 18 jl è 6.242 ´10 eV ø n= = = 2.92 » 3photons needed hc (6.626 ´10 -34 J × s)(2.9979 ´108 m / s) (b) If three photons are absorbed,

c

KEelectron = h n - f = (6.626 ´10

l

-34

æ 2.9979 ´108 m / s ö æ ö 1J -20 J × s)ç ÷ =1.93´10 J ÷ (3) - (4.67eV )ç -9 18 è 6.242 ´10 eV ø è 776.5´10 m ø

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Instructor‟s Manual

1 KEelectron = 1.93´10 -20 J = mv 2 2 v= 9.40.

2(1.93´10 -20 J) = 2.06 ´10 5 m / s -31 9.109 ´10 kg

By combining the equations E = h and c = , we get an expression that can be used hc directly: E  . For the wavelength of 10 m:



(6.626 ´10-34 J × s)(2.9979 ´108 m/s) = 1.99 ´10-26 J per photon, 10 m 1.99 ´10-26 J per photon ´ 6.022 ´1023 / mol = 0.0120 J/mol photons E=

For a wavelength of 10.0 cm = 0.100 m: (6.626 ´10-34 J × s)(2.9979 ´108 m/s) =1.99 ´10-24 J per photon, 0.100 m 1.99 ´10-24 J per photon ´ 6.022 ´10 23 / mol = 1.20 J/mol photons E=

For a wavelength of 10 microns = 0.00001 m: (6.626 ´10-34 J × s)(2.9979 ´108 m/s) =1.99 ´10-20 J per photon, 0.00001 m 1.99 ´10-20 J per photon ´ 6.022 ´1023 / mol = 12000 J/mol photons E=

For a wavelength of 550 nm = 5.5010−7 m: (6.626 ´10-34 J × s)(2.9979 ´108 m/s) = 3.61´10-19 J per photon, 5.50 ´10-7 m 3.61´10-19 J per photon ´ 6.022 ´10 23 / mol = 217,500 J/mol photons E=

For a wavelength of 300 nm = 3.0010−7 m: (6.626 ´10-34 J × s)(2.9979 ´108 m/s) = 6.62 ´10-19 J per photon, 3.00 ´10-7 m 6.62 ´10-19 J per photon ´ 6.022 ´10 23 / mol = 398,700 J/mol photons E=

For a wavelength of 1 Å = 10−10 m: (6.626 ´10-34 J × s)(2.9979 ´108 m/s) = 1.99 ´10-15 J per photon, 10-10 m 1.99 ´10-15 J per photon ´ 6.022 ´10 23 / mol = 1.20 ´ 20 9 J/mol photons E=

72

Chapter 1

9.42. Equation 9.34:  C2   J  s 2 2 2  n h J  s2 Jm =m r o   kg  m m e e 2 kg(C 2 )

Recognizing that a J is a

kg  m 2 , everything cancels out but one meter unit on the top. s2

me v 2 e2  9.44. Equation 9.27 is . Mass has units of kg, velocity has m/s, r is a distance r 4 0 r 2 in meters, electrical charge has units of C, and the permittivity of free space has units of C2/Jm. Substituting these units for all quantities: kg  (m/s) 2 C2  2 m (C / J  m)(m 2 )

kg m 2 J m C 2 kg m J  2 2 which reduces to 2  , both of which equal the unit N. 2 ms C m m s 9.46. Equation 9.35:

E tot  

mee4 8 o n 2 h 2 2



kg  C 4 2

 C2    (J  s) Jm

kg  m 2  J s2 2

9.48. Angular momenta are simply integral units of h/2: for n = 4, L = 4(6.62610−34 Js)/2 = 4.2210−34 Js.

For n = 5: L = 5(6.62610−34 Js)/2 = 5.2710−34 Js.

For n = 6: L = 6(6.62610−34 Js)/2 = 6.3310−34 Js.

9.50. For the Lyman series electrons fall from higher energy levels to n=1:

The Brackett series diagram will be similar, except electrons from n>4 will fall to n=4.

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Instructor‟s Manual

 0n2h2 9.52. (a) The formula should be r  2 . (b) For n = 100 and Z = 91: Z me e 2 r=

9.54.

o e0 n2 h 2 (8.854 ´10-12 C2 / Jm)(100 2 )(6.626 ´10-34 J ×s)2 -11 = = 6.39 ´10 m or 0.639 A .

Z 2 p mee2 912 p (9.109 ´10-31 kg)(1.602 ´10-19 C)2

Starting with n= 2r, substitute for wavelength using de Broglie‟s relation  

n

h : mv

h  2r , which rearranges to nh  2mvr . Finally, dividing both sides by 2 yields mv

nh , which is Bohr‟s postulate that the angular momentum, mvr, is a quantized 2 multiple of h/2. mvr 

9.56.

 1.602  10 19 J  1  . If we substitute in the mass of an mv 2  (100  10 3 eV) 2 1eV   −31 8 electron, me=9.109x10 kg, v = 1.88x10 m/s.

KE electron 



h (6.626  10 34 J  s)   3.89  10 12 m 31 8 m  v (9.109  10 kg)(1.88  10 m / s)

Chapter 10 Introduction to Quantum Mechanics 10.2.

Acceptable wavefunctions must be bounded or finite (that is, non-infinite over continuous ranges), single-valued, continuous, and integrable (that is, we must be able to evaluate the integral of the function).

10.4.

(a) No, the function is not quadratically integrable. (b) No, the function is not acceptable because not finite over the range given. Notice that over a certain range, the function is imaginary. This does not disqualify it as a possible wavefunction! (c) No, the function is not acceptable because it is not continuous. (d) No, the function is not quadratically integrable. (e) No, the function is not acceptable because it is not bounded.

10.6.

74

(a) The operator is the multiplication by 2.

Chapter 1

(b) The operator are either

1

()

or

( ).

5

(c) The operator is the natural logarithm function. (d) The operator is the sine function. (e) The operator is the exponential function. (f) The operator is the first derivative operation.

10.8.

 (a) Aˆ p  (4 x 3  2 x 2 )  12 x 2  4 x 3 x (b) Cˆ q 

1  2  0.5

 2x  (c) Bˆ s  sin   3 

(d) Dˆ q  10 0.5 

1  0.3162277...

(10) 0.5

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Instructor‟s Manual

  1  1  (e) Aˆ (Cˆ r )   2  x  45 xy  45 x 2 y 2 (f) Â(D̂q) 

10.10. (a)

  10 0.5  (0.3162277...)  0 x x





d x  x sin  cos . Because you are not getting the original sine function back, this dx 2 2 2 is not an eigenvalue equation.

d2 x d   x  2 x (b) sin   cos    sin . We are getting the original sine function 2 2 dx  2 2 4 2 dx times a constant back, so this is an eigenvalue equation with eigenvalue of –2/4.

 x  x sin  i cos . Because you are not getting the original sine function x 2 2 2 back, this is not an eigenvalue equation.

(c)  i

 imx e  i(im)e imx  me imx . We are getting the original exponential x function times a constant back, so this is an eigenvalue equation with eigenvalue of  m .

(d)  i

(e)

 (e  x 2 )  2x . The original function is not present. Therefore, this is not an x eigenvalue equation.

(f)  2 d 2  2x 2 d 2 2x 2x  2 4 2 2x 2x   0 .

5 sin   sin  0 .

5 sin   sin  0.5 sin   2 2 3 2m dx 3 3 2m 9 3 3  2m dx 

 4 2  2  2x  0.5 sin Factoring the sine function:   . We are getting the original 3  18m  exponential function times a constant back, so this is an eigenvalue equation with 4p 2 2 eigenvalue of + 0.5.

18m (g)

2   y2 e  2 ye  y . The original function is present, but it‟s now multiplied by x another function,-2y. Therefore, this is not an eigenvalue equation.

10.12. When multiplying a function by a constant, the result is simply the original function times a constant. Since this is what‟s required by an eigenvalue equation, multiplying a function by a constant always yields an eigenvalue and the original function.

76

Chapter 1

10.14. In terms of classical mechanics, K = ½ mv2, or in terms of linear momentum p = mv, p2  . Substituting for the definition of the linear momentum operator pˆ  i : K 2m x 1    1  2 2  2 2   , which is the one-dimensional kinetic Kˆ   i         2m  x  2m  2m x 2 x 2  energy operator.

2

 1 im   1 im    1 im   e   i(im) e   m e  . Since the original function is   2   2   2  returned multiplied by a constant, the function is an eigenfunction and the value of the angular momentum is m .

10.16.  i

10.18. 1.23me = 1.23(9.10910-31 kg) = 1.1210-30 kg

10,000 m  6.626  10  x1.12  10 -30 kg   s 4  

34

J s

x  4.71  10 9 m .

 1.055  10 34 J  s    9.77  10  26 kg  m / s 10.20. x  p  ; p  m  v  12 2x 2(540  10 m) 2 v 

9.77  10  26 kg  m / s 9.77  10  26 kg  m / s   1.07  10 5 m / s 31 me 9.109  10 kg

10.22. First, we need to convert the 1.00 cm-1 to energy: c 3.00  10 8 m/s 1  1m      3.00  1010 s 1  0 .

01 m      0.01 m 1.00 cm -1  100 cm 

Now calculating the equivalent E: E  h  (6.626  10 34 J  s)(3.00  1010 s -1 ) 

E = 1.9910-23 J Calculating the t from the expression in the exercise:

(1.99 ´10-23 J)Dt ³

6.626 ´10-34 J × s t ³ 2.6510-12 s.

4p

This is just under 3 picoseconds.

10.24. (a) * = 4x3 (b)  *  e i (c) * = 4 – 3i (d)  *  i sin

3x 2

(e)  *  e  iEt / 

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10.26. Since all parts of the exercise use the same integral, just with different units, let us solve the integral first and then substitute the appropriate limits. *

upper limit

 2 x   2 x  2 x 2 x 1 2x  P    d    sin   sin dx   sin 2 dx    sin a  a a a a a  2 4( / a) a  lower limit  a *

(a) From 0 to 0.02a: 0.02 a

2 x 1 2x   sin  a  2 4( / a) a  0 



2  0.02a a 2 (0.02a)   0 a 2 (0)   sin sin      a  2 4 a a    2 4

2 a  0.01a  sin(0.04 )  0  0  2(0.01  0.0099737)  0.0000526  a 4 

(b) From 0.24a to 0.26a: 0.26 a

2 x 1 2x  2  0.26a a 2 (0.26a)   0.24a a 2 (0.24a)   sin    sin  sin     a  2 4( / a) a  0.24 a a  2 4 a 4 a   2 



2  a a    sin(0.52 )    0.12a  sin(0.48 )   0.13a   a  4 4   

 2(0.01  0.07942  0.07942)  0.02

(c) From 0.49a to 0.51a: 0.51a

2 (0.51a)   0.49a a 2 (0.49a)  2 x 1 2x  2  0.51a a  sin    sin  sin     a  2 4( / a) a  0.48 a a  2 4 a 4 a   2 



2  a a    sin(1.02 )    0.245a  sin(0.98 )   0.255a   a  4 4   

 2(0.01  0.0049967  0.0049967)  0.0400

(d) From 0.74a to 0.76a: 0.76 a

2 (0.76a)   0.74a a 2 (0.74a)  2 x 1 2x  2  0.76a a  sin    sin  sin     a  2 4( / a) a  0.74 a a  2 4 a 4 a   2 



2  a a    sin(1.52 )    0.37a  sin(1.48 )   0.38a   a  4 4   

 2(0.01  0.07942  0.07942)  0.02

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(e) From 0.98a to 1.00a: 1.00 a

2 x 1 2x  2  a a 2a   0.98a a 2 (0.98a)   sin    sin  sin     a  2 4( / a) a  0.98 a a  2 4 a   2 4 a  

2 a  0.01a  0  sin 1.96   2(0.01  0.0099737)  0.0000526  a 4 

A plot of the probabilities would show very low probabilities at the sides and a relatively high probability in the middle of the system.

2

2

0

0

10.28. (a) To normalize: N 2  (e im ) * e im d  1 . N 2  e im e im d  1 2

N 2  1  d  1 N 2   0  1 N2(2 – 0) = 1 N 2  2

0

1

Therefore, the normalized wavefunction is 2 / 3

 1 im   e  (b) P    2    0

*

1 2

e im  d 

2

1 2

N

1 2

e im .

2 / 3

2 / 3

1 1 e im   e im  d  d  2 0 2 0

1   02 / 3  1  2  0   1 . The probability is 1/3. This makes sense because 2 2  3  3 the region of interest is 1/3 of the ring. The probability does not depend on m, so any value of m would have the same value of probability over this interval. 

10.30. First, we need to normalize the function: a

a

a

N  (kx) kx dx  1 . N k  x dx  1 2

*

2

0

2

2

0

1  N k  x3   1 3  0

3 1  N 2 k 2  a 3  0   1 N2k2a3 = 3 N 2  2 3 k a 3 

Therefore the normalized wavefunction is  

2

2

N 3 ka

3/ 2

3 ka

3/ 2

kx 

3 a

3/ 2

x . Again, note how the k

cancels.

For the first third of the interval: *

a/3  3   1 3 3 3 1 a/3 3  1 a3 .

P    3 / 2 x  3 / 2 x dx  3  x 2 dx  3  x 3  3   0   a 0 a 3 0 a  3 27  27 0 a  a a/3

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For the last third of the interval: *

a  3  3 3 3 1 3a 3  1 3 1 8a 3  19 2    a   .

P    3 / 2 x  3 / 2 x dx  3  x dx  3  x  3 27  27 a 2a / 3 a 3 2 a / 3 a 3  3 2a / 3  a  a a

In this case, the particle has a higher probability of being in the last third of the interval, and the probability distribution is not equal like it is in the previous problem.

10.32. These are similar to Example 10.7. None of theses have imaginary parts so  * ( x )   ( x ).

1

(a)  ( Nx ) 2 dx  1 0

1

x3 1 N  N 2    1; N  3 3 0 3 2

 ( x )  3x 2

(b)  ( Nx ) 2 dx  1 0

2

x3 3 8 N  N 2    1; N  3 0 8 3 2

(x) 

3 x 8



(c)  ( N sin(2x )) 2 dx  1 (using the integral table in Appendix 1): 0



x  1  N   sin(2  2 x )  N 2    1; N  2  2 4(2) 0 2

(x) 

2 

2 sin(2 x ) 

2

(d)  ( N sin(2x )) 2 dx  1 0

2

x  1  2  N   sin(2  2 x )  N 2    1; N   2   2 4(2) 0 2

(x) 

80

1 sin(2 x ) 

1 

Chapter 1

2

2

3

(e)  ( N sin 2 ( x )) 2 dx  N 2  sin 3 ( x )dx  1 0

0

2

1   N  cos x  cos x   N 2 (0)  1 3  0 2

10.34. The wavefunction must be finite. If the integral diverges, it is impossible to normalize it.

10.36. The Schrödinger equation contains a specific expression for the kinetic energy because 1 p2 kinetic energy has a specific formula in classical mechanics ( mv 2 , or - the two are 2 2m equivalent) which can be transferred into a kinetic energy operator. However, there is no single expression for the potential energy of a system. The expression of the potential energy depends on the definition of the system. Therefore, in the Schrödinger equation, there is only a general expression for the potential energy operator. 2 2 (k )  0  k  Ek Since the derivative of a constant is zero, we have 2m x 2 0 + 0 =Ek Therefore, E must be 0.

10.38. 

10.40. The only property of Hermitian operators described in the text is that Hermitian operators have real (i.e. non-imaginary) eigenvalues. While not a strict proof, we can argue that if the Hamiltonian operator yields energy as an observable, and since we know that energy is a real quantity, we submit that the Hamiltonian must produce real eigenvalues and so be a Hermitian operator. (There are other mathematical requirements for Hermitian operators, but this text doesn‟t cover them.

2

2

0

0

10.42. To normalize: N 2  (e iKx ) * e iKx dx  1 . N 2  e iKx e iKx dx  1 2

N 2  1  dx  1 N 2 x 0  1 N2(2 – 0) = 1 N 2  2

0

Therefore, the normalized wavefunction is

1 2

1 2

N

1 2

e iKx . Now, substituting this into the

Schrödinger equation:  2  2  1 iKx   2  1 iKx   2 K 2  1 iKx  2    e   ( iK )    2m  2 e   2m  2 e  . Therefore it is an 2m x 2  2       2K 2 . The eigenvalue doesn‟t change with 2m inclusion of a normalization constant.

eigenfunction with an eigenvalue of

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Instructor‟s Manual

10.44. 

2 2  2 nx   2 n 2 2  2 nx  n 2  2 2  2 x       Therefore, the sin  sin sin 2  2 2 2m x  a a  2m a  a a  a  2ma  a

n 2  2 2 h eigenvalues are . Since   , we can substitute and simplify to get 2 2 ma n 2 h 2 2 . The 2 terms cancel, and multiplying the constants in the denominator, we 2 2 4  2ma n2h2 get , which is the expected expression for the energy eigenvalue.

8ma 2

10.46. n 2h 2 ; a  1.00nm 8ma 2 (1) 2 (6.626  10 34 J  s) 2 E1   3.28  10  23 J  27 9 2 8(1.673  10 kg)(1.00  10 m) En 

(2) 2 (6.626  10 34 J  s) 2 E2   1.31  10  22 J  27 9 2 8(1.673  10 kg)(1.00  10 m) E3 

(3) 2 (6.626  10 34 J  s) 2  2.95  10  22 J  27 9 2 8(1.673  10 kg)(1.00  10 m)

10.48. E n 

n 2h 2 ; If a doubles, E decreases by a factor of 4.

8ma 2

10.50. First, we should convert the wavelength of the photon into an equivalent value in joules: c 3.00  10 8 m/s  1.38  1015 s -1 . The  = 2170 Å = 2.1710-7 m. Therefore,     2.17  10 -7 m energy of a photon of this frequency is E = h = (6.62610-34 Js)(1.381015 s-1) = 9.1410-19 J. Now we use the expression for the energies of the particle-in-a-box wavefunctions:

(3 2  2 2 )(6.626  10 34 J  s) 2 We have all of the 8(9.109  10 -31 kg) a 2 information except the length of the „box‟, a. Solve algebraically: a = 5.7410-10 m = 5.74 Å.

9.14  10 19 J  E (n  3)  E (n  2) 

10.52. The drawings are left to the student. See Figures 10.6 and 10.7 in the text.

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Chapter 1

*

0.505 a  2 x   2 x  2 x  10.54. For n = 1: P    sin   sin dx  sin 2 dx  a a  a a a 0.495 a a 0.495 a  0.505 a

0.505 a

2 x 1 2x   sin  a  2 4( / a) a  0.495 a 

2  0.505a a   0.495a a   sin 2 (0.505)     sin 2 (0.495)    a  2 4 4   2 

P  2(0.005  0.00250  0.00250)  0.02 *

0.505 a  2 2x   2 2x  2 2x   dx  For n = 2: P    sin sin sin 2 dx     a a   a a  a 0.495 a a 0.495 a  0.505 a

0.505 a

2 x 1 4x   sin  a  2 4(2 / a ) a  0.495 a 

2  0.505a a   0.495a a   sin 4 (0.505)     sin 4 (0.495)    a  2 8 8   2 

P  2(0.005  0.002498  0.002498)  0.000008 *

0.505 a  2 3x   2 3x  2 2 3x   dx  For n = 3: P    sin sin sin dx    a  a a a a a 0.495 a  0.495 a    0.505 a

0.505 a

2 x 1 6x   sin  a  2 4(3 / a) a  0.495 a 

2  0.505a a a   0.495a   sin 6 (0.505)     sin 6 (0.495)    a  2 12 12   2 

P  2(0.005  0.002496  0.002496)  0.01998 *

0.505 a  2 4x   2 4x  2 4x     For n = 4: P    sin sin dx  sin 2 dx     a a   a a  a 0.495 a a 0.495 a  0.505 a

0.505 a

2 x 1 8x   sin  a  2 4(4 / a ) a  0.495 a 

2  0.505a a a   0.495a   sin 8 (0.505)     sin 8 (0.495)    a  2 16 16   2 

P  2(0.005  0.002493  0.002493)  0.00028

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Instructor‟s Manual

The probabilities are relatively high for the first and third wavefunctions (as they should be, as the middle of the box is the point where the wavefunctions have the highest magnitude), and relatively low for the second and fourth wavefunctions (as they should be, as the middle of the box is where these wavefunctions have a node).

10.56. (a) E n  (b)

n 2 h 2 12 (6.626  10 34 J  s) 2   3.94  10 72 J 8ma 2 8(0.145kg)(310m) 2

1 mv 2  3.94  10 72 J 2 v

(c) KE 

(3.94  10 72 J )  2  7.37  10 36 m / s 0.145kg

1 1 mv 2  (0.145kg)(44.7m / s) 2  145J 2 2

n 2 (6.626  10 34 J  s) 2  145J 8(0.145kg)(310m) 2 n  6.07  10 36

Note that these values don‟t make sense for a macroscopic situation.

10.58. At high quantum numbers, the positive peaks of the probability waves effectively blend together, mimicking a straight line of constant probability across the box. This is what would be expected for a classical particle bouncing back and forth between two walls of a box, which is consistent with the correspondence principle.

10.60. If the limits are –a/2 to +a/2, let us assume that we still have the same general form for the wavefunction,   A sin Bx  C cos Dx . In this case, however, we have to satisfy the boundary conditions that  = 0 at –a/2 and +a/2. This is probably most easily done by recognizing that the boundaries are simply shifted by half of the system width, making the sine function inappropriate in this case (since the sine function goes to zero at the center of our new, shifted box. Therefore, the simplest solution will be to assume that A = 0 and that the sine function cancels:   C cos Dx . In order for the cosine function to go  3 5 to zero at –a/2 and a/2, Dx must be half-integral values of : , , , etc. Thus, we 2 2 2 a  (2n  1) have D  (2n  1) , or D  .

2 2 a

84

Chapter 1

(2n  1)x , where n is any integer (including a zero). All we need to do now is normalize: Therefore our wavefunction is   C cos

(2n  1)x   (2n  1)x   2 2 ( 2n  1)x dx  1 N   cos   cos dx  1 N  cos a a a    a / 2  a / 2 *

a/2

a/2

2

a/2

a  x 1 2(2n  1)x   a N   sin  1 N 2   0      0   1  a  4 4   2 4(2n  1 / a)  a / 2 2

a 2 2 (This is the same normalization constant for the 1 N2  N 2 a a other particle-in-a-box system.) Therefore, the complete wavefunction is 2 (2n  1)x , n 0, 1, 2, 3, …  cos a a N2

2 nx can be very easily shown to not be eigenfunctions of sin a a the position operator. The position operator x̂ is defined as “x”; that is, multiplication by the x variable. This generates a new function, the sine function times the x variable. Since this is a new function and not the original function times a constant, the wavefunctions are not eigenfunctions.

10.62. The wavefunctions  

10.64. Since the answer must come out in units of length, we are going to assume that the  1  constant part of  should be:  3   a  

 1  (a) r    3  0  a 

1/ 2

e

r / a

 1   r  3   a 

1/ 2

1/ 2

e  r / a  4r 2 dr

 4  3!  24a  1    4 3   r 3 e  2 r / a dr  3  a  2 / a 4  16  a  0

(b) 7.94x10-11m

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Instructor‟s Manual

*

a  2 x   2 x 2 x  x    sin   i sin dx  i  sin sin dx a a x a a a0 a x a 0 a

10.66.

px

 i

2 x x sin cos dx  aa0 a a

 i

2  1 x  2  a 2  a   sin 2   i 2  sin   sin 0  i 2  (0  0)  0 2  a 0 a  / a a  a   

a

a

10.68. 1 ( x )   2 x

2 x 2 2x sin ; 2 ( x )  sin a a a a  2 x

a

  1 ( x ) p 1 ( x )dx

(a) p

0

 2 x  2  2  2 x    dx    sin    sin 2   0 a a a a  x     a

x    x  2   a  cos dx      2     sin  0 a  x  a  a a 





 2  a  x  x  2     2  2    sin  sin dx a  a  a a  0

 

a

 2 2  2   x 1  2 2  2  a    2  2  2x        2     sin  3 3  2 4  a a a      2   a  0 



a

(b) p 2x   2 ( x ) p 2x 2 ( x )dx 0

a 2 2x  2  2  2 2x     dx     sin sin 2  0 a  a  x  a  a

2x    2x  2  2  a    cos      2     sin  dx a  x  a  a  a 0





2 2x  2x   2 2    a       4 2    sin  sin dx 0 a  a  a  a  

 

a

 8 2  2   x 1  8 2  2  a   4 2  2  4x            sin  3  a  0  a 3  2   a 2   a  2 4

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Chapter 1

10.70. A normalization constant is not going to affect the value of the eigenvalue, so let us simply evaluate the operator/wavefunction combination without normalizing:  3 i e  i(3i )e 3i  3e 3i . Therefore, the eigenvalue is 3 , that being the value  of the angular momentum. To determine the average value, set up the expression as follows, using N as the normalization constant:  i

2



p   Ne 3i 0

2

  i  Ne d  i  3i  Ne 3i

*

3i

Ne 3i d . If the wavefunction is

0

properly normalized, the value of the integral is simply 1. Therefore, we have p  3 , which is the same value as the eigenvalue for angular momentum, as it should be.

1 d2 2mE X as  2 allows us to ultimately rewrite the expression into a form 2 X dx  that mimics a one-dimensional Schrödinger equation. If we had defined the second derivative as simply E, the connection to the one-dimensional Schrödinger equation would not have been as obvious. (See also the next exercise.)

10.72. Defining

10.74 

n y y n x n z  2  2 2  2  8  2  2  2   sin x  sin  sin z  2m  x a b c  y z   abc

n yy n yy  n xx  n xx   2 n z z n z 2  8  2 8   sin z  2 sin   sin   sin  sin   2 sin  2m  abc  x a  b c abc a  y b  c 

n yy   2 n x n z   8  2 sin z   sin x  sin abc a b  z c 

 2  nx  2   2m  a2 2

nz  2  c2 2

n yy n x n z n y  8 sin x  sin  sin z  abc a b c b2 2

2

n yy n x n z 8 sin x  sin  sin z abc a b c

n yy n x n z  8 sin x  sin  sin z  abc a b c 

2 2 2 2 2 ny nz  2  nz   2  nx  2 n y  h 2  nx             , where the energy 2m  a 2 8m  a 2 b2 c 2  b2 c 2  2

2

2 2 ny nz  h 2  nx eigenvalue is  2  2  . This verifies that the wavefunction is an  8m  a 2 b c  2

eigenfunction of the three-dimensional Schrödinger equation.

