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Student Solutions Manual
1
for
Physical Chemistry
David W. Ball
Cleveland State University
THOMSON
*
BROOKS/COLE
Australia­ Canada> Mexico· Singapore> Spain· United Kingdom. United States
CHAPTER 1. GASES AND THE ZEROTH LAW OF THERMODYNAMICS
1.1. The drawing is left to the student. The calorimeter, water bath, and associated equipment
(thermometers, ignition system, and so forth) are the system, while the surroundings are
everything outside the apparatus.
(b) 4SOC + 273.15 = 318 K
(c) 1.055 atm x 1.0~
bar x 10;~:
Pa = 1.Q~)Os
Pa
(d) 1233 mmHgx ltorr x 1atm x 1.01325 bar
1mmHg 760 torr
1atm
= 1.644 bar
l cm"
(e) 125mLx­­ = 125cm 3
'lmL
(f) 4.2 K­273.15 =­269.0°C
1bar
(g) 25750Pax
= 0.2575 bar
.
100,000 Pa
.
~
1.5. In terms of the zeroth law ofthennodynamics, 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.
0.0250L. =!J)4 xl 0­ 4 ~
(33.0+273.l5)K
.he temperature.) tfthe volume is going to be 66.9 cnr' =
For this sample orgas under these conditions, F(p)
.....~
(Note the conversion to. .:'
0.0669 L: 0.0669 L
.
.
//' lOC ­4
.04 x 10
,~
1.9. There are many possi
8.314
34~' :
hL/
. \.0 4,-x II) ~
{c::...
K.
.>'f7;/
'.
0 versions.
=_
{
L
.
Using the fact that 1 cal = 4.184
J
x 1cal = 1.987 cal
mol.K 4.184J
mol.K
,­
. ­"
~.
..
-~'_
J~'
_­­­-II_
i \;
.._.....
1.11. Calculations using STP and SATP use different numerical values of R because the sets of
conditions are defmed 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
•
lb
1.13. The partial pressure ofNz = 0.80x 14:71b =11.8.
m.
m.
7
. I pressure 0 f 0. z = 0 .20 x14.
2 9 ­.
lb.
The partia
. lb =.
.
m.
m.
1.15. (a) This is an equation for a straight line with slope = 5, so at x
simply 5.
d
(b). The slope of this function is given by its first derivative: ­ (3x 2
dx
5, the slope is 6(5) ­ 5 = 25. At x = 10, the slope is 6(10) ­ 5 = 55.
(c) The slope of this function is given by its first derivative:
·
s1ope IS
­ ­7 ;
25
1.17. (a)
at x = 10, thee slone
s ope iIS
(BVJ
Bp
T,n
= ~(nRTJ
BT) =~(PV)L BV nR nR
(BV
(d)
BT r Rn=)T (~=
»
(:) = ­(
(e) .
er v
tJ
p
r, =..!!.­,
and using data from Table 1.6, we have:
bR
.
for CO2 : Tn =
.
3.592 e a ~
mol.
(0.04267 Llmol{0.08205 Latm)
\.
molK
. ..
= 1026 K
1.360 I!a~
TB =
= 521K
mol
(0.03183L1mol{0.08205 Latm)
\.
molK
2
.
TR=) n(~=
(.Bp)
van
Jv.
dp dV
. (~dTJ dV· dp vJ or (!!-(dT)
2.. Using
/
/
:2).
At x
= 5, the
.
1.19.
p
.
5x + 2) = 6x ­ 5. At x =
....
2T
(Bp)
p,n
­
nRpan.
(b) (BV)
r,» = ~(nRTJ an p = RTp
=­
Bp p
7
100
­ ­
~
= 5 and x = 10, the slope is
T,V
(c)
an v
v
1.390ea~
forN z: Tn =
mol
= 433 K
(0,03913 Llmo/ 0.08205 Latm)
'\
molK
.
1.23. The C term is
or
e.
;2 .
In order for the term to be unitless, C should have units of (volumej',
The C' term is C'pz, and in order for this term to have the same units as p V (which would
be Latm), C' would need units of ~ .
(The unit bar could also be substituted foratm ifbar
atm
units are used for pressure.)
1.25. Gases that have lower Boyle temperatures will be most ideal (at least at high
temperatures). Therefore, they should be ordered as He, Hz, Ne, N z, Oz, AI, CI­Lt, and COz..
rn Let us assume standard conditions of temperature and pressure, so T= 273.15 K andp =
\~
atm. Also, let us assume a molar volume of22.412 L = 2.2412
have for hydrogen:
=1 + B = 1+
15 cm3:o~
RT
V
2.2412xl0 em Imol
compressibility. For HzO, we have:
pV
X
104 cnr'. Therefore, we
=1.00067 , which is a,0.067% increase in the
(O,~
J~.
,-~
.t. 0
0 ";;;
3/mol·
=
pV = 1 + ­=
B ·1'·
. th e
+ -1126cm
4
3
=09497
'.
, whi Ch'IS a 50°/
. 10 d ecrease In
RT
V
2.2412xl0 em Imol
compressibility with respect to an ideal gas.
~
­
~Y
comparing
th~ tw~
expressions from
+(0 ­ ­!!:­)
1 +f( b) + ~ .. and Z = 1 +'~
RT V
V
the~\
C
C + ..,\
v' V 2
~
it seems straightforward to suggest that, at the first"approximation, C = bZ• . Additional terms
involving V 2 may occuririlater terms of the first expression, necessitating additional
corrections to this approximation for C.
f
1.31. In terms ofp, V, and T, we can also write the following two expressions using the cyclic
rule:
=
(av)
er
p
C~fl ap )
(
av
and
T
(avJ
ap
=
T
(:1
(aT)
avo
There are other constructions possible that would
p
be reciprocals of these relationships or the one giveri in Figure 1.11.
3
1.33, Since the expansion coefficient is defined as
"
-.!-(8V)
, a will have units of.
V M
. p
1
volume
1
­­­ ­­­­­ =
, so it will have units ofK­ 1, Similarly, the isothermal
volume temperature temperature
is defime d as ­ ­1 (av)
its of
0
­
, so K. WI'II h ave units
compresslibili
I ity IS
V
EJp
T
1 . volume
volume pressure
=
I
pressure
,
or atm­I or bar'] ,
1.35. For an ideal gas,
.
K
= _-.!-(av) = _.!-~(nRT)
V ap
T
V EJp P
=...!.. nRT , Since nRT = V, this last
V p2
P
,T
ion b ecomes ­
1­
V = ­I lor
c. an Iidea1 gas, Th e­"expresston
expressIOn
-a.IS eva 1uated as
Vp p
P
TTl
(av) T ­
a (nRT) T­nR, For an Ideal
, gas, the Ideal
, gas law can be
-a
pV P
p =­­
p V ­er =­
pV et ­p­ =­
' ­nR =­,
V
b
. .IS
rearranged to give
so we
su '
stitute to get th at thiIS Iast expression
p
:v ,
~
T
p
which = ~
Thus, the two sides of the equation ultin.Iately yield the same expression and
so are equal.
",
­ RT
1.37. For an Ideal gas, V = ­ . Therefore, the expression for density becomes, substituting for
p
the molar volume, d =
temperature is
(:a. ~-=
(ad)
aT
p.«
M" = pM . The derivative of this expression with respect to
RTf
p RT
= ­ P~
. Using the definition of V,. this can be rewritten as
RT
4
CHAPTER 2. THE 'FIRST LAW OF THERMODYNAMICS
=F· s = IFIlsl cosO
work = 30 N . 30 m· cos 0° = 900 Nrn = 900 J.
work = 30 N· 30 m· cos 45° = 900·0.7071 Nrn =
2.1. work
(a)
(b)
.
2.3.
.
= -Pext8V = ­­c(2.33 atm)(450.mL ­ 50.mL)x
W
640 J.
1L
1000mL
.' 101.32 J
.
= 94.4 1.
lL·atm·
= 0.932 L· atm x
2.5. (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.
(d) The work would be greater if the process were irreversible.
o
(kli7~'"
£,rGtf\.!'/lN-~I"
It­I lifz T-(,/·t~
lk't;l#z.u­
I'
.
2.7. These three compounds experience hydrogen bonding between their molecules. Because it
requires more energy to overcome the effects of this hydrogen bonding, these compounds have
hi=:ecific heat capacities than other, similar­mass molecules.
I '?!l1,./q~
,
5­1 ~
1"IlP~ ;
""'\11­{­11?~';:Wf 1l" -~t°
1".< ~1/' lVL
01
i,.&t\;. ­u­1;:.nIi. OZf· r« =t­­L­ .tocbV'~
2.9. First, calculate the energy needed to warm the water:
q = m·c·I:1T= (1.00xl0 5 g)(4.l8 J/g·K)(L.Q.Q.K.1, ~ l8xl 05 J
y,~tl-1
?f!AL
~,}/i'"
rVL~f1_;
5 "r~
~
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)(9.8l m/s 2)(2.00 m) = 392.4 J. Therefore,
4.18xl0 5 J
# drops =
= 1070 drops.
392.4 J/drop
2.11. The verification of equation 2.8 follows from Boyle's law, which says PYi
can be rearranged to give Vf
= Pi
Vi
Pf
W rev
= -nRT~
= P[Vf
.
This
. Substituting this into equation 2.7:
Vf becomes Wrev = -nRTln Pi (which is what we are supposed to verify);
Vi
Pf'
2.13. Equation 2.10 is not a contradiction of equation 2.11 because equation 2.11 is applied for
systems in which the total energy does change. This can happen for open or closed systems.
. Equation 2.10 only applies to isolated systems.
V
.
OOOIL
-nRTln---:L = ­(0.245mol)(8.3l4J/mol·K)(95.0+ 273.15 K)ln­'­Vi
!.OOOL
W= 5l80J
2.15.
W=
5
.
~.t;
2.17. 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!
vt".:l.,..
2.19. Temperature is a state function because an overall change in temperature, is determined
solely by the initial temperature and the final temperature, not the path a series oftemperature
changes took.
.2 . First, we should determine the number of moles of gas in the cylinder. Assuming the ideal
aw holds:
pV
(172atm)(80.0L)
"
"
PV =nRT can be rearranged to n =­ =
RT (0.08205 L· atm)(20.0+ 273.15 K)
mol·K
n = 572molN 2 gas
(a) The final pressure can be determined using Charles' law:
,
PiTf
r, =T=
(172 atm)(140.0+273.15 K)
(20.0+ 273.l5K)
Pi __ PI
1'; Tf
Pf=242atm
~
(b) w = 0 since the volume"of the tank does not change.
q = n-c-S'I> (572 mol)(21.0 J/mol·K)(140.0°C ­ 20.0°C) = 1.44x106 J
I1U= q + w = 1.44x106 J + 0 = 1.44x106 J.
.23. w=-nRTl~(0.5mo)8314
+:
" J
VI'
17;
" 0 10L '
)(5.0+273.15K)1h '=+2690J
mol­K
1.0L
.
q = ­1270 J (given)
I1U= q + w = +2689 J ­1270 J = +1420 J
Mi = I1U + 11(pV)Since the process occurs at constant temperature, Boyle's lawapplies and
"
11(pV)= O. Therefore, Mi= +l420 J.
."
0!j'
,
In terms ofpressure and volume: dU
Up ,dp +
(au)
av
dV .
p
(aH) ,dp + (aH)
-
For enthalpy: dH = ­
~
(au)
=
ap
II
av
dV.
p
In order for each term to have units of J/mol·K for each term, the first term has units
\1bR!I­K; the second term has units J/mol·K2 ; and the third term has units J·K/mol.
6
Q(OHJ ==(O(U + PV»)
~
op T . Op ~(T
which is zero at constant
~
2 3.
1
~=
)T
+(OPV) ==O+(OPV) ==(OPV) ==(ORT)
OPT
OPT OPT OpT
v~·
temp ra u ~
Therefore,
.(~H) op
is zero .
T
This derivation is given explicitly in the text in section 2.7 .
. ForRe:
e.
T == ­2a ==
(0~825
Rb
ForH2:
2a
T == ­ ==
~
Rb
~
2(0.244 L2 • arm/mol")
. '.
== 224 K (compared to 202 K from text)
,
(0.08205 L· atm/mol­ K)(0.0266 Llmol)
.
.
~,)bl.
.
: Because the presu~
change isn.'t !90 dras~ic,
our ~swer
to exercise 2.34 is probably
hin a few degrees ofbemg correcr­­tf a truly isenthalpic process can be arranged.
2.~
"
~
, . 2(0.03508
arm/mol")
36 I K (compared to 40 K fr'om text)"
==.
L· atm/mol­ K)(0.0237 Llmol)
"P'(i6\
,
.
..
-:
~.41?
..
b~,
1M;," ~ \ t :
.
2.37. Because strictly speakmg, heat capacities are extensiveproperties; they depend on the
am t of matter in the system.. Thus, the form in equation 2.37 is the most general expression
/ t re "" the two ~s.
..