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Instructor‟s Manual

2 2  h 2  n x 2 n y n   2  2  z2  10.76. (a) E smallbox   1 1   8m  1

2 2  h 2  n x 2 n y n   2  2  z2  E l arg ebox   2 2   8m  2 2 2  h 2  n x 2 n y n    2  2  z2  1 1  1  8m  1 E smallbox    4; E l arg ebox  h 2  n 2 n y 2 n 2  1   x2  2  z2  4 8 m 2 2    2

The smallbox has higher energy levels. (b) Yes. The degeneracies of each set of quantum numbers will be the same and will follow Figure 10.13. The energy levels in the small box will not be equal to the energy levels of the large box.

10.78. Figure 10.13 can be consulted to answer this exercise. The first set of quantum numbers for which degeneracy occurs is (1,1,2), whose energy is the same as the sets (1,2,1) and (2,1,1). The first appearance of degeneracy for sets of different quantum numbers (as opposed to simply the rearrangement of the same quantum numbers) is (5,1,1) and its iterations (1,5,1) and (1,1,5) with the set (3,3,3).

2  2 2   2  2  , the wavefunctions are 10.80. For the two-dimensional box: Hˆ   2m  x y  

2 2 n yy ny  n xx 4 h 2  nx sin sin , and the quantized energies are  2 .  ab a b 8m  a 2 b 

10.82. We must evaluate the following integral: *

 8  8 x y z  x y z  x    sin  sin  sin   x   sin  sin  sin dx dy dz a b c  a b c   abc  abc

We can simplify this by separating the wavefunction (including the normalization constant) into three parts, one part for x, one part for y, and one part for z: *

*

*

b c  2  2  2  2 x  x  y   2 y  z   2 z  x    sin   x   sin  dx    sin   sin  dy    sin   sin  dz a a  a  b b   b b  c c   c c  0 0 0  a a

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Chapter 1

The reason we do this is so we can now recognize that the second and third integrals represent normalized one-dimensional wavefunctions, so the integrals are simply 1.

Therefore, we get *

 2  2 x  x  x    sin   x   sin dx . We solved this integral in Example 10.12; it is a a a 0  a a b x  . Finally, rather than repeat everything, by extension we can deduce that y  2 2 c and z  .

2 a

2 2x 2 3x sin and 3 ( x )  sin . If they are a a a a 2 2x  2 3x   dx will equal 0. Using the integrals sin sin a a  a a 

10.84. For the 1-D particle in the box, 2 ( x )  a orthogonal, the integral   0  in appendix 1, a 2 2x  2 3x  0  a sin a  a sin a dx

a

   2 3     2 3       x   5x     x  sin    x   sin  sin   sin   a   a   a    a  2    a  2   a          a     2   2 3    10    a   2 2  3   2            a  a   a  a   0  a  a     0 a

2   0  0  0 a 10.86. Although there are two different wavefunctions in the integral there is also the position operator, x̂. This equation is not identical to equation 10.27.

10.88. (a) Because the wavefunctions are the same, and should be normalized, the value of the integral is exactly 1. (b) Because the wavefunctions are different, and should be orthogonal, the value of the integral is exactly 0. (c) Since the wavefunction is an eigenfunction of the Hamiltonian operator, and the two wavefunctions are the same, the value of this integral is simply the energy of the 16h 2 fourth level, which for a particle-in-a-box is .

8ma 2 (d) Although 2 is an eigenfunction of the Hamiltonian operator, the two wavefunctions inside the integral are different. Therefore, the value of this integral is exactly 0.

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Instructor‟s Manual

(e) Let us assume that these wavefunctions are for the three-dimensional particle-in-abox (because there are three quantum numbers given). All three quantum numbers are the same, so the two wavefunctions are the same wavefunctions, which should be normalized, so the value of the integral should be exactly 1. (f) Let us assume that these wavefunctions are for the three-dimensional particle-in-abox (because there are three quantum numbers given). Not all quantum numbers are the same, however, so these two wavefunctions are different and should be orthogonal. Therefore, the value of this integral is exactly 0. (g) Since the wavefunction is an eigenfunction of the Hamiltonian operator, and the two wavefunctions are the same, the value of this integral is simply the energy of the h2  1 1 1 particle-in-a-box having quantum numbers 1, 1, and 1:  2  2  2 .

8m  a b c  (h) Although 223 is an eigenfunction of the Hamiltonian operator, the two wavefunctions inside the integral are different. Therefore, the value of this integral is exactly 0.

Et Et   10.90. In terms of sine and cosine, e iEt /    ( x)   cos  i sin    ( x) . (Here, we are using     the fact that cos(-x) = cos(x).)

Chapter 11 Quantum Mechanics: Model Systems and the Hydrogen Atom 11.2.

11.4.

n=

n=

1 2p

1 2p

2 2 k so k = ( 2pn ) m = 2 × p × 0.277s-1 × 500.0kg =1514kg / s 2 m

(

)

æ Nö ç866 ÷ è mø

k k so m = = = 0.443kg 2 -1 2 m 2 pn ( ) 2p × 7.04s

(

)

æ1ö æ1ö 2p ç ÷ ( kg) 2p ç ÷ ( kg) 2pn m 1 èsø èsø 11.6. a = = = = 2 ; Yes, in Eqn. 11.8, the units in the exponent must 2 kg × m J ×s m ×s 2 s cancel each other. X is in meters so they do.

90

Chapter 1

11.8.

 2 d 2   2 2 2 mx 2    E , substitute the definition to  to get Starting with  2  2m dx  2 2 2 2 2   d  x   2     E  .

Dividing both sides of the equation by yields    2 2m  2m  2m dx

 d2 2mE 2 2  2   x     2  . Finally, bringing all terms to one side of the equation and factoring   dx  d 2   2mE  out the  from two terms, we get   2   2 x 2   0 , which is our ultimate result.

2 dx     2mE  11.10. (n  1)(n  2)c n  2  2nc n   2   c n  0   

c n2 

2mE   2mE c n  c n c n    2n  2  2      (n  1)(n  2) (n  1)(n  2)

2nc n 

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Instructor‟s Manual

11.12. For any two adjacent vibrational levels labeled n and n+1, the difference in energy is: E(n+1) – E(n) = (n+1+½)h – (n+½)h = nh + h + ½ h – nh – ½h. The two nh terms cancel, as do the ½h terms. All that remains is the h term; therefore, we can E = h for the transition between any two adjacent vibrational levels. E = (6.62610−34 Js)(6.0001013 s−1) = 3.97610−20 J.

11.14. (a) E = h

3.00  10 8 m/s  5.00  10 6 m (c) This wavelength is in the infrared region of the  6.000  1013 s -1 spectrum. (d) Even in the early 20th century, technology existed to be able to detect infrared electromagnetic radiation. c

(b)  



11.16. In constructing the harmonic oscillator wavefunctions, let us simply use N2 and N3 to denote the collection of constants for the second and third wavefunctions, respectively. Thus,

2  N 2 H 2 ( 1 / 2 x)e x / 2 and 3  N 3 H 3 ( 1 / 2 x)e x / 2 . Therefore, we have to evaluate 2

 N H ( x)e



1/ 2

2

x 2 / 2

2

2

 N H ( x)e *

1/ 2

3

x 2 / 2

3

dx . Since there is no imaginary term in the



wavefunction, the complex conjugate has no effect and all functions are simply multiplying each 

other. Therefore, we can simplify this integral to N 2 N 3  H 2 ( 1 / 2 x) H 3 ( 1 / 2 x)e x dx . Looking 2



at the integrals involving Hermite polynomials in Table 11.2, we see that this integral has the same form, and because there are two different Hermite polynomials in the expression, the integral‟s value is exactly 0. Therefore, the two wavefunctions are orthogonal. 

11.18.

x N

2

  (x )  x̂   (x )dx . For the quantum harmonic oscillator, all even numbered ‟s are *



even and all odd numbered ‟s are odd. Since (odd)x(odd)=even and (even)x(even)=even, the product of (x) and *(x) is always even. x̂ is defined to be multiplication by x, which is odd, so we have the product of an even function and an odd function integrated over symmetric limits.

This integral is 0.

 xe



11.20. N

2

x 2 / 2

 xe *

x 2 / 2



dx  1

N



2

x e 2

x 2

dx  1 Now we apply one of the integrals in the



integral table from the appendix:

1   N   2  3  2

1/ 2

1

 3  N  2   

wavefunction is 1  2

92

1/ 2

N 2

2

1/ 2

 3     

1/ 2

 3     

1/ 4

Therefore, the complete

1/ 4

xe x / 2 , which can be shown to be algebraically equivalent to 2

Chapter 1

   1  1/ 2 x 2 / 2 , which is the form of the wavefunction obtained when using     (2 x)e    2 equation 11.19.

1/ 4

1/ 2

11.22. (a) This integral is identically zero because the two different wavefunctions are orthogonal. (b) This integral is identically zero because the overall expression inside the integral is an odd function. (c) This integral is not necessarily identically zero. (d) This integral is identically zero because the two different wavefunctions are orthogonal. (e) There are limits on the integral. This integral is not necessarily zero. (f) The value of this integral is indeterminate, because it depends on the exact mathematical form of the potential energy operator.

11.24. Drawings are left for the student. As n increases, the particle has a very high probability of being found near the turning points and a very low probability of being found in the middle. At higher n‟s, the probability will become increasingly continuous.

2 1  11.26. For a simple harmonic oscillator, 0    e x / 2 , and V̂  kx 2 . To calculate the average 2  value of V, we must evaluate the following integral:

1/ 4

1/ 4      1 / 4  1  2 2     * V   0 V̂ 0 dx      e x / 2  kx 2    e x / 2 dx    2           

1   k  2 

1/ 2 

2  x  x e dx 2



We can use an integral from appendix 1 in the textbook, but note that the limits on the integral are different. If we recognize that f(x) is an even function, then we know that the integral of f(x) from −∞ to ∞ is equal to 2  the integral of f x from 0 to ∞.

1/ 2 

1   k  2 

x e



2  x 2

1  dx =  k  2 

Since k 4π2ν2m and  

1/ 2



2 x e 2

0

 x 2

 dx k  

1/ 2

 1   1 / 2       4    

2m 1 ,we can determine that V = hn which is one−half of the zero  4

point energy.

11.28. The number of nodes in a harmonic oscillator wavefunction is just the quantum number, n.

11.30. K 

1 1 2 2 m1 x 1  m 2 x 2 ; Using equations 11.21 and 11.22: 2 2

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Instructor‟s Manual

2

2

 m1  2 1  m2  2 1  q  m 2   q K  m1  2  m1  m 2  2 m  m 2   1 

2 2 m m  1  m1 m 2  m1  m 2  1  m1 m 2  2 1 2  m1 m 2    q q   1 2   q 2  2  m1  m 2 m1  m 2  2  m1  m 2  m1  m 2  2  m1  m 2 

(2.435  10 24 kg)( 2.995  10 22 kg) 11.32. = 2.9591022 kg.

24 22 2.435  10 kg  2.995  10 kg 11.34. Since we know that the frequency of vibration is inversely proportional to the square root of the reduced mass, we can make the following ratio:



1



and  * 

1

*

leads us to the following ratio:

1/   *   . Now we need only   * 1/  *

calculate the reduced masses of the O−H and O−D bonds. Since the expression we derived is a ratio, it doesn‟t matter what ultimate units the reduced mass is in, so let us simply use values in grams: (OH) = (1)(16)/(1+16) = 0.941, while (OD) = (2)(16) (2+16) = 1.778. Therefore, we 3650 cm -1 1.778 get e   1.374 Solve for  * :  *  2656 cm -1 . * 0.941  11.36. Equation 11.39 shows that the quantized values for angular momentum are not dependent on mass, only a quantum number and Planck‟s constant divided by 2. (Note that the m variable in the equation is the quantum number, not mass!)

11.38. m 

1

2 equal to 0.

2p

æ 1

e im  ; To demonstrate orthogonality, we need to show that the following integral is

öæ 1

ö

1 2 p 2i -i

1 2p i

ò çè 2p e j ÷øçè 2p e j ÷ødj = 2p ò e j j dj = 2p ò e j dj = 2ip [cosj + isin j ] -i

2i

0

=

0

1 [(1+ 0) - (1+ 0)] = 0.

2ip

They are orthogonal.

94

0

1

2p 0

Chapter 1

11.40. The angular momenta of the first five energy levels are simply 0, 1, 2, 3, and 4 . The energies are given by the formula E 

m2 2 : 2I

0 2  (6.626  10 34 J  s) 2 E (m  0)  0 (2 ) 2 (2)(9.109  10 31 kg) (1.51  10 10 m) 2 E (m  1) 

12  (6.626  10 34 J  s) 2  2.68  10 19 J 2 31 10 2 (2 ) (2)(9.109  10 kg) (1.51  10 m)

E (m  2) 

2 2  (6.626  10 34 J  s) 2  1.07  10 18 J (2 ) 2 (2)(9.109  10 31 kg) (1.51  10 10 m) 2

3 2  (6.626  10 34 J  s) 2 E (m  3)   2.41  10 18 J 2 31 10 2 (2 ) (2)(9.109  10 kg) (1.51  10 m) E (m  4)  11.42 (a) (b) E 

4 2  (6.626  10 34 J  s) 2  4.28  10 18 J 2 31 10 2 (2 ) (2)(9.109  10 kg) (1.51  10 m)

mh m(6.626 ´10-34 J × s) 2 = 600 kg × m /s = 2p 2p

\m » 5.69 ´1036

(5.69  10 36 ) 2 (6.626  10 34 J  s) 2  113 J (2 ) 2 (2)(25 kg)(8 m) 2

L2 (600 kg  m 2 / s) 2   113 J , the same 2I 2(25 kg)(8 m) 2 amount – but this shouldn‟t be surprising, since we calculated the „quantum−mechanical‟ amount of energy using the same values but in a more roundabout fashion.

Classically, the child would have an energy of

11.44. (a) E (m  1)  E (m) 

(m  1) 2  2 m 2  2 (2m  1) 2   2I 2I 2I

(b) If E(1) – E(0) = 20.7 cm−1, we can use the formula from part a to determine I, then calculate E(2) – E(1) using that value of I. (2  0  1)(6.626  10 34 J  s) 2 20.7 cm -1  I  2.686  10 70 J 2 s 2 /cm -1 (we are not reducing 2 (2) (2)I the units, since we‟re simply using this quantity to recalculate an energy difference in cm−1 units) Now, calculating E(2) – E(1):

(2  1  1)(6.626  10 34 J  s) 2 E (2)  E (1)   62.1 cm -1 .

2  70 2 2 -1 (2 ) (2)(2.686  10 J s /cm )

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Instructor‟s Manual

(c) This is not a great agreement with the experimental value of 41.4 cm−1, suggesting that the 2−D rotational model may not be the best model for HCl. 11.46. For 2−D rotational motion, E 

m 2 2 ; If m=0, E=0.

2I

11.48. The quantized angular momentum is independent of mass (see eqn 11.5). This is counter to the concepts of classical mechanics.

11.50. he square root of equation 11.45 cannot be taken analytically. The first and last terms are second derivatives and can be considered perfect squares, but the second term isn‟t the appropriate cross term to allow us to take the square root of the entire expression.

11.52. The average value of r, <r>, would not have any meaning for Y2−2 because the spherical harmonic does not contain r as a variable.

  2  1  2  105   sin 2  cos e  2i  11.54. (a) Lˆ2 3, 2   2  2  cot  2 2   sin    4    Let us determine each derivative singly, then combine the three terms and simplify.

The first term:



 105   105 105 3, 2    2 sin  (cos  ) cos   sin 3  e  2i  2 sin  cos 2   sin 3  e  2i  4 4  4 





2 105 3, 2  2 cos 3   4 sin 2  cos   3 sin 2  cos  e  2i 2 4 





The second term: cot 

 105 105 3, 2  cot  2 sin  cos 2   sin 3  e  2i  2 cos 3   sin 2  cos  e  2i  4 4









The third term:

 1 2 1  105 105 2  2 i 2    4   sin  cos  e  (  2 i ) cos e  2i 3 ,  2 2 2 2   4 sin   sin   4  Combining these terms:

 105    2  2 cos 3   4 sin 2  cos   3 sin 2  cos   2 cos 3   sin 2  cos   4 cos  e  2i   4 



This simplifies to

96



Chapter 1

 105    2  4 cos 3   8 sin 2  cos   4 cos  e  2i   4 





Factoring a cosine out of the first two terms:

 105    2  cos  4 cos 2   8 sin 2   4 e  2i   4 





If we recall that sin2 + cos2 = 1, then we can substitute the 4 inside the parentheses with 4sin2 + 4 cos2 but don‟t forget to distribute the minus sign:

 105    2  cos  4 cos 2   8 sin 2   4 sin 2   4 cos 2  e  2i   4 





The 4cos2 terms cancel, and we can group the sin2 terms together to get  105   105    2  cos   12 sin 2  e  2i   12 2  sin 2  cos e  2i   12 2 3, 2  4   4 





Therefore, this wavefunction is an eigenfunction of the square angular momentum operator with an eigenvalue of 12 2 – as expected from the eigenvalue formula. (b) This determination is more straightforward because there is only one term in the wavefunction that depends on : the exponential function. We have:

 i

  105     105 2  2 i 2  21    i (2i )   23, 2 3, 2  i sin  cos  e sin  cos  e   4     4    Therefore, this wavefunction is an eigenfunction of the 2−D angular momentum operator with an eigenvalue of  2 – as expected from the eigenvalue formula. (c) To determine the energy eigenvalue, technically we have to evaluate the entire three−part Lˆ2 derivative in equation 11.43. However, if we recognize that Hˆ  , we can use the answer 2I to part a because we have already evaluated the eigenvalue to the L2 operator. Thus, we have





1 ˆ2 1 12 2 Hˆ 3, 2  L 3, 2  12 2  – as expected from the eigenvalue formula.

2I 2I 2I (  1)(  1  1) 2 ()(  1) 2 (2  2) 2   11.56. (a) E (  1)  E ()  .

2I 2I 2I

(b) Use the formula from part a to determine the value of I, then use this value to determine the energy difference between the new quantum levels. We have:

(2  0  2)(6.626  10 34 J  s) 2 I  5.370  10 70 J 2  s 2 / cm -1 (we are not reducing 2 (2 ) (2) I the units, since we‟re simply using this quantity to recalculate an energy difference in cm−1 units) 20.7 cm -1 

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Instructor‟s Manual

(2  1  2)(6.626  10 34 J  s) 2  41.4 cm -1 . This 2  70 2 2 -1 (2 ) (2)(5.370  10 J s /cm ) is exactly the same as the experimental value of 41.4 cm−1, suggesting that the 3−D rotational model is a good model for HCl. Now, calculating E(2) – E(1): E (2)  E (1) 

11.58. For the four l values in exercise 11.57: For = 5: L2 = l l+1)h2/(2 = 5(5+1)(6.62610−34 Js)2/(2 = 3.3410−67 J2s2. Therefore, the angular momentum is the square root of this: L = 5.7810−34 Js. ml=±5, ±4, ±3, ±2, ±1, 0

Lz = ml = ±5 , ±4 ,±3 ,±2 ,±1 , 0

For l = 6: L = l l+1)h /(2 = 6(6+1)(6.62610−34 Js)2/(2 = 4.6710−67 J2s2. Therefore, the angular momentum is the square root of this: L = 6.8310−34 Js.

2



2

ml=±6, ±5, ±4, ±3, ±2, ±1, 0 Lz = ml =±6 , ±5 , ±4 ,±3 ,±2 ,±1 , 0 For l = 7: L2 = l l+1)h2/(2 = 7(7+1)(6.62610−34 Js)2/(2 = 6.2310−67 J2s2. Therefore, the angular momentum is the square root of this: L = 7.8910−34 Js. ml=±7, ±6, ±5, ±4, ±3, ±2, ±1, 0

Lz = ml =±7 , ±6 , ±5 , ±4 ,±3 ,±2 ,±1 , 0

For l = 8: L2 = l l+1)h2/(2 = 8(8+1)(6.62610−34 Js)2/(2 = 8.0110−67 J2s2. Therefore, the angular momentum is the square root of this: L = 8.9510−34 Js. ml=±8, ±7, ±6, ±5, ±4, ±3, ±2, ±1, 0 Lz = ml =±8 , ±7 , ±6 , ±5 , ±4 ,±3 ,±2 ,±1 , 0 2 11.60. If Lz=2  , cos    .816;   35.3 6 If Lz=  , cos  



If Lz=0  , cos  

0

6

 .408;   65.9 

6

 0;   90 

If Lz= −1  , By geometry, =90°+ (90−65.9)°= 114.1° If Lz= −2  , By geometry, =90°+ (90−35.3)°= 144.7° 11.62. (a) +2 (b) +5 (c) +25 (d) +61 (e) +53 (f) +86 (g) +91 (h) +105.

11.64. V  (6.673  10

11

(9.109  10 31 kg)(1.672  10 -27 kg) N  m / kg )   1.92  10 57 J -10 (0.529  10 m) 2

2

11.66. The only difference between the two definitions of the Bohr radius is that one uses the mass of the electron while the other uses the reduced mass of the hydrogen atom. Therefore, we need the (1.674  10 27 kg)(9.109  10 -31 kg) reduced mass of the hydrogen atom, . It is = 9.10410−31  27 -31 1.674  10 kg  9.109  10 kg kg. So, the two values for the Bohr radius can be calculated as

98

Chapter 1

4 0  2 4 (8.854  10 12 C 2 / J  m)(6.626  10 -34 J  s) 2 a0    5.29294  10 11 m 2 2 31 -19 me e (2 ) (9.109  10 kg)(1.602  10 C) a

4 0  2 4 (8.854  10 12 C 2 / J  m)(6.626  10 -34 J  s) 2   5.29585  10 11 m 2 2 31 -19 e (2 ) (9.104  10 kg)(1.602  10 C)

As you can see, they are not too different (about 0.05% difference).

11.68. Because D has a slightly larger reduced mass (9.10710−31 kg) than H does (9.10410−31 kg), the positions of the Balmer series will be slightly higher by that proportion. First, let us convert the positions into cm−1 units: 1 10 7 nm   15,230 cm -1 656.5 nm 1 cm

1 10 7 nm   20,560 cm -1 486.3 nm 1 cm

1 10 7 nm   23,030 cm -1 434.2 nm 1 cm

1 10 7 nm   24,370 cm -1 410.3 nm 1 cm

Next, we multiply each value by the fraction

9.107  1.000329 : 9.104

15,230 cm−1  1.000329 = 15,240 cm−1

20,560 cm−1  1.000329 = 20,570 cm−1

23,030 cm−1  1.000329 = 23,040 cm−1

24,370 cm−1  1.000329 = 24,380 cm−1

These shifts are not large, but should be detectable experimentally.

11.70. E 

 e 4  (1.602  10 19 C) 4 (9.104  10 31 kg)   2.18  10  22 J 2 2 2 8 0 h n  2 C2   6.626  10 34 J  s 100 2 8 8.854  10 12 Jm 





11.72. a) for the f subshell, l=3. The total angular momentum =

(  1)  12

b) Since this is still for the f subshell, the total angular momentum is the same as in part a.

11.74. N 2    e r / a  r 2 sin drdd  1 ; This is a triple integral, where the limits on  are 0 to 2, the 2

limits on  are 0 to , and the limits on r are 0 to ∞. We can separate this out: 2







N  d sin d r e 2

2

0

0

2

r / a

 dr  1; Using one of the integrals in Appendix 1 for the r integral,

0

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Instructor‟s Manual

     2a 3  2!  2  2 2    1 N  0  cos  0  N (4)  2 3  8        a   N

1 a 3

This term is consistently found in all of the wavefunctions in table 11.4.

11.76. For a single H atom:

Z 2e4  (1) 2 (1.602  10 19 C) 4 (9.109  10 31 kg)    2.179  10 18 J . Therefore, 2 2 2 -12 2 2 34 2 2 8 0 h n 8(8.854  10 C / Jm) (6.626  10 J  s) 1 for a mole of atoms, we multiply this number by Avogadro‟s number: E

−2.17910−18 J  6.0221023 = −1.312106 J = −1312 kJ. For a single He+ ion:

Z 2e4  (2) 2 (1.602  10 19 C) 4 (9.109  10 31 kg)    8.716  10 16 J Therefore, 2 2 2 -12 2 2 34 2 2 8 0 h n 8(8.854  10 C / Jm) (6.626  10 J  s) 1 + for a mole of He ions: −8.71610−18 J  6.0221023 = −5.249106 J = −5249 kJ. E

The difference is caused by the +2 charge in the helium nucleus.

1/ 2

1/ 2

 1000   1000   e 10 r / 0.529 A    e 10 r / 0.529 A  r 2 sin drdd 11.78. The setup is: P    3  3    (0.529A)    (0.529A)  This is a triple integral, where the limits on  are 0 to 2, the limits on  are 0 to , and the limits on r are 0 to 0.1Å. Thus, we have 2   1000 0.1A 2 20 r / 0.529 A  r e P   d   sin d   dr As shown in the text (see example 11.24), 3     (0.529A)  0 0 0 the  and  integrals collectively equal 4. Thus, we have

 1000 0.1A 2 20 r / 0.529 A  r e P  4   dr 3     (0.529A)  0 Using the integral table in the appendix, we can solve the integral and evaluate:

100

Chapter 1

0.1A

     20 r / 0.529 A  1000 r2 2r 2   P  4   e    3   (20 / 0.529A) (20 / 0.529A) 2 (20 / 0.529A) 3    (0.529A)     0



P  4 2150.221A

-3



  20 ( 0.1) / 0.529  (0.1A) 2 (0.529A) 2(0.1A)(0.529A) 2 2(0.529A) 3      e   20 400 8000   

 2(0.529A) 3    e 0  0  0  8000   P = 4(2150.221A−3)[0.022806(−0.0002645A3 – 0.0001399A3 – 0.00003701A3)] + 0.00003701A3 P = 0.7280, or 72.80%. The difference is that the +10 charge on the neon nucleus pulls the single electron much closer to the nucleus than the +1 charge on the hydrogen nucleus, resulting in a much higher probability that the electron is within 0.1Å of the nucleus.

11.80. No. The degeneracies of the shells do not change. The energies of the shells do change.

11.82. (a)  has an angular node when the value of 3cos2−1=0. This occurs when  1   . The angular nodes occur at 54.7° and 125.3°. = cos 1   3  (b) The drawing should look like the dz2 boundary plot in Figure 11.22.