L~al
2 . First, calculate
pressure and assume that this is the exte
(a good approximation; since t
co stant throughout
~== nRT == (0.145mol)(0.08205L·atmlmol·K)(273.15K) == 0.650
P
V
5.00L
~ : res ion
Now we can calculate work as -Pext~V:
.
.
w == ­(0.650 atm)(3.92L ­5.00 L)x
101321·
== +71.1 J
l Lvatm.
To determine ~U,
we need to calculate q first. We need the final temperature, which can be
determined by Charles' Iaw (since pressure is constant):
V; == Vf
1'; Tf
Using q = n·c~T:
~u
5.00L _ 3.92L
,273.15K
Tr '>
T ==214K
f
_____
q =(O.l41\r~;zuJ2#mg-59
Thismeant~=(214-73.l5)9'K
K) == ­178 J.
= q + w == ­178 J + 71.1 J = ­107
.)
2.41. Starting with ­ R In V (I v; == C v lnT (I T,: both logarithm terms can be evaluated similarly.
Since In(a) -In(b) = In (alb), this
The logarithm on the left is evaluated as ­R(ln(Vr) ­In
(Vi»'
7
simplifies to -R·ln(VtlVi), which is the left side of equation 2.44. The right side gets evaluated
and simplified similarly.
R Performing a similar substitution as in
r
C
=
C
p ­
y
c,
7
5
-R--R
= 2
'2
~R
2
2
-R
2 = ­",­,
2/2 =­
2 f or ann 1id ea1diiatomic
. gas.
=­5­
-R 5/2 5
2
2.4 . If the melting process occurs at standard pressure, then Mf = qp = 333.5 J'{from Table
To correct for the volume change, we need to calculate the volumes of both water and ice
at 0 C:
1
,
1mL '
1.00016 mL
for water: 1gram
,
0.99984g
1mL
for ice: 1gram x
= 1.0907mL Therefore, fl.V = 1.00016 ­­: 1.0907 = ­0.0905 mL
0.9168g
.
~
Using the equation fl.U = Mf + fI.(p,)1) = Mf + pfl.V (for constant pressure):
JMO: -Dl)~6ptf
1L' )(10132J)
AU = 333.5 J + (1 atm)(­0.090,mL),
m1.
.
= 333.5 ­ 0.0090 J ~ 333.5 J
.,
1000
1Lvatm
x=
This shows that fl.U and Mf can be very close, ifnot v
phase processes.
~
m~
2.47. Steam bums hurt
ofheat as the heat of vaporization.
y
e same, for many condensed­
.
water bums because steam gives up a considerable amount
' e h eat of fusioh
~1\. given up by th e fr eezmg
. water can b e transferre
~d
in part
part) to th e
(at least m
2.49. Th
citrus fruit, keeping them warmer and (hopefully) keeping the fruit itself from freezing.
Arx"H = LfI. [H(prods)-LfI. [H(rcts)
=(2mol)(26.5kJ/mol)­ 0­ 0 = 53.0kJ
aRe03 (s) ­7 Na (s) + Yz H2 (g) + C (s) + 3/202 (g)]
2 Na (s) + C (8) + 3/202 (g) ­7 Na2C03 (s)
C (s) + 02 (g) ­7 C02 (g)
H 2 (g) + Yz O2 (g) ­7 H 20 (1)
-2xfl.rH= +1901.62 kJ
fl.rH= ­1130.77 kJ
fl.rH= ­393.51 kJ
ArB = ­285.83 kJ
.I
r--:
This yields the overall reaction (you can verify that), and the overall fl.rxJl is the sum of the
values on the rightfl.rxnH = 91.51 kJ.
2.55. Since process is constant­volume, qv = AU = ­31,723 J.
.
----
>
8
w ~ 0 since the process is in a constant­volume calorimeter. To determine MI, we need to know
9
the balanced chemical equation for the combustion ofbe~d:
.CJI,COOH(s) +
~+
For every mole of benzoic acid combust~er
1.20 grams of benzoic acid are ~gx
(
l
:a
~-'
3H,OOj
q ­ 15/2) ~es
~
isa Ch~ .sof
(~
!-~.Q
~ ~ : ~ ~ ; n rmbe:F-clino~s)X(
mo~fbenz01c
.~,
~0.5)
of g ~
..............-aCI
.= .-0,~492Il1;()f
. Therefore, the
gas.__ :qsing the
= AU +.6(pV) =.J1U + t:.(nRT) = t:.U + (.6ri)RT, we can determine the Mi of the process:
Ml= ­31,723 J + (­0.00492 mol)(8.314 J/mol·K)(24..6 +273.151<.) = ­31,723 ­ 12.2 = ­31,735 J.
. Mi
J
R'
2.57. This problem is very similar to Example 2.19, so we will follow that example, taking data
from Table 2.1.
­./ VI)
~
The heat needed to bring the reactants from 500°C (or 773 K) to 298 K is:
Mil = q = (2 mol)(2 glmol)(l4.304 J/g·K)(-475 K) + (1 ~0.918
J/g­K)(­475 K)
!~
=­41131J
The heat of reaction is 2x(AfH[HzO(g)]) = 2 mol x ­241.8 kJ/mol = ­483.6 kJ = tJlh
,
The heat needed to bring the products from 298 K to 500°C (or 773 K) is: .
Mi3 = q = (2 mol)(l8.0 glmol)(l.864 J/g·K)(475 K) = +31,870 J
Q
t:.
verall t:.rxnH is the sum of these three parts. Converting all energy values to kJ:
=­41.131 kJ ­ 483.6 kJ + 31.87 kJ =­491.9 kJ
.
.
.
.
{j( t4./M/
9
.
.
CHAPTER 3. THE SECOND AND THIRD LAW OF THERMODYNAMICS
3.1. (a) Spontaneous, because ice's melting point is DOC. (b) Not spontaneous, because ice's
in
melting point is DOC. (c) Spontaneous, because potassium compounds are generally so~le
water. (d) Not spontaneous; an unplugged refrigerator should warm up. (e) Spontaneous,
because ofthe effect of gravity on the leaf. (f) Spontaneous, because both Li (s) and F2 (g) are
rat~el.
elements. (~)
Not spontaneous, because water does not break apart into hydrogen
and oxygen without some mput of energy.
3.3. e=- wcyc/e =_(­334­115+72+150)1 =_ ­227 ~ 0 . 2 6 7 = 2 6 . 7 %
850J
q\
.. e ·
35
= l _ ~ ow
Thigh
­:,>~
850
0.440 = 1------'=-(150 + 273.15) K
0.560= T/ow
423K
T.low = 237K=­36°C
Superhea\el steam has the advantage of a higher temperature, so (hopefully) there will he a
a heat engine using superheated steam.
higher efic~yor
m~tion
machine is a machine that creates more energy than it
ne definition of a p~
ses. Ifit were to do so in an iso1atea'ystem, it would violate the first law of thermodynamics.
.erpetual motions of this sort are not known (and probably never will be).
i
~l
defint~s
~:rln
~:ch
~ac9mpnled
e~y
IS
~
_ _~ .
o~
are apl~JP
real-g~
m{{ependent oft]}elype of~ tl
as well ~ ideal gases.
involved m a p~s.
A better statement of the secondlaw of thermodynamics includes the conditions under
the second .law is ~W7tly
apyVf,able: for an isolated system, a ~<t¥1eous
change is
by anincrease m entropy.
2, 1
.....v I ~
.
units are standard, so let us simply substitute into the proper expression
3.17. The number of moles of air being breathed in depends, of course, on the temperature of the
295 K. (You may assume a slightly
air. Let's assume a normal room temperature o~
different temperature, but the final answer probaoyon't be too far off.) Under those.
.
10
./
conditions, 1 liter of air at 1 atm pressure is
n = pV
=
(1 atm)(1 L)
= 0.0413 mol air. The entropy change of 0.0413 mol
RT (0.08205 L . atmlmol K)(295 K)
of gas undergoing a pressure change from 760 torr to 758 torr is
­4
Pf '
758 torr
M = -nR In­ = ­(0.0413 mol)(8.314 J/mol· K) In
= +9.05 x 10 JIK.
Pi
760 torr
3.19. Since the sample isa real gas, the change in entropy is probably greater than it would be
than if it were an ideal gas. Therefore, let us calculate the entropy change assuming it is an ideal
gas and state that the true entropy change must be greater than this. We separate the total
entropy change into two parts, a change­in­pressure part and a change­in­volume part:
M. = ­(1mol)(8.314J/mol·K)In 1 atm·,.~52JIK
for the change in pressure. '
230atm'\,j
,
,
195 em"
,
M 2 =(1mol)(8.314J/mol· K)In ~ 3 =+43.8 JIK for the change in volume.
lcm V
ore, for ~real
gas we can
The overall entropYfQaIige is­A.S'.­==::d5.2 + 42:8 ~
~nfltug!.
.4 J .
­
'­.,.,
'-\)~
~.'
~P-.R:i§Jltcy
~
of an.ideal gas under conditions of co stant pressure. However,
the process in Example 3.3 is not a constant­pressure J2rocess. Therefore, while the value ofthe
heat capacity
may latn~irepx a i oce l
valu; for an ideal gas it probably wouldn"t
gIven
be
~
"
3.23. 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) -7 Ar, He (6.50.L; 298 K., 1.50
atm)
Ar, He (6.5,0 L, 298 K, 1.50 atm) -7
AI, He (20.0 L, 298
11
~.
)
~,-l_
M mix = ­(8.314 J/mol· K)[(0.245 mol)ln(0.6l) + (0.153 mol)(ln(0.384)] = +2.20 JIK
The entropy change for the expansion step is:
V
. 20.0L
M
=nRln-.L=(0.398mol)(8.3l4J/mol·K)ln
=+3.72JIK
exp
V
. 6.50L
I
Therefore, the total entropy of the process is M= +2.20 + 3.72 = +5.92 JIK.
3.25. (a) According to ~hf:um.Jw,
the energy of the isolated system remains constant.y .
Therefore, any energyIQsl by onepart of the system will be gained by another part of the system.
of
In this case; the hotcopper will lose energy/and the cooler water will gain energr:_.1n te~§
the second law; t1}is_~()Il.t~! 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.
E)
~ost=
~
,
.
J
.
<m-e ·/),T = -on-c ·(Tf ­1';):: -(5.33g)(0.385-)(Tf ­372.85 K) = -2.052Tf + 765.107
~
g·K
Heat gained =
+ m- c­ /),T = m- c· (Tf
.'
J
-
1';) = (99.53 g)(4.l8-)(Tf
g·K
-
295.75 K) :: 416.03Tf -123,042.470
Equate the two quantities and solve for Tf
-2.052Tf +765.107= 416.03Tf -123,042.470
.
"
­­­
Tr 296.13 K = 23.0°C.
29613K
.
T=­0.473JIK=entropylossofCu
(c) M::mcln-L=(5.33g)(0.385J/g·K)ln .'
~
7;-'
372.85 K
I
123,807.577 = 418,082T/
(d) M
=meln Tf = (99.53 g)(4.18 Jig ·K)ln 296.13K =O.534JIK = entropy gain of water
T;
295.75K
(e) Th=­­e­to­tal­=­­entropy change for the system is M = ­0.473 + 0.534 = 0.061 JIK.
(f) Because the over~n.e tropy
change of the isolated system is positive, so we would expect
equalization oftemperatures ­ would be spontaneous.
­,".
that the process ­ ~
'"=,
~
~en rgy,
.
.
3.27. Since (for an isolated system) the first law of thermodynamics prohi~ts
the crJalk,n of
the concept of'vyou can't win" maybe used to cW\veXY­tah-yle ruc~a.ilf
fact.
Since the second law ofthermodynamics requires an efficiency ofless than 100%, yon will
always get less energy out of a process than the energy going into that process. Thus, ''you can't
even break even" may be a way to convey that idea.
3.29. If S = kin 0, where 0 is the number ofpossible combinations, the number of
combinations can never be less than 1 for any real system. The logarithm of I is zero, so it may
be possible that the entropy of a system is zero. However, numbers greater than 1 have positive
(which it is), then the absolute amount of entropy S can
logarithms, and if k­is a positive c o ~
never be negative. (This does not preclude that changes in S might have negative values.)
12
\,~{-l
l.Q;
\~ l
Ytt-.
~
3.31. (a) the dirty kitchen (b) the blackboard with lriting on it (c) 10 gram of ice (d) If
perfectly crystalline, both have the same entropy (zero) (e) 10 grams of ethanol at 22°C.
~
3.33. ~inde
a liquid at absolute zero, it would not have ~ropy.
because according to the third law oftbennodvnamics, only a perfect ryst
have zero elW:opy at absolute zero
That's
(that is, a solid) can
3.35. Unlike the listings of !lfH and !IN, S for elements are not zero because the condition isn't
what's required for Sto equal zero: a perfect crystal at 0 K. Much of the tabulated
thermodynamic data is for something close to room temperature, like 25°C (298 K).
3.37. The balanced chemical reaction is 2 Al (s) + Fe203 (s)
!lS = [50.92 + 2 27.3)] ­ [2(28.30) + 87.
'i39 For the formation of H 20 (1), L\S= 9~
~. 0 ation of H20 (g), !lS = 188.83 ­ 130.. ­
e formation of the gas has a /).S that is hi
the different phase ofthe product H20.
3.41. L\S = nCln Tf = (800 Ib)(2200
1';
Ah03 (s) +' 2 Fe (s)
~
.-uQ,6~
(205.14)= ­163.29 JIK. For the
1 2 (205. ~
­44.42 JIK. The difference is that
y 118.87 JIK. The reason for this difference is
gl~)(0.45
JIg· K)ln 923 K = 9.09 X 105 JIK
293K
3.43. At 37°C = 310 K, the number of moles of gas is
n = pV =
RT
(1 atm)(1 L)
= 0.0393 moles. The change in entropy is then
(0.08205 L· atm/mol­ K)(31 OK)
.
!lS = ­(0.0393mol)(8.314J/mol·K)ln 590mmHg = 0.0827 JIK
.
760mmHg
--'--' --.....
--.,.,-........<,
'f~1-
~
e
I,
J­;
.... ~r"
__ •. ~.
-" --",,-"-, •.....
-:
~b
,~:( j. ")
\-~
­t
'2.(1{~)J
--.-
+- ~\
;"'1.-"
­ ­._.. ­­­._­­­­.­...-_.
'l.
__ .'.
­
'- '. ~
~
_"
~-
......---)
(1­ )
(2­ S 1,?6
__ .... .._... _. __.-­_.. ........
.
o. -l~
.. b ~
­ nl
('.P)
~"
...........
~
I
tI- ]
-.
G~ ,q ,
13
lO5\'(~
-~
)
I--k..-o t')"') ( ~ e.9§j
-)
1 'ho
+ 8'7
__._._- -,.
V\
Qn 7'
@
CHAPTER 4. FREE ENERGY AND CHEMICAL POTENTIAL
4.1. 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 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 Ml must be zero.
4.3, The total change in U or H is required to be negative for these spontaneity conditions, not
anyone component (as given by a partial derivative).
4.5. 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 ::; O. Under conditions
of constant Tand V, the second and third terms are zero, so this spontaneity condition simplifies
to
(dA)T,v::; O.
­r­'
4.7. 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.
4.9. The reaction can do up to ~3
an la~osi
/4.1
(vl,llll/tU,
\
kJ of work for every mole of H 2 0 formed.
U= 1lH= 0 (for
process). To determine the pressure at the bottom ofthe
if the water pressure­­iiicreases by 1 atrn for every to.55 meters of depth and the depth is
0 30 meters, then the pressure increase is 10,430 m x 1atm = 989 atrn. If this is the pressure
­,
~.
~)01
"inCreaSe, then the pressure at that depth must be ~?"_:'=j.-fitrn
Therefore:
.
~_.·990atIn
.
Jlmol·
K) In
­15;1001 .
w~
d'R~=-:<."18!4
K)(~2:
Since thisis a reversible J?o.ce~ls
M ~R
~
.. -
.~;
In p',
i
=~
.
'1";.
~qu:jlseLFi!Y,
for!':.S:
990 atrn
mol)(8.314 J/mol· K) In 1 atm
~np-
.
=+57.3 J / K . .
.
.
. .
...
4.13. For NaHC0 3 (s) ~ ~a+
(aq) + HC03­ (aq),!':.G = [­261.88 + (­586.85)] ­ (­851.0) =
+2.3 kJ.
For Na2C03 (s) ~ ­2 Na+ (aq) + col- (aq),!':.G = [2(­261.88) + (­386.0)] ­ (­1048.01) =
+138.3 kJ.
4.15. 6.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 25°C, not DoC.
Solid H20 is not thermodynamically stable at 25°C, so a lot of data is not given for that phase
under that condition. However, since we know that O°C is the normal melting point ofH 20
under standard pressure, we expect that the value of I1G would be exactly O.
14
4.17. M is zero for thecomplete Camot cycle, since A is a state function and,the cycle returns
to its original conditions.
4.19. Analogous fo equation 4.26:
A=U-TS
:.U:=A+TS
One of the relationships we can determine about A is (::) y
= ­S .. We can substitute this for
the entropy variable in the second term: U:= A ­ T( aA ) . The differential form of the equation
aT y
.
is
aA ) y dT for our
. final answer. We don't use the chan~_f
dU = dA ­ ( aT
1: tceps!~iw
to
volume.
4.21. To show that
(au)
av
:=
-p has consistent units on both sides of the equation, let us look at
s
.
.the units distribution through the equation. U has units of energy (J), V has units of L, and p has
units of atm. Thus, we get
.!­L =atm, which does not appear to make much sense.
But recall that
we can convert J to L'atm, so we can substitute Latm for J in the numerator:
Lvatm atm
. . .
--= = atm, so the umts are consistent.
L
1
4.23.
:J) =(~
eo (~
( ; ( : )) = ( ; (
8(X;
8(X; y»)) = (~
(1))= 0
y) ) ) = ( ; (1)) = O.Since the two derivatives
are equal, they are exact
differentials.
(~
(b)
:))
=( (~ 8(X'; Y'»)) ~(
( ; (: ) ) = ( ;
(2Y)) =0
(8(X'; Y'»)) = ( ; (2X)) = O. Since the two derivatives are equal, they are -
exact differentials.
(~
(c)
1­7" .
:))
-r)~"
~i
o V~
.z<. 'S"" h /!-
"'' 1:
=( (~ 8(X;;'»))
=
(~ (nx"y'-' l) = n'x'-'Y'-'
( ; (: ) ) = ( ; (
B(X;:'))) =(; (n['y' l)
=
n'x·-'y'-'. Since the two derivatives are
equal, they are exact differentials.
(d)
(~
:))
=( (~
B(X;;')
))= ~(
(nx"y'-'
l) =mnx"-'y'-'
(mx--' y' l) = mnx"?'y'-'.
( ; ( : )) = ( ; ( B(X;:')) }= ( ;
Since the two derivatives are
equal, they are exact differentials.
e) ( ; (: ) ) = ( ; (
(
(~
:)) =
B(YS:(XY)))) = ( ; (y' eos(xyl)= 2yeos(xy)- y'xsin(xy).
(~ (B(YS:(XY)))) = (~
~t
CoH'i.'i)
(sin(xy) +xy eos(xy)))= yeos(xy) + ycos(xy)-xy' sin(xy)
Since the two derivatives are different, they are not exact differentials.
4.25. Starting with dH = TdS + Vdp, divide both sides by dp and hold T constant:
aH)
( ­Op T
T as
-)
=Tas- + V = V(1 + Op
All we need to do
V ap
.
IS
show that a.T =
1 dS
.
­­
­ . Using
the
V dp
definition of a.:
..!..(av)
=..!..(_ as)_ ,where we have used a Maxwell relation as a substitution.
Op
.
V aT
V
Multiplying by T: aT = T (-- as) ,which is the desired relationship. Therefore,
V Op
"a =
~
p
C:l
T
=V(I-aT)
4.27. Simply put, by multiplying through by the denominator in the partial derivative, we get.
that d(llU) = -(tyJ)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 II .
o demonstrate the cyclic rule ofpartial derivatives, let us evaluate the derivatives listed in
e 1.11 using the ideal gas law:
"
=nRV
(av) = ~(nRT)
aT
aT p
nRT and substitute:
.
P = ­7
(aapv) = Opa(nRT)
=~(nRT)
v
aT
p
T
16
"
=
nRp
_ap_ ) _ (aVlaT)p
( et v ­ (aV lap)T
YerJ~}g
_nR_=_ (nRlp)
V
-(nRTl p2)
V
-L [: s­ )r
4.31. Subs ituting for the definitions ofq and K:
~
K
ev
=
as T
(l/V)
(JP .(av)
~V
__________/'
I
T
~
V,.l
the cyclic rule.
(IIV)
orpV=nRT,
whichreducesto­nR_=+p
ap
.
,Y,
v ,/
-:..
,t
(~
-;r'/
Using a Maxwell relati?nship, substitute for the fin,al partial
as)T
~
JY:
01
T'
<l
r"
r:
~
(f!-)
<lJ7. '(
e, and rewrite the denominator as its reciprocal in the numerator (note that the 1 Vs
(ajJ) (aT
( av)
aT p av T Op
.
,
Inspection ofthe three partial derivatives shows that the three variables
V
are organize . such a way as to give the cyclic rule ofpartial differentiation, so that the product
d~ eriVat vesi - l:
+1)~ .
',')/, //
iJ'(( (/
( \, \ (',
\ \","
:. ~< "
~-'
',,0- or '
to a Maxwell relationship,
(fJp)
as
<
er
/ V)
fl
'-]<v
- b)
2
'~
V
, ,
= __
p
P;;-_
'or a van der
nI!J
~als
=_(aT)
av
~
fY' ,,\,\
.
gas:
For an ideal gas,
p :
~
1=.~
=­__
1 [(­ 2~n
Therefore: _(aT)
J(w ­
J]. Here,
nb)+ (p + an:
V
V
o take the derivative of the two V­containing terms using the chain rule.
nR
av
'11l{! "
p
4.35 ..The derivation of a Gibbs­Helmholtz expression for A is exactly the same as that given hi
section 4.7 in the text, except that A is substituted for G and U is substituted for H. Rather than
redo the entire derivation, here we provide the final answer:
:r(,)~
6
.37
2-U~ =
"
~(A)
,aT T p
U
=_ 2 or
T '
.
= nRTln A,
fu
RT
i'
(0.988'mol)(8.314 J/mol.K)(350 K)ln 25.0 L = "':'967 J .
V,:,...,
35.QL
Thel ubstitution for volumes instead of pressures was made using Boyle's law.
J9.
Starting with a natural variable expression for A, we know that A varies with volume as
oA ) = _p.
( oV
T
We rewrite this as dA =
-pdV.
17
Integrating both sides: M = ­
JpdV .
Substituting for p from the ideal gas law: M
.
M
=­ Jn~T
dV, which integrates to
V
=-nRT In ­!..., which is the desired expression.
Vi
4.41. The change in the chemical potential is zero because the molar Gibbs free energy does not
change for a single­phase, single­component system if the amount ofthe component changed.
The molar Gibbs free energy. or chemical potential, is an intensive property. The total Gibbs
free energy changes (because it is an extensive property), but J..l does not.
~
4.43. dG =-SdT + Vdp + JiN2dnN2 + Jio2dn02' The first two terms can also be broken down into
th e ind.,ividual entropies of the two components an.d the change in the pressures of the two
comp ents (assuming that dT and V are the same for the two gases).
,
\,~
".'.
. 1.
fiL__.//
-
bar equals 0.987 atm. Therefore, going from 1.00 atm to 0.987 atm:
= RTln p f =(8.314 Jzmol­ K)(273.15 K) In 0.987 = -29.~
J. This may seem like a
Pi
1.00
lot. but most chemical processes occur with energy changes in the thousands ofjoules, so the
difference is relatively trivial.
~
#;niriat
4.47. (a) .1 mol of H20 (g). (b) 10.0 g of Fe at 35°C. (c) the compressed air.
4.49. Oxygen should deviate from ideality more than helium at any pressure. although as the
pressure decreases the behavior ofboth gases should approach that of an ideal gas.
18
CHAPTER 5. INTRODUCTION TO CHEMICAL EQUILIBRIUM
5.1. A battery that has a voltage is not at equilibrium because there still exists a driving force for
a reaction to occur. On the other hand, a completely dead battery is at equilibrium because there.
is no driving force for a chemical reaction to occur.
5.3. (a) Rb+ and OH­ and H2 would be the prevalent equilibrium species. (b) NaCI (xtal) would
be the prevalent equilibrium species. (c) H+ (aq) and cr (aq) would be the prevalent
equilibrium species. (d) C (graphite) should be the prevalent equilibrium species, but from
experience we know that C (diamond) is chemically very stable once formed.
~
The minimum value of ~ is 0 (as it is for any reaction). The maximum value can be found
which reagent is the limiting reagent:
.
~et rmin g
O.l500Lx 2.25 molHCI = 0.3375mo1HC1
..
100 g Zn x 1molZn = 1.53 mol Zn
65.4g
.
L
The reaction requires a 1:2 molar ratio of Zn to HC1, but we have a 1:0.221 mole ratio, so HCI
Therefore, according to the definition of~:
, will be t h e ~
~
= 0 -~.375
= 0.1688
will be the maximum value of~.
5.7. The heme + CO reaction lies farther towards products. In fact, one can show that the
equilibrium constant for the reaction heme­O, + CO ~
heme­Cf) + 02 is
18
4
23/9.2x
10 = 2.5x10 , indicating a strong preference by heme for CO ­ explaining its
2.3x 10
potential toxicity.
5.9. False. pO is the standard pressure for gaseous substances in a system, defined as either 1 atm
or 1 bar.