11.84. a =

4pe0 2 4p (8.854 ´10-12 C2 / J × m)(6.626 ´10-34 J × s)2 = = 5.29585´10-11 m 2 2 -31 -19 2 me (2p ) (9.104 ´10 kg)(1.602 ´10 C)

11.86. According to equation 11.67, 2 p x 

1 2



2 p 1



 2 p1 . Using the forms of the wavefunctions

from Table 11.4, we have 1  1  2Z 3   3 2 p x  2  8  a 

1 1  2Z 3   3  2 8  a 

1/ 2

1/ 2

Zr  Zr / 2 a 1  2Z 3   i e sin e   3  a 8  a 

1/ 2

 Zr  Zr / 2 a e sin e i   a 

Zr  Zr / 2 a e i  e  i e sin  e i  e i Now we use the fact that cos   and a 2

1  2Z 3   3 substitute: 2 p x  4 2  a 



1/ 2



Zr  Zr / 2 a e sin  cos  . a

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Instructor‟s Manual

Also according to equation 11.67, 2 p y  

i 2



2 p 1



 2 p1 . Using the forms of the

wavefunctions from Table 11.4, we have 1/ 2 1/ 2  i  1  2Z 3  Zr  Zr / 2 a 1  2Z 3  Zr  Zr / 2 a  i  3  2 p x   e sin e   3  e sin e i   a 8  a  a 2  8  a  

i 1  2Z 3   3  2 8  a 

1/ 2

Zr  Zr / 2 a e i  e  i  e sin  e i  e i Now we use the fact that sin   and a 2i



1  2Z 3   3 substitute: 2 p y  4 2  a  negative sign canceled.

1/ 2



Zr  Zr / 2 a e sin  sin  . Note how the two i terms and the a

 e2 . To calculate the average 4 0 r value of the potential energy of the 1s orbital, we must calculate the following integral:

11.88. The potential energy operator for the hydrogen atom is: V̂ 

  1 1 / 2  r / a   e 2   1 1 / 2  r / a  2  r sin drdd . This is a  V     1s V̂ 1s d       3  e e   a   4 0 r   a 3         triple integral, where the limits on  are 0 to 2, the limits on  are 0 to q, and the limits on r are 0 to ∞. We can separate this out: *

2

   e 2  1  2   1  3   d sin d r 2   e  r / a dr; Using an integral from appendix 1, we find V   r  4 0  a  0 0 0 that:       e 2  4  1!   e2  3  V    2   4 0  a   2   4 0 a a      This is equal to −4.36x10−18J. The total energy of the 1s orbital of the hydrogen atom is −2.178x10−18J. Therefore, the potential energy equals twice the total energy.





11.90. The combinations are:





3d 2  

i 3d 1  3d 1 2





3d 4  

i

3d1 

1 3d 1  3d 1 2

3d 3 

1 3d 2  3d 2 2

3d 5  3d 0

102



2



3d  2



 3d  2



Chapter 1

The numerical labels on these composite wavefunctions are arbitrary; they are actually given Cartesian coordinate labels that reflect their distribution in three-dimensional space. Note that one of the wavefunctions is a pure eigenfunction, rather than a combination of two eigenfunctions.

11.92. Graphs are left to the student. As n increases the most probable distance of the electron from the nucleus also increases.

Chapter 12 Atoms and Molecules 12.2.

A particle with a magnetic moment moving through an inhomogeneous magnetic field will have a force exerted on it and will be deflected. Because the Ag ions were deflected in two directions, it is understood that the Ag particles have two distinct magnetic moments. The magnetic moment is proportional to the angular momentum. But this experiment shows that the Ag atoms have two distinct angular momentums. Eventually this angular momentum was understood to be an intrinsic property of the electron itself. Just like the z−component of the angular momentum from the motion of the electron can go from –l to +l in integral steps, the z−component of the spin angular momentum can go from –s to +s in integral steps. Since the number of allowed states of the z−component of the spin angular momentum is equal to 2s+1 and this is equal to 2 states, s must be ½.

12.4.

If s=3/2 for an electron, we would expect a magnetic field to split a beam of silver atoms into 4 parts corresponding to ms values of +3/2, +1/2, −1/2, −3/2

12.6.

For a 3d orbital of the hydrogen atom, n=3 and l=2. Possibilities for ml are −2,−1,0,1,2. All possible combinations of quantum numbers are: (3,2,−2,+1/2) (3,2,−2,−1/2) (3,2,−1,+1/2) (3,2,−1,−1/2) (3,2,0,+1/2) (3,2,0,−1/2) (3,2,+1,+1/2) (3,2,+1,−1/2) (3,2,+2,+1/2) (3,2,+2,−1/2)

12.8.

One would think that the two spin functions  and  would be orthogonal. Consider the two wavefunctions 1s and 1s. The spatial parts of the wavefunctions are the same, but the spin parts are different. The only way these two wavefunctions can be orthogonal – as different wavefunctions are required to be – is if the spin wavefunctions are themselves orthogonal; otherwise, the normalization requirement is satisfied.

12.10. Yes. The  means that it is a “spin−up” electron. This is intrinsic to the particle and does not affect the symmetry of the wavefunction.

12.12. (a) The potential energy is negative because the attraction between the positive and negative charges leads to a decrease in energy. (b) The potential energy is positive because the repulsion between two positive charges leads to an increase in energy.

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(c) The potential energy is negative because the attraction between opposite magnetic poles leads to a decrease in energy. (d) The potential energy is negative because the attraction between the two masses leads to a decrease in energy. (e) The potential energy is positive because the rock is higher than the base of the cliff and it has extra gravitational potential energy.

12.14. Hˆ  





2 2 3e 2 3e 2 3e 2 e2 e2 e2  e #1   e2# 2   e2#3       2 4 0 r1 4 0 r2 4 0 r3 4 0 r12 4 0 r13 4 0 r23

The last three terms are what make this Hamiltonian non−separable.

12.16. Acceptable wavefunctions are linear combinations (i.e. sums and/or differences) of individual spatial wavefunctions because this way the eigenvalue energy is the correct value. If the wavefunction were constructed as the product of spatial wavefunctions, then the resulting energy eigenvalue would be the product of the energies of the individual wavefunctions, which is not the correct energy eigenvalue for the atom.

12.18. Let us start with the acceptable wavefunction for He:

 (1,2) 

1 2

(1s1 )(1s 2  )  (1s1  )(1s 2 ) . If we exchange the two electrons, then the “1”

and “2” labels switch to get  (2,1) 

1 2

(1s 2 )(1s1  )  (1s 2  )(1s1 ) . If we switch the −

and −containing functions in each term (which we can do because multiplication is commutative), we get  (1,2) 

1 2

(1s1  )(1s 2 )  (1s1 )(1s 2  ) . The first term in this

wavefunction is the same as the second term in the original wavefunction, while the second term in this wavefunction is the same as the first term in the original wavefunction.

Algebraically, since (a – b) = –(b – a), we have shown that (2,1) = -(1,2), as required by the Pauli principle.

12.20. The wavefunction in example 12.6 is antisymmetric. None of the rows or columns are identical to each other. Therefore it does not violate the Pauli Exclusion Principle.

12.22. The Slater determinant for the Li− is isoelectronic to that of Be:

1s1 1s1  (Li - ) 

2s1

2s1

1 1s 2  1s 2  2s 2  2s 2  24 1s 3  1s 3 2s 3  2s 3 1s 4  1s 4  2s 4  2s 4 

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Chapter 1

1s1 1s1 12.24. (a)  (Be) 

2s1

2s1

1 1s 2  1s 2  2s 2  2s 2  24 1s 3  1s 3 2s 3  2s 3 1s 4  1s 4  2s 4  2s 4  1s1

 (B) 

1 120

1s1 

2 s1

2 s1 

2 p x ,1

1s 2 1s 2 

2 s 2

2s 2 

2 p x , 2

1s3 1s3 

2 s3

2s3 

2 p x ,3 . Keep in mind that, because the

1s 4 1s 4 

2 s 4

2s 4 

2 p x , 4

1s5 1s5 

2 s5

2s5 

2 p x ,5

six p orbitals are degenerate, the last column could have been labeled 2px, 2py, 2py, 2pz, or 2pz. (b) There are six possible determinants for the C atom that have the two unpaired p electrons having the same spin. There are also six possible determinants for the F atom, depending on which spin−orbital is not used (analogous to the wavefunction for Be in part a above).

12.26. The 1s and 1s orbitals are spherically symmetric. When these are operated on by the L̂2tot operator (Equation 11.45 − which looks for changes in  and  only), the resultant answer is 0.

So, equation 12.9 is an eigenfunction of equation 11.45 with an eigenvalue of 0.

12.28. If an electron had three possible values of ms, , , and , the Slater determinants for Li and Be would be:

 (Li) 

1 6

1s1 1s1 1s1  1s 2  1s 2  1s 2  1s 3  1s 3 1s 3 

1s1 1s1 1s1   (Be) 

2s1

1 1s 2  1s 2  1s 2  2s 2  The final column could also be  or .

24 1s 3  1s 3 1s 3  2s 3 1s 4  1s 4  1s 4  2s 4 

12.30. The ground state electronic configurations are as follows: (a) Li: 1s2,2s1 (b) N: 1s22s22p3 (c) Ne: 1s22s22p6 (d) Sc: 1s22s22p63s23p64s23d1 For s orbitals, ml=0, ms= ±1/2 For p orbitals, ml= −1,0,+1, ms= ±1/2 For d orbitals, ml= −2,−1,0,+1,+2, ms= ±1/2

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Chapter 1

12.32. (a) Li will have 6 possibilities for its unfilled valence shell: 2px1, 2px1, 2py1, 2py1, 2pz1, and 2pz1. (b) C will have six possibilities for its unfilled valence shell: 2px12py1, 2px12pz1, 2py12pz1, 2px12py1, 2px12pz1, and 2py12pz1. (c) K will have six possibilities for its unfilled valence shell: 4px1, 4px1, 4py1, 4py1, 4pz1, and 4pz1. (d) Be will only have one possibility for its excited state, since both electrons have been excited to an s orbital: 3s1 3s2. (e) U will have 6 possibilities for its unfilled valence shell: 7px1, 7px1, 7py1, 7py1, 7pz1, and 7pz1. Note the similarity in the answers for Li, K, and U. 12.34. The ground−state harmonic oscillator wavefunction is found in Chapter 11. *

1/ 4 1 / 2    1 / 4 x 2 / 2    4   x 2 / 2 4 x 2 e  cx  e dx  c x              e dx .         



According to the table in the Appendix, there is an appropriate form to use if the integral limits are 0 to  . Since the function in the integral is even, we can split the interval and multiply the value of the integral by 2. We get:

   2  c   



1/ 2

 1  3     3 2    2     

1/ 2

The two square root terms cancel. Simplifying:

3c , where c is the anharmonicity constant.

4 2

12.36. According to perturbation theory, the energy of the particle is: E  E ( 0)  E (1) . We will assume that Ĥ'  kx 2 . To calculate the average energy of the particle using the lowest−energy wavefunction we need to evaluate the following integral: a  2 nx  2  2 nx   kx  dx E (1)    sin sin   a  a a a 0   

 

      3   a 2k 2 2k  x 1  2nx x 2nx   x2 2 nx  x sin dx    sin  cos 3  2 a 0 a a  6   n  a a  n  n       4  8     4     a   a a         

a

0

 2k  a 3 a3 a 3   ka 2   0  2 2  0  0  2 2   a  6 3 4n   4n   

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Instructor‟s Manual

12.38. To determine a3, follow the same procedure used in Example 12.10 but use 1 and 3 for the particle-in-a-box: a

 3  kx  1dx *

a3 

0

E1  E 3

a



 0

2 3x 2 1x sin  kx  sin dx a a a a  12 h 2 3 2 h 2     2 2  8 ma 8 ma  

2k 3x 1x x sin sin dx  a 0 a a a



 h2     2  ma  

Again, we use the trigonometric identity sinaxsinbx = ½[cos(a−b)x – cos(a+b)x] and substitute:

2kma 1  2x 4x     x cos  x cos dx This expression can be integrated: 2 2 0 a a  h a

a3  

a

kma  a 2 2x ax 2x a2 4x ax 4x    2  2 cos  sin  cos  sin  Evaluating at the limits: 2 a  a a 4 a 0 h  16

a3  

 a2  kma  a 2 a2 a2 0 a2 (1)  (0)  (1)  (0)  (1)  0  (1)  0    0 .   2 2 2 2 2 h  4  16 4 16  4 

All of the terms cancel, so the value of a3 is zero.

12.40. The integral we need to evaluate is

1 2     8  a 3  r , ,

1/ 2

r r / 2a  1  e cos   eEr cos    3  a  a 

1/ 2

e  r / a  r 2 sin drdd (Note that we have

the 2pz orbital as one wavefunction and the 1s orbital as the other wavefunction.) Everything is multiplicative, so we can separate this integral into three integrals as follows: 2





2eE  d   cos 2  sin d   r 4 e 3r / 2 a dr . The integral over  is simply 2. The integral 8a 4 0 0 0 over  is easy to evaluate as a power function of cosine: 

1

 cos  sind   3 cos  2

0

3



1 2   (1  1)  0 3 3

The integral involving r can also be evaluated using a formula from the appendix: 

4 3 r / 2 a  r e dr  0

108

4!

24  32a 5 256a 5 .   243 81 (3 / 2a) 5

Chapter 1

Combining the three integral solutions with the collection of constants, we get that

2eE 2 256a 5 128 2eaE  2     .

3 81 243 8a 4

E1 

(c   c  ) Hˆ (c   c  )d  12.42. Starting with , we distribute the operator and  (c   c  ) (c   c  )d *

a ,1

1

a,2

2

a ,1

1

a,2

2

*

a ,1

1

a,2

2

a ,1

1

a,2

2

multiply through to get the individual integrals:

 (c   c  ) ( Hˆ c   Hˆ c  )d =  (c   c  ) (c   c  )d *

a ,1

1

a,2

2

a ,1

1

a ,1

1

a,2

2

*

a ,1

1

a,2

2

a,2

2

(c  ) Hˆ c   (c  ) Hˆ c   (c  ) Hˆ c   (c  ) Hˆ c  d    (c  ) c   (c  ) c   (c  ) c   (c  ) c  d *

a ,1

*

1

a ,1

1

a,2

a ,1

*

a ,1

1

*

2

1

a ,1

*

a ,1

1

a,2

2

*

1

a,2

2

a,2

2

*

a ,1

1

a ,1

a,2

2

*

1

a,2

2

a,2

2

a,2

2

The integral sign can be distributed through to all of the terms:

 (c  ) Hˆ c  d   (c  ) Hˆ c  d   (c  ) Hˆ c  d   (c  ) Hˆ c  d   (c  ) c  d   (c  ) c  d   (c  ) c  d   (c  ) c  d *

a ,1

*

1

a ,1

1

a,2

a ,1

*

a ,1

*

2

1

a ,1

*

1

a ,1

1

a,2

2

*

1

a,2

2

a,2

2

*

a ,1

1

a ,1

a,2

2

*

1

a,2

2

a,2

2

a,2

2

Now the constants can be brought outside the integral signs:



2 * * * 2 * c a ,1  1 Hˆ 1 d  c a ,1 c a , 2  2 Hˆ 1 d  c a ,1c a , 2  1 Hˆ 2 d  c a , 2  2 Hˆ 2 d

c a ,1  1 1 d  c a ,1 c a , 2  2 1 d  c a ,1c a , 2  1 2 d  c a , 2  2 2 d 2

*

*

*

2

*

At this point the definitions from equation 12.28 can be applied, and the expression becomes

c a ,1 H 11  c a ,1 c a , 2 H 12  c a ,1 c a , 2 H 12  c a , 2 H 22 2



c a ,1 H 11  2c a ,1 c a , 2 H 12  c a , 2 H 22

2

c a ,1 S11  c a ,1 c a , 2 S12  c a ,1 c a , 2 S12  c a , 2 S 22 2

2

2



2

c a ,1 S11  2c a ,1 c a , 2 S12  c a , 2 S 22 2

2

This is equation 12.29.

12.44. First, we need to normalize the function. The integral is

N 2  (e  kr )(e  kr )r 2 sin drdd  1 The  and  parts integrate to yield a combined total of r , ,

4. Now we need to integrate the r integral: 4N

 2

r e 2

 2 kr

dr  1. This integral has a known

0

form (see the appendix), and we can substitute:

 2  4N 2  1 3  ( 2k ) 

4N 2 

2 1 8k 3

N2

 k

1 3

N2 

k3



N

k3



Now that we have a normalized constant, we can substitute into the expression for the average energy, solve for E, take the derivative of E with respect to k, and minimize it to determine k.

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Since this wavefunction only has r as a variable, the derivatives in the Hamiltonian operator with respect to  and  are zero. Thus, we present a more simplified expression: 2  k 3 kr   2  1   2   k 3 kr       e e r e       2  r 2 r  r    4 0 r r , ,    

 e kr r 2 sin drdd   

k3

We evaluate the derivatives with respect to r stepwise from the inside (and using N to represent the normalization constant for clarity):

 Ne  kr   kNe  kr r

r2

 Ne  kr   kr 2 Ne  kr r

  2   Ne  kr   k 2 r 2 Ne  kr  2krNe  kr r r  r 

1   2  2k   r Ne  kr    k 2 Ne  kr  Ne  kr Substituting this into the Schrödinger equation 2  r r  r  r  yields:

    Ne   2  k Ne 2

 kr

2





r , ,

 kr



 2k e2  Ne  kr   Ne  kr r 2 sin drdd r  4 0 r 

Multiplying all the constants through gives

  k  Ne   2 Ne 2

 kr

2

 kr



r , ,



 2 2 k e2 Ne  kr  Ne  kr r 2 sin drdd 2r 4 0 r 

Now we multiply through the wavefunction on the left through the three terms and separate into three integrals: 2  2k 2   kr  2  kr  2 k   Ne r sin drdd   Ne  Ne  kr r 2 sin drdd  2   2r  r , ,

 Ne   kr

r , ,





 e2    Ne  kr  Ne  kr r 2 sin drdd  4 0 r  r , ,





We can remove the constants (including the normalization constant) to outside the integrals.

Notice also that we lose one term in r in the second and third integrals:



  

  

N 2 2k 2 2N 2 2 k  kr  kr 2 e e r sin  drd  d   e kr e kr r sin drdd 2 r ,, 2 r ,,



  

N 2e2 e kr e kr r sin drdd 4 0 r ,,

The exponentials can be multiplied together to make e−2kr, and now we can integrate. All three integrals have the same functions in  and , and the integration over the two angles yields a factor of 4 for each one. Thus, we now have:

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

4N 2  2 k 2 2  2 kr 8N 2  2 k 4N 2 e 2  2 kr r e dr  re dr  re  2 kr dr    2 2 4 0 r r r

Note that the last two integrals are the same, so we really only have two integrals to evaluate. 

From the appendix, we see that

r e n

br

dr 

0

n! , which we can apply to all three integrals. b n 1

The equation above simplifies to (including canceling out some terms from numerators and denominators):

2N 2  2 k 2  2!  4N 2  2 k  1!  N 2 e 2  1!         3  2  2     ( 2 k ) ( 2 k ) ( 2 k )       0 At this point, we should now replace N with the normalization constant, appears as N2, we can replace it with

k3



k3



. Since N always

:



2k 3  2 k 2  2!  4k 3  2 k  1!  k 3 e 2  1!        Simplifying, this becomes 3     (2k ) 2   0  (2k ) 2   ( 2k ) 



2k 2 2k 2 ke 2   2  4 0

We can combine the first two terms to get

2k 2 ke 2   This is our final evaluation of the integral. In order to determine what value of 2 4 0 k gives the minimum value of the energy eigenvalue, we take the derivative of this expression with respect to k, set it equal to zero, and solve for k:

  2k 2 ke 2   2 k e2      0 k  2 4 0   4 0

2k





e2 4 0

k

e 2 4 0  2

If you check the original definition of  from Chapter 11, we find that  

4 0  2 . Therefore, e 2

we see that k is simply 1/! The proper wavefunctions, as determined using variation theory, is



1



3

e r / a – which is the same wavefunction we found for H in Chapter 11.

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12.46. An “effective nuclear charge” approach to determine appropriate wavefunctions for H atoms is unnecessary for two reasons. First, since we can solve the Schrödinger equation analytically for the H atom, we have no need of approximation methods. Second – and perhaps more to the point regarding Example 12.11 – hydrogen atoms only have one electron, so there are no additional electrons acting to “shield” the nucleus from other electrons.

12.48. (a) Using equation 12.31, the non−trivial solution is found by evaluating the determinant

 15  E  1

 2.5  E  0

 2.5  E  0

 4  E 1

 0 . Because the basis wavefunctions are orthonormal, we

know that S11 = S22 = 1 and S12 = S21 = 0. We can simplify the determinant to

 15  E

 2.5

 2.5

4E

 0 . We expand the determinant to get the quadratic equation E2 +19E

+ 53.75 = 0. Using the quadratic formula, we can determine the two roots for E:

E

 19  391  4(1)(53.75) 2

E  3.46 or  15.54

(b) The energies of the real system are still rather close to the ideal energies, but are farther away than in Example 12.12. The reason is the larger values of H12 and H21. As those values get larger, the energies of the real system deviate more and more from those of the ideal system.

12.50. (a) The secular determinant is

H 11  ES11

H 12  ES12

H 13  ES13

H 14  ES14

H 21  ES 21

H 22  ES 22

H 23  ES 23

H 24  ES 24

H 31  ES 31 H 41  ES 41

H 32  ES 32 H 42  ES 42

H 33  ES 33 H 43  ES 43

H 34  ES 34 H 44  ES 44

 0 . It

requires knowing the values of 32 different integrals and will ultimately yield 4 values of energy and 4 real wavefunctions. (b) Secular determinants get complex quickly as the number of ideal wavefunctions grows. If there are n ideal wavefunctions, there are n2 Hij integrals and n2 Sij integrals, but you only get n real wavefunctions and n values of energy.

12.52. For the perturbation method, the first approximation we made was that the Hamiltonian was separable into an ideal part and a perturbation part. For real systems, that is probably not completely accurate. For the variation method, the first approximation we make is when we use any trial wavefunction that is not the analytically correct wavefunction of the system.

12.54. The Born−Oppenheimer approximation is likely to produce less error for Cs2 rather than H2. The cesium atom is so much more massive than the electron that its motion is so much slower that the nuclei can be considered effectively motionless. While H nuclei are also much larger than electrons, the ratio of masses is not as extreme as for Cs atoms.



12.56. For R = 1.00Å: S12  e (1.00 A / 0.529 A) 1 



112

1.00A (1.00A) 2    0.616  0.529A 3(0.529A) 2 

Chapter 1

1.00A   H 12  (13.60 eV)(0.616)  2(-13.60 eV)e 1.00 A/0.529A 1    20.25 eV  0.529A 

 13.60 eV  (20.25 eV)  20.95 eV and 1  0.616 (13.60 eV)  (20.25 eV) E2   17.32 eV . Using equation 12.43, the wavefunctions are 1  0.616

Therefore, E1 



 (H2+,1) = 0.556(H(1) + H(2)) and  (H2+,2) = 1.141(H(1) − H(2)).



For R = 1.15Å: S12  e (1.15 A / 0.529 A) 1 



1.15A (1.15A) 2    0.540  0.529A 3(0.529A) 2 

1.15A   H 12  (13.60 eV)(0.540)  2(-13.60 eV)e 1.15 A/0.529A 1    17.16 eV  0.529A 

 13.60 eV  (17.16 eV)  19.97 eV and 1  0.540 (13.60 eV)  (17.16 eV) E2   7.74 eV . Using equation 12.43, the wavefunctions are 1  0.540

Therefore, E1 



 (H2+,1) = 0.570(H(1) + H(2)) and  (H2+,2) = 1.043(H(1) − H(2)).



For R = 1.45Å: S12  e (1.45 A / 0.529 A) 1 



1.45A (1.45A) 2    0.403  0.529A 3(0.529A) 2 

1.45A   H 12  (13.60 eV)(0.403)  2(-13.60 eV)e 1.45 A/0.529A 1    12.04 eV  0.529A 

 13.60 eV  (12.04 eV)  18.28 eV and 1  0.403 (13.60 eV)  (12.04 eV) E2   2.61 eV . Using equation 12.43, the wavefunctions are 1  0.403

Therefore, E1 



 (H2+,1) = 0.597(H(1) + H(2)) and  (H2+,2) = 0.915(H(1) − H(2)).



For R = 1.60Å: S12  e (1.60 A / 0.529 A) 1 



1.60A (1.60A) 2    0.344  0.529A 3(0.529A) 2 

1.60A   H 12  (13.60 eV)(0.344)  2(-13.60 eV)e 1.60 A/0.529A 1    10.00 eV  0.529A 

 13.60 eV  (10.00 eV)  17.56 eV and 1  0.344 (13.60 eV)  (10.00 eV) E2   5.49 eV . Using equation 12.43, the wavefunctions are 1  0.344

Therefore, E1 



 (H2+,1) = 0.610(H(1) + H(2)) and  (H2+,2) = 0.873(H(1) − H(2)).

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The plots are left to the student.

12.58. The hydrogen molecule ion must be treated in light of the Born−Oppenheimer approximation, in that the two nuclei are treated as having fixed relative positions so that the wavefunction of the single electron can be approximated at that distance. Understanding the H2+ system thus requires us to make such approximations at a series of fixed distances. The helium atom has only one nucleus, which we can treat as unmoving with respect to the two electrons.

Understanding the helium atom wavefunctions requires us to approximate the potential energy interaction between the two electrons, but the Born−Oppenheimer approximation need never be invoked.

12.60. Li2: (1s)2 (*1s)2 (2s)2 Be2: (1s)2 (*1s)2 (2s)2 (*2s)2 B2: (1s)2 (*1s)2 (2s)2 (*2s)2 (2px,y)2 C2: (1s)2 (*1s)2 (2s)2 (*2s)2 (2px,y)4 N2: (1s)2 (*1s)2 (2s)2 (*2s)2 (2px,y)4 (2pz)2 O2: (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2px,y)4 (*2px,y)2 F2: (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2px,y)4 (*2px,y)4 Ne2: (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2px,y)4 (*2px,y)6 12.62. No, it shouldn’t, because the additional two electrons will occupy antibonding orbitals, leading to a bond order of zero.

Chapter 13 Introduction to Symmetry Quantum Mechanics 13.2.

(a) E, C2 (the principal axis), 2 C2’ (perpendicular to the principal axis), h, 2v, i, S2. (b) E, C2, 2v. (c) E, Cf , ∞v. (d) E, Cf , ∞v.

13.4.

If all of the points on the jacks were the same they would have Oh symmetry and include the symmetry elements: E, C3, C2, C4, C2’, i, S6, h,S4, and d

13.6.

For an S1 operation, you rotate an object by 360º/1 = 360º to return it to its original position, then reflect through a  plane perpendicular to the axis of rotation. The net result is simply a reflection, . For S2, you rotate an object by 360º/2 = 180º. That has the effect of changing the sign of two of the three spatial coordinates of every point of the object. Then, in performing the  reflection perpendicular to the axis, the third spatial coordinate changes sign. Thus, ultimately

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all three spatial coordinates have their signs changed, corresponding to the inversion operation i.

13.8.

If a molecule has a C6 axis, it automatically also has a C3 and a C2.