~
(a) The equilibrium constant for this expression isK =
1 1/2' S03 doesn't appear
Pso, 'Po2
in the expression because it's in a condensed phase. (b) From the data in the appendix: l!Go = ­368 ­ [­300.13 + Y2(0)] = ­68 kJ. (c) /)'Go = -RTIn K. ­68,000 J = ~(8.314
Jzmol­ K)(298.15 K)
InK
\Y'
InK=
-68,000J/mo1
(-8.314J/mol· K)(298.15 K)
InK=27.432
K=8.2x10 1l • (d) Thereaction
would move toward the direction of more products.
5.13. For example, consider the following gas­phase reaction:
2 H 2 (g) + O 2 (g) ~
2 H 20 (g)
Its equilibrium constant expression is K =
P~20
2
PH2 • P02
H2 (g) + Y2 O 2 (g) ~ ~
19
•
Now, divide all coefficients by 2. We get
H 20 (g)
For this reaction, the equilibrium constant expression is K =
PH20 1/2 • The exponents on all
PH 2 • P0 2
of the partial pressures are half ofwhat they were in the first expression, meaning that all of the
terms in this expression are the square root of the terms in the original expression. Therefore, K
= (K)1/2 when the reaction itself is halved.
5.1 . The reaction will reverse when I1G equals O. Let P be the pressure at which I1G = 0: .
~=
2,800+RTln4. Simplify and solve forp: 32,800 == (8.314)(298.15)lnp-2
p.p
IIp-2
32,800
= 13.232
(8.314)(298.15) .
reaction to reverse.
=
lnp == ­6.6161
P = 1.34 X 10­ 3 atm or bar
for the
5.17. 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 ofthe gases involved
in the reaction do not change.
5.19. I1 rxnOO = [­41.9 ­73.94] ­ [2(51.30~
228.61] = 10.2 kJ.
10,200 J/mol = ­(8.314 J/mol· K)(298.15 K) In K
~
lnK=~4.1l 8
InK
=_
10,200 J/mol
(8.314 J/mo1· K)(298.l5K)
... K= 1.63xl0·2 .
9
Using the 11(? from the previous problem: I1Go = -RTln a dia , and substituting from
apa
equation 5.14:
2900 J/mo1 = - RT[ Vdia (p ­1) _Vgra (p ­1)]. The RT terms cancel, leaving us with
RT
RT
2900 Jzmol == -[Vdia (p ­1) ­ Vgra (p ­1)]. Now we need to substitute the molar volumes of
diamond and graphite. Using 1 mol C = 12.01 g:
20
1 3
12.01gx cm x
1L 3
2.25 g 1000 em
= 5.34x 10­ 3 L
for the molar volume of graphite.
= 3.42 x 10­ 3 L
for the molar volume of diamond. Substituting:
3
12.01 g x 1cm x
1L 3
3.51 g 1000cm
2900Jlmol=­[p.42x10­3 Umol)(p-1)-(5.34x10- 3 Umol)(p-1)]1
.
~
p
\
-H--
\+:
:
14,900 atm.
..".­_.,.
5.25. (a) ~Go=-(8.314J/ml·K)295nx0Z
~Go=
10.96kJ.
~ Assuming that the salt is completely soluble, [Na+] = 0.010 m. To determine the .
. .
concentrations ofHS04­, H+, and so,", we first assume that HS04­ starts at a concentration of
0.010 m,and that some amount -x- dissociates, leaving O.OIO­x. The amounts of
and
W are therefore +x. We therefore set up the following expression:
sol-
­K = 1.2x 10- z = o.O~x_and
solve for x using the quadratic formula: x = ­0.0185 or 0.00649.
e reject the negative root: x = 0.00649.­ Therfo;-tmalcOi~ns
.00649 m and [HS04­] = 0.010 ­ 0.00649 = 0.00351 m.
~
are [:H+] = [SOl-]
5 . 2 . Use equation 5.20 and letK2 = 2KI:
. I' = -100,000J/mol( 1 - - 1
KI
8.314 J/mol· K 298 K t;
'.
J
'ln2­
- ­100,000 J/mol ( 1 - - 1
8.314 J/mol· K 298 K 1;
J S 1ve .~ lor T'
0
2.
. (1n2)(8.314) '=_1__ ~
T =293 K.
­100,000
298 Tz
2
To determine a temperature for the equilibrium constant is 10 times the original value, let K 2 =
lOKI and perform the same calculations:
1n lOKI = -100,000J/mo1( 1
KI
8.314 J/mol· K 298 K
(ln10)(8.314) = _1__ ~
­100,000
298 t;
-~J'
t;
~J
1n10 = ­100,000 J/mol ( 1
8.314J/mol·K 298K
t;
Solve for T2:
T = 282 K.
2
For a reaction whose Mi = ­20 kJ, repeat both sets of calculations using this value of Mi.
Letting K 2 = 2KI :
1n 2KI - - 20,000 J/mol (
KI
1
8,.314J/mol·K 298K
­
~J
1n2 ­ -20,000J/mol( 1
1
­ 8.314 J/mol· K 298K ­ Tz
t;
J S(jlv~
for T2:
(ln2)(8.314) =_1
1 T2 =274 K.
­ 20,000
298 T;
To determine a temperature for the equilibrium constant is 10 times the original value, let K2 =
10K! and perform the same calculations:
I lOKI
n~=
1
- 20,000J/mol ( 1
8.314J/mol.K 298K ­ Tz
J
In10 = ­ 20,000 J/mol ( 1
8.314J/mol·K 298K
21
_1
T2
J Solve for T2: .
-'~
(ln10)(8.314)
­ 20,000
=­ 1 ­ ­1
298
T2 = 232 K.
T2
5.29. The easiest way to show that equations 5.18 and 5.19 are equivalent is to take the
differential of liT and substitute:
d (l I T) =-(1 I T 2 )dT. Substitute this into equation 5.19:
dIn;
-(lIT )dT
or
=_ Ml
and bring the -(l/r) term to the other side: dInK
. dT
.
R
=_(_1)(_
Ml)
T
R
2
dInK
MI ,whiICh iIS equation
' 5 . 18.
.dT = RT
2
5.31. If 1.0 mol of glycine is made into 1.00 L of solution, let us assume that the original pH is
7.00 and calculate the amount of protonated glycine is in the solution. According to equation
5.21:
K 1 = [gly][H+]
.. [glyH+]
=10­2.34 •
.
If [gly] = 1.00 M and
[W] =
. 0 f[ g IyH+]: 10 ­234
concentration
. = -'"~_:. (1.00)(1.00 x 10­
.
.
. ,_
[glyH+]
Similarly, according to equation 5.21:
.. K 2
= [gly­][H+] =10­9.60 •
[gly­]:
. [gIy]
1O~9.60
1.00xlO­7 M, we can calculate the
7
)
Again, assuming [W] == 1.00xlO­7 M, we can calculate the resulting
== [gly­](1.00xlO­
(1.00)
.
7
)
[gly'] == 2.51xlO­3 M.
22
CHAPTER 6. EQUILIBRIA FOR SINGLE COMPONENT SYSTEMS
6.1. (a) 1 component. (b) 2 components. (c) 4 components. (d) 2 components. (e) 2
components.
6.3. FeCb and FeCb are the only­chemically stable, single­component materials that can be
.>:
made from iron aiOCh~
6.5. The water is boiling because the v~e
of.the.water equals the
,­­­.ambient pressure
inside the ~e.
By drawing back the plunger of the syringe enough, we can reduce the
pre~sw:
on the w~ter
sUf~iently
sc(that.­the vapo;rressure and ambient pressure are:~!,_)
which IS the physical requirement fO~-'2I
. 'tr.::f\,
....._­.
", ~;q:" )
,,,l,!L;I\; "
6.7. Any substance has only one normal boiling point. The normal boiling point is the boiling
point when the ambient pressure is 1.00 atm, and all substances only have one temperature where
their vapor pressure is 1.00 atm.
6.9. (a). M:/-shouldbep~v
Mf sh~ld
be n~e
because e!~.ID'pasl0
go.into a solid in order to sublime it. (b)
because energy has to be removed from a gas to condense it.
r1\'vI.
6.11. The heat offusion given up by the freezing water can betransferred (at least in part) to the
citrus fruit, keeping them warmer and (hopefully) keeping the fruit itself from freezing.
'.
6.13.
AC'
LW
= ~ Q'+30,700J
='
T
Trouton's rule.
(80.1+273.15)K
. edby
= +86.9 JIK , fairl
Ian y c1ose to w h at wou ldbe predict
.
.
6.15. Using the defmition of entropy for thisisothennal change: 6S = ~
1K=
510,400J/mol
sub SItU
. mo·
----t · tin'g: 1247JI
T
q
T
=
H
!!.
fw
T
and
T = 4093 K or 3820°C.
6.17. The derivation of equation 6.12 from 6.11 assumes that the molar enthalpy change, Mf ,
and the molar volume change, !!.V , do not vary with temperature.
,
6.19. First, we will need to calculate the molar volumes ofrhombic and monoclinic sulfur:
l cm'
1L
.
256.48 g x
= 123.9 cnr' x
3 = 0.1239 L for the molar volume for rhombic sulfur
2.07 g
1000 cm
".~
~
3
256.48 g x 1cm = 130.9 em' x
1L 3 = 0.1309 L for the molar volume for monoclinic sulfur
.'
1.96g
1000cm
In going from monoclinic to rhombic sulfur, the change in molar volume is 0.1239 ­ 0.1309 L =
!',.p
M
!',.p' 1.00 Jzmol­ K 1L· atm
=
x­­­0.0070 L. Using the Clapeyron equation: ­ ~ ­= ~ r­
!!.T ·!!.V ~ . 5
K
­ 0.0070 L
101.32 J/,
23
/
~I
«
j<-: ~
-.
/ 7!.(~y,·
j\·t~
.
1~C=;:fi3-atm.Th@euI,
presu-b¥_ao~tm,
should.be
to make rhombic sulfur the stable fonn at 1Oo~
en
.
~.-')
..
~
6.21. (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. (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.
6.23. Using the fact that 6V was ­0.0070 L for the phase change in exercise 6.19, and the fact
that the monoclinic­to­rhombic phase change must have a Mf of ­0.368 kJ/mol:
)In
!¥J = (Mf)ln Tf
= ( ­368 J/mol
(100+ 273.15) K x 1L· atm !¥J =;:; 6.3 atm, so that
AV
~.
­0.0070Umol
(95.S+273.15)K lO1.32J
increasing the pressure by 6.3 atm to 7.3 atm should make the rhombic phase more stable. This
is the same answer we got in exercise 6.19.
6.25. No, the behavior of chemical hot packs cannot be described using the Clapeyron equation
or the Clausius­Clapeyron equation because supersaturated solutions are not equilibrium
systems, nor does the process involve a phase change (it involves a solubility change).
·t
6.27. Ifwe use dp = Mf
as the form ofthe Clausius­Clapeyron equation, we can substitute
.
. ~RT
.directly the values given in the problem:
5
.dp = (71,400 J/mol) . (7.9 x 10­ bar) = 7.8 x 10­6 barlK.
dT (8.314 Jzmol­ K)(22.0 + 273.15 K)2
6.29. Since, according to equation 6.16, the vapor pressure is related to the exponential of the
temperature, if the temperature is increased linearly, then the vapor pressure will increase
exponentially. Thus, at high temperatures nearing the boiling point, even small changes in
temperature can lead to large changes in vapor pressure.
6.31. (a) ill order for the term rdA to have units of energy, and dA has units of'rrr', r must have
units of J/m2 • (b) The derivative of A, in terms of r; is dA = 87tr dr and the derivative of V in
terms of r is dV = 41tr2 dr. Substituting into the right side of the equation:
2dV = 2(4nr
­­
r
r
2
dr)
=8nrdr =d.
A
()
ri ,
C
By sunp
. 1y rearrangmg
. th e expression,
.
we can get
dV :;: dA . r . (d) According to the expression dA ::: 2dV , droplets with smaller radii will
2
r
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.
24
6.33. For equation 6.18,
(OJ!)
aT
= -8, the units of I! are J/mol
and the unit on Tis K, so the
p,n
overall unit on the left side is Jzmol­K, which are the units for molar entropy. For equation 6.19,
(aapJ! J
=
V, the units of I! are J/mol and the unit onp is atm or bar, while the unit on the molar
t:»
volume is Llmol. However, if we remember that the unit J can be written in terms of Latm, then
the units of I! can be written as Latm/mol, which when divided by the unit atm equal units of
Llmol, which are the units of molar volume.
6.35, Figure 6.6 shows 12 phase boundaries.
6.37. If there were such a thing as a single­axis phase diagram, equation 6.
probably be
rewritten as 'degrees of freedom = 2 ­ P', because there would be 0 1 s variable n eded to
6\:0 \"~
determine the exact state of the system.
6.39. The critical point represents the point in the phase diagram beyond w ioh only single
fluid phase, a supercritical fluid, exists. Since P = 1, according to the Gibbs phase rule, degrees
of freedom = 3 ­ 1 =,2, requiring two variables to be specified.
~
6.41. The answer will depend on the Figure and the line chosen. All of them are
aT
f)p
.. derivatives, but the phases involved vary with the line.
6.43. 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.
25
.