13.10. (a) Since Sn involves Cn followed by reflection in a plane perpendicular to the axis, the matrix would be the same as Cn but with a –1 in the lower right corner:

 cos  S n   sin   0

sin  cos  0

0 0  .  1

0  1 0   (b) Inversion involves changing the sign on every coordinate. Therefore, i  0  1 0 .    0 0  1 13.12. There is a bulleted list of these in section 13.3.

13.14. The number of classes can be determined by simply counting the number of rows or columns of characters from the appropriate character table. The order is the total number of symmetry elements in the point group. (a) 4 classes, order = 4. (b) 8 classes, order = 8. (c) 12 classes, order = 24. (d) 4 classes, order = 4. (e) 2 classes, order = 2.

13.16. (a) The following shows a multiplication table for the C3v point group (which you can verify by using a molecule like NH3 as an example): E C3 C32 v v’ v” 2 E E C3 C3 v v’ v” C3 C3 C32 E v’ v” v 2 2 C3 C3 E C3 v” v v’ E C32 C3 v v v” v’ C3 E C32 v’ v’ v v” C32 C3 E v” v” v’ v Since only the original symmetry elements appear in the multiplication table, we can state that the closure property is satisfied. (b) Using the above table to substitute sequentially: v(EC3) = vC3 = v’.

Also, (vE)C3 = vC3 = v’. Thus, the associative law is satisfied.

13.18. (a) In C2v, C2v = v’. (b) In C2h, iC2 = h. (c) In D6h, C6h = S6. (d) In D2d, C2C2’ = S4. (e) In Oh, iS4 = C43.

13.20. (a) C32. (b) h. (c) C53. (d) i 13.22. Porphine has D2h symmetry as a neutral molecule. However, if a metal ion is substituted into the center of the porphine ring, with the simultaneous loss of the two H atoms on the ring nitrogens, the symmetry becomes D4h.

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13.24. The tetrahedron has Td symmetry: E, 8C3, 3C2, 6S4, and 6d. The cube and the octahedron have Oh symmetry: E, 8C3, 3C2, 6C4, 6C2’, i, 8S6, 3h, 6S4, and 6d.

13.26. Using the scheme in Figure 13.14: (a) H2O2 has C2 symmetry. (b) Allene has D2d symmetry. (c) dGlycine has C1 symmetry. (d) -Glycine also has C1 symmetry.

13.28. Using the scheme in Figure 13.14: (a) H2Se has C2v symmetry. (b) HDS has Cs symmetry. (c) The chair conformation of cyclohexane has D3d symmetry. (d) The boat conformation of cyclohexane has C2v symmetry.

13.30. 1,4-cyclohexadiene has D2h symmetry. 1,3-cyclohexadiene has C2v symmetry.

13.32. CH3Cl has C3v symmetry. CH2Cl2 has C2v symmetry. CHCl3 has C3v symmetry. CCl4 has Td symmetry.

13.34. (a) The wavefunctions of HCl have C v symmetry. (b) The wavefunctions of sulfur dioxide would have C2v symmetry. (c) The wavefunctions of SO3 would have D3h symmetry.

13.36. (a) Dichloromethane will have a permanent dipole moment. (b) Chlorobenzene will have a permanent dipole moment. (c) Ammonia will have a permanent dipole moment. (d) Carbon dioxide will not have a permanent dipole moment.

13.38. Of the listed species, the following will not have a permanent dipole moment: phosphorus pentachloride, boron trifluoride, and diborane.

13.40. Molecules with a center of inversion have high symmetry. Since a dipole moment reflects uneven distribution of electron density in a molecule, an inversion center guarantees that a molecule won’t have a dipole moment. Any electron density at one point on the molecule is exactly replicated at the exact same point on the other side of the inversion center.

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 x N  1  y  0  N   z N  0     x H1  0  y H1  0     z H1  0  13.42. E  x H2  0     y H2  0  z  0  H2    x H3  0     y H3  0  z H3  0  x N  1  y  0  N   z N  0     x H1  0  y H1  0    z 0   H1    x 0  H2    y H2  0  z  0  H2    x H3  0     y H3  0  z H3  0

0 0 0 0 0 0 0 0 0 0 0  x N  1 0 0 0 0 0 0 0 0 0 0  y N  0 1 0 0 0 0 0 0 0 0 0  z N    0 0 1 0 0 0 0 0 0 0 0  x H1  0 0 0 1 0 0 0 0 0 0 0  y H1    0 0 0 0 1 0 0 0 0 0 0  z H1  0 0 0 0 0 1 0 0 0 0 0  x H2    0 0 0 0 0 0 1 0 0 0 0  y H2  0 0 0 0 0 0 0 1 0 0 0  z H2    0 0 0 0 0 0 0 0 1 0 0  x H3    0 0 0 0 0 0 0 0 0 1 0  y H3  0 0 0 0 0 0 0 0 0 0 1  z H3  0 0 0 0 0 0 0 0 0 0 0  x N  1 0 0 0 0 0 0 0 0 0 0  y N  0 1 0 0 0 0 0 0 0 0 0  z N    0 0 1 0 0 0 0 0 0 0 0  x H1  0 0 0 1 0 0 0 0 0 0 0  y H1    0 0 0 0 1 0 0 0 0 0 0  z H1  0 0 0 0 0 0 0 0  1 0 0  x H2    0 0 0 0 0 0 0 0 0  1 0  y H2  0 0 0 0 0 0 0 0 0 0 1  z H2    0 0 0 0 0 1 0 0 0 0 0  x H3    0 0 0 0 0 0 1 0 0 0 0  y H3  0 0 0 0 0 0 0 1 0 0 0  z H3 

13.44. By multiplying the corresponding characters together, one can show that the following relationships exist for the D2 point group: A1A1=A1

A1B1=B1

A1B2=B2

A1B3=B3

B1B1=A1

B1B2=B3

B1B3=B2

B2B2=A1

B2B3=B1

B3B3=A1

Therefore, the closure requirement is satisfied.

13.46. The specific solution depends on which pair of irreducible representations you select. The following illustrates one pair for each point group. (a) For C2: AB = (111 + 11-1) = 0 (b) For C2v: A1B1 = (111 + 11-1 + 111 + 11-1) = 0

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(c) For D2h: AgAu = (111 + 111 + 111 + 111 + 11-1 + 11-1 + 11-1 + 11-1) = 0 (d) For Oh: T1uT2u = (133 + 800 + 3-1-1 + 61-1 + 6-11 + 1-3-3 + 800 + 311 + 6-11 + 61-1) = (9+0+3-6-6+9+0+3-6-6) = 0 (e) For Td: ET1 = (123 + 8-10 + 32-1 + 601 + 60-1) = 6+0-6+0+0 = 0 Since all of the direct products equal zero, the irreducible representations are orthogonal. You should convince yourself that any two irreducible representations are orthogonal, not just these pairs.

13.48. This irreducible representation is not normalized.

A?  A? 

1 1  1  1  1  1  1  1  1  0  1  1  0  1 4 2

Since this value is not equal to one, the irreducible representation is impossible.

13.50. Using C4v for which h=8: 12  12  12  12  2 2  8 13.52. In both C v and Dh , any value of  is possible, not just discrete values. Therefore, we use a mathematical expression for some of the characters, rather than a specific value.

13.54. (a) Sin has E, (xy), S2 (the y axis), and i, among others. (b) Cos has E, C2 (the y axis), and 2 ’s (the xy and the yz planes).

13.56. All wavefunctions of linear molecules must belong to either the C v point group or the Dh point group.

13.58. The drawing is left to the student. Benzene has D6h symmetry, and the irreducible representations of the lowest- and highest-energy  orbitals can be determined by the following analysis. For the lowest-energy  orbital: Its character for i should be −1 (since the phases of the p orbitals will change), the character of h should be -1 (again, because the phases of the p orbitals should change), and the characters of any vertical or dihedral plane should be +1 (because the phases of the p orbitals will not change). This is sufficient to label the lowestenergy  orbital with the A2u irreducible representation. For the highest-energy  orbital: Its character for i should be +1 (since the phases of the p orbitals will not change), the character of h should be -1 (because the phases of the p orbitals should change), and the characters of any vertical plane should be +1 (because the phases of the p orbitals will not change; vertical planes will cut through an orbital, whereas dihedral planes will pass between orbitals). This is sufficient to label the highest-energy  orbital with the B2g irreducible representation.

13.60. (a) In C3v, the product A1A2 yields the character set (1, 1, -1), which is the A2 irreducible representation. Thus, A1A2 = A2. (b) In C6v, the product E1E2 yields the character set (4, -1, 1, -4, 0, 0). Using the GOT to reduce this representation, we find that it equals B1 + B2 + E1.

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(c) In D3h, the product A2’A1”E” yields the character set (2, -1, 0, 2, -1, 0), which is the E’ irreducible representation. (d) In D6h, the product B2gB2u yields the character set (1, 1, 1, 1, 1, 1, -1, -1, -1, -1, -1, -1), which is the A1u irreducible representation.

13.62. We know that the sum of the irreducible representations has to equal the reducible representation. Remember to take into consideration the number of operations in each class.

Since this is the C2v point group, each class only has one operation making the problem easier.

This  is the sum of 5 A1 irreducible representations. All of the other irreducible representations of C2v contain -1’s which would prevent the sum from adding up to 5 5 5 5.

13.64. (a) For D6h, B1g  B1g  A1g which is the totally symmetry representation. This integral does not have to be 0. (b) For Td,

E  T1

E 6

8C3 0

3C2 -2

6S4 0

6d 0

This is not an irreducible representation. Using the GOT, we must determine if this contains the A1 irreducible representation:

a A1 

1 1  6  1  0  3  2  1  0  0  0 24

Since this does not contain the A1 totally symmetric representation, the integral must be 0.

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(c) For Td,

T2  T2

E 9

8C3 0

3C2 1

6S4 1

6d 1

Using the GOT, we can determine that this does contain the A1 irreducible representation.

This integral does not have to be 0. (d) For Oh,

E g  T2g

E 6

8C3 0

3C2 -2

6C4 0

6C2’ 0

i 6

8S6 0

3h -2

6S4 0

6d 0

When we reduce this we find that it does not contain the A1 irreducible representation. This integral must be 0.

13.66. Because s orbitals are spherically symmetric, they have the Dg(0) irreducible representation in the Rh(3) point group. If the operator has the irreducible representation Du(1), the integral of interest has an overall symmetry given by the product Dg(0) Du(1) Dg(0). The product of these three irreducible representations is equal to Du(1). Since this is not the all-symmetric irreducible representation (represented in this point group by Dg(0)), the integral is exactly zero.

13.68. Using the GOT, we can reduce this representation to E + T2. This means that two d orbitals will be doubly degenerate and the other three will be triply degenerate.

13.70. For f orbitals, l=3. In Rh(3) the irreducible representation for f orbitals is Du(3). Using the formulas for the characters for the symmetry operations we can calculate the characters for D4h symmetry. 

E 7

2C4 -1

C2 -1

2C2’ -1

2C2’’ -1

i -7

2S4 1

h 1

2v 1

2d 1

Using the GOT we can determine that the linear combination of irreducible representations of the D4h point group is: A2u Å B1u Å B2u Å 2Eu 13.72. If the core orbitals of S (that is, 1sS, 2sS, 2px,S, 2py,S, and 2pz,S) were included in our SALC analysis of H2S, there would be more SALC molecular orbitals because the number of independent molecular orbitals equals the number of atomic orbitals used to construct the SALCs. Also, the 1s and 2s orbitals would appear in every linear combination where the 3s orbital appears and the same-axis 2p orbitals would appear in every linear combination where the corresponding 3p orbital appears. That’s because the 1s and 2s orbitals have the same symmetry properties as the 3s orbital does, and the 2p orbitals have the same symmetry properties as the corresponding 3p orbitals do.

13.74. Most of the molecular orbitals from Example 13.13 are actually atomic orbitals, so they are certainly orthogonal to each other. Molecular orbitals having different symmetry species are also orthogonal to each other, because the product of two different symmetry species in C2v will not be A1. The only questionable wavefunctions are the group of 4 A1 wavefunctions, the products of which also have A1 symmetry. The molecular wavefunctions that are identical to oxygen atomic wavefunctions are orthogonal to each other, since the atomic wavefunctions

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should be mutually orthogonal. The only ambiguity is the orthogonality of the oxygen atomic wavefunctions with the linear combination  = ½(1sH1 + 1sH2). Because the wavefunctions are on different atoms, there is no immediate requirement that these wavefunctions must be orthogonal. At this point, the best we can do is state that the wavefunctions should be orthogonal.

13.76. Molecular orbital theory assumes that orbitals on molecules are linear combinations of atomic orbitals. Valence bond theory assumes that orbitals are products of atomic orbitals and their linear combinations. The text lists other differences.

13.78. The four hybrid orbitals will be different but will still be four independent, acceptable linear combinations. To illustrate this, we can substitute the definitions of 2px and 2py into the expression for an sp3 orbital and see how it reduces (recall that 2pz can also be labeled 2p0). In the derivation below, we are dropping the “” and are simply using the wavefunction labels for clarity:

1 

1 s  p x  p y  p z   1  s  1 2 p1  2 p1   i 2 p1  2 p1   2 p0  2 2 2 2 

This simplifies to

1 

1 s  p x  p y  p z   1  s   1  i 2 p1   1  i 2 p1  2 p0  2 2  2 2 2  2 

A hybrid orbital made from the pure hydrogenic wavefunctions would look like

1 

1 s  2 p1  2 p1  2 p0  -- which is definitely not the same hybrid orbital! However, 2

hybrid orbitals can be defined in terms of pure hydrogenic wavefunctions and can perform the same functions.

13.80. Let us take two sp3 orbitals and show that they are orthogonal to each other. Using 1 and 3:

 2 s  p  p  p  2 s  p  p  p d 1

*

x



y

z

1

x

y

z



1 * s sd   s * p x d   s * p y d   s * p z d  4

  p x sd   p x p x d   p x p y d   p x p z d *

*

*

*

  p y sd   p y p x d   p y p y d   p y p z d *

*

*

*

  p z sd   p z p x d   p z p y d   p z p z d *

*

*

*



Taking advantage of the orthonormality behavior of the individual atomic orbitals, we know that any integral with different atomic orbitals is exactly 0, while any integral with the same atomic orbitals is exactly 1. Substituting:

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

1 1  0  0  0  0  1  0  0  0  0  1  0  0  0  0  1  0 4

Thus, these two hybrid orbitals are orthogonal to each other. The same conclusion can be made about any two sp3 hybrid orbitals.

13.82. The character for E is 3, as all three orbitals operate onto themselves. The character for C3 would be 0, as no orbital operates onto itself. The character for C2 would be 1 (for the axially coincident p orbital), for h would be 3, for S3 would be 0, and for v would be 1. Therefore, the set of characters for the three sp2 hybrid orbitals would be (3, 0, 1, 3, 0, 1). Using the GOT, this representation reduces to A1’ + E’.

13.84. Using the appropriate model, one can show that the characters for the sp3d2 hybrid orbitals in octahedral symmetry are: 

E 6

8 C3 0

3 C2 2

6 C4 2

6 C2’ 0

i 0

8 S6 0

3 h 4

6 S4 0

6 d 2

Using the GOT, this representation reduces to A1g + Eg + T1u.

13.86. The drawings are left to the student.

13.88. The three sp2 orbitals collectively represent D3h symmetry. Consider the symmetry operations in the D3h point group and how they affect the lone p orbital. For E, its character would be 1; for C3, its character would be 1 (because the p orbital is co-linear with the rotation axis); for C2, the character would be –1; for h, the character would be –1; for S3, the character would be –1; and for v, the character would be 1. This set of characters describes the A2” irreducible representation.

Chapter 14 Rotational and Vibrational Spectroscopy 14.2. (a) This integral must be zero since the product AuB2uAu = B2u, which is not the completelysymmetric representation in D2h. (b) This integral must be zero since the product A1A1A2 = A2, which is not the completelysymmetric representation in C3v. (c) This integral can be non-zero since the product g+g−g− = g+, which is the completelysymmetric representation in Dh. (d) The irreducible representations multiply to get the character set (6, 0, −2, 0, 0). Using the GOT to see if A1 is a part of this reducible representation, we get 12.56 L 

122

1000 mL 1 cm 3   12,560 cm 3  1.256  10 4 cm 3 . Therefore the integral must be zero.

1L 1 mL

Chapter 1

14.4. (a) (b) (c) (d) 14.6. The irreducible representations multiply to get the character set (18, 0, 2, 0, 0, −18, 0, −2, 0, 0).

Using the GOT to see if A1 is a part of this reducible representation, we get (ignoring all of the zero terms) V



R 

nT p

. Therefore the integral for the transition

must be zero and the transition is forbidden.

14.8. Since c = , we can rewrite this into an expression for 1/: R



pV nT

Thus, we have shown that the wavenumber is directly proportional to , with the proportionality constant equal to 1/c.

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14.10. First, we must determine the masses of each atom in kg units: R

(2.66 bar)(27.5 L) L  bar  0.0830 (1.887 mol)(466.9 K) mol  K

x atm 64.0 m



1 atm 10.1 m

Now we can determine  and, after that, the moment of inertia I:

(1.04 atm )(67 L )  9 .4 L (7.38 atm ) I = r2 = (1.6510−27 kg)(1.61710−10 m)2 = 4.3110−47 kg·m2 Now, using equation 14.7, we can determine the energies of the first four rotational levels:

1.82  1013 g SO 2 

nRT p  V

1 mole SO 2  2.84  1011 moles SO 2 64.07 g L  atm )(273.15  17.0K ) mol  K  7  10 10 atm 21 8  10 L

(2.84  1011 moles SO 2 )(0.0821

0.0821

L  atm 5 / 9 K L  atm   0 .

0456 mol  K 1 R mol  R

0.80 

14.7 lb lb  11 .

8 in.2 in.2

14.12. Since 2J+1=5, we know that J=2. 0.20  14in..7 lb  2.9 in.lb , so MJ could equal −2,−1,0,1, or 2.

2

2

14.14. (a) oblate symmetric top (b) linear (c) linear (d) linear (e) asymmetric top (f) asymmetric top 14.16. Both SF6 and UF6 are octahedral, spherical-top molecules. Therefore, their moments of inertia are independent of the axis of rotation. Therefore, we can choose a simple-to-envision rotation and use that to calculate the moments of inertia, and from that, the value of B. Let us select rotation of the molecule along one of the F−S−F or F−U−F axes. That way, the three atoms along the axis do not contribute to the moment of inertia because they are not rotating. What is

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rotating are four F atoms a particular distance away from the rotational axis. The moments of inertia can be calculated simply, by considering it as four atoms rotating about an axis. For SF6: I = 4(0.0190 kg/6.0221023)(1.56410−10 m)2 = 3.08710−45 kg·m2 For UF6: I = 4(0.0190 kg/6.0221023)(1.99610−10 m)2 = 5.02810−45 kg·m2 Now to calculate the values of B: 4.50torr (44.01g / mole)  0.01706 g / L  1.71 10 5 g / cm 3 m pM m pV   RT;   L  torr M V RT   (62.36 )(273.15  87K) mol  K

3y

2



y 2 z 3 w

When you consider that the central atoms are so different in mass, these values of B are not that much different from each other!

2

3

2

3

w z xy z  14.18. 332 so: y w 2

3 3

w z 2xyz 6xy  2  32 y w

3

14.20. There are 2J + 1 MJ levels that are degenerate. Also, for each of these levels, the positive and negative values of K that have the same magnitude also yield the same energy, because energy is dependent on K2. Therefore, for most energy levels, the degeneracy is 2(2J + 1). When K = 0, there’s not another value of K that yields the same energy, since –0 = 0. Therefore, the degeneracy of those energy levels is dictated solely by the various MJ values – 2J + 1 of them.

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14.22. Ethane is a prolate symmetric top with the two lower rotational constants being the same:



3 y 2 z 2 w



3 y 2 z w

2

The lowest rotational energy level has E = 0. The second energy level comes from the J = 1, (K=0) value of the lower of the two rotational constants: 3

2 y z w 2

2

The third energy level comes from the J = 1, (K=1) value of the lower of the two rotational constants:   

n V

   T, P



R

p  T

The fourth energy level comes from the J = 2, (K=0) value of the lower of the two constants:    

T p

    V ,n



n

V  R

Finally, the fifth energy level comes from the J = 2, (K=1) value of the lower of the two constants:   

n T

   P, V



- pV RT 2

14.24. (a) Dimethyltriacetylene will not have a pure rotational spectrum. (b) Cyanotetraacetylene will have a pure rotational spectrum. (c) Nitric oxide will have a pure rotational spectrum. (d) Nitrogen dioxide will have a pure rotational spectrum (e) SF4 will have a pure rotational spectrum (f) SF6 will not have a pure rotational spectrum.

14.26. (a) not allowed (b) allowed (c) not allowed (d) allowed 14.28. A spectrum plotted in units of wavenumbers should consist of a series of equally−spaced lines, since the wavenumber unit is directly proportional to energy. However, if the spectrum is replotted using units of wavelength, the series of lines will no longer be equally spaced because wavelength is not directly proportional to energy.

14.30. If the rotational spectrum consists of lines spaced by 15.026 cm−1, this value represents 2B, so B = 7.513 cm−1. Calculating the reduced mass of LiH:  p     n  T , V

126



RT V

Chapter 1

Using the wavenumber forms of the equation for the rotational constant:

V (T , p )



nRT p

Solving for r2, then for r:  V   V  dV   dT   dp   p    T  p  T

r = 1.6010−10 m = 1.60 A. 14.32. If the rotational spectrum consists of lines spaced by 15.026 cm−1, this value represents 2B, so B = 7.513 cm−1. Converting this into frequency units:    Latm   Latm     1 mole  0.0821  1mole  0.0821  350K     nRT   nR  Kmole   Kmole    0.10atm  dp   dT   10.0K    dV   2 2    p 1 .

08 atm p 1.08atm             

 V       p  n , T



  2V   nRT   ;  2  p2  p  n ,T

Calculating the frequency: 

2 nRT p3

In terms of energy: E = (6.62610−34 J·s)(2.2521011 s−1) = 1.49210−22 J. Using equation 14.23: nR   2 p   p    ; 0   (a) 2  V   T  n , V  T  n , V



a bR

(b)

TB

(c)

L2 atm mol 2 TB   1026 K  0.04267 L/mol  0.08205 L atm  mol K   3.592

L2 atm mol 2   521 K  0.03183 L/mol  0.08205 L atm   mol K   1.360

14.34. T

B

L2 atm mol 2   433 K  0.03913 L/mol  0.08205 L atm   mol K   1.390

TB

The spectrum consists of evenly spaced lines 2B apart from each other.

14.36. (a) The rotational spectrum should split into two individual lines. (b) The rotational spectrum should split into (up to) 6 individual lines. (c) The rotational spectrum should split into (up to) 12 individual lines.

14.38. CH4 has no permanent dipole moment. It will not exhibit a pure rotational spectrum.

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Instructor‟s Manual

14.40. First, we need to convert B to cm−1.

C 2 V

14.42. As the masses of the two atoms get larger, the values of both B and D get smaller. This makes sense for B because as the masses of the atoms increase, so does the moment of inertia of the molecules and, because I is in the denominator of the expression for B, the value of B gets smaller. This makes sense for D because as the masses of the atoms get larger, it (presumably) gets harder and harder to distort the molecule rotationally because, again, of the increased inertias of the larger atoms.

14.44.

1m 3 0.0237 L  mole 1000 L



2.37  10  5

m3 mole

14.46.

Total deg. Freedom (3N) 9 9 15 12 6 60 60

(a)H2S (b)OCS (linear) (c)SO42− (d)COCl2 (e)Cl2 (linear) (f)linear 20 atom molecule (g)non-linear 20 atom molecule

Vib. deg. freedom 3 4 9 6 1 55 54

14.48. (a) 3 (b) 4 (c) 9 (d) 6 (e) 1 (f) 55 (g) 54 3

3

mole 6.0210 atoms

atom

14.50. Starting with the equation 2.37105 m  1mol23e  3.941029 m , we multiply the first fraction on the right by m2/m2 and the second fraction on the right by m1/m1:

4 m3 V  r 3  3.94  10 29 ; r  2.11 10 10 m or 2.11. Now we can take the reciprocal of both sides of the equation to get 3 atom an equivalent expression for the reduced mass:

128

0.03049L . mole

Chapter 1

14.52. For O2,

0.0638L mole

14.54. The ratio of the reduced masses from the example is 0.7170. All we need to do is show that the ratio of the two frequencies in cm−1 units has that same ratio: which is close enough to the ratio of reduced masses.

14.56. We need to determine the reduced masses of 56FeH and 54FeH, assuming a mass of 1 for the hydrogen: 2

a  1   b  Z  1 b      RT  V  V 

V

Z  1

B C   V V 2

2

14.58. The drawing is left to the student.

14.60. (a) CO2 will have a pure vibrational spectrum. (b) H2 is a homonuclear diatomic molecule and will not have a pure vibrational spectrum. (c) NaCl(g) will have a pure vibrational spectrum (d) SO32− will have a pure vibrational spectrum (e) S2 is a homonuclear diatomic molecule and will not have a pure vibrational spectrum.

14.62. Deviations from ideality are more likely to be seen at higher energies, because lower-energy vibrations typically result from higher-mass atoms or less-distorted motions, either of which tends to follow classical harmonic oscillator behavior.

14.64. A Morse oscillator has two significant differences from a harmonic oscillator. First, the width of the potential energy curve increases as the oscillator deviates further and further from the equilibrium separation, ultimately leading to a dissociation of the oscillator at some point. Second, the quantized energy levels get closer and closer together as the internuclear separation increases. There are no quantized energy levels above the dissociation limit.

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Instructor‟s Manual

14.66. Using equation 14.39 to determine the energy of the first six energy levels of HCl: p 

RT  V  b

  

a T V ( V  b)

p V

 RT  ( V  b) 2

   T ,n

a T V ( V  b)

2



a



2

T V ( V  b)

an 2 (p  )( V  nb)  nRT V2

 RT

pV RT p  ,p  V

(0.0821

L  atm e p  (0.0821 )(273.15K ) K  mol   

   

V T

V p

1 V

Latm    (10.91atm L2 ) /  ( 22.41L )( 0.0821 )( 273.15 K )  K mol  

  

   T

L  atm )( 273.15K ) K  mol  1.00atm 22.41L  0.981atm

(22.41L  0.0401L) 



p



  



      

     

T p T V

p T p V

  V   T

   V    p

V T

  

p

1 volume 1   volume temperatur e temperatur e 

1 V

   

V p

   T

The energy differences are: E1 – E0 = 8.67610−20 – 2.94410−20 J = 5.73210−20 J = 2886 cm−1 difference

130

0.0007%

Chapter 1

E2 – E0 = 1.42010−19 – 2.94410−20 J = 1.12610−19 J = 5669 cm−1

0.02% difference

E3 – E0 = 1.95210−19 – 2.94410−20 J = 1.65710−19 J = 8344 cm−1

0.03% difference

−19

E4 – E0 = 2.46310

−20

– 2.94410

J = 2.16910

−19

J = 10920 cm

−1

E5 – E0 = 2.95410−19 – 2.94410−20 J = 2.66010−19 J = 13390 cm−1 14.68. For this, we need equation 14.38: 

0.03% difference 0.05% difference

1 volume 1   volume pressure pressure

 1  V    V  P  T



14.70. From 14.69,



1  1   nRT  1  nRT  1  1 1             V  2   p  V  V V p  p  p     p 

1  V  1   nRT  1 nRT   14.72. Starting with equation 14.39,        , we can construct a general V  p  T V p  p  V p 2 expression for the change in energy between adjacent energy levels:

nRT p



V

Multiplying out the binomials and collecting terms which won‟t be shown here , we get 1 V

V p



−1

(either in s or cm T p



−1

1 p

. If we want to express this change of energy in frequency units directly , we divide all terms by Planck‟s constant:

. For the fundamental absorption, v = 0 and this becomes

T T 1  V  T   nRT  T nR       p p V  T  p pV T  p  pV p

. We know (see example 14.12) that we can use the ratio of reduced masses as a weighting factor to predict the shift of a vibrational frequency upon isotopic substitution. The frequency of the heavier isotope, e*, is shifted by a factor

nR V  . Defining this factor as , we have p T

that e* = e. This variable also appears in the anharmonicity term as well. However, considering equation 14.40, we also find that the anharmonicity constant xe is also defined in terms of e, so xe also reduces by the same factor of . We can thus state that xe* = xe. Assuming that, for the T V 1 isotopically−substituted molecule, pV T , which  p , we can substitute to get: V



RT p

, which is the expression from exercise

14.71.

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14.74. The first vibration has g+ symmetry and is IR-inactive. The second vibration has u+ symmetry and is IR-active.

14.76. (a) E and C3 are proper rotations; the vs are improper rotations. (b) The E, C3s, and C2s are proper rotations; the S4s and ds are improper rotations. (c) The proper rotations are E, the C6s, the C3s, and the three classes of C2s. The improper rotations are i, the S3s, the S6s, and all planes of symmetry. (d) The proper rotations are E and C2, while the two S4s are improper. (e) The proper rotations are E, the C(), and the infinite C2s. The improper rotations are i, S(), and the infinite vs. (f) All of the five classes of symmetry elements in O are considered proper rotations.

14.78. (a) H2O2 has C2 symmetry. A check of the C2 character table shows that both irreducible representations, A and B, will be IR-active. Therefore, H2O2 will have 6 IR-active vibrations. (b) Oxalic acid, (COOH)2, has Cs symmetry. A check of the Cs character table shows that both irreducible representations, A’ and A”, will be IR-active. Therefore, (COOH)2 will have 18 IRactive vibrations. (c) Sulfur trioxide, SO3, has D3h symmetry, and not all of the irreducible representations are IRactive. Therefore, we must go through the steps to determine the character of the vibrations: Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 4 0º 3 12 3 3 6

2 C3 1 120º 0 0 0 0 0

3 C2 2 180º −1 −2 −1 −1 0

h 4 180º −1 4 −1 1 4

2 S3 1 60º 2 −2 2 −2 −2

3 v 2 180º −1 2 −1 1 2

Using the GOT, this reduces to A1‟ 2E‟ A2”. According to the character table, E‟ and A2” modes are IR-active. Therefore, we expect SO3 to have three IR-active vibrations. (d) Formaldehyde is a C2v molecule. Not all of the irreducible representations are IR-active.

Therefore, we have to go through the steps to determine the characters of the vibrations:

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Chapter 1

Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 4 0º 3 12 3 3 6

v 4 180º −1 4 −1 1 4

C2 2 180º −1 −2 −1 −1 0

v‟ 2 180º −1 2 −1 1 2

Using the GOT, this reduces to 3A1+2B1+B2, all of which are IR-active. Therefore, CH2O will have 6 IR-active vibrations. (e) Acetone is a C2v molecule. Not all of the irreducible representations are IR-active. Therefore, we have to go through the steps to determine the characters of the vibrations: Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 10 0º 3 30 3 3 24

v 6 180º −1 6 −1 1 6

C2 2 180º −1 −2 −1 −1 0

v‟ 2 180º −1 2 −1 1 2

Using the GOT, this reduces to 8A1+4A2+7B1+5B2. Of these, the A1, B1, and B2 vibrations are IR-active, so acetone will have 20 IR-active vibrations.

14.80. If KrF4 were ever synthesized, one could tell if it were square planar or tetrahedral by determining the number of IR-active (and Raman-active) infrared absorptions the molecule has. If the molecule were planar, it would have 3 IR-absorption bands, while if it were tetrahedral, it would have 2 IRabsorption bands. Some of the bands contain degenerate vibrational modes.

14.82.

Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 16 0º 3 48 3 3 42

8C3 4 120º 0 0 0 0 0

3C2 0 180º −1 0 −1 −1 2

6C4 0 90º 1 0 1 1 −2

6C2‟ 0 180º −1 0 −1 −1 2

i 0 180º −1 0 −1 1 0

8S6 0 120º 0 0 0 0 0

3h 0 180º −1 0 −1 1 0

6S4 0 90º 1 0 1 −1 0

6d 8 180º −1 4 −1 1 8

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Since only T1u is IR-active, all we need to do is determine how many T1us there are in this reducible representation. Using the GOT, we find that there are 3T1us as part of the irreducible representation.

Therefore, there are three (triply-degenerate) IR-active vibrations. This represents 9 of the 42 possible vibrations of C8H8.

14.84. Combination bands are typically (but not always) weak because they are not usually allowed by the quantum-mechanical selection rules.

14.86. Both dioctyl sulfide and hexadecane are largely composed of long hydrocarbon chains, the only difference being that dioctyl sulfide has a sulfur atom in the middle. Thus, their vibrational spectra should be very similar.

14.88. SiH4 is a spherical top whose rotational energy levels are given by the expression d   d    p .n

whose spacing between rotational levels is 2B =  T   

M pM  and RT / p RT

pM Since 2B = 16.72 cm−1, we convert this RT 2

to joule units: 2B = 3.3210−22 J. Solving for I:

V Now we need to determine a formula for the moment of inertia. If we assume that the molecule is spinning about an Si−H bond, then there are three equivalent H‟s making a triangle whose angle is determined by the tetrahedral bond angle, 109.45º. The angle that the bond makes from the rotational axis is half that, or 54.725º. The component of this bond that is rotating is related to the sine of this angle, or 0.8164. Finally, there are three hydrogens spinning. Thus, we have:  d   T

   p .n





M VT

Solving for r:

r = 1.0010−10 m = 1.00 Å.

This information could not be determined from a pure rotational spectrum because SiH4 does not have a permanent dipole moment, so does not have a pure rotational spectrum.

14.90. If centrifugal distortion were negligible, there would be no DJ terms in either equation 14.41 or 14.42. They would simplify to: 

E = h – 2xee + (B1 + B0)(Jlower + 1) + (B1 – B0)(Jlower + 1)2 and



E = h – 2xee – (B1 + B0)Jlower + (B1 – B0)J2lower

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Chapter 1

14.92. We are interested in the difference in the energy gap between the kg  m 2 s2

p e



Mgh RT

 g  m    m  Mgh mol  s 2    J RT    K   K  mole 

absorption and the

absorption. Using equation 14.41 and neglecting the centrifugal distortion term, we

have:

Mgh RT

2   kg  m    2    mol  s   J    K   K  mol 

14.94. (a) For CH4: Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 5 0º 3 15 3 3 9

8 C3 2 120º 0 0 0 0 0

3 C2 1 180º −1 −1 −1 −1 1

6 S4 1 90º 1 −1 1 −1 −1

6 d 3 180º −1 3 −1 1 3

Using the GOT, this reduces to A1+E+2T2. Of these, the point group shows that the A1, E, and T2 vibrations are Raman-active, so all four vibrations should appear in a Raman spectrum. (b) For CH3Cl: Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 5 0º 3 15 3 3 9

2 C3 2 120º 0 0 0 0 0

3 v 3 180º −1 3 −1 1 3

Using the GOT, this reduces to 3A1+3E, all of which are Raman-active. Therefore, CH3Cl will show 6 Raman-active vibrations.

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Instructor‟s Manual

(c) For CH2Cl2: E 5 0º 3 15 3 3 9

Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

C2 1 180º −1 −1 −1 −1 1

v 3 180º −1 3 −1 1 3

v‟ 3 180º −1 3 −1 1 3

Using the GOT, this reduces to 4A1+A2+2B1+2B2, all of which are Raman-active. Therefore, this molecule will have 9 Raman-active vibrations. (d) CHCl3 will have the same number of Ramanactive vibrations as CH3Cl: six. (e) CCl4 will have the same number of Raman-active vibrations as CH4: four.

14.96. Assuming XeF4 were tetrahedral, we could understand its vibrations as follows: Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 5 0º 3 15 3 3 9

8 C3 2 120º 0 0 0 0 0

3 C2 1 180º −1 −1 −1 −1 1

6 S4 1 90º 1 −1 1 −1 −1

6 d 3 180º −1 3 −1 1 3

Using the GOT, this reduces to A1+E+2T2. Of these, only the T2 vibrations are infrared-active, so XeF4 should show only two IR-active vibrations. Also, the A1, E, and T2 vibrations are Ramanactive, so all four vibrations should appear in a Raman spectrum. If, on the other hand, XeF4 were square planar (with D4h symmetry): Nstationary  1+2cos ±N(1+2cos) r, (1+2cos) tr, ±(1+2cos) vib

E 5 0º 3 15 3 3 9

2 C4 1 90º 1 1 1 1 −1

C2 1 180º −1 −1 −1 −1 1

2 C2 ‟ 3 180º −1 −3 −1 −1 −1

2 C2 ” 1 180º −1 −1 −1 −1 1

i 1 0º 3 −3 −3 3 −3

2 S4 1 90º 1 −1 1 −1 −1

h 5 180º −1 5 −1 1 5

2 v 3 180º −1 3 −1 1 3

2 d 1 180º −1 1 −1 1 1

Using the GOT, this reduces to A1g+B1g+B2g+A2u+B2u+2Eu. Of these, the A2u and the Eu vibrations are infrared-active, and the A1g, B1g, and B2g vibrations are Raman-active. Thus, for the square planar geometry of XeF4, we would expect 3 infrared absorptions and 3 Raman absorptions. Since this is exactly what we get experimentally, we can conclude that XeF4 probably has square planar geometry.

Chapter 15 136

Chapter 1

Introduction to Electronic Spectroscopy and Structure 15.2. In integrals that contain two wavefunctions and an operator, the combination of two of the functions must yield a resulting symmetry species that is the same as the third (section 13.8). In this case we know that T2  A1g  T2 . The allowed electronic excited states must have a symmetry of T2.

15.4. First, we need to determine the reduced masses of D and T. For D:



(9.109  10 31 kg)(3.344  10 -27 kg)  9.1065  10 31 kg 9.109  10 31 kg  3.344  10 -27 kg

 For T:

(9.109  10 31 kg)(5.008  10 -27 kg)  9.1073  10 31 kg 31 - 27 9.109  10 kg  5.008  10 kg

Thus, the new Rydberg constants are:

(1.602  10 19 C) 4 (9.1065  10 31 kg) R(T)   2.1783  10 18 J  109,660 cm -1 12 4 2 34 2 8(8.854  10 C / J  m) (6.626  10 J  s) R(T) 

(1.602  10 19 C) 4 (9.1073  10 31 kg)  2.1785  10 18 J  109,670 cm -1 12 4 2 34 2 8(8.854  10 C / J  m) (6.626  10 J  s)

These are very small – but detectable – changes.

15.6. The drawing is left to the student.

15.8. (a) not allowed,   1 (b) allowed (c) allowed (d) not allowed   1 15.10. The possible values of MS for a hydrogen nucleus are –1/2 and +1/2, the same as for a single electron. For a deuterium nucleus, the possible values of MS are –1, 0, or +1 – illustrating that MS values are spin-dependent and may not be a simple few values.

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15.12. (a)  = 2, m = -2, -1, 0, 1, 2. Two possible values of j: j = 5/2, with possible mj = -5/2, -3/2, -1/2, 1/2, 3/2, 5/2; also, j = 3/2, mj = -3/2, -1/2, 1/2, 3/2. (b)  = 3, m = -3, -2, -1, 0, 1, 2, 3. Two possible values of j: j = 7/2, with possible mj = -7/2, -5/2, -3/2, -1/2, 1/2, 3/2, 5/2, 7/2; also, j = 5/2, mj = -5/2, -3/2, -1/2, 1/2, 3/2, 5/2. (c)  = 4, m = -4, -3, -2, -1, 0, 1, 2, 3, 4. Two possible values of j: j = 9/2, with possible mj = 9/2, -7/2, -5/2, -3/2, -1/2, 1/2, 3/2, 5/2, 7/2, 9/2; also, j = 7/2, mj = -7/2, -5/2, -3/2, -1/2, 1/2, 3/2, 5/2, 7/2.

15.14. (a) With a single s electron in its valence shell, a Li atom has the term symbol 2S1/2. (b) With a single p electron in its only unfilled subshell, the Al atom has the term symbols 2P1/2 and 2P3/2. According to Hund‟s rules, the 2P1/2 will be the lower-energy ground state term symbol. (c) With a single d electron in its only unfilled subshell, the Sc atom has the term symbols 2D3/2 and 2D5/2. According to Hund‟s rules, the 2D3/2 will be the lower-energy ground state term symbol.

15.16. According to a more mundane version of Hund‟s rules, electrons will spread out among available orbitals in a subshell before they pair in a single orbital, and will do so having the same orientation of spin. This “same orientation of spin” is what yields a maximum multiplicity of the ground state. In such a situation, however, the overall L – as determined by the vector sum of all of the ml values – is zero, as they all cancel. Thus, half-filled subshells will always have their highest multiplicity for an S term symbol.

15.18. No. In order to get a multiplicity of 5, S must be 2 which implies 4 unpaired electrons. N does not have four unpaired electrons in the ground state.

15.20. Table 15.1 gives us the allowed term symbols for this configuration. Given the first, we see that 2 P has S=1/2 and L=1. So J can be 3/2 or ½ giving 2 P3 / 2 and 2 P1 / 2 . We need to calculate the degeneracy (2J+1) for each of these states which is equal to 4 and 2, respectively. For each of the 2 D term symbols we get 10 available states. 2F has 14. 2G has 18. 2H has 22. 4P has 12, and 4F has 28 for a total of 120 states. We can also calculate this using the formula:

G! , where G is equal to the maximum number of electrons allowed in the subshell e!G  e ! and e is equal to the number of electrons in the subshell. For this exercise that is equal to 120.

N

15.22. See exercise 15.20 above. We expect 210 available states for this configuration. The two 1S‟s have 1 each. The two 1D‟s give 5 each. 1F has 7. The 1G‟s have 9 each. 1I has 13. The 3P‟s have 9 each. 3D has 15. The 3F‟s have 21 each. The 3G has 27. 3H has 33, and 5D has 25 for a total of 210.

138

Chapter 1

15.24. There are many term symbols for this configuration. L can range from 5+5 to 0. S can equal ½ + ½ or 0. Term symbols are: 3 M11 , 3 M10 , 3 M 9 ,1 M10 , 3 L10 , 3 L 9 , 3 L 8 ,1 L 9 , 3 K 9 , 3 K 8 , 3

K 7 ,1 K 8 , 3 J 8 , 3 J 7 , 3 J 6 ,1 J 7 , 3 I 7 , 3 I 6 , 3 I 5 ,1 I 6 , 3 H 6 , 3 H 5 , 3 H 4 ,1 H 5 , 3 G 5 , 3 G 4 , 3 G 3 ,1 G 4 , 3 F4 , 3 F3 , 3 F2 ,1 F3 , 3

D 3 , 3 D 2 , 3 D1 ,1 D 2 , 3 P2 , 3 P1 , 3 P0 ,1 P1 , 3 S1 , and 1 S0 . Many of these are not allowed due to the Pauli Principle. The ground state will be a triplet state of highest L value. Since the shell is less than half filled, the ground state will be: 3 M 9 .

15.26. Since the two electrons are in the same s orbital, the spins must be paired. Therefore S=0 and this is a singlet state. L is also equal to 0 since the electrons are both in s orbitals. So the term symbol is 1So. 15.28. Since S = 0, any allowed excited state will have S = 3/2. L = 0 or 1, so allowed excited states can be either D, F, or G terms. Finally, since J = 0 or 1, allowed excited states can have a value for J of 7/2, 9/2, or 11/2. Combining these three items, the possible term symbols are 4D7/2, 4 D9/2, 4D11/2, 4F7/2, 4F9/2, 4F11/2, 4G7/2, 4G9/2, or 4G11/2. Some of these aren‟t possible due to the relationship between L, S, and J: only 4D7/2, 4F7/2, 4F9/2, 4G7/2, 4G9/2, or 4G11/2 are valid term symbols.

15.30. The wavefunctions of a heteronuclear diatomic molecule do not have a center of inversion.

Therefore, we don‟t need a “u” or “g” label to indicate the wavefunction‟s symmetry with respect to an i symmetry element.

15.32. A  state means that its  value is 1 so possible excited states can have  values of 0, 1 or 2.

Since the ground state has a multiplicity of 2, excited states must have a multiplicity of two.

Available states are: 2, 2  and 2 .

15.34. = 0 since it is a  state. S=1/2 since it is a doublet state.

15.36. If the molecular orbitals of O2 are considered, the ground state is a triplet it‟s actually 3g-.)

Transition to a 1 state represents a forbidden transition, since it violates the S = 0 selection rule. Thus, the transition back to the ground state might be slow because the transition is formally not allowed.

15.38. The transition that has the greatest Franck-Condon factor is the transition that is (a) vertical, and (b) connects two wavefunctions that have approximately the same amplitude and linear momentum (i.e. either both near the turning points or both near the center of the potential well).

15.40. (a) Cr2O72- has unpaired electrons in Cr‟s d subshell, so it is highly likely that the ion will absorb visible light and have some color. (It is, in fact, yellow-orange.) (b) There are no unpaired electrons in O22-, so we expect it to be colorless. (c) Acetylide ions have unpaired electrons in its molecular orbitals (see the answer to exercise 15.37), so we expect it might have a color. It‟s grayish-black in color.)

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15.42. According to the point group character table in the Appendix, the electric dipole operator has the irreducible representations A1, B1, or B2 (depending on which three-dimensional axis is involved). In order to guarantee that the A1 irreducible representation is part of the product upper ˆ lower , an excited state must also have the irreducible representations A1, B1, or B2. Thus, possible excited states will have irreducible representations 1A1, 1B1, or 1B2.

n 2h 2 15.44. For a particle in a box, E  . So the energy difference between two levels would be: 8ma 2

6.626 ´10-34 J × s) ( h2 2 2 E4 - E3 = 7 = 4.35´10-19 J (4 - 3 ) = 2 ( ) -31 -9 8ma 2 8 ( 9.109 ´10 kg) ( 0.985´10 m) 2

-34 8 hc ( 6.626 ´10 J × s) ( 2.9979 ´10 m / s) DE = ;so l = = = 457nm l DE 4.35´10-19 J

hc

This is a reasonable approximation to the experimental value especially since the hexatriene isn‟t quite linear!

15.46. S11, S22, S33, and S44 are overlap integrals in which the wavefunctions are located on the same atom. Thus, what we really have are wavefunctions overlapping with themselves. If they are individually normalized (which proper wavefunctions are), then the value of the overlap integral is 1.

E  0   E  0 0 15.48. For cyclobutadiene: 0  E   0  E

E  0 0   E  0 0 For cyclopentadiene: 0  E  0 0 0 0  E   0 0  E

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Chapter 1

15.50. (a) The energy levels of butadiene are shown in Figure 15.19. All of the electrons occupy low lying molecular orbitals, thus stabilizing the molecule. (b) Using an online Huckel determinant solver, we find that the energy levels for cyclobutadiene are different. Two electrons fall into a stabilizing molecular orbital and the other two occupy separate molecular orbitals (unpaired!) at the  level. Since the expected energy for cyclobutadiene is higher than that of its linear counterpart, it is called anti-aromatic.

Experimental data suggests that cyclobutadiene does not behave as an aromatic compound, but rather behaves as if it has two independent double bonds.

15.52. Cyclopentadiene easily accepts an electron because when it does, it has six  electrons and is aromatic.

15.54. (a) Ideally, neutral cyclopolyenes will be aromatic if they have 4n+2 carbon atoms with  electrons in the ring. (b) Ideally, cyclopolyenes having a single positive charge if they have 4n+3 carbon atoms with  electrons in the ring. (c) Ideally, cyclopolyenes having a double positive charge if they have 4n+4 carbon atoms with  electrons in the ring.

15.56. Fluorescence is a very fast process on a molecular time scale. This indicates that it is spinallowed. Phosphorescence is much slower since it is actually a “forbidden” transition.

15.58. Phosphorescence involves intersystem crossings that technically violate the S = 0 selection rule. Such transitions are usually rather slow on an atomic or molecular time scale. As such, it may be possible to use phosphorescence-related transitions to construct a population inversion in a system, possibly leading to laser action.

15.60. Refer to Figure 15.24. Because there is some internal dissipation of energy before fluorescence, the fluorescent photon typically has a lower energy than the excitation photon. Since blue light is higher energy than red light, it is unlikely that blue fluorescence will result from red excitation – more likely, it would be the other way around. (There is a phenomenon known as frequency doubling, in which two photons “combine” to make a single photon having twice the energy – and thus twice the frequency – but this is a different phenomenon.)

15.62. (a) 10.6 m 



c





1m  1.06  10 5 m 10 6 m

2.9979  10 8 m/s  2.83  1013 s -1 -5 1.06  10 m

E = h = (6.62610-34 Js)(2.831013 s-1) = 1.8710-20 J per photon 300,000 J 1 photon   1.60  10 25 photons per second - 20 s 1.87  10 J Therefore, .

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(b) 1.60  10 25

 1E photons  E    26.6 23 s s  6.02  10 photons 

æcö 8p h ç ÷ 3 A 8p hn è l ø 8p h 15.64. = = = 3 3 B c c3 l 3

15.66.

(a) E=nh=  8 m  2.9979  10  nhc s  23 34   6.022  10 photons 6.626  10 J  s  3.00  10 5 J 9  405  10 m 34 A 8h 8 6.626  10 J  s kg ( b)  3   2.51  10 13 3 B ms  405  10 9 m





 







15.68. Consider equation 15.33, which gives the ratio of spontaneous emission coefficient A to A 8h 3  stimulated coefficient B: . Note the “frequency-cubed” dependence in the numerator.

B c3 That means that the higher the frequency, the greater (by a cubed factor) A will be with respect to B; that is, the greater (by a cubed factor) spontaneous emission will be with respect to stimulated emission. X-rays are very high-frequency radiation, and the rate of spontaneous emission will be so much greater than stimulated emission that achieving a population inversion and laser action will be extremely difficult.

Chapter 16 Introduction to Magnetic Spectroscopy 16.2.

Since there are six wires, the maximum total magnetic field is the field of one magnetic field, multiplied by six: B  6

0 I (4  10 7 T  m/amp)(10 amp)  6  6  10 6 T  0.06 gauss 2r 2 (2 m)

In reality, due to the vector properties of magnetic fields, it may be less (and in fact may be exactly zero if certain symmetry constraints are met, like equal spacing).

16.4.

According to the text, a tesla equals a kg/(coulombs). Thus, we have:

T

142

kg m 2 /s 2 J J J  2 2   2  2 2 2 Cs m /s C  m  s/s m C/s m  amp .

Chapter 1

Therefore, we have shown that a tesla also equals a J/m2amp.

16.6. The drawing is left to the student; see Figure 16.6. Without a magnetic field, there is only one transition. However, in the presence of a magnetic field, the 1P state splits into three evenly−spaced energy states, and the spectrum will be a series of three equally−spaced (if plotted in units proportional to energy) lines.

16.8. (a) Of the three states in the 1S  1P transition, one state decreases in energy, one state increases in energy by the same amount, and one state doesn‟t change in energy at all. The change in energy is given by equation 16.9: E = BMLB: 



E = (9.27410−24 J/T)(1)(2.35 T) = 2.1810−23 J. This corresponds to 1.10 cm−1.

(b) For the 1P  1D transition in the presence of a magnetic field, there will be 15 possible transitions, as the 1P state as three possible ML values while the 1D state has 5 possible ML values. However, many of the energy differences will be the same, and the selection rules in equation 16.8 will limit the number of allowed transitions. In the end, there will be only three lines having, not surprisingly the same E as for the 1S  1P transition: E = 2.1810−23 J. 16.10. We need the gJ value for the 2P3/2 term symbol. Using equation 16.13: (3 / 2)(3 / 2  1)  (1 / 2)(1 / 2  1)  (1)(1  1) gJ  1  1.3333...

2(3 / 2)(3 / 2  1)

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Now we use equation 16.11 and the fact that the maximum MJ value is +3/2 and the minimum is –3/2 to calculate the energy difference between these two states (the difference being equal to 3): 

E = (1.3333)(9.27410−24 J/T)(3)(0.00006 T) = 2.2310−27 J. This is equivalent to 1.1210−4 cm−1, a difficult energy difference to differentiate.

16.12. Using equation 16.12:

 (4)(4  1)  (2)(2  1)  (2)(2  1)  gJ  1   (2.0023  1)  1.50115 2(4)(4  1)   Using equation 16.13:

 (4)(4  1)  (2)( 2  1)  (2)( 2  1)  gJ  1    1.5 2(4)(4  1)   The difference is less than 0.08%.

16.14. Since this is a triplet state, S=1. L=0. Using equation 16.13:

 (0)(0  1)  (1)(1  1)  (0)(0  1)  gJ  1   . This involves dividing by 0! We never need worry 2(0)(0  1)   3 about this unusual value because the S0 state can‟t exist. J  L  S  L  S so J can only have a value of 1.

16.16. In several places, the text suggests that the frequency of microwave radiation used in ESR spectroscopy is on the order of 10 GHz. Let us use that frequency to determine an approximate energy per photon: E = (6.62610−34 J)(10109 s−1) = 710−24 J/photon.

Of course, the energy depends on the frequency of microwave radiation used. This can be compared to 310−19 – 510−19 J for a single photon of visible light.