CHAPTER 7. EQUILIBRIA IN MULTIPLE COMPONENT SYSTEMS
).1. There would be three degrees of freedom. They could include, for example, the mole
fractions of alcohol, liquid water, and olive. The mole 'fraction of the fourth component can be
determined by subtraction.
.
7.3. 0 = C ­ P + 2 0 = 3 ­ P + 2 and solve for P: P ~ 5. So you would need 5 separate phases
in equilibrium with each other to have zero degrees of freedom.
7.5. In this reaction, there are only three independent components (the amount of the fourth can
be determined from the equilibrium constant) and 4 phases. Therefore, there are 3 ­ 4 + 2 = I
degree of freedom.
PV =
(23.76 torr)(5.00 L)
1atm. 6 39 10­ 3
1 f
x
=. x
rno 0
7..
7 For water: n =­
RT (0.08205 L· atmlmol· 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: n=pV =
(l25.0torr)(5.00L)
x latm =3.36xl0­ 2molof
RT (0.08205 L· atm/mol­ K)(298.15 K) 760 torr
methanol is needed to ensure that there is a liquid and gas phase. This equals 1.08 grams.
. 7.9. Since, according to equation 7.11,
Q
j
= Pi. , and since p/ =
760 torr, we can calculate aj:
Pi
a. = 748.2 torr = 0.984
I
760 torr
7.11. We start with the expressions similar to those in equation 7.18 in the text, but in terms of
Y2:
X;P2
.' Next, we recognize that Xl = 1 ­ X2, so we substitute in the
X2P2 + XtPt
•
denominator: Y2 =
• Xzpz
•. This rearranges algebraically (see exercise 7.10) into
.
X 2Pz + (1- xz)Pt
=
Yz
= • X2~
•
•
, which is the equation of interest.
Pt + (pz ­ Pt )x 2
7.13. p(ethanol) = (0.0006)(115.5 torr)
7.15. Equation 7.19 is Y. =
denominator: Y1
= • XtP~
•
XI~'
= 0.0693 torr = 0.07 torr (1 sig fig)
•
•
•
P2 +(Pl ­ pz )x t
P2 + X,Pt ­ xtlh
.
•. Now, multiply the denominator over to the other side
26
\
First, let us multiply through the XI in the
of the equation: Yt (P2* + XtPt* ­ X1P2 .)= XtPt*, or YtP2* + YtXtPt' ­ Yt XtP2 * = XtPl*. Collect
all terms having XI in it on one side of the equation: YtP2* =xtPt* -YtXtPI* +YtXtP2' and
factor the XI out of all three tenus on the right side: YtP2* = X t (Pt * - YIPt' + YtP2 *). Now,
divide the parenthetical terms over to the other side of the equation (which we will flip):
•
Xt
=
Xl
=
•
YtP'2
•. Now factor out the YIS from the last two terms in the denominator:
Pt ­ YtPt + YtP2
.
•
Yt~2
•
•
P, + (P2 ­ Pt )Yt
, which is our ultimate expression.
.
• •
7.17. Consider the equation P tot = • P 2 .PI •
. IfYl goes to zero, then the entire second
. PI + (P2 ­ PI )YI
• •
term in the denominator is zero, and the expression becomes P2 ~I
,which reduces to P2*, as it
Pt
should when the system' contains only component #2. IfYI goes to 1, then the expression forptot
• •
• •
• , which becomes P2 ~t
= Pt· , as it should when the system contains
becomes • P2 ~t
Pt + P2 ­ Pt
P2
only component #1.
7.19.
~mixS
~mixG
= (2 mol)(8.314 J/mol·K)(293.15 K)[(0.5)(ln 0.5) + (0.5)(1n 0.5)] = ­3380 J
= ­(2 mol)(8.314 Jzmol­K) [(0.5)(ln 0.5) + (0.5)(1n 0.5)] = + 11.5 JIK.
7.21. If an initial composition of Xl = 0.1 were used, the tie lines that you would draw would
eventually lead you to the.minimum­boiling azeotrope, near the middle of the phase diagram.'
Thus, the azeotrope will be your ultimate product.
7.23. 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, ifit is two (or more) components, then each component should freeze at
distinctly different temperatures.
.
7.25. The mixture of water and ethylene glycol has both a higher boiling point and lower
freezing point than pure water alone.
7.27. 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 ofHCl will act ideally. HCl is very polar and unlikely to act ideally.
7.29. A mole fraction of 4.17xlO­s implies that there are 4.17xlO­s moles ofCChF2 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
27
\.\
.
0.9999583
mo I x 18.01g x 1.00em
1mol
1.00 g
3
= 18.0 em 3 , or 0.0180 L.
.
c:
t IS
Therefore
thee molari
mo anty 0 f thi
smol
· IS
. 4.17xlO­
. . th e Henry ' s Iaw constant, use 1 atm =
= 000232M
.
. T 0 detennme
so Iution
0.0180L .
101,325 Pa as the pressure:
101,325Pa=Kjx(4.17x10's)
Ki=2.43x109Pa.
~
7.31. (a) The mole fraction ofnitrogen in water is about 90% ofthe 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. (b) If air is 20% oxygen and the mole fraction
of air in water is 1.388x 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 mol~
fraction of oxygen in water would be about (0.20)(1.388x'l 0,5)
6
=2.78x10· • (c) If oxygen is 20% of 1 atm = 101,325 Pa, then the partial pressure of oxygen is
x, =7.29xl09 Pa.
(0.20)(101,325 Pa) = 20,265 Pa. Therefore: 20,265 Pa = Kjx(2.78x10-6)
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 IG. should be the same. However, Table 7.1
shows that the actual value of K, is somewhat less than this (4.34x 109 Pa), indicating that
nitrogen and oxygen do not dissolve in water to an extent proportional to their composition of
arr.
/
7.33. If 87.0 grams of phenol can be dissolved in 100 mL of water, to calculate the molarity we .
.need to know the moles ofphenol and the total volume ofthe solution. The number ofmoles of
phenol are 87.0 g x 1mol = 0.925 mol phenol. The volume ofthephenol is given by
94g
.
ImL
.
87.{) g x ­ ­ = 82.1 ml., Ifwe assume that the volumes are additive, then the total volume of
1.06g
the solution is 100 + 82.1 mL = 182.1 mL = 0.1821 L. Therefore, the molarity of the solution is
0.925 mol = 5.08 M.
.
.
0.1821L
7.35. (a) The calculated mole fraction ofnaphthaIene in toluene is 0.311, so there are 0.311 mol
ofnaphthalene in 0.689 mol oftoluene. The molecular weight of toluene (CJiSCH3) is 92.0
g/mol, and using the density of toluene as given, we can calculate the volume of toluene used:
92.0g
1mL
.
0.689 mol x
x
== 73.2 mL of volume. We do the same for naphthalene (ClOHs):
.
1mol 0.866 g
0.311 mol x 128.0 g x 1 mL = 38.8 mL of volume. Thus, the total volume is 73.2 + 38.8 =
1mol 1.025 g
112.0mL=0.1120L. Determining the molarity: M
= 0.311 mol =2.78M.
(b) Because the
0.112L
ideal solubility is calculated using only properties of the solute, the calculated mole fraction
solubility of naphthaiene in n­decane is the same as in toluene: Xsolute = 0.311. However, the
28
concentrations expressed in other units will be different. To get the solubility in grams per 100
mL of solvent:
0.311 mol 128.0gC IOH g
29.7g
I mol
0.730g 39.8 g naphthalene 100/134
­­­x
x
x
=
x
=­­­­''0.689 mol
mol
142.0gdecane
1mL
134mLdecane
100/134 100mL
In terms of molarity, again we need to determine the total volume ofthe two components:
0.689 mol x 142.0 g x 1mL = 134.0 mL of volume. We do the same for naphthalene (ClOHg):
1mol
0.730g
1280g
1mL
.
. '
0.311molx
.
x
.
=38.8 mLof volume. Thus, the total volume is 134.0+38.8=
1mol 1.025 g
.
0311mol . .
172.8 mL = 0.1728 L. Thus, the molarity is: M = .
= 1.80 M .
0.1728L.
.
7.37. Of the four solutions listed, the C20142 in cyclohexane is probably the closest to ideal. .The
sodium chloride/water and sucrose/water solutions deviate because ~Olar
interactions between
solute and solvent, and the water/carbon tetrachloride combines a polarsolute with a nonpolar
solvent. Therefore, the calculated properties of the C2ol42/cyclohexane solution will probably be
closest to actual properties.
_1­) _
7.39. In(8.0x 10­3 ) = ­ 14,900 J/mol (
I
4.828314... = ­1792( 1
8.314J/mol·K 298.15K T
298.15
l/T= 0.006600.... T= 1515 K = 1242°C.
0.002694 = (0.003354­ l/T)
1)
T
7.41. We can tell by looking.at a phase diagram of the NaCI­H20 system and see if the
temperatures and relative concentrations involved point to a eutectic or to the colligative
property. But as mentioaed in the chapter, the percent ofNaCI in the salt/water eutectic is 23%
salt (that is, about 1 part NaCI to 3 parts H20). 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 freezing­point depression.
7.43. The drawing is left to the student.
1
­
1
7.45. lux!t =­ 2,600J/mol (
soue
8.314 J/mol·K 273.15 K . (97.8 + 273.15) K
J Inx=­0.30184 .:
x=0.739.
7.47. Molarity includes the concept of partialmolar volume because it is defined using the
numbenof liters ofsolution. Thus, even if partial molar volumes may change during a range of
compositions, only the total volume is used to define m o l a r i t y . '
7.49. First, let us calculate Kf and Ks:
K
f
::: MsolvRTM/ = (18.02 g/mol)(8.314J/mol· K)(273.15 K)2 = 1.86 Klmola1
1000AfusH
(1000 glkg)(6009 J/mo1)
,
29
K; = MsolvRTB/ = (18.02 glmol)(8.314 J/mol· K)(373.15 K)2
1000Ll vapH
(1000 glkg)(40,660 J/mo!)
= 0.513 Klmolal
For the freezing point depression: LlT= (2 particles)(1.86 Klmolal)(1.08 molal) = 4.02 K.
Therefore, the freezing point goes down by 4.02°: FP = ­4.02°C.
For the boiling point elevation: LlT= (2 particles)(0.513 Klmolal)(1.08 molal) = 1.11 K.
Therefore, the boiling point goes up by 1.11°: BP = 101.11°C.
For the osmotic pressure, we need the mole fraction of the solute. Ifthere is 1.08 moles of .
''NaCI'' per liter of solution (yielding 2.16 mol of particles per liter), and a liter of solution
contains approximately 1900 mL of water (assuming negligible volume change from the solute),
that's 55.5 moles of H 20 . Therefore, the mole fraction of solute is 2.16 =0.0375. Themolar
57.65
volume of the solution is approximately the same as the molar volume of water, or 0.01802 L.
Using the van't Hoff equation, equation 7.56:
nv == xRT .TI(0.01802 L) = (0.0375)(0.08314L· bar/mol­ K)(298.15K) .
Solve for ll: 51.5 bar.
7.51. Ifx = 0.739, we can use the van't Hoff equation:
nv ==xRT TI(0.0152L) = (0.739)(0.08314L· bar/mol­ K)(273.l5K)
Solve for FI: n = 1210 bar.
7.53. K· = MSOIvRTM /
.
/
1000LlfusH
= (159.8g1mol)(8.314llmol·K)(273.15-7.2 K)2
=
8.89KJmolal
(1000 g!kg)(l0,570 llmol)
2
.
2
K; = MsolvRTBP == (159.8 glmol)(8.314 llmol· K)(273.15 + 58.78 K) = 4.95 KJmolal
1000,6 vapH
(1000 glkg)(29,560 Jzmol)
.
n
n
Xi
30 Pa
1bar
7.55. m =-.- ~
=_.
m=
x­"­­­­­­#kg L water RT
(0.08314L· bar/mol·K)(310K) 100,000Pa
5
m = 1.16x10­ molal. For 1 kg of solvent (water);
­5
185000 g
.
1.16xl0 molalxlkgx
'
=2.15gpolymer.
mol
30
~-
~
CHAPTER 8. ELECTROCHEMISTRY AND IONIC SOLUTIONS
0.0225N=
8.3. (a) F=
q1(1.00C)
ql=2.50x10-8C
4n-(8.854 xl 0. 12 C 2/J . m)(1 00.0 m)"
2q:
qtq2 2 1.55xlO­6N=
1
Solveforq:
4n-(78)(8.854 xl O' 2 C IJ· m)(0.06075 m)?
41l"&or
q = 4.98x10"9 C, so one particle has that charge and the other has twice that, or 9.96xlO·9 C. (b)
The electric fields for the two particles are equal to the force divided by their charges.
Therefore, the electric field on the first particle is 1.55 xl 0­: N = 311 J/C . m , while the electric
4.98x10' C
. I . 1.55x10­{;N
field
e oth
0 er partie e IS
9
= 156 JIC· m. '
e on the
9.96x10" C
8.5. F =
19
19
(+ 1.602 xl 0­ C)(­1.602 xl 0. C)
4n-(8.854x10·12 C 2/J . m)(5.29 x 10.11 m)"
, F = .8.24 X 10­8 N. This may not be
much force, but it's huge compared to the sizes of the proton and electron!
8.7. "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.
'
9 T h e two half reactions are:
Mn02 + 2 H 20 -7 Mn04" + 4 H+ + 3e"
4 e' + O 2 + 2 H 20 -7 4 OH"
E = ­1.679 V .
E = 0.401 V
To balance the electrons,' we multiply the first reaction by 4 and the second reaction by 3.
Combining the H+ and Off ions to H 20 and consolidating the water molecules on both sides, we
get
E= ­1.278 V
4 Mn02 + 3 02 + 2 H 20 -7 4 Mn04' + 4 W_
--To calculate LSG':
AGO = -nFE = ­(12 mol)(96,485 C/mol)(-1278 V) = +1.480 x 10 6 J = 1480 kJ
(b) The two half reactions are:
'­­­
Cu" -7 Cu2+ + e'
E= ­0.153 V
e' + Cu+ -7 Cu
E=0.521 V
Because both half reactions have one electron, they can be combined directly without
multiplying through either reaction. The overall reaction is
2 eu+ -7 Cu + Cu2+
E= 0.368 V_
To calculate AG':
fj,G = -nFE ;= -(1 mol)(96,485 C/mol)(0.368 V) = ­3.55 x 10 4 J =­ 35.5 kJ '
(c) The two half reactions are:
E= 1.087 V
Br2 + 2 e' -7 2 Br
2 P" -7 P2 + 2 e"
E= ­2.866 V
(
D
31
Because both half reactions have one electron, they can be combined directly without
multiplying through either reaction. The overall reaction is
.
Br, + 2 F­ ~ Fz + 2 Br' E = ­1.779 V
To calculate tJ.G':
= -nFE = ­(2 mol)(96,485 C/mol)(­1.779 V) = +3.433 X 105 J =343.3 kJ
. (d) The two half reactions are:
.
HzO z + 2 W + 2 e­ ~ 2 HzO
.
E = 1.776 V
2Cr~lz+eE=-1.358V
Because both half reactions have two electrons, they can be combined directly without
multiplying through either reaction. The overall reaction is
2 HzO z + 2H+ +2Cr ~ 2H 0 + Clz
E=0.418V
To calculate tJ."G':
tJ.Go = -nFE = ­(2 mol)(96,485 C/mol)(0.418 V) = ­8.07 X 10 4 J =
<;j
8.11. The two half reactions are:
E = ­0.771 V
Fez+ -7 Fe 3+ +' ez+
E=-0,447 V
Fe + 2 e­ ~ Fe
To balance the electrons, we multiply the first reaction by 2. For the overall reaction, we get
.
3Fez+~2
E=­1.218V
Since the overall voltage is negative, the reaction is not spontaneous. To calculate ~G':
AGO = -nFE = ­(2mol)(96,495 C/mol)(­1.218 V) = 2.350x10 s J =235.0kJ
'8.13. For the standard hydrogen electrode, the spontaneous process would be
2 Li + 2 W -7 2 Lt + Hz
The voltage ofthis process is 3.04 V. For the standard calomel electrode, the process is
2 Li + HgzClz ~ 2 Lt + 2 Hg + 2 cr
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
2Ag + 2W -7 2Ag+ + Hi
and the voltage is 0.7996 V. With the calomel electrode, the spontaneous reaction is
.
2 Ag + 2 Hg + 2 cr -7 2 Ag+ + HgzClz
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.
.
8.15. Because these half­reactions 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 tJ.G) of the process. Although all alkali metal ions have a +1 charge, the
.smaller, higher­charge­density lithium ion interacts more strongly with water molecules,
. .
increasing the energy of the process.
8.17. E = EO ­ RT In [Zn2+]
nF [Cu 2+]
EO for this reaction is 1.1037 V, so we have
32
2
1 OOOV = 11037V
o
0
[Zn 2+]
-
(8.314JIK)(298.l5K) I [Zn +] Th'
n
IS rearranges to
(2 mo1)(96,485 C/mol) [Cu 2+]
[Zn 2+]
3
2
=3.21x10 • Unfortunately, we can't mathematically
[Cu "l
. [Cu +] ..
determine specific concentrations without more information; we can onlyspecify the ratio.
In
2
8(a)
=8.07279...
E'3~
0 V for any concentration cell, since both half reactions are the same but opposite.
(b) Q = [Fe ]= 0.001 . (c) E = 0 V _ (8.314JIK)(298.15 K) In 0.001
. [Fe 3+]
0.08
(3 mol)(96,485 C/mol)
0.08
= 0.0375 V.
(d) The
.
.
opinion is left to the student.
8.21. In order to use the equation
E~ EO +(MO
]I1T , we will need the entropy change of the
nF
.
.
.
.
reaction. The entropy oflr (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) 144.96130.6S = 14.28 JIK. The number of electrons transferred is 2 so we have:
o~
+(
14.28 JIK
)I1T Solving for I1T: I1T= 595 K. Therefore, if we
(2 mol)(96,485 C/mol)
raise the temperature from 298 K to (298 + 595) = ~893
K, the voltage ofthe reaction should be
about O.
.
­0.044 V
8.23. Since heat capacity (at constant pressure) is defined as
(OH)
,we can take the derivative
oT
. .
p
.8.30
.
of equation
WIth respect to temperature:
.
'.
(O(MI)) = -nF(OEO
+ [OEO
oT
oT.
p
2
ED]J.'
0 .. -t
+ T-.-
oToT
.
oE D
o2 E O]
which simplifies to I1Cp =-nF( 2 - + T - 2- •
oT
.oT
8.25. E = 0 V ­ (8.314JIK)(298.15 K) In 0:0077
(2 mol)(96;485 C/mol)
0.035
=0.0194 V
8.27. Following Example 8.7: the reaction can be written in terms of twohalf reactions:
AgCI (s) + e­ ~ Ag (s) + cr (aq)
E = 0.22233 V
Ag (s) ~. Ag+ + e­
E = ­0.7996 V
The overall voltage of the combination of the two reactions is ­0.5773 V. Using equation 8.32: .
_ 0.5773 V = (8.314 JIK)(298.15 K) InK
(l mol)(96,485 C/mol)
sp
Solve for K : K = 1.74xl0·10 •
sP
33
sP
8.29. First, we need to determine the overall reaction. Using Table 8.2, fmding the Mn04­lMn2+
half reaction, and combining it with the hydrogen electrode reaction, we find a 1a­electron
overall reaction:
2 Mn04­ + 6 H+ + 5 H2 ~
2 Mn 2+ + 8 H20
.eo = 1.507 V
However, because some ofthe concentrations are not standard, equation 8.35 can't be used
directly. It is probably best to use the complete Nemst equation:
1.200 V =1.507 V _ (8.314 JIK)(298.15 K) In
(0.288)2
Solving for IH+]:
. (10 mol)(96,485 C/mol) (0.034)2[H+ t (1)5
7.88x10 51 =
[If] = 4.6x10­ 9
(0.288)2
(0.034)2[H+ (1)5
r
Therefore, the pH = ­log (4.6xlO·9 ) ::= 8.34.
8.31. TheKsp for Hg2Ch was determined in exercise 8.28 and is 1.29xlO· 18 . If'x moles per liter
of Hg 2Ch dissociates, one gets x M Hgz z+ and 2x M cr. Therefore, we have:
1.29xlO­18 = 4x 3 X = 6.86xl0·7 Since the equilibrium concentration
1.29xlO­ 18 ::= (x)(2xi
of'Cl' is twice this, [Cn = 1.38xlO­6 M.
8.33. Using the definition of ionic strength in equation 8.47:
(a) 1= !(0.0055m)(+1)2 + (0.0055m)(­1) 2 0.0055m
2
)=
(b) 1=
~
(0.075m)(+1)2 + (0.075m)(­1)2)= 0.075m
(c) 1= !(0.0250m)(+2)2 + (0.0500m)(-I) 2
2
(d) 1=
~
(0.0250m)(+3)2
)= 0.0750m
+(0. 75 m) -l ~ =
0.150m
8.35. In the equation Hz (g) + 12 (s) ~ 2 If" (aq) + 2 T (aq), the overall enthalpy of reaction
is ­110.38 kJ. Using the concept ofproducts­minus­reactants to determine Ml, 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 l:1iHs are zero. By
convention, the I:1rH ont (aq) is also zero, so the only non­zero I:1rH is that for r. Therefore, we
have
I:1rH [f] =­55.19 kl/mol.
­11O.38kJ = (2 mol)(l:1rH[f])
8.37. The reaction is HF (g) ~ If" (aq) + F (aq). Using the thermodynamic values from the
appendix, we have:
Ml= (­332.63 + 0) ­ (­273.30) = ­59.33 kJ
M = (­13.8 + 0) ­ (173.779) = ­187.6 JIK
I:1G = (­278.8 + 0) ­ (­274.6) = ­4.2 kJ
(b) Using I:1Go = -RTInK: ­4200 J/mol = ­(8.314 J/mol·K)(298.15 K) InK·In K = 1.6943.... K = 5.44. This is rather far off from the 3.5xlO­4 value as measured. The
difference is that the equilibrium constant refers to HF in the aqueous phase being the reactant,
34
rather than the gas phase. The predicted value for K should be much closer to the measured
value if the hydration of HF step were included.
8.39.· 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 molar ll2 .
8.41. 0.9% NaCI implies 0.9 g NaCI in 99.1 g water. Assuming that the volume of the solution
is 100 mL = 0.100 L = 0.100 kg: (0.9 g)(1 mol/58.5 g)
.
0.100kg
strength is 1=
~
= 0.154 molal.
(0.154 molal)(+1)2 + (0.154 molal)(­1)2)= 0.15~
Therefore, the ionic
molal.
8.43. (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 ionicstrengths WQ~19_
Il~ 9:. t 12
.
recalculated. Ifsulfate were the counterion, the salt's formula is Fe2(S04h and the sulfate·
concentration is 3/2 of the iron ion concentration:
I (Fe 3+ soIn) = Y;·[(0.100)(+3i + (0.150)(-2i] = 0.750 molal
If sulfate were the counterion, the other salt's formula would be CUS04 and the sulfate
concentration would equal the copper ion concentration:
I (Cu2+ soln) = Y2.[(0.050)(+2i + (0.050)(-2i] = 0.200 molal
Using equation 8.52 for each ion:
. (1.171 molaf l12)(+3)2 (0.750 molalyl2
In ) =_
". YFe •
1 + (2.32 X 10 9 m "molal'!" )(9.00 X 10­10 m)(O.750 molal)'?
Y (Fe 3 ) = 0.0388
Iny Fe). == ­3.250
Therefore, the activity of Fe3+ is (0.0388)(0.100 m)
=0.00388 m.
Similarly, for the Cu2+:
lny
l.
Cu
=_
(l.i71 molal
l12)(+2)z'(0.200
molalyl2
1+ (2.32 X 10 m "molal?" )(6.00 x 10­ m)(0.200 molalj'?
9
Inycul • = ­1.291.
Y (Cu l ) = 0,275
10
Therefore, the activity ofCu2+ is (0.275)(0.050 m) =
0.0138 m.
Substituting these activities into the Nernst equation:
E
=0.379 V +
(8.314 J/K)(298.15K) In (0.00388)2
(6 mol)(95,485 C/mol)
(0.0138)3
=0.379 + 0.0075 = 0.386 V
. .
When you compare this to the voltage from the example (0.372 V), we see how the ionic
strength ofthe solution, as influenced by the counterion, can have an influence on the voltage
even though the counterionsdon't participate in the reaction.
J. ~.
j
A.
Equation 8.5 shows that E has units of
8.45. Equation 8.61 is I = e 2 ·1 z 12 .( N
V
61r''lrj .
N/C, e has units of C, z is unitless (it is simply the magnitude of the charge on the ion), the
fraction (N/V) has units of 11m3 , A has units of m', and in SI units, viscosity has units of kg/m­s.
The numbers 6 and 1t have no units. The radius r has units of m. Combining all of these units:
35
C
2
1 )
2
C 2 ­rn' ­Nvm­s
N/C.
This can be rearranged to get
. One of the C
(kg/m­sjun)
C·m 3 ­rn­kg
units cancels; as do the three m units in the numerator and three of the four munits in the .
•
( ­m'
•
m
•
denominator. This results in C· N ­s . Ifwe break down the newton unit into kg­m/s', we have
kg­rn
C ~ kg . m . s . The kg and m units cancel, as do one of the second units in the numerator and
s ­kg­rn
denominator. What's left is Cis, and one coulomb per second is an ampere, the unit of current.
8.47. For a galvanic cell, oxidation (the loss of electrons) occurs at the cathode and is considered
the negative electrode. Therefore, L 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 cathode, so L is still the current towards the cathode and 1+ is the current,towards the anode.
For any given cell reaction, however, the identities ofthe cathode and anode are switched for
galvanic and electrolytic cells.