16.18. The text mentions 9.5, 24, and 35 GHz as specific frequencies used in ESR spectroscopy. Let us use equation 16.16 to determine the magnetic field strengths needed to achieve resonance for an electron in each of these regions: (2.002)(9.274  10 24 J/T) B 6.626  10 34 J  s

B  0.339 T

(2.002)(9.274  10 24 J/T) B 24  10 s  6.626  10 34 J  s

B  0.857 T

9.5  10 9 s -1  9

-1

35  10 9 s -1 

144

(2.002)(9.274  10 24 J/T) B 6.626  10 34 J  s

B  1.25 T

Chapter 1

 e  ge   B  1.602  10 19 C 2.002  0.335T   e B  2m e  9     16.20.  L    9.39  10 Hz 31 2 2  2 9.109  10 kg  2 









16.22. (a) This pattern would be caused by a single unpaired electron on an atom with an I=1. (b) In this case 2NI+1=3 so NxI=1. N=2 and I=1/2. (c) This pattern would be caused by a single unpaired electron on an atom with an I=2 (d) 2NI+1=5 so NxI=2. I could equal ½ and N could equal 4.

16.24. Since the carbon atoms have I = 0, they do not contribute to the hyperfine coupling of the electron. However, the five hydrogen atoms all have I = ½, and the possible combinations of those spins can lead to MI values of +5/2, +3/2, +1/2, −1/2, −3/2, and –5/2. Thus, we would expect 6 ESR signals for the cyclopentadienyl radical.

16.26. 7.20410−24 J = (2.0023)(9.27410−24 J/T)(B)

h res 16.28. B   g e B

16.30. Using  

6.626  10

B = 0.388 T



1  J  s 1.165  10 9  s   0.042T   24 J  2.002 9.274  10  T  34

c , 

   2.9979  10 8 m  s  6.626  10 34 J  s   1 m     4.05cm h res  100cm    ge    1.906 B B  1  10  4 T   24 J   9.274  10 2776G   T  1G 



16.32.  res 

g e B B  h



2.0023 9.274  10 24 J 10.2T  T  34 6.626  10 J  s





 2.86  1011

1 s

 8 m  2.9979  10  c  s    1.05  10 3 m 1  2.86  1011 s

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16.34. With six lines in the ESR spectrum, an atom in the radical can have a spin of up to 5/2 (assuming that there are other atoms in the radical that are contributing to the signal); if it has a spin higher than that, then that particular atom probably isn‟t in the molecule. Checking the table of nuclear spins, we find that (a) 42K (I = 2), (b) 35Cl (I = 3/2), (c) 37Cl (I = 3/2), (d) 67Zn (I = 5/2), (e) 47Ti (I = 5/2), and (f) 32S (I = 0) are all possible in the radical. In the cases of 67Zn and 47Ti, no other atoms in the radical should have a non−zero nuclear spin; in all other cases, other atoms with non−zero I are necessary in order to produce the given ESR spectrum.

1   16.36. We expect: 2  3   12  1  3  1  28 signals in this ESR spectrum. The three equivalent H 2   atoms split the pattern into a quartet. The 10 B splits the pattern into a heptet. We would expect a heptet of quartets.

16.38. Yes. Although we would expect the spectra to be similar, the g value would be different due to different local environments for the electron.

16.40. The E between adjacent spin states is the same no matter what the value of MI is (assuming a constant g). The E is proportional to the resonance frequency given in equation 16.24. Because of the selection rules governing NMR, this frequency is independent of MI. 16.42. Nuclear spins and electron spins both have a quantized value of the total spin and a quantized value for the z component of the total spin. All individual electrons are fermions and a nucleus can either be a fermion or a boson. Also when placed in a magnetic field, the E‟s of the electron states and the nuclear states are significantly different.  6 1   6.7238  10 3.22T   NB  sT  16.44.  L    3.45MHz 2 2

16.46. (a) We would expect three signals (including the alcohol H). (b) We would expect five signals (including the alcohol H).

16.48. For methylacetate we would expect two distinct signals, both singlets and having the same integration. The methyl group attached to the carbonyl carbon should appear around 2.2 ppm and the other methyl group should appear around 3.8 ppm.

16.50. If a H is split by three adjacent hydrogen atoms, the options are:

        This results in four distinct transitions that create a 1:3:3:1 pattern.

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16.52. The top spectrum is diethyl ether since it should only have two distinct signals. The hydrogens on the carbon near the O are shifted more downfield and split into a quartet by the methyl hydrogens. The methyl hydrogens are split into a triplet. The other spectrum is 2−butanol.

16.54. The 35Cl nucleus, with I = 3/2, will have four different orientations with respect to the magnetic field, and so will show four different changes in energy. They will be: E = (0.5479)(5.05110−27 J/T)(−3/2)(3.45T) = −1.43210−26 J. E = (0.5479)(5.05110−27 J/T)(−1/2)(3.45T) = −4.77410−27 J. E = (0.5479)(5.05110−27 J/T)(+1/2)(3.45T) = +4.77410−27 J. E = (0.5479)(5.05110−27 J/T)(+3/2)(3.45T) = +1.43210−26 J. 16.56. Boron NMR is more complicated than proton or 13C NMR for two reasons. First, naturally−occurring boron is a mixture of two isotopes, boron−10 (20 %) and boron−11(80 %), both of which have non−zero spins and so are NMR−active. Second, their spins are very different. Boron−10 has a nuclear spin of 3, while boron−11 has a nuclear spin of 3/2 (which you can verify by consulting a table of nuclear spin properties). A boron NMR spectrum of naturally−occurring boron is thus much more complicated than 1H or 13C (both of which have a nuclear spin of ½).

Chapter 17 Statistical Mechanics: Introduction 17.2. There are four ways of putting one of three balls in each of four boxes. It doesn‟t agree with equation 17.1 because we are assuming that the balls are indistinguishable. If they are distinguishable, then there are 24 different ways, which is consistent with equation 17.1. If there are no restrictions on the number of balls in each box, then there are 20 different ways of putting the balls in the boxes.

17.4. According to Stirling‟s approximation, the natural logarithm of 1,000,000! is ln (1,000,000!) = (1,000,000)ln(1,000,000) – 1,000,000 = 12,816,000. Therefore, 1,000,000! is about e12,816,000, or about 105,566,000.

17.6. Recognizing that N!=N(N-1)(N-2 …,

1000 mL 1 cm 3 12.56 L    12,560 cm 3  1.256  10 4 cm 3 1L 1 mL For large N this is approximately NlnN-N. 17.8. The logarithm of the expression is

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Instructor‟s Manual

1.055 atm 

1.01325 bar 100,000 Pa   1.069  10 5 Pa Using properties of logarithms, 1 atm 1 bar

this reduces to

1 torr 1 atm 1.01325 bar    1.644 bar . To evaluate ln(5000!), we set 1 mmHg 760 torr 1 atm N to 5001 and substitute: 1233 mmHg 

1 cm 3 125 mL  1 mL

 125 cm 3

Evaluating each term: ln(5000)! = -5001 + 42,591.22 + 0.9189 + 1.666310-5 = 37,591 This is one of the values that‟s listed in the table in chapter 17, so this expression is much more accurate than Stirling‟s approximation.)

17.10. The first method to calculate the average involves summing all of the values and dividing by the total number of grades: 25750 Pa 

1 bar  0.2575 bar 100,000 Pa

For the second method we need to recognize that someone has a 1/15 probability of making a 2 on the quiz, a 4/15 probability of making a 10, etc.

V



R



nT p

17.12. (a) For 10 objects and two objects in each system, we need five systems. Therefore, we have to pV determine R  . nT (b) For 3 objects and one object in each system, we need three systems. Therefore, we have to determine R 

( .2 66 bar)(27.5 L) L  bar  .0 0830 .

(1.887 mol)(466.9 K) mol  K

(c) For 6 objects and three objects in each system, we need two systems. Therefore, we have to x atm 1atm  . (d) For 6 objects and two objects in each system, we need three 64.0 m 10.1 m (1.04 atm)(67 L)  9 .4 L . systems. Therefore, we have to determine (7.38 atm)

determine

17.14. For a microcanonical ensemble, equations 17.4 – 17.6 would be 1.82  1013 g SO 2 

1 mole SO 2  2.84  1011 moles SO 2 64.07 g

The first two of these are the same as for a canonical ensemble; the third equation is new.

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Chapter 1

17.16. In order to obtain a total system energy of 5EU‟s, one particle must be in the 3EU level and two particles must be in the 1EU level. Since the particles are distinguishable, there are three ways to make this distribution:

nRT p  V

L  atm )(273.15  17.0K) mol  K  7 1010 atm 8 1021 L

(2.84 1011 moles SO2 )(0.0821

This is the same answer as for 17.15.

17.18. The most probable distribution for 17.15 is to have 2 particles in the 2EU level and one in the 1EU level.

17.20. Having all gas molecules in one corner of a room is only one possible distribution out of countless possible distributions. The chances that this one distribution would actually occur is infinitesimally small.

17.22.

0.0821

L  atm 5 / 9 K L  atm    0.0456  mol  K 1 R mol  R

17.24. Use equation 17.21: 0.80 

14.7 lb lb . If we assume the degeneracies are the same, the g1/g0  11 .

8 in.2 in.2

term cancels. Converting 200 cm-1 to J units:

0.20 

14.7 lb lb  2.9 2 in. in.2

E = h = (6.62610-34 Js)(5.9961012 s-1) = 3.97310-21 J  m  m pM 4.50torr (44.01g / mole) 5 3 Now we can use equation 17.21: pV   M RT; V  RT  (62.36 L  torr )(273.15  87K)  0.01706 g / L  1.7110 g / cm . mol  K

Thus, there are almost three times as many atoms in the ground state as there are in the first excited state.

17.26. The only difference in the two equations is in the second terms, so if we show that they are equivalent, we‟ve shown that the two equations as a whole are the same. The easiest way to show that equations 17.29 and 17.30 are equivalent is simply to substitute 1/kT for  in the second term of equation 17.29:

3y 2



y 2 z 3 w

. In the second step, the k is

removed from the denominator because it is a constant. The canceling of the k terms shows that the two partial derivative terms are, in fact, equal.

17.28. We assume that the energy levels are singly degenerate:

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Instructor‟s Manual

z3 xy 2 z 3  32 y w2

At 250 K: 3w

2

6 xy  At 350 K:



2 xyz 3 w

3y 2 z 2 w



3 y 2 z w

 At 500 K:

w3 z 3 32 y 2

2

3y 2 z 2 w2

p  n     R  T  V  T, P 17.30. Partition function values are constant for a given set of conditions, including temperature, pressure, volume, etc. If temperature, volume, etc., were to change, then the specific value of the partition function would change. So yes, q is a constant, but does vary with conditions. Thus, the derivatives of q with respect to conditions like T and V are not zero.

17.32. If we assume the convention that our ground energy state is equal to 0, as T approaches 0, q approaches the degeneracy of the ground state. This makes sense because at T=0 we would expect only the ground state to be occupied.  T  V  , the restriction is that the amount of any other substance in   p  V ,n n  R

17.34. For equation 17.46, 

the system, nj (j  i), must remain constant. This is the same restriction we placed on the chemical potential when it was introduced earlier in the text.

17.36. (a) From a statistical perspective, we can argue that there are more energy states available when a gas expands, so q would be larger, meaning that S would increase. (b) At 5 °C, there is enough energy available to allow the molecules to access a larger number of energy states, suggesting an increase in q and a concurrent increase in S.

 n    T  P, V

17.38. We need to show that 



- pV . The first term is easy.

RT 2

Consider the summation. The first term in the summation is g0, because 0 in the exponent is zero (as the ground state) and the limit of that term as T goes to zero is just g0. For all of the

150

Chapter 1

remaining terms, as T goes to zero, the exponentials become e-, which is zero. So for the first term, we get k ln g0 as T goes to zero. The second term is more problematic, and we will need to apply ‟Hopital‟s rule. Let us rewrite the second term by factoring out the exponential with respect to temperature out of every term in the summation:

 p     n  T , V



RT V

By factoring the e-1/kT out of each term, we are taking advantage of a property of exponentials. nRT All we need to do now is determine the limit of V(T, p)  as T approaches 0. We can‟t simply p substitute T = 0 into the expression because we will have trouble with infinities and zeros.

However, we can use ‟Hopital‟s rule and take the derivative of the numerator and denominator:

 V   V   dV   dp .  dT      T  p  p  T Evaluating this expression at the limit of T rule again to get   Latm     Latm    1mole 0.0821    1mole 0.0821 350K   nR   nRT  Kmole   Kmole    0.10atm dV   dT   2 dp    10.0K     1.08atm2   p   p   1.08 atm      

0 is still problematic, but we can apply ‟Hopital‟s

, which is an expression that can be evaluated at T = 0. The fraction

equals 0 at T = 0, so the entire second term goes to zero as T goes to zero, and we are left with  V 2nRT the fact that  Vp    nRT . ;   p  p  p 2



2

 n ,T

2



3

 n ,T

3

1/ 2     3/ 2   3/ 2    mkT  3 / 2 1   3 / 2 8 3 / 2 m 3 / 2 Vk 3 / 2 T 3 / 2  2 3  2mkT        17.40. q    V  V  2 3  h    2  h2 23 h 3 / 2  h        2/3   8mV kT  

 

17.42. For a particle in a 1-D box, TB  L2 atm mol 2  1026 K  0.04267 L/mol  0.08205 L atm  mol K   L2 atm 1.360 mol 2 TB   521 K  0.03183 L/mol  0.08205 L atm  mol K  

a where L is the length of the box. For a particle in a bR

3.592

2-D box,

TB 

where A is the area of the box (assuming a square box). Since

,

L2 atm mol 2 L atm  0.03913 L/mol   0.08205  mol K   1.390

TB



 433 K

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C 2 V 17.44. The original definition of the partition function was based on the energy states of the particles in the sample. The individual particles are atoms or molecules (whose energy states can be understood with quantum mechanics), so we use the mass of the individual atom or molecule in the calculation of the thermodynamic properties via statistical thermodynamics.

17.46. Using the definition of S, note that

, we substitute for q and evaluate. First, we

0.0237L 1m3 m3   2.37 105 . Evaluating the second form of the expression: mole 1000 L mole

3 m3 1mole  29 m 2.37 10   3.94 10 mole 6.02 1023 atoms atom 5

Now, dividing this by q:

4 3 29 m3 10 . Complex as this looks, everything cancels except 0.03049L . Substituting back V  r  3.9410 ;r 2.1 10 mor2.1 mole 3 atom into the original expression:

0.0638L mole

Now we substitute the expression for q into the second term:

. The final step is recognizing that V/N can be rewritten in terms

of the ideal gas law, which on the atomic scale uses k Boltzmann‟s constant instead of R, the ideal gas law constant: V/N = kT/p. Substituting: , which is the Sackur-Tetrode equation.

2

a 1 b  17.48. Z  1   b        RT  V  V 

152

Chapter 1

17.50. (a)

Z

 1

B V



C V 2



2

V Solving: S = 164.9 J/K (b)

p 

RT  V  b

a T V ( V  b)

 p      V  T ,n



S = 210.1 J/K (c)

 RT  ( V  b) 2

a T V ( V  b) 2



a 2

T V ( V  b)

an 2 (p  )( V  nb)  nRT V2 S = 174.9 J/K.

17.52. Assuming a gas phase:

pV  RT

Since the mass of C-13 is greater than that of C-12, its partition function is also greater.

L  atm

17.54. Using q = msT, p  RT , p  (0.0821 K  mol )(273.15K )  1.00atm 22.41L

V

Using equation 17.56: L  atm e p  (0.0821 )( 273.15K ) K  mol

Latm    (10.91atm L2 ) /  ( 22.41L )( 0.0821 )( 273.15 K )  K mol  

( 22.41L  0.0401L)

 0.981atm

17.56. First, we can calculate the de Broglie wavelength of electrons going 0.01c (= 2.9979106 m/s):  V   T

  

 p



 p   T p    V

  V   T

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Instructor‟s Manual

Next, we use this as the wavelength in the thermal de Broglie equation and solve for T:    

V p

   T





 T   p   T   V

   V    p

T = 94,100 K Let us use the volume-containing form of the Sackur-Tetrode equation: 3/ 2 5  2mkT  V  S  Nk   ln   . If we consider ΔS as S2 – S1, we can label the different 2  h  N   2 conditions in the Sackur-Tetrode equation and subtract two expressions: 3/ 2 3/ 2 5 5  2mkT  V2   2mkT  V1  S  S 2  S1  Nk   ln     Nk   ln  2 2  h  N   h  N   2  2

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Chapter 1

The two 5/2 terms cancel, and using the properties of logarithms we can write the subtraction of two logarithms as a fraction:  2mkT  V2   h2  N  S  Nk ln . All of the terms in the fraction cancel except for the volumes, 3/ 2  2mkT  V1   2  h  N V V leaving S  Nk ln 2 . In molar terms, NAk = R, so we have S  R ln 2 .

V1 V1 3/ 2

Chapter 18 More Statistical Mechanics 18.2.

The effective nuclear partition function is the degeneracy of the ground state, g1, which is calculated using the equation: g(I)=2I+1. The degeneracies are: (a) 3 (b) 4 (c) 2 (d)1.

18.4.

(a) A check of the electronic states of N2 shows that the first excited state is at 69290 cm−1, which is high enough that only the first term of the electronic partition function will be significant. Thus, we can estimate qelect with only one term. (b) A check of the electronic states of O2 shows that the first excited state is at 7918 cm−1, a second state at 13195 cm−1, and a third at 36096 cm−1. Thus, we can estimate qelect with only one term. Therefore, we suggest that a one-term partition function will probably be OK, but including a second and even third term in the expansion, to cover the first two excited states, may be warranted.

18.6.

For both E and S, the nuclear contribution is zero. That’s because both E and S are determined as a derivative of the partition function. In our approximation, qnuc is a constant (equal to the degeneracy of the ground nuclear state), so the derivative of this constant is simply zero.

18.8.

The minimum value of qelect is 1, because this is the minimum degeneracy of any electronic ground state.

     27.2cm 1   =7.14 18.10. q elec  1  7 exp 1   cm  298K     0.695 K     18.12.

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Instructor‟s Manual





1 J  1mole   1 34 13 1 D e  D o   h   918  10 3    6.626  10 J  s 4.76  10 s 23 2 mol  6.022  10 molecules  2  















1 1 6.626  10 34 J  s 1.08  1014 s 1  6.626  10 34 J  s 1.12  1014 s 1  1.61  10 18 J 2 2

 J  1mol  3   497  10 mol  6.022  10 23 molecules       3 exp  59800K  18.14. q elec  3  exp    T        23 J  1 .

381  10  T     K     18.16. (a) Assuming that the 89.8 J is the value for De (to be accurate, we would need to include the zero−point vibrational energy):

89.8 J 1 mol   1.49  10 22 J Substituting into qelect: 23 mol 6.02  10 molecules

q elect  1  e1.4910

22

J/(1.38110 -23 J/K)(4.2 K)

 13.1

(b) Room temperature is enough thermal energy (= RT = 2.48 kJ/mol) to break such a weak bond, so He2 probably won’t exist at such “high” temperatures.

18.18. The ratio of qvib(H2) to qvib(D2) can be written using equation 18.20:

1 qvib (H 2 ) (kT / h H2 )  D2 . Since     2 qvib (D 2 ) (kT / h D2 )  H2

k



, where k is the force constant and  is the

reduced mass, this ratio of partition functions becomes

reduces to

q vib (H 2 )  q vib (D 2 )

H

2

D

2

q vib (H 2 ) (1 / 2 ) k /  D 2 , which  q vib (D 2 ) (1 / 2 ) k /  H 2

. We can substitute the reduced masses for H2 and D2 into this

expression or, recalling that deuterium has twice the mass of hydrogen, recognize that deuterium will have twice the reduced mass of hydrogen. Thus, the ratio simplifies and we get our final answer:

18.20. q vib =

q vib 

156

q vib (H 2 )  q vib (D 2 )

1 .

2

-34 13 -1 hn ( 6.626 ´10 J × s) (1.15´10 s ) = = 552K J k 1.381´10-23 K

1 1 e

 vib / T



1 1 e

552 K / 300 K

 1.19

Chapter 1

18.22. For 250 K: q vib 

 e  / 2T   e 1360 / 2250  e 2330 / 2250  e 2330 / 2250           / T  1360 / 250   2330 / 250   2330 / 250  1 1 e  1 e  1  e  1  e 

3 N 6

 e 4800 / 2250  e 4880 / 2250  e 4880 / 2250        4800 / 250   4880 / 250   4880 / 250  1 e  1  e  1  e  qvib = (0.06616)(0.009467)2(0.00006773)(0.00005772)2 = 1.33810−18 For 500 K: q vib 

 e  / 2T   e 1360 / 2500  e 2330 / 2500  e 2330 / 2500           / T  1360 / 500   2330 / 500   2330 / 500  1 1 e  1 e  1  e  1  e 

3 N 6

 e 4800 / 2500  e 4880 / 2500  e 4880 / 2500        4800 / 500   4880 / 500   4880 / 500  1 e  1  e  1  e  qvib = (0.2748)(0.09823)2(0.008230)(0.007597)2 = 1.25910−9 For 1000 K: q vib 

 e  / 2T   e 1360 / 21000  e 2330 / 21000  e 2330 / 21000           / T  1360 / 1000   2330 / 1000   2330 / 1000  1 1 e  1 e  1  e  1  e 

3 N 6

 e 4800 / 21000  e 4880 / 21000  e 4880 / 21000        4800 / 1000   4880 / 1000   4880 / 1000  1 e  1  e  1  e  qvib = (0.6815)(0.3455)2(0.09147)(0.08783)2 = 5.74010−5 3

2

 e  / 2T   e 1870 / 2298   e 2180 / 2298   e 4170 / 2298  e 4320 / 2298           18.24. q vib     / T  1870 / 298    2180 / 298    4170 / 298   4320 / 298  1 1 e  1 e  1 e  1 e  1  e  3 N 6

3

qvib = (0.04347)3(0.02581)2(0.0009149)(0.0007114)3 = 1.80210−20 18.26. For a gas-phase molecule, the minimum value of qnuc is 1, assuming that all nuclei have singlydegenerate nuclear states. Similarly, the minimum value of qrot for a molecule is 1, which would be at absolute zero (and thus the molecule would be in the J = 0 rotational state for all possible rotations). At any temperature above T = 0 K, the qrot would be greater than 1. The minimum value of qvib is zero (see the examples in exercise 18.13) if either T were very low (i.e. 0 K) or the vibrational frequencies were very large. However, for any nonzero temperature and finite vibrational frequency, qvib is greater than zero.

18.28.  r 

2 2Ik

0.116K 

1.054  10 J  s ; I  3.47  10 2I1.381  10 J / K  34

 23

2

 45

kg  m 2

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Instructor‟s Manual

18.30. Using the expanded form of equation 18.34 (in which the definition of r is given explicitly): 2 2 I H2  H2 r 2  H2 1 q rot (H 2 ) 8 I H 2 kT / 2h      . q rot (D 2 ) 8 2 I D 2 kT / 2h 2 I D 2  D 2 r 2  D 2 2

18.32. (a) For 14 N 2 (bosons) the ratio of the number of molecules in odd rotational states to the number of molecules in even rotational states is:

(2I 2  1) 3  . We would expect I  12I  1 6

about ½ as many 14 N 2 molecules in odd rotational states compared to even states. (b) For 15 N 2 (fermions) the ratio of the number of molecules in odd rotational states to the number of molecules in even rotational states is:

I  12I  1  3 . We would expect (2I 2  1)

1

three times as many 15 N 2 molecules in odd rotational states compared to even states.

18.34. Using equations 18.31 and 18.32, q nuc  (2I  1) 2  9 . q rot 

T T  2 r 22.86K 

18.36. As J increases, there is increasing centrifugal distortion in the molecule. The atoms get farther away from the center of mass, increasing r and therefore I. Since I is in the numerator of the expression for qrot, we conclude that as J increases, so does qrot. Therefore, because r is inversely proportional to qrot, J increases, r decreases.

18.38. PH3 is a symmetric top molecule. We can calculate qrot using equation 18.38: 1/ 2

   47 2  23 J  47 2  23 J  kg  m  1.381  10 kg  m  1.381  10 273 K  2  6.645  10 273 K  1/ 2  2  5.478  10  K K      q rot  2 2   3  34 34 1.0546  10 Js 1.0546  10 Js      









=147.3 18.40. Using equation 18.40:

 1 / 2  T3   q rot  2  0.590K 0.624K 11.5K  

1/ 2

=2182 (for 295K) and 3446 (for 400K)

18.42. Cp will always be greater than Cv, due to the additional Nk term in it. Note that this is consistent with the conclusion from phenomenological thermodynamics, which states that Cp and Cv differ

by R.

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Chapter 1

18.44. This problem is an exercise in computation! It is easiest to work using a spreadsheet. Assuming one mole: E

 v

3

NkT  ND o  NkT

 2T

2



v / T    NkT  /T e v 1

 3720 J  397000 J  17600 J  2480 J  373000 J / mole H

5 2

 v

NkT  ND o  NkT

 2T



 2mkT     h 2  

G   NkT ln 

3/ 2 



v / T    NkT  6200 J  397000 J  17600 J  2480 J  371000 J / mole  /T e v 1

   / T   v  V T     NkT1  ln  ND  NkT  ln  1  e v   2T    o N      r 

 39600 J  397000 J  17600 J  4897 J  424000 J

where the ideal gas law has been used to calculate the volume of one mole of gas at 1atm and 298K. For S, the degeneracy of the electronic ground state of HCl is assumed to be 1.

3/ 2       /T   / T   2mkT   5 T v        NK ln S   Nk ln   kT / p    NkT ln g  Nk  ln  1  e v  Nk i   h 2    2  v / T    r 1 e     

 

 154 J / K  0  0.0000873 J / K  33.1J / K  187 J / K

18.46. From Tables 18.1 and 18.3, we get v and r for H2: 6215 and 85.4 K, respectively. By calculating the reduced masses of H2, HD, and D2, we can determine the following relationships between them: H2)/(HD) = 0.5/0.66666…= 0.75 and (H2)/(D2) = 0.5/1 = 0.5. From these ratios and knowing the relationships between the v and r and , we get 5382 K and 64.0 K for v and r of HD (respectively), and 4395 K and 42.7 for v and r of D2 (respectively).