36
CHAPTER 9. PRE­QUANTUM MECHANICS
9.1. The kinetic energy for a mass falling in the z direction is .!.mi 2 and the gravitational
2
potential energy is mgz. Therefore, the Lagrangian is L
=~
mi.2
-
mgz . Using the formula
. teo
h Ii 11owmg
. deri
.
d (aL
- -J=oL
- , we d etermme
envatives:
-
dt
oi
az
0
ai. = oi
(aL)
(1 .
2
2 mz ­ mgz
)
•
= mz
!!-(OL)
= !!-(mi) = mz
dt Oi
dt
: =~
(~
mi
2
­
=-mg
mgz)
Combining the last two lines, we have for the Lagrangian equation of motion: mi = -mg.
9.3. Inthiscase,qisz, if is i,pis-mi,and pis-mz (these last two are because an object
falling in the z direction is decreasing its z value). According to equations 9.14 and 9.15,
OH.) = i and (OH)
= mz . The first differential is easy to demonstrate:
(Ornz
OZ
.
1 .­
0. (1-mz; + mgz =­1 (mz. + 0 =) 'z, as reqUIre.
. d The deri
.
.(OH)
- - . = -1 (OH)
- . =­
envative
2
Ornz
m Oi
)
mOi 2
m
.
0f
th e
second term is zero because i does not show up as a variable in that term. The second
differential is a little tricky. Ifwe perform the differentiation, we get
(0::)
=
~ (~
mi 2 + mgz)
= (0 + mg) = mg , which does not appear to be the required
expression. However, remember what g is: it is the acceleration due to gravity experienced by
the mass falling down. Therefore, g IS Z. Therefore,. we do in fact have
(0:) =
mz = -
p, as
required by Hamilton's laws ofmotion.
9.5. The drawing is left to the student.
9.7. Iftwo spectra of two different compounds have some lines at exactly the same wavelength,
then by Kirchhoff's and Bunsen's proposition, they must share one or more constituent element.
9.9. To determine the series limit, assume that lin?
=
wavenumber 109,700
~
cm-tl~
wavenumber 109,700 em
­
0)= 109,700 cm· is the series limit. For the Brackett series:
i
­t;, ­ ~
0)
6856 cm' is the series limit.
37
:;o.",­­, ••
:­.J:"'.'.:
li;.c.
= 11 00 2 = O. For the Lyman series:
9.11. First, we need to convert the wavelengths to wavenumbers:
­5
1
1
1m
100 em
656.2nrnx 9
x
=6.562xlO em
-==
5
= 15,240cm'l
10 run
1m
A 6.562xlO­ em
486.1nmx
1m x100cm=4.861xlO­5cm
.!..=
1
=20,S70cm·1
10 9 run
1m
A 4.861 x 10­ 5 em
1m
100 em
­5
1
1
434.0 nrn x 9
x ­ ­ ­ = 4.340 x 10 em :. ­ =
5
=23,040 ern"
10 run
1m
A 4.340xl0­ em
Using these wavenumber values, we can calculate R for nz == 2, assuming that the first three lines
correspond to n2 = 3,4, and 5, respectively:
1,)
15,240 em"
3
15,240 cm' = R(0.1388888. __l
= 109,730cm­'
~
20,570cm·1
.'
..
R(;, ­
=R(~-)
23 040cm·1 =
,
R
1
R(~
2
22
4
20,570cm· =R(0.1875)
R =109,710cm·1
__
I) 23,040cm·1 = R(0.21) R == 109,71Ocm·1
52
­I
. 109,730+109,710+109,710 109720
The average 0 fthese three IS
3
=,
em.
9.13. (a) A.sjugle~of,I
kg. Ifwe 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~1.6 2xI0·27
~}.9'i1Q27)
kg =Q ~.
The ratio of these two masses will tell us how-~ybeta
27
. Ies. to
.
le
.
ak
I
h
f
1
l'
h
.
I
6.694
x
l
0­
­31 :::: 7349 b eta partie
partie es It t es to equa t e mass 0
a p apartie e:
9.109xlO
.
mass the same as one alpha particle. (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.
4)
•
2
W
­8
W
9.15. For a flux of 1.00 W/m: 1.00­
)(T
Solvmgfor T:T = 65 K
2 = (elj~Qs._:'1{4
m
m
.
.
2
W
.4)
s W
2
4)(T
Forafluxof10.00W/m: 10.00­
(5.6705xl0­
SolvingforT:T=115K
=
m2
mK
2
W'
­SW)4
Fora fl uxo f 100.00W/m: 100.00­'­2 ::::(5.670SxlO
24 (T)
m
mK
Solving for T :T == 205K
9.17. (a) power per unit area == (5.6705 x 10.8 W/m 2K4)(5800K)4 == 6.42 x 107 W/m 2
'(b) 6.42xl07 W/m2 x 6.087xl012 m 2 = 3.91x1020 W
38
365 d 24 hr 3600 s
(c) 3.91x102oW=3.91xl02o J/s x ­ ­ x ­ ­ x
=1.23xl0 28 Jperyear.
1 year
1d
1 hr
9.19. (a) Using Wien's law: Iv max·5800 K =­,89§ !JIW:~
Ivmax = 0.4997 urn = 4997 A.
(b) 5000 A = 0.5000J.!m, so using Wien's law: 0.5000 J.!m·T= 2898J.!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.
9.21. Planck's law is written s dE
= 2~C
(e hC/
L­1 )dA. In order to use the given integral,
we're going to have to re efine the variable for frequency, not wavelength. Since c = IvV, we
have A =!?. and dA ­ ---;'dv. Substituting:
v
v
2
dE ­ 2wc
- AS
dE = ­
2~
(
v,
e
/»:J
2
)dA ­ 21ihc (
1
)(­ .s,
d v)
-1
- (clv)s ehc/(c/v)kT -1
v2
(ehV/~T
dv = kT h- dx and v
_1)dV. Ifwe now substitute x
=
This expression simplifies into
=h v I kT so that dx = hi kT . dv , or
xkTIh, we substitute again:
w(xkT I h)3 ( 1 ) k T · .
+ ­ ­ : ­ " ' 2 ­ ­ ' ­ ­ ­ ­ -dx. Collecting terms: dE=
. c
eX -1 h
21leT
3 2
h c
4
3
x
' ­ ­ . Now we can
e' ­ 1
ate from x = 0 to x = (you should verify that when v = 0, x = 0, and when v = , x = :
2JTk4T4
x3
2JTk4T4 co x 3
21ik4T4 1f4
21fseT4
co
dx
=JdE=J-: h3c2 'ex_ldx=
h3c2
=
h3c2· '15=­ 15h3c2 which is
ex _ l
J
o
0
the Stefan­Boltzmann law. (The negative sign indicates that energy is being givenoff.)
9.23. Substituting the values of the various constants:
23
4
s
21f (1.;81 x ~ 0­ JIK)
5.6­7'6:'i;;;;;;;)which is almost exactly the value
34
3
15(2.9979xl0 mls) (6.626xl0- Js)
. . . . ""..­"'"
..
ofthe Stefan-BQlzIm£2.J,§~ird
in the text.
19
3·46 xO
1 ­19 J .
9.25. A work functi
ction of2. 1 6
eV 'IS equa I to 2. 16 eV x 1.602x10· J =.
leV
Determine the energy equivalent of the wavelengths of light, then subtract 3.46x10·19 J from that
energy value. Any remaining energy is converted to kinetic energy of motion, ~ mv".
8
1m =5.50xl0-7 m v == 2.9979xl0 mls = 5.45xl014 S·I
(a)
10 9 run
5.50xlO·7 m
E=(6.626x 0­ 34 Js)(5.45xl0 14 S · 1 )
= 3.61xl0­ 19
If 3.46x 10' J are used in overcoming the work function, there is 0.15xl0'19 J l~:ft.
formula for kinetic energy and the mass of the electron:
550nrnx
39
Using the
B
19m =4.50x10­ 7m v = 2.9979x107 mls = 6.66x10 14
(b)
4.50xlO: m
10 run
1
4
I
E = (6.626 X10­34 Js)(6.66x10 S· )
E= 4.41x10­ 19 J.
450 nm x
S·I
If3.46xlO'19 J are used in overcoming the work function, there is 0.95x10'19 J left. Using the
formula for kinetic energy and the mass of the electron:
0.95x10­19 J =.!­(9.109X 10­31 kg)v 2 v=457,000mls.
2
.
8
14­ 1
7
­'350
­
. x 10­ m v ­_2.9979x10 7 rn/s_857
­ . x 10 s
10 nm
3.50xlO' m
E = (6.626 x 10­ 34 Js)(8.57 x 1014s·1)
9
(c) 350 nm. x 1m
E= 5.68xlO·19 J.
If3.46x10'19 J are used in overcoming the work function, there is 2.22x10· 19 J left. Using the
formula for kinetic energy and the mass ofthe electron:
2.22x 10­19 J =.!-(9.109x 10­3 1 kgjv"
2
9.27. By combining the equations E
..directly: E
= he .
A
V
= 698,000mls.
= hv and e = AV, we get an expression that can be used
For the wavelength of 10m:
8
.
602·2 x 10231 mo I == 00120JI
= (6.626xl0­34 j .s)(2.9979xI0 mls) =.199 x 10­ 26 J per photon,
x.
.
moI photons
10m
For a wavelength of 10.0 em = 0.100 m:
E
E = (6.626xl0­34 J.s)(2.9979xl0
8rn/s)
0.100m
199 x 10­ 24J per ph oton, x 6022
120JImo I photons.
==.
o.
x 1023/ moI =.
For a wavelength of 10 microns = 0.00001 m:
E
= (6.626 x 10­
34
8m1s)
J.s)(2.9979xl0
=.199 x 10­ 20J perph oton,x.6022 x 10231 mo1=12000JImo1photons
O.OOOOlm
For a wavelength of 550 DIn = 5.50x 10'7 m:
E = (6.626xl0
-~
8
.
J·s)(2.9979xl0
mls) == 361
6022 x 10231 mo1== 217500JI
. x 10­19 J per photon, x.
,
mo1ph0 t ons
7
5.50xl0' m
For a wavelength of300 DIn = 3.00x10'7 m:
E == (6.626 x 10­
34
8
mls)
J .s){2.9979xl0
7
3.00x'10· m
For a wavelength ofl
ec
662
6022 x 10231 mo J== 398700JI
•
x 10­ 19 J per photon, x.
,
mo1ph otons
.
A = 10­10 m:
(6 .626xl 0­34 J·s)(2.9979xl08 rn/s);"199 10­ 15J
231
602210
E -­
­. x
perph oton,x.
x
mol-12020'JI
­. x
mo I ph otons
10.10 m
9.29. Using equation 9.34, for n = 4:
40
r= son h =,i8.54x10-I2g;kUn)(~<s
2
1lm e e
For n = 5:
2
2
=8.47xl0­IOm=8.47 A
.
Jr(9~g)n2
2 2
12 2IJm)(5 2)(6.626xlO­34
J 'S)2 =1.32x10­9 m=13.2A
r = son h = (8.854xl0­ C
Jr(9.109x10­31 kg)(1.602x10·19 C)2
mn ee 2
For n = 6:
2
2
r = son h = (8.854x 10­
mn ee 2
Jr(~.109x-3
12 2IJm)(62)(6.626xlO­34
C
J 'S)2 = 1.91x10­9 m=19.1A
1
9
kg)(1.602x10· C)2
9.31. Angular momenta are simply integral units of h/2n: for n
L = 4·(6.626x 10­34 J.s)l2n = 4.22x 10.34 J.s.
For n = 5: L = 5·(6;626x 10.34 J·s)l2n = 5.27xlO­ 34 J·s.
For n = 6: L = 6.(6.626xlO'34 J.s)l2n = 6.33xlO·34 Jvs.
=
4,
e2
(b) For n = 1,
2s onh
­19
2
.
6
(1.602
x
10
C)
. about 0 .73% 0 f the speed
v=
12 2
34
= 2. 18 x 10 mIS ThiIS IS
2(8.854x10­ C /Jm)(1)(6.626xl0' Js)
oflight. (c) L = mvr = 9.109xl0·31 kg· 2.19x106 m/s­ 5.29xlO·11 m = 1.06x10.34 Ls, which is
equal to h/2n.
9.33. (a) v =
9.35. Starting with n'A = 27tr, substitute for wavelength using de Broglie's relation A = ~:
n~
mv
= 21CY , which rearranges to
mv
nh = Znmvr . Finally, dividing both sides by 2n yields
mvr =;; , which is Bohr's postulate that the angular momentum, mvr, is a quantized multiple of
sn«
9.37. A =
A=~ 1.0 x -IO
~
= 1.00 x 10':'10 m =
mv
m=
mv
~
(6. 2 x1~7-34
(6.626 x 1~:34
Js)
(9.109x10' kg)(v)
Js)
41
v = 7.27 X 10 6 mls for an electron,
v=3.96x10 3 mlSfOr
..
aA
~
CHAPTER 10. INTRODUCTION TO QUANTUM MECHANICS
10.1. From the text, the following statements were given as postulates. The state of a system
can be described by an expression called a wavefunction; all possible information about the
observables of the system are contained in the wavefunction. Forevery physical observable,
there exists a corresponding operator, and the value of the observable can be determined by an
eigenvalue equation involving the operator and the wavefunction. Allowable wavefunctions
must satisfy the Schrodinger equation. The average value of an observable can be determined
from the appropriate operator and wavefunction using the expression given by equation 10.13.