In terms of the changes in H and S, the only contributions will be from rotational and vibrational partition functions, since the translational and electronic contributions are the same, while nuclear contributions are ignored. In addition, for H, the rotational contributions are the same as well, leaving only changes in vibrations. Thus, we have for H:  5382 K 5382 K / 298 K  H  2(8.314 J/mol  K)( 298 K)  5382 K / 298 K  1 1  2(298 K) e    6215 K 6215 K / 298 K   (8.314 J/mol  K)( 298 K)  6215 K / 298 K  1 1  2(298 K) e    4398 K 4398 K / 298 K   (8.314 J/mol  K)( 298 K)  4398 K / 298 K  1 1  2(298 K) e  

H = 49701 – 28313 – 20760 = 628 J/mol = 0.628 kJ/mol

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Instructor‟s Manual





 5382 K / 298 K   298 K  S  2  (8.314 J/mol  K) 5382 K/298 K  ln 1  e 5382 K/298 K   2  (8.314 J/mol  K) ln  1 1  e   64.0 K    6215 K / 298 K   298 K   (8.314 J/mol  K) 6215 K/298 K  ln 1  e 6215 K/298 K   (8.314 J/mol  K) ln  1 1  e   85.4 K  





 4395 K / 298 K   298 K   (8.314 J/mol  K)  4395 K/298 K  ln 1  e  4395 K/298 K   (8.314 J/mol  K) ln  1 1  e   42.7 K 





S = 42.21 – 18.70 – 24.47 J/molK = −0.96 J/molK

  ln q N  1 q   NkT 2 : the expression for the translational q T  T 

18.48. Using the expression E  kT 2 

energy is derived in chapter 17 (see equation 17.56 and the expressions leading up to it), so won’t be repeated here. For the electronic contribution to the energy, we have:





D  g1e De / kT 1 E elect  NkT  NkT 2  g1   e2 e De / kT De / kT De / kT T kT g1e g1e 2

1

The degeneracies cancel, the exponentials cancel, the kT2 terms cancel. The only terms that don’t cancel are N and –De, giving us Eelect = −NDe. For (polyatomic) rotations, we have

  1 / 2  T 3 1 / 2          A B C   1   E rot  NkT 2 1/ 2 1/ 2 3  T    T      A B C   NkT

2

1

 1/ 2  1     1/ 2    A B C  

 1/ 2  T 3      A B C 

1/ 2

3  T 1/ 2 2

All of the terms (, , s) cancel, and what we have left is (after collecting all of our terms in T): 1/ 2 3 3 1 3 2 T E  NkT  NkT 2  NkT 3/ 2 2 2 T 2 T

160

Chapter 1

18.50. Using equation 18.52 and assuming one mole: 2 2  3  1900K  e 1900 K / 298 K e 1980 K / 298 K    1980K   23 J  3 C v  N  1.381  10        K  2 2  298K  1  e 1900 K / 298 K 2  298K  1  e 1980 K / 298 K 2  









e  2330 K / 298 K J  2330K   26.21   2 K  mol  298K  1  e  2330 K / 298 K 2





Using a similar procedure for 400 K, we find that Cv = 28.90

J K  mol

18.52. If the reaction deals only with isotopes, the translational and electronic partition functions cancel, and we will assume that the nuclear partition function doesn’t affect the equilibrium (which may not be the case, since different nuclei are involved. However, we will make the assumption). Thus, the only contributions to the equilibrium constant come from the vibrational and rotational partition functions. For the given reaction, we would estimate K as

q vib q rot mixed 2 K . q vib q rot  N q vib q rot  N 14

Using the high-temperature limit expressions for each q,

15

2

2

this becomes

T T  2      r  v  mixed K  T T   T T         2 r  v  14 N  2 r  v  15 N 2

The T variables cancel. On the assumption that the v 2

and r values are similar, they too would cancel and all that would remain is

K

1 4 (1 / 2) 2

Thus, isotope exchange reactions are predicted to have a limiting value of K of 4.

18.54. It might be tempting to think that the values for H and G are reasonably close to the actual values of these energies, much like the statistically-determined entropy S values are very close to the experimentally-determined values. However, one part of the molecular partition function is based on an arbitrary energy value: the electronic partition function, which is defined in terms of the dissociation energy De, which is itself defined in terms of an arbitrary zero point. Note that according to Table 18.5, S does not have that problem. Finally, for atomic systems, the total energies are equal to what is predicted from the kinetic theory of gases and does not include any contributions from electronic or nuclear energies, so H or G cannot be construed as the “total energy” of the atoms.

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Instructor‟s Manual

Chapter 19 The Kinetic Theory of Gases 19.2.

A single mercury atom has a mass of about 3.3310−25 kg. Using the formula for kinetic energy: 12.56 L 

1000 mL 1 cm 3   12,560 cm 3  1.256  10 4 cm 3 for one atom. For a mole of atoms, the 1L 1 mL

kinetic energy is 6.6610−21 J  6.021023/mol = 4010 J/mol.

19.4.

We know that

1.01325 bar 100,000 Pa 1.055 atm    1.069  10 5 Pa 1 atm 1 bar 19.6.

19.8.

1 torr 1 atm 1.01325 bar 1233 mmHg     1.644 bar 1 mmHg 760 torr 1 atm To calculate the average energy of one molecule, we can use equation 19.9:

We can use equation 19.11 to calculate the temperature:

1 bar 25750 Pa   0.2575 bar 100,000 Pa 19.10. Using equation 19.8: (a) V

(b)

162

R



R 



nT p

pV nT

Chapter 1

19.12. (a) The “pressure” of hydrogen in interstellar space can be determined by the ideal gas law:

p = 3.6810−21 atm = 3.7310−16 Pa (b) Now we can use equation 19.8 to determine the average speed of the hydrogen molecules, whose total mass is 3.3510−26 kg:

This is a fairly slow velocity, but the temperature is also rather low.

19.14.

19.16. False.

, not 5.

Latm (2.84101 molesSO )(0. 821 )(273.1517.0K)

19.18. In equation 19.23, the units on K are:pnRT 2 molK 7 1010atm. In equation 19.31, the units are: V 8 0 L1

21

Latm 5/9K Latm 0 . 0 8 2 1    0 .0 4 5 6  . molK 1 R mol R

14.7 lb lb in. in.

Recognizing that a J= 0.80  2  11.8 2 we can see that the units on K in both equations are identical.

19.20. The g functions are the one−dimensional velocity probability distribution functions.  is the three-dimensional velocity probability distribution function, which is the product of the three one-dimensional functions. G is the non-vector speed probability distribution function.

19.22. Showing that K is also equal to 0.20 

 m  m pM 4.50tor (4 .01 /g mole) 14.7 lb lb and pV MRT; V  RT  Ltor 0.01706g/L 1.7110 /g cm follows the same steps as presented  2 .

9 (62.36 )(273.1587 )K molK in.2 in.2 5 3

in the text, starting with equation 19.16 and ending with equation 19.21. The only difference is that the derivatives in the steps are taken with respect to vy or vz instead of vx. Rather than repeat the steps, the reader is referred to that section of the text.

19.24.

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Instructor‟s Manual

Using one of the integrals in the appendix, we find that:

w3 z 3 2 xyz 3 19.26. The most probable speed is: 6 xy   2 w 32 y 19.28. (a) Rubidium atoms have a mass of 85.5 g/mol or 0.0855 kg/mol. Using the formula for the most probable velocity:



3 y 2 z 2 w

(b) Arguments are left to the student.

19.30. The relative values of the three different average velocities will always be the same. In fact, their ratios are also invariant (as shown by exercise 19.25).

19.32.



3y 2 z 2  w  will always be >1 since Cp>Cv, butwill vary based on temperature (see section 2.8). For a diatomic ideal gas (air?) with no vibration, Using this value for , all of the ratios are less than one which means that the sound wave can‟t travel faster than the speed of the molecules.

This makes sense since sound waves propagate by collisions of molecules in air.

19.34. The pressure from exercise 19.6 is 3.6810−21 atm. In millitorr: 3 y 2 z w 2

2

p  n     19.36. Rearranging equation 19.40, R T  V  T, P 19.38. Equation 19.41 is derived explicitly in the text, starting with equation 19.40. Rather than repeat it here, please refer to that section of the text.

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Chapter 1

19.40. Although argon is an atom, hydrogen is a molecule. The fact that there are two atoms of hydrogen bonded together gives H2 an effective molecular diameter commensurate with those of other atoms.

19.42. The average collision frequency depends on the density (which relates how many gas particles there are per volume) and the temperature (which relates how fast the has particles are moving).

19.44. 

 = 6.421015/m3. That is, there need to be 6.421015 atoms of Xe per every cubic meter. Since 1  n    P, V

mole of Xe has 6.021023 atoms, we need  T 



- pV . That‟s RT 2

the volume of a cube that is 454 meters (just under half a kilometer) on a side.

19.46. (a) The collision frequency can be calculated using the following equation: 3





2  molecules  10dm   10  23 J  4 8.06 1019   5.00 10 m 1.38110 228K  3 2 K dm 4d kT 1   1m   z   2.84 10 7 s m kg    7.97 10 26  molecule  

(b) To calculate the number of collisions per molecule we need to know a length of time to count collisions in. Assuming 1s, we can use the answer from part a: 2.84 107 collisions. (c The total collisions per second per unit volume can be calculated using equation 19.42:

 V   V  dV   dT   dp   p    T  p  T

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Instructor‟s Manual

19.48. (a) At (roughly) 80% N2 and 20% O2, the average molecular weight of air is (0.80)(28.0 g/mol) + (0.20)(32.0 g/mol) = 28.8 g/mol. We can use the rearranged ideal gas law to determine the density of air:

  Latm     Latm    1mole 0.0821    1mole 0.0821 350K   nR    nRT    Kmole   Kmole    0.10atm dV   dT   2 dp  10.0K    2     p 1 .

08 atm p 1.08atm             If 80% of that is nitrogen, then the density of nitrogen is 2.6910250.80 = 2.151025 molecules/m3, while the density of oxygen is 0.202.691025 = 5.381024 molecules/m3.

Using equation 19.44:

 V   nRT   2 V  2nRT      ;   p   p 2  p2 p3   n ,T   n ,T That is, the mean free path between a N2 molecule and an O2 molecule is 4.610−7 m. The other mean free path is

nR   2 p   p    ;   0   2   V  T  n , V  T  n , V (b) For the average collision frequencies, we will need the reduced mass of the N2−O2 system: (28.0)(32.0) 1kg m= g´ ¸ 6.02 ´10 23 = 2.48´10-26 kg 28.0 + 32.0 1000g Now, using equation 19.45:

L2 atm 3.592 mol 2 TB   1026 K L atm   0.04267 L/mol  0.08205  mol K   and

L2 atm 1.360 mol 2 TB   521 K L atm   0.03183 L/mol  0.08205  mol K  

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Chapter 1

(c) For the total number of collisions, there is only one value:

L2 atm mol 2   433 K L atm   0.03913 L/mol  0.08205  mol K   1.390

TB

19.50. What the question is asking is which quantity is larger, zHeHe , zArAr , or zHeAr ? Since the two gases have equal concentration, we expect that the density value used to calculate z should be the 1 same. Since za d 2a , and the atomic diameter for He and Ar are almost similar, we can take m only consideration of the masses. Because Ar is more massive than He, we expect that zHeHe will be larger, therefore the number of collisions between He and He will be higher.

19.52. First we need to calculate the density of the air: Next we need to calculate 1/z:

3

3

0.0237L 1 m 5 m   2.37  10 mole 1000 L mole 19.54. From equation 19.51, we know that: 3 m3 1mole  29 m 2.37 10   3.94 10 mole 6.02 1023 atoms atom 5

19.56. Using equation 19.51:

3

4 3 29 m 10 V  r  3.94 10 ; r  2.1110 m or 2.11 3 atom That is, 2.021014 mercury atoms are diffusing through per second.

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Instructor‟s Manual

19.58. Here, we use equation 19.51 but solve for the vapor pressure. First, let us recalculate the rate of mass loss in terms of atoms per second:

0.03049L mole Now, using equation 19.51:

0.0638L mole Solve for p: p = 663 Pa = 0.00654 atm 19.60. If air is 20% O2, then the oxygen partial pressure is 210−13 torr. In the effusion equation 19.51, use 1 cm2 = 110−4 m2 as the area:

19.62. First, we need to determine the densities of the gases. If we‟re under standard conditions, let us take advantage of the approximate molar volume of a gas, 22.4 L per mole:

6.02 ´10 23 particles = 2.69 ´10 22 particles = 2.69 ´10 22 particles 3 L m 22.4L a  1  b  (a) Z  1    b    RT  V  V 

Z  1 (b)

168



B C   V V 2

2

V p 

2

RT  Vb

a T V ( V  b)

Chapter 1

 p    V  T ,n

19.64. Diffusion stops because the term    , the concentration gradient, is zero when a minor component is evenly distributed throughout the mixture.

19.66. Using equation 19.53(a) as the temperature increases, D increases. (b) as the molar mass increases, D decreases, (c) as the density of the gas increases, D decreases, (d) For an ideal gas, PM . Substituting in to equation 19.53, we find that D1/P. D should decrease.  RT 19.68. Using equation 19.56:

an 2 ( p  2 )( V  nb)  nRT V For one day, the 3D displacement will be

p V  RT

We can estimate the total distance traveled by calculating the average speed of Helium at 298K:

p

RT ,p  V

(0.0821

L  atm )(273.15K) K  mol  1.00atm 22.41L

So, in one second, the He atom would travel 1256m. In one day it would travel 109,000,000m!

Clearly the He atom is experiencing MANY collisions as it diffuses!

19.70. (a) If the ammonia were diffusing through helium, diffusion will likely be faster. That‟s because the value of (r1 + r2)2 will likely be smaller (since helium atoms are smaller than air molecules), and the reduced mass  of NH3−He is smaller as well. Both of these terms are in the denominator of the expression for D12 in equation 19.54, so D12 itself will be larger, implying faster diffusion. (b) If ammonia were diffusing through SF6, the arguments are reversed. The value of (r1 + r2)2 will be larger, and so will . Therefore, D12 will be smaller and diffusion slower.

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Instructor‟s Manual

Chapter 20 Kinetics 20.2.

If 1.00 mmol of H+ are consumed in 2 min 33.8 s (= 153.8 s), then the invariant rate of the 1 1.00  10 3 mol  4.06  10 7 mol/s . reaction is 16 153.8 s

20.4.

Since HNO2 disappears at a rate of 0.045mol/s, HOCH2CH2OH disappears at a rate of −0.045/2 = 0.0225 mol/s. NO2−CH2CH2NO2 appears at a rate of 0.0225mol/s, and H2O appears at a rate of 0.045mol/s.

20.6.

This reaction is 0th order with respect to SO2 and first order with respect to Cl2. It is first order overall.

20.8.

We can begin by comparing the first and second reactions where the concentration of H2O2 is held constant:  3.43  10 2   0.740       3   0.219   3.01  10  n

We can quickly see that n=2 by guessing small whole numbers. Next, we can look at the first and third reactions where [Mn2+ ] is held constant:  3.43  10 2 M / s   0.556M   In this case, n is also equal to 2. So the rate law for this     2  0.662M   4.86  10 M / s  reaction is: rate=k[Mn2+]2[H2O2]2. To find the rate constant, we can plug in data from any of the 1 2 2 experiments: 3.43  10 2 M / s  k 0.740M  0.556M  ; k  0.203 s  M3 n

20.10. No useful information can be obtained by comparing the first and third sets of data because two different variables are changing, the concentration of A and the concentration of B. In order to extract useful information form any two trials, only one variable can be changed so its effect on the rate of the reaction can be isolated. (However, these trials would be useful if other, additional trials have been performed.)

20.12. In this exercise, we need to find the value of k for the first condition, then use that value to determine the concentration of A for the second specified rate. Determining the value of k: 2.44  10 4 M/s  k (1.67 M) 2

k  8.75  10 5 M -1s -1

Using this value of k, determine [A] for the new rate: 1.5510−6 M/s = (8.7510−5 M−1s−1)[A]2

170

[A] = 0.133 M

Chapter 1

20.14. If a rate is given in units of M/s, in order for that to be the ultimate unit, k would have to have units of M2s−1.

20.16. We can use the integrated rate law for first order kinetics:

At  Ao e  kt A1000s  3.4  10 M e

1   1.0610  5  1000 s  s 

A7500s  3.4  10 M e

1   1.0610  5  7500 s  s 

5

5

 3.36  10 5 M  3.14  10 5 M

Ao  kt  lnA  lnA multiplying through by −1: o t At At  kt   lnAo  lnAt  ln Ao

20.18. ln

20.20. Equation 20.15 is not written in the form of a straight line, despite the fact that the equation has the y variable on the left side and the x variable on the right side. A straight−line equation has the form y = mx + b, whereas this equation is of the form y = ex, which would not plot as a straight line.

20.22. First, we need to determine how many grams of HgO needs to decompose in order to make 1.00 and 10.0 mL of O2 gas at STP. Using the fact that at STP there are 22.4 L of gas volume: 1.00 mL 

1L 1 mol 2 mol HgO 216.6 g HgO     0.0193 g HgO 1000 mL 22.4 L 1 mol O 2 mol HgO

must decompose in order to form 1.00 mL of O2. Of course, for 10.0 mL, ten times that amount, or 0.193 g HgO, need to be decomposed. If we start with 1.00 g, then there are 1.00 – 0.0193 = 0.9807 grams left over (= [A]t) after 1.00 mL of oxygen are made, and 1.00 – 0.193 = 0.807 g of HgO left over [= [A]t) after 10.0 mL of oxygen are made. Using equation 20.15: −4 s−1)t

(a) 0.9807 g = 1.00 g  e−(6.0210

−4 −1

ln 0.9807 = ln 1.00 – (6.0210 s )t −(6.0210−4 s−1)t

(b) 0.807 g = 1.00 g  e

ln 0.807 = ln 1.00 – (6.0210−4 s−1)t

Taking the logarithm of both sides: t = 32.4 s Taking the logarithm of both sides: t = 356 s

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Instructor‟s Manual

d [A]  k  dt . Integrate both sides, with the left side‟s limits being [A] 2 [A]0 through [A]t and the right side‟s limits are 0 to some time t:

20.24. Start with equation 20.19: 

d[A] t = ò k × dt 2 [A] [A]0 0 [A]t

-ò

The integral on the left side is simple 1/[A], evaluated at the limits, while the integral on the right is simply kt, integrated at its limits. We have: [A] t

1 t  kt 0 [A] [ A]0

Evaluating at the limits, we get 1 1   kt [A] t [A] 0

(which is equation 20.20).

d [A] d [A]  k  dt , which  k[A] 3 . We can rearrange this to get  [A] 3 dt 1 1 ultimately integrates to   kt .

2 2 2[A] t 2[A] 0

20.26. (a) The rate law is 

(b) If we rearrange the integrated expression to

1

 kt 

1

, we see that we would 2 2[A] t 2[A] 0 have to plot 1/2[A]t2 versus time t to get a straight line having slope k and intercept 1/2[A]02.

2

20.28. For a zeroth order reaction, the integrated rate law is written as [A]0 – [A]t = kt, which can be rewritten as [A]t = −kt + [A]0. This shows that if [A]t were plotted versus time, the slope would be –k and the intercept would be [A]0.

20.30. (a) The third life is constant for a first order reaction t 1 / 3 

ln3 / 2 k

(b) If one-third of the original amount of reactant is gone, we are left with 2/3 of the original amount. For a 0th order reaction:

Ao  At  kt Ao  2 Ao  kt 1 / 3 3 Ao t1/ 3  3k

After 3 third−lives the reaction will be complete.

20.32. For a 0th order reaction:

172

Chapter 1

Ao  At  kt Ao  1 Ao  kt 1 / 2 2 Ao t1/ 2  2k

Yes. Part b of Exercise 20.31 calculates the half−life for the reaction. It is 4.81s.

20.34. A process that has constant rate is a zeroth-order reaction.

20.36. The [HNO2]>>[C6H5OH] so we can estimate that the [HNO2] is essentially constant at the beginning of the reaction. The pseudo first-order rate law is rate k‟[C6H5OH] and k‟ k[HNO2]o =0.043s−1 20.38. Four experimentally-determined parameters can be (but are not limited to) initial amount of ethyl benzoate, initial amount of sodium hydroxide, the ratio of ethyl benzoate and sodium hydroxide, and the temperature of the system. Proper selection and understanding of such parameters is necessary when studying the kinetics of any chemical reaction.

20.40. The units on k are different based on the order of the reaction. Since the rate must be in M/s, for 1 1 1 1st order: k is in . 2nd order k is in . 3rd order k is in , etc. s M s M2 s 20.42. Rate laws are typically defined for initial reaction conditions. As the reaction progresses and eventually approaches equilibrium, the original description of the kinetics for the process probably won‟t apply. In fact, as a reaction approaches equilibrium, its net rate decreases and eventually becomes zero. As such, it‟s unlikely that a zeroth-order reaction will continue at a constant rate for two complete half-lives.

20.44. We need to determine the rate constants for the forward and reverse reactions, then take their ratio. First, let us consider the forward reaction, with only A and B present. Using the first and second trials:

1.081  10 5 M/s k (0.660) a (1.23) b  6.577  10 -5 M/s k (4.01) a (1.23) b This reduces to 0.164 = 0.165a. It should be obvious that a = 1. Using the second and third trials:

6.577  10 5 M/s k (4.01) a (1.23) b  6.568  10 -5 M/s k (4.01) a (2.25) b

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This reduces to 1.001 = 0.5467b. In this case, b = 0. Now using a set of data to determine the value of k: 1.08110−5 M/s = k(0.660)1(1.23)0

k = 1.6410−5 s−1

Next we consider the reverse reaction, with only C and D present. Using the first and second trials:

7.805  10 7 M/s k (2.88) c (0.995) d  1.290  10 -6 M/s k (2.88) c (1.65) d This reduce to 0.6050 = 0.6030d. It should be obvious that d = 1. Using the second and third trials:

1.290  10 6 M/s k (2.88) c (1.65) d  1.300  10 - 6 M/s k (1.01) c (1.65) d This reduces to 0.9923 = 2.851b. In this case, c = 0. Now using a set of data to determine the value of k: 7.80510−7 M/s = k(2.88)0(0.995)1

k = 7.8410−7 s−1

Now that we have the rate law constants, we can use equation 20.36:

K

k f 1.64  10 5 s -1   20.9 k r 7.84  10 7 s -1





[A] 0 k r  k f e ( kf  kr )t . If the reverse reaction is negligible, that (k f  k r ) suggests that kr is negligible and can be ignored in the above expression. If so, equation 20.33 [A] 0 becomes [A] t  k f e  kf t , which reduces to [A] t  [A] 0 e  kf t , which is the integrated rate law kf for a normal first−order reaction.

20.46. Equation 20.33 is [A] t 





20.48. This is a parallel reaction and the initial ratio is equal to the appropriate ratio of the rate constants(Equation 22.43), in this case k2 divided by k1 (because the product A−B is made by the k 3.95  10 4 s -1  8.98 second reaction): ratio  2  k1 4.40  10 5 s -1 The ratio is valid for any time t, as long as the reverse reactions (if any) are negligible.

20.50. First we need to solve for [A]o for the [B]eq and [C]eq equations 20.44.

( K1 + K 2 +1) [ B]eq K1

[ B]eq K1 = [C ]eq K 2 174

= [ A] o =

( K1 + K 2 +1) [C ]eq K2

Chapter 1

20.52. Straightforward linear plots of equations 20.41 and 20.42, with t on the x axis, won‟t be easy. We can try to take the logarithm of both sides of the equations (as is a common tactic), but because the logarithm isn‟t a linear function, it does not distribute itself through the two-term bracketed function that contains the exponential.

20.54. (a) Taking the derivative of [B]t in equation 20.47 with respect to t:

 k [A]   k1 [A] 0 k1t e  e k2t   1 0  k1e k1t  k 2 e k2t  t  k 2  k1  k 2  k1









Set this last expression to 0 and solve:





k1 [A] 0  k1e k1t  k 2 e k2t  0 k 2  k1

 k e

 k1t

1



 k 2 e  k 2t  0

k1e  k1t  k 2 e  k2t

Take the natural log of both sides: ln k1 – k1t = ln k2 – k2t

ln k1 – ln k2 = k1t – k2t

ln k1 – ln k2 = (k1 – k2)t

Solving for t: t

ln k1  ln k 2 k1  k 2

(b) Using k1 = 1.6010−6 s−1 and k2 = 5.7910−8 s−1 from example 20.7 and the expression from part a:

ln(1.60 ´10-6 ) - ln(5.79 ´10-8 ) 3.319 t= = = 2.15´10 6 s (about 25 days) -6 -1 -6 -8 -1 (1.60 ´10 - 5.79 ´10 ) s 1.542 ´10 s Using the three expressions in equation 20.47 and assuming that [210Bi]o=1.00M: -6

-1

[ 210 Bi] = (1.00 M)e-(1.60´10 s )(2.15´10 s) = 0.0321 M 6

(

-6 -1 6 -8 -1 6 (1.60 ´10 -6 s-1 )(1.00 M) e-(1.60´10 s )(2.15´10 s) - e-(5.79´10 s )(2.15´10 s) -8 -1 -6 -1 5.79 ´10 s -1.60 ´10 s = 0.883 M

[ 210 Po] =

)

æ æ ö -6 -1 6 -8 -1 1 ç ç ÷é( 5.79 ´10-8 s-1 ) e-(1.60´10 s )(2.15´10 s) - (1.60 ´10-6 s-1 ) e-(5.79´10 s )(2.15´1 [ Pb] =1.00M 1+ -6 -8 -1 ë ç ç éë1.60 ´10 - 5.79 ´10 ùû s ÷ ø è è 206

= 0.0850 M

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20.56. From Exercise 20.7, k1= 1.60  10 6 s 1 and k2= 5.79  10 8 s 1 .

Using equations 20.47 where A=Bi−210, B=Po−210, and C=Pb−206, -6 -1 6 é ù éë 210 ù 83 Biû = (1.000M ) exp- ë(1.6 ´10 s ) (1.00 ´10 s )û = 0.202M

æ -6 1 ö ç1.6 ´10 ÷ (1.000M ) sø éë 210 ù è exp- éë(1.6 ´10 -6 s -1 ) (1.00 ´10 6 s)ùû - exp- éë( 5.79 ´10-8 s) (1.00 ´10 6 s)ùû 84 Poû = 1 1 5.79 ´10 -8 -1.6 ´10-6 s s = 0.770M é æ -8 1 ö -6 -1 6 ù æ -6 1 ö -8 -1 6 ùù é é 1.00 ´10 s - ç1.6 ´10 exp- 5.79 ´10 s 1.00 ´10 s ú ÷ exp- ë 1.6 ´10 s ÷ ê ç 5.79 ´10 û è ë û sø sø éë 206 ù ê è ú 82 Pbû = (1.000M ) 1 + -6 1 -8 1 ê ú 1.6 ´10 - 5.79 ´ 10 êë úû s s = 0.029M

(

)

(

)(

)

(

)(

)

These three concentrations add up to 1M as they should.

20.58. (a) When k1 >> k2, the second rate constant is negligible with respect to the first. The equations reduce to

[A] t  [A] 0 e  k1t [B]t 

(no change)











k1 [A] 0 k1t e  e k2t  [A] 0 ~ 0  e k2t  [A] 0 e k2t  k1



    1 1 [C]t  [A] 0 1  k 2 e  k1t  k1e  k2t   [A] 0 1  ~ 0  k1 e  k 2 t   k1   k1 







  1  [A] 0 1   k1e  k2t   [A] 0 1  e  k 2t  k1 











(b) When k1 << k2, the first rate constant is negligible with respect to the second. The equations reduce to

[A] t  [A] 0 e  k1t [B]t 

(no change)









 

k1 [A] 0 k1t k [A] k [A] e  e k2t  1 0 e k1t  ~ 0  1 0 e k1t k2 k2 k2

    1 1 [C]t  [A] 0 1  k 2 e  k1t  k1e  k2t   [A] 0 1  k 2 e  k1t  ~ 0    k2   k2   k1t  [A] 0 1  e





176









Chapter 1

20.60. This problem needs to be answered graphically, using the techniques illustrated in example 20.8.

Your final answer will depend on the accuracy of the graph, but you should get a pre−exponential factor around 410−10 cm3/molecules·s.

20.62. If 295 K is the nominal base temperature, then a 10 degree increase would make the second temperature 305 K. At 295 K, if the rate constant were k, then at 305 the rate constant is 2k.

Using equation 20.52: ln

EA k 1   1     2k 8.314 J/mol  K  295 K 305 K 

The k‟s cancel; solving for EA:

EA = 51,850 J/mol = 51.85 kJ as the activation energy.

20.64. Using the Arrhenius equation: 9.910−12 cm3/s = Ae−1900 J/mol/(8.314 J/molK)(1153K) Solving for A: A = 1.210−11 cm3/s 20.66. ln

k 20000 J/mol  1 1      6 8.314 J/mol  K  373 K 298 K  1.77  10 /M  s

Solve for k: k = 8.9710−6 /Ms 20.68. (a) To determine the energy of the photon (which represents the activation energy):



c





2.9979  10 8 m/s  3.997  1014 s -1 7.50  10 -7 m

Therefore: E = h = (6.62610−34 Js)(3.9971014 s−1) = 2.6510−19 J On a molar basis, this is (2.6510−19 J)(6.021023/mol) = 159,442 J/mol = 159 kJ/mol (b) 3´1011 s-1 = Ae-159,442 J/mol/(8.314 J/mol×K)(310 K) Solve for A: A = 2.21038 s−1 (c) ln

k 159, 442 J/mol æ 1 1 ö = ç ÷ 3´1011 /s 8.314 J/mol× K è 271 K 310 K ø

Solve for k: k = 4.1107 s−1 20.70. (a) The first step is probably the breaking of the Br−Br bond to make two Br atom radicals.

Then, a Br atom abstracts an H atom from CH3CH3. Then, a Br atom combines with the ethyl radical to make a product, and the cycle continues. (b) The reaction would probably go faster than chlorination, since the bromine molecule is more easily broken than Cl2 is.

20.72. The “consecutive reaction” approach doesn‟t work because it doesn‟t implicitly recognize that the first reaction comes to equilibrium and that the concentration of B is essentially constant.

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20.74. If the mechanism of the chlorination of methane presented in the text is correct and the second step is the RDS, then the rate law is rate = k2[Cl][CH4]. Invoking the steady-state approximation, [Cl]2 and if the first step is truly at equilibrium: K  . Therefore, [Cl]  ( K[Cl 2 ])1 / 2 . Substituting [Cl 2 ] 1/2 into the rate law: rate = k(K[Cl2]) [CH4] = k‟[Cl2]1/2[CH4].

20.76. If the second step is the RDS, then the rate is rate = k[CH4Cl]. If the first step is in equilibrium, [CH 4Cl][Cl·] we can use the fact that K = and solve for [CH4Cl]: [Cl2 ][CH 4 ]

[CH 4Cl] =

K[Cl2 ][CH 4 ] Substitute into the original rate law: [Cl·]

é K[Cl2 ][CH 4 ]ù k'[Cl2 ][CH 4 ] as the final rate law. Note that this one still has the rate = k ê ú= [Cl·] [Cl·] ë û concentration of a relatively unstable intermediate, the Cl· radical.

20.78. We will assume that reaction one is in equilibrium. The forward reaction 1 will have a rate constant of k1, the reverse reaction 1 will have a rate constant of k−1. The second reaction will have a rate constant of k2, etc. Since the second reaction is the RDS, we can write the experimental rate law as: Rate = k2[CH3OH2+]. We can also assume that [CH3OH2+] is steady state.

d éëCH 3OH 2+ ùû dt

= -k2 éëCH 3OH 2+ ùû - k-1 éëCH 3OH 2+ ùû + k1 [CH 3OH ] éë H + ùû = 0

k [CH 3OH ] éë H + ùû éëCH 3OH 2+ ùû = 1 ( k2 + k-1 ) é k [CH OH ] é H + ùù k1k2 1 3 ë ûú rate = k2 ê = kobs [CH 3OH ] éë H + ùû where kobs = êë ( k2 + k-1 ) úû ( k2 + k-1 ) 20.80. The equivalency of the two equations is shown explicitly in the text between equations 20.64 and 20.68. Rather than repeat the discussion, readers are referred to that section of the text.

178

Chapter 1

20.82. Since V = k2[E0], let us substitute into the denominator of the second term in equation 20.68:

1 1 K 1    rate V V [S]

Multiplying the first term on the right by [S]/[S]:

1 1 [S] K 1 K  [S]     rate V [S] V [S] V [S] V [S] rate  expression: .

K  [S]

Now we take the reciprocal of both sides to get the final

20.84. When K = [S], the Michaelis−Menten equation simplifies to

1 1 [S] 1 1 1 2 V       . Taking the reciprocal: rate  . So when the constant K rate V V [S] V V V 2 equals the substrate concentration, the rate equals one half of the constant V. This is an experimental (rather than graphical) way to determine V. 20.86. k = Ae−EA/RT = (2.41011 M−1s−1)e−(142,000 J/mol)/(8.314 J/molK)(400K) k = 6.910−8 M−1s−1 20.88. The extent of reaction depends on the relative concentrations of 235U and the product neutrons, as product neutrons can participate in additional reactions. If few 235U nuclei are present, the reaction will not proceed quickly. If there is the proper balance of 235U nuclei and product neutrons, the reaction maintains a chain-reaction process. If there is an abundance of 235U present, more and more neutrons can induce more and more reactions, ultimately leading to a runaway reaction (and, in the proper circumstances, an explosion – which disperses the 235U so that the chain reaction cannot be sustained).

20.90. For Mechanism 1, the rate laws for the three steps are: rate = k1[A]

rate = k2[B][C]

rate = k3[C]

For Mechanism 2, the rate laws for the three steps are: rate = k1[A][B]

rate = k2[B][C]

rate = k3[C]

Note that the only difference is for the first elementary process.

20.92. Using the form of the Eyring equation, equation 20.80:

8.5  10 1 M -1s -1 

(1.381  10 23 J/K)(450 K)  G /( 8.314 J/mol K)(450 K) e (1 M)(6.626  10 -34 J  s)

Rearranging: 9.063  10 14  e  G / 3741.3

Take the logarithm of both sides :

G 3741.3 J/mol

G  112,300 J/mol  112.3 kJ/mol

- 30.032  -

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Instructor‟s Manual

J  J   e1.381  10  23 335K  75 .

0 3  ekT * K Kmol   1.57  1014 m exp  20.94. A  o e S / R    mol  mol  s c h  34  8.314 J  1000 3  6.626  10 J  s Kmol   m  





20.96. Refer to equation 20.85: the pre-exponential factor A is related to the relative entropy of the transition state. If A for the chlorine-containing reaction is lower than A for the brominecontaining reaction, equation 20.85 suggests that the proposed transition state, HCl2, is lower in entropy with respect to the original reactants than HBr2 is to its original reactants.

20.98. Although the expression for k in equation 20.84 contains energy and entropy terms, the energy and entropy terms relate to the transition state, not directly to the reactants or products.

Therefore, we can‟t use equation 20.84 to justify any laws of thermodynamics. However, we can use equation 20.84 to suggest conditions that would increase the value of k, thereby suggesting a faster reaction. For the EA term, a lower EA (energy of activation) will increase the overall value of k. For the S* term, a higher S* would increase the overall value of k. So while “lower energy” and “higher entropy” are concepts that can be applied to equation 20.84, they do not directly relate to the laws of thermodynamics.

Chapter 21 The Solid State: Crystals 21.2.

BN has the structure of diamond, a covalent network solid.

21.4.

It is impossible for objects with 10−fold symmetry to pack together to fill all space either in two dimensions or three. Drawings are left to the student.

21.6.

This structure is also a face-centered cubic unit cell.

21.8.

Triclinic: 1 Monoclinic: 1,2 (in order of the chart) Orthorhombic: 1,2,4,2 Tetragonal: 1,2 Rhombohedral: 1 Cubic: 1,2,4 Hexagonal: (1/6)*12+(1/2)*2=3

21.10. Let us consider the face−centered cubic method of close−packing. For a face−centered cubic unit cell, there are four radii going from corner to opposite corner in a face. That means that 4r =

180

Chapter 1

(2)1/2a, where a is the unit cell parameter. Thus, r 

2a . The volume of a sphere that has 4

4 3 4 (2) 3 / 2 a 3 2 3 2a  a . There are four atoms total in an fcc unit cell, so the r is r  3 192 24 4 2 3 2 3 a  a . Numerically, this volume is total volume taken up by the atoms is 4  24 6 0.74048…a3, compared to the a3 volume of the cube. Thus, the close−packing of atoms in a unit cell uses up 74.048…% of the volume of the cube.

21.12. In simple cubic unit cells, the maximum radius is ½a. Therefore, the volume of that sphere is 4 4 3  (1 / 2 a) 3  a  0.5235...a 3 . Thus, in simple cubic unit cells, a maximum of only 3 24 52.35…% of the volume is taken up by the atoms.

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Instructor‟s Manual

21.14. First, determine the mass of a single unit of ZnSe. Using a periodic table, we find that ZnSe has a mass of 65.39 + 78.96 = 144.35 amu per formula unit. Now, let us determine the volume of the unit cell. For a cubic unit cell, the volume is (5.669 Å)3 = 182.2 A3. Knowing that mass = density  volume, we calculate for a single unit cell:

mass  (5.263 g/cm 3 )  182.2 A 3 

1 cm 3 1 amu   577.5 amu 24 3 10 A 1.6605  10 -24 g

Thus, there are 577.5 amu per unit cell. At 144.35 amu per ZnSe unit, this means we have 577.5  4 ZnSe units per unit cell. The type of cubic cell that has four units is face-centered 144.35 cubic.

21.16. First, we will have to determine the volume of the unit cell using equation 21.4:

volume  (5.287 A)(9.158 A)(18.95 A)  1 - cos 2 90  cos 2 99.50  cos 2 90  2 cos 90 cos 99.50 cos 90 volume = 917.53 0.972759… 1/2 = 904.95 Å3 The mass of one formula unit of talc is 3(24.3) + 4(28.1) + 12(16.0) + 2(1) = 379.3 amu. Four times this (since there are four formula units per cell) is 4379.3 = 1517.2 amu. Thus, the density of talc is:

d

1517.2 amu 10 24 A 3 1.6605  10 24 g    2.78 g/cm 3 amu 904.95 A 3 cm 3

The experimental density of talc is 2.7 – 2.8 g/cm3, in excellent agreement with the calculated value.

21.18. With one angle of 120º (implying that the other internal angle is 30º), you can fit three of the unit cells into a hexagonal shape. Such three−piece collections can fill three−dimensional space much like a honeycomb can fill space. Therefore, while the unit cell itself isn‟t hexagonal in shape, hexagonal−shaped combinations can be found in such crystals.

21.20. The formula weight of turquoise is 813. Next we can calculate the volume of the unit cell using equation 21.4:

V  7.424Å 7.62Å 9.910Å  1  cos 2 68.61  cos 2 69.71  cos 2 65.08  2 cos68.61 cos69.71 cos65.08  461Å 3 d

813  1.6605  10  27 kg  10 24 Å 3  1000g  g      2.93 3 3  3  1 461Å  cm  1cm  1kg  21.22. When n = 0, the process is simple reflection, which is wavelength−independent. Diffraction (even diffraction through transparent media) is a wavelength−dependent process.

21.24. If a first order diffraction occurs at 30º or larger, then we can show that the wavelength used is equal to or greater than the d spacing:

182

Chapter 1



 = 2dsin

 = 2dsin(30º)

  d (since sin 30º = 0.5)

At the limit where  = d, what angle will the second order diffraction diffract at? Substitute into the Bragg equation: 

 = 2dsin 

sin  

2  1 if  = d.

2d

This would correspond to a diffraction angle of 90º. If  were greater than d, sin  would be greater than 1, according to the Bragg equation. However, the sine function can‟t have any value greater than 1, leaving us to the conclusion that there won‟t be any second−order diffraction if the first−order diffraction is greater than 30º.

21.26. Assuming first−order diffraction: 

 = 2dsin

1.54056 Å = 2(3.615 Å)sin 

sin = 0.21307

 = 12.3º

Of course, with a face−centered cubic unit cell, there will be other diffractions from other planes of atoms, but this will be the diffraction caused by the atoms at the corners of the unit cells.

21.28. Miller indices (100), (010), and (001) describe planes that contain two of the three crystal axes.

Each set contains a different pair of crystal axes.

21.30. Each of the atoms in the plane represent half of a cell axis (for all three directions). Thus, the coordinates of the atoms are (½, ½, ½). Since Miller indices are reciprocals of the plane‟s intersections with the unit cell axes, the Miller indices of the indicated plane are (222).

21.32. If the lower right corner were selected as the origin, then the plane would be intersection the a axis at 1, the b axis at 1, and the c axis at ½. The reciprocals of these intersections yield the Miller indices (112). This is the same answer as was found in Example 21.7, suggesting that the Miller indices for a given plane are independent of which point is selected as the origin (as long as the identities of the a, b, and c axes remain the same).

21.34. We use equation 21.9 and the Miller indices given: (a) 1.544 A  2  (b) 1.544 A  2  (c) 1.544 A  2 

2.88 A 12  0 2  0 2 2.88 A 12  12  0 2 2.88 A 2  12  0 2 2

sin 

sin   0.26805

  15.5

sin 

sin   0.37909

  22.3

sin 

sin   0.59939

  36.8

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Instructor‟s Manual

21.36. First we need to determine d and 1/d2 for the two angles: 13.48° 15.62°

d

1/d2

3.1690 2.744

0.0996 0.133

0.0996  0.75 . The crystal is face−centered cubic.

0.133

21.38. If Br has the sodium chloride structure, then it‟s a face−centered cubic lattice. Thus, we use the last column of Table 21.3 to determine which sets of Miller indices will diffract X rays.

According to the table, the (111), (200), (220), (311), (222), and (400) planes will be the first six diffractions. Therefore, the angles will be, using equation 21.9: sin  

 h2  k 2  2 2a



(1.54056 A) 12  12  12  0.20245 2(6.59 A)

(1.54056 A) 2 2  0 2  0 2 sin    0.23377 2(6.59 A)

  13.5

sin  

(1.54056 A) 2 2  2 2  0 2  0.33060 2(6.59 A)

  19.3

sin  

(1.54056 A) 3 2  12  12  0.38767 2(6.59 A)

  22.8

sin  

(1.54056 A) 2 2  2 2  2 2  0.40491 2(6.59 A)

  23.9

sin  

(1.54056 A) 4 2  0 2  0 2  0.46754 2(6.59 A)

  27.9

  11.7

21.40. Copper atoms and zinc atoms are almost the same size. (In fact, they are adjacent in the periodic table). It would be expected that their abilities to scatter X rays are very similar. Thus, the diffraction pattern of a CuZn alloy might look as if it were caused by a pure substance having a simple lattice, rather than a two−component mixture composed of similar elements.

21.42. Use Table 21.4 to obtain the ionic radii of the ions involve, and take the ratio of the smaller to larger ions.

rsmaller 1.33 A   0.6786 . Therefore, KBr should have the rlarger 1.96 A sodium−chloride structure, face−centered cubic.

(a) For potassium bromide:

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rsmaller 1.33 A   0.7964 . Therefore, CsF should have the rlarger 1.67 A cesium−chloride structure, simple cubic.

(b) For cesium fluoride:

rsmaller 1.31 A   0.9776 . Therefore, BaO should have the rlarger 1.34 A cesium−chloride structure, simple cubic.

(c) For barium oxide:

21.44. In its normal elemental state, sulfur exists as cyclic S8 molecules. As such, it is more difficult for it to adopt a space−efficient, close−packed crystal structure.

21.46. Consulting Figure 21.7, it should be easy to determine that the coordination number in the cesium chloride structure is 8.

21.48. Despite the fact that in diamond the carbon atoms are making four strong bonds to neighboring atoms, the graphite phase is more stable under normal conditions. This suggests that the relative stability of an allotrope is not easily related to the unit cell of the allotrope.

21.50. Lattice energy is a potential energy because it is an energy of position: it is the energy decrease due to the proximity of positive ions to negative ions. It is certainly not energy of motion!

21.52. (a) According to Table 21.4, the ions in CsCl are (1.67 + 1.81) = 3.48 Å apart. Therefore, according to Coulomb‟s law, the energy of attraction is (1.602  10 19 C)(-1.602  10 19 C)  6.63  10 19 J  6.02  10 23 /mol 4 (8.854  10 -12 C 2 / J  m)(3.48  10 -10 m) E  399,000 J/mol  399 kJ/mol E

According to equation 21.13: (6.02  10 23 /mol)(1.7627)(1.602  10 -19 C) 2 (1)(1)  0.331 A  1   3.48 A  4 (8.854  10 -12 C 2 / J  m)(3.48  10 -10 m)  E  636,000 J/mol  636 kJ/mol E

(b) According to Table 21.4, the ions in ZnS are (0.74 + 1.84) = 2.58 Å apart. Therefore, according to Coulomb‟s law, the energy of attraction is (2  1.602  10 19 C)(2  -1.602  10 19 C)  3.58  10 18 J  6.02  10 23 /mol 4 (8.854  10 -12 C 2 / J  m)(2.58  10 -10 m) E  2,153,000 J/mol  2,153 kJ/mol E

According to equation 21.13: (6.02  10 23 /mol)(1.6381)(1.602  10 -19 C) 2 (2)(2)  0.289 A  1   2.58 A  4 (8.854  10 -12 C 2 / J  m)(2.58  10 -10 m)  E  3,132,000 J/mol  3,132 kJ/mol E

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Instructor‟s Manual

(c) According to Table 21.4, the ions in TiO2 are (0.68 + 1.31) = 1.99 Å apart. There are also two Ti−O interactions per formula unit. Therefore, according to Coulomb‟s law, the energy of attraction is (4  1.602  10 19 C)(2  -1.602  10 19 C)  1.85  10 17 J  6.02  10 23 /mol -12 2 -10 4 (8.854  10 C / J  m)(1.99  10 m) E  11,160,000 J/mol  11,160 kJ/mol E  2

According to equation 21.13: (6.02  10 23 /mol)(2.408)(1.602  10 -19 C) 2 (2 2 )  0.250 A  1   1.99 A  4 (8.854  10 -12 C 2 / J  m)(1.99  10 -10 m)  E  11,750,000 J/mol  11,750 kJ/mol E  2

21.54. Using Equation 21.13 and recognizing that there are two Ti−O interactions per formula unit: 2 1  2  23 19  6.022  10   2.408  2  1.602  10 C  mole   0.250A  Lattice Energy  2   1   2 1.99A    12 C  10 4  8.854  10  1.99  10 m  Jm  

E  11,750,000J / mol  11,750kJ / mol

The experimental lattice energy is −12150kJ/mol so they are comparable (3.3% difference).

21.56. Generally, the higher the charge density on the ion(s), the greater the lattice energy.

21.58. The presence of H or H2 in certain vacuum systems would present a major contamination problem. Hydrogen atoms and molecules are so small that they can be easily incorporated into the lattice of solid elements (i.e. making a defect), altering their physical and/or chemical properties.

21.60. The drawing is left to the student. For an n−type semiconductor, germanium (or any other Group IVA element) can be substituted for Ga, and selenium (or any other Group VIA element) for the As. For a p−type semiconductor, calcium (or any other Group IIA element) can be substituted for Ga, and germanium (or any other Group IVA element) can be used to substitute for As. 21.62. Holes conduct electricity by serving as sites to host moving electrons. The flow of electrons in one direction is equivalent to the flow of holes in the opposite direction.

Chapter 22 Surfaces 22.2.

186

Since force is equal to mass x acceleration:

Chapter 1

1000 mL 1 cm 3 12.56 L    12,560 cm 3  1.256  10 4 cm 3 which is the unit of work.

1L 1 mL dyn 1N 100 cm N ´ 5 ´ = 4.355´10-1 cm 10 dyn 1m m

22.4.

435.5

22.6.

Assuming that the change takes place around 20°C, and using the information in Table 22.1:

1233 mmHg  22.8.

1 torr 1 atm 1.01325 bar    1.644 bar 1 mmHg 760 torr 1 atm

Alternately, surface tension can be defined as

1 cm 3 125 mL   125 cm 3 (In this last derivative, the variable A is used to r 1 mL epresent both the Helmholtz energy and the area!)

22.10. The surface energies of gases are so small that we can safely ignore them. Since surface energy and surface tensions are based on unbalanced forces between particles, surface tensions and surface energies of gases are miniscule under most conditions, as forces between gas particles are very small – certainly smaller than those forces in liquids and solids.

22.12. Let us use some of the results from Example 22.3. The amount of gravitational work done by the falling razor blade is 2.8710−8 J, and the area involved is 7.74 cm2. Therefore, the approximate limit to the surface tension needed to float a razor blade is

25750 Pa 

1 bar  0.2575 bar , which is a fairly low surface tension. However, we 100,000 Pa

conceded that this is an estimate, and for all practical purposes the limit may be substantially higher.

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Instructor‟s Manual

22.14. (a) At first approximation, raindrops should be spherical due to surface tension. (b) Raindrops would experience a terminal velocity because the drag of the atmosphere is balancing the force due to gravity. That compensating force is acting on the bottom of the droplet, pointing up. Thus, the resulting shape of the drop would be a distorted sphere, with a flattened bottom, ultimately resembling an oblate spheroid.

22.16. You won’t get the Laplace−Young equation because simply writing the area of a sphere in terms of its volume and evaluating

¶A is simply an exercise in geometry. The Laplace−Young ¶V

equation was derived by realizing that the first law of thermodynamics holds for surface energies as well as any other type of energy. The derivative in terms of volume and area is a consequence of the energies involved (i.e. p−V and −A), not due to the shape of the system.

22.18. You would think that we would be able to minimize evaporation by increasing the external pressure. The external pressure is pII, and as pII increases the pressure difference decreases.

However, equations 22.14 and 22.15 give expressions for p that are ultimately dependent only on the surface tension and the radius of the droplet, suggesting that the external pressure may not affect the p across the interface.

22.20. Using equation 22.14 (because we’re considering a droplet, not a bubble), and converting the radii into centimeters: (a)

R



pV nT

This is about one−hundredth of an atmosphere. (b)

R 

( 2.66 bar)(27.5 L) L  bar  0.0830 (1.887 mol)(466.9 K) mol  K

This is about 9 and one−half atmospheres.

22.22. Using equation 22.14:

x atm 1 atm  64.0 m 10.1 m 22.24. Looking at equation 22.15 we see that the pressure differential is proportional to the surface tension. Since the surface tension for pure water (~70 erg/cm2) is greater than that of soapy water (~24.3 erg/cm2), we expect a soapy bubble of the same radius to last longer since the pressure differential is less.

22.26. The drawing is left to the student.

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Chapter 1

22.28. We know from equation 22.17 that:

(1.04 atm)(67 L)  9 .4 L (7.38 atm)

(a) If we are trying to increase the wettability of the surface,  needs to be close to 0 so cos should approach one. We need to adjust the surface tension of the liquid such that 1mole SO 1.82 10 g SO   2.84 10 moles SO .

64.07 g 13

2

11

2

2

(b) If we want to decrease the wettability of the surface,  needs to be close to 180° so cos L  atm should approach negative one: p  nRTV  (2.8410 molesSO )(80.082110 Lmol  K)(273.15 17.0K)  710 atm.

11

2

10

21

22.30. Rearranging equation 22.19: 0.0821

L  atm 5 / 9 K L  atm    0.0456  . Substituting with the given data, and using known mol  K 1 R mol  R

values for water:

0.80  22.32.

0.20  N/m2 = 1 Pa)

14.7 lb lb  11 .

8 in.2 in.2

14.7 lb lb  2 .

9 in.2 in.2

(where we have used the fact that 1

22.34. Pressure times volume (essentially the left side of equation 22.21) yields energy (or energy per mole) as the ultimate unit. In order for the other side to have energy (or energy per mole) as the ultimate unit, the temperature variable must be multiplied by the conversion factor between energy and temperature. At the atomic level, this conversion factor is Boltzmann’s constant, k.

At the molar level, it’s the ideal gas law constant R. 22.36. In a cubic crystal, atoms are closest together in one of the planes that contains two of the three axes.

22.38. The data in Table 22.2 suggest that for similar ionic compounds, as ions get smaller, the surface energy increases.

22.40. On earth, there are so many gas−phase species present that are adsorbed onto a surface that two connecting surfaces are actually separated by layers of adsorbed species. In space, without a large ambient pressure, little such gas−phase species are present, and two surfaces have the opportunity to bond more directly with each other.

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Instructor‟s Manual

22.42. One monolayer corresponds to 1 L – one−millionth of a torr of pressure exposed to a surface for one second. Thus for a pressure of 1.010−4 torr, a monolayer builds up in m pV   RT; M

m pM   V RT

4.50torr (44.01g / mole) (62.36

L  torr )( 273.15  87 K ) mol  K

 0.01706 g / L  1.71  10 5 g / cm 3

time = 0.01 s Thus, in about one hundredth of a second, a “clean” surface will have a monolayer of adsorbed molecules on it. It won’t be clean for long!

22.44. (a) heterogeneous catalysis (b) heterogeneous catalysis (c) homogeneous catalysis (d) homogeneous catalysis (e) heterogeneous catalysis (unless the transition metal catalyst is soluble in the carbon, making an alloy; this is unlikely) 22.46. Early attempts were probably based on simple physisorption, since poly(tetrafluoroethylene) is rather chemically inert. Chemisorption would provide a better−adhering film. Today, a surface is roughened up by sandblasting and a primer is applied. The Teflon is then embedded into the primer. Thus, ultimately Teflon coatings are still physisorbed onto a surface, but the surface is better prepared so the physisorption is more durable.

2

22.48. If we look at equation 22.48, a plot 3 y 2  y z

3

should give a straight line if this data

w

follows a Langmuir isotherm. In this case the amount of adsorbent is varying and it is a little more complicated to calculate the values for the y axis. Since m is the change in mass of the adsorbent due to the adsorbate this is proportional to the concentration change of the urea from initial exposure to equilibrium. We must divide this change in concentration by the amount of adsorbent used. So to determine if this experiment follows a Langmuir isotherm we must

z3 xy 2 z 3 . For this data set, you do indeed get a line, although  32 y w2

plot: 3w

2

the r−squared value is only around 0.9. The adsorption process may be more complex.

190