Other sources may list postulates differently; your answer may vary.
10.3. (a) Yes, the function is an acceptable wavefunction. (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) Yes, the function is an acceptable wavefunction. (e)
No, the function is not acceptable because it is not bounded. (f) Yes, the function is an
acceptable wavefunction. (g) No, the function is not acceptable because it is not single­valued.
(h) No, the function is not acceptable because it is not continuous. (g) No, the function is not
acceptable because it is not single­valued.
10.5. (a) Evaluation of the operation yields the value 6. (b) Evaluation of the operation yields
the value 9. (c) Evaluation of the operation yields the function 12x 2
10.7. (a) Px(4,5,6)=(-4,5,6)
.(c) Px~(5,0)
= (5,0,0)
(d) Py~(n,
/2,0) = (-n~
­
7­
~
x
.
(b) PyPz(O,-4,-l)=(O,4,l)
(the two operations cancel each other out)
/2,0)
(e) Yes, PXPy should equal PyP for any set of coordinates, because the x coordinate and the y
.
coordinate are independent of each other.
X
10.9. 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.
10.11. In terms of classical mechanics, K= 1/2. mv", or in terms oflinear momentum p
K = p2 . Substituting for the definition of the linear momentum o p e r a t o ~
2m
A
K
=­ 1
2m
'f:. 8)2
( -In
-
ox
2
2
2
ax
2m
ax
= mv,
~
. I kin etic
. energy
=­ 1 ( ­ *­2 ­82) =­ 1i­ ­82 ,whiICh'IS th e one­ d'tmenSlOna
2m
rt
operator.
42
.
,
10.13. ­ ih.
:¢ C,."~
e;";)
~
-ih(im{
.J."~
e'·; )
= mh( .J."~
e'.'). Since the original function
is returned multiplied by a constant, the function is an eigenfunction and the value of the angular
momentum is mb .
10.15. 1.23me = 1.23(9.109x10'3! kg) = 1.12x10·3o kg
J
34
&(1.12 x 10.30 kg. 10,000 m :2: 6.626 X 10­ J . S
'S
4Jr
&:2: 4.7lx10­9 m.
10.17. First, we need to convert the 1.0Q em'! to energy;
J
1
( .1 m
= 0.01 m ="t,
v = ~ = 3.00x 10 mls = 3.00x 1010 S­l
1.00 cm' 100,cm'
A.
0.01 m
Now calculating the equivalent M;; M; = h v = (6.626 X 19­34 J . s)(3.00 X 1010s·1 )
23
M; = 1.99x 10. J
Calculating the t:.t from the expression in the exercise:
8
34
(1.99x 10­ 23 1)t:.t= 6.626 X 10­ J·s
4Jr
This is just under 3 picoseconds.
­ n~
is»
t:.t=2.65x10·!2 s.
~
ptGO .
j
.,l
2n
2n
10.19. (a) To normalize: N 2 f(e il7l¢)' eim¢d¢ = 1. N 2 Ie-'im¢ e im¢ d¢
o
=1
.0
1
N==-
.J2Jr
Therefore, the normalized wavefunction is ~
e il7l¢ .
.
v2Jr
2n/3()"
(b)P= I _1_. eimt}
o .J2Jr
1 r",]2n/3
=2Jr LV' 0
2n/3
2n/3
_l_ eim¢d¢=_l Ie- im¢ .eim¢d¢=_l Id¢
.J2Jr
27r. _ ~ _
2Jr.o
oJ
1 (2Jr
= 2Jr
3 -' = 3"1 . ~/-b" .L~.a
b·l··
.
. -:-'1/3
~_
1 Ity IS
S
do.~ l1o.! ~
interest is 1/3 ofthe ring'. !he prQba?il~1:)'
the same value of probability over this interval.
ak
b
he rezi
f
m es sense ecause t e region 0
so any value of m would have
­
'.
10.21. First, we need to normalize the function:
a
N
2
I(l<x)'" kx dx = 1.
o
N
2
­OJ =1
e (.!.a
3
3
N
2
Jla = 3
3
Therefore the normalized wavefunction is 'l' =
cancels.
43
N
2
;2
=~
k 2 a3
kx
= a.Jl x..' Again, note how the k
For the ftrst third of the interval:
3
J'..J3
.
3 a/3
3 1 al3
3 (1 a
J 1
a3/2Xdx=7 Ix dx=7'3 X 10 =7 327- 0 =27'
a/3(.J3
p= I a 3/2 X
2·
3
For the last third of the interval:
p=
J' a.J33/2 xdx==73 asX dx=7'jX
3 1 31 a
3 (1
1 8a
2al3 =7 ja ­j
(..J3
as
. a3/2 X
2a/3
2
3
3 )
27
2al3
19
== 27'
In this case, the particle has a higher probability of being in the last third ofthe interval, and the
probability distribution is not equal like it is in the previous problem,
10.23. (a) N I(x
2
)'
x 2 dx = 1
o
N 2(l/5 ­ 0)
N=.J5
If = 5
=1
wavefunction is 'I' =
.J5x
2
I ~I
'X~[ N
1
2
Therefore, the normalized
,
6
2
(b) N J(lI X)' (11 x)dx = 1
s
N'( ~- -(1=)~. wavefuncti
nction IS
N
\T1
T
=.J30
Therefore, the normalized
.J3O
=­
­,
X
1</2
1<12
N 2 Icos 2 xdx = 1 Using the integral table in the
-1<12
(c) N 2 J(cosx)' (cosx)dx = 1
-1</2
appendix:
.,1<12
2
N [ ; -+- : sin 2x
I
1.
sm 2(- 7£ 12)]] =1
2+ 41.sm 2(7£12)) ­ [-trI2
N 2[(7£12
2 +"4
=1
- -tr/2
7£ 1
] =1
N 2 [7£
­+­+­(0­0)
4 4 4
N2
N
7£
the normalized wavefunction is 'I' =
~
co
cos X
= ~ ­.
Therefore,
7£
•
co
(d) N 2 J(e- rla)' (e- rla)4nr 2dr = 1
47rN 2 Jr2e-2rla dr
o
the appendix:­ 47rN 2 [ 2!
(11 a)
.
=­2
=1
Using the integral table in
o
3] = 1
normalized wavefunction is 'I' =
N
=
~
1
~
v81Ta
3
.
Therefore, the-
e -r I a •
8mz3
(e) The tactic for this wavefunction is to recognize that since the­exponent has the variable
squared, the wavefunction in the interval from ­ co to 0 has the same as it has in the interval 0 to
44
Therefore, we can use the interval 0 to
Therefore, we have
00.
00
and multiply the value of the integral by 2.
00
00
2N 2 J(e- r2Ia)" (e-r2/a)4nr2dr
8nN 2 Jr2e-2r2/adr = 1 Using the integral table in
=1
o
o
the appendix: 8nN
2
(',
I
2 . (2/ a)
,.
~
1£
(2/ a)
J= 1
11N 2
a~
1m
2
=1
2 )1/4
..
( 2 )1/4
N = ( n 3a 3
Therefore, the normalized wavefunction is \f'= n 3a 3
«'>.
I 0.25. The Schrodinger equation contains a specific expression for the kinetic energy because
kinetic energy has a specific formula in classical mechanics (
~
mv 2 ,or ~
­ the two are
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 Schrodinger equation, there is 011y a general
expression for the potential energy operator.
"
.
il 2 8 2
I 0.27. ­ - - - (2k ) + O· k = Ek
Since the derivative of a constant is zero, we have
2m 8x
0+ 0 =E-kTherefore, E must be O.
10.29. The only property of Hermitian operators described in the text is that Hermitian operators
have real (i.e. non­imaginary) 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 isa 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.)
.
2H
2H
o
0
10.31. To normalize: N 2 J(e iKx )* eiKxdx = 1. N 2 Je-iKxeiKxdx = 1
2 ..
1
N=2n
45
.. -
;;.:.,,­
1[.:
1 N=. J2ff
10.33. First, we should convert the wavelength of the photon into an equivalent value injoules:
7
c 3.00x 10 8 mls
7
= 1.38 X 1015 s". The energy of a
A = 2170 A = 2.17x10· m, Therefore, v = - =
A 2.17xlO· m
photon of this frequency is E == hv = (6.626x1 0.34 J·s)(1.38x 1015 S·I) = 9.14x10· 19 J. Now we use
the expression for the energies ofthe particle­in­a­box wavefunctions:
9.14x 10­ 19 J = E(n = 3) ­ E(n = 2)
= (3
2
2
2)(6.626x
34
10­ J . S)2
. 8(9.109 x 10-3 1 kg)a 2
­
We have all of the
information except the length of the 'box', a. Solve algebraically: a = 5.74x10· 1O m
10.35. The drawings are left to the student. See Figures 10.6 and 10.7 in the text.
10.37. For n = 1: P = o,sT (
0.4950
f2 sin 11XJ' ( f2 ~is
V-;; a V;
~ oo5r
~
11XJdx =
sin 2
a
a 0.4950
a
2 [x
1
. 211X]0.sosa
­;; 2 ­ 4(1r / a) Sl~
049Sa
=
~[(0.5a
-~Sin21r(0.5)
a
2
41r
P = 2(0.005 + 0.00250 + 0.00250) = 0.02
_ (0.495a _ .!!.­.sin 21r(0.495))]
2
41r
/
oso.
o.sosa(l'
211XJ'(l-:-211XJ
2°.s
(211X
2
==dx .
Forn=2: P=
-Slll- S l l l - dx=- f· sm
0.49Sa a
a
a
a
a 0.4950
a
J
2 [x
. 411X]0.50sa
1
­;; '2­ 4(21r / a) sm~
~ ~
[(
0
0.4950
5~a
_ 8: sin 4,,(0505)) ­ (
0.4~5a
;". sin 4,,(0.495»)]
P = 2(0.005 ­ 0.002498 ­ 0.002498) = 0.000008
f . 0.49Sa a
For n = 3: P = 0.5050 ( l sin 311X
-
J' ( l­sin 311XJdx =­2 005050Jsin 2 ::!!:!..ax
3
a .
a .
a
a 0.495 a.
. a
2 [x
1
. 611X]0.50S0
a 2' ­ 4(31r / a) sin ~
0.4950
= 2 [(0.505a _~Sin61r(0.5 )_(0.495a
a
2
121r
2
.
P = 2(0.005 + 0.002496 + 0.002496) = 0.01998
For n = 4: P == o'ST(
0.49Sa
-~Sin61r(0.495»)]
121r
~
0
f2 sin 411XJ'( (2 sin 411XJdx = 2. 0.51 sin? 411Xdx
V; a V-;; a
a 0.49Sa
a
46
.
.
= 5.74A.
2
X
-;; "2 [
0.505a
1
. 8nx
4(4n­ / a) sm ­­­;­]
=
--
~[(
0.505a
2
a
~
0.495a ,
16n-
sin 8n-(0.505)) _ (OA95a ­
,2
.E: sin 8n-(0.495))]
16n-
P ­ 2(0.005 ­ 0002493 ­ 0.002493) ­ O.OOQ2S
The probabilities are relatively high for the first and third wavefunctions (as they should be, as
the middle ofthe 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).
'
10.39. At high quantum numbers, the positive peaks of the probability waves effectively blend
together, mimicking a straight line of constantprobability 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.
10041. The wavefunctions 'f' =
'1{2
-;; sin nnx
a
can be very easily shown to not be eigenfunctions of
the position operator. The position operator .x 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.
Q{ (2 sin nxJ* ~ni-'
10043. (Px) =
1lV­;;
a
. nx nx
=- I n2-n­- aJSln­cos­
'f:.
a a
°
a
f2 sin nx dx =-:in~
ax'1­;;
a
JSin
a
°
nx nis~
a
ax
nx dx
a
'
dx'
a
== ­iii 27r[
__1_ Sin 2 nx]Q == ­iii 27r2 [!:(SinK­:­SinO)] == ­iii 2n­2 [!:(O­O)] == 0
a2 n l a
a
a tt
a n-
°
10.45. A normalization constant is not going to affect the value ofthe eigenvalue, so let us
simply evaluate the operatorlwavefunction combination without normalizing:
­ in ~
e3i ; == ­in(3i)e 3i ; == 3ne3i;
. Therefore, the eigenvalue is 3n, that being the value of the
8¢
angular momentum. To determine the average value, set up the expression as follows" using N
as the normalization constant:
2;
(p;) = J(Ne
3i
;
r.
8
21<
­in _Ne 3i ; d¢ = ­in· 3i Ne­Ji,p ·Ne3i; d¢. If the wavefunction is properly
I
°
8¢
°
normalized, the value of the integral is simply 1. Therefore, wehave
same value as the eigenvalue for angular momentum, as it should be.
47
,j
(p;) == 3n, which is the
2
~2
'
,
X as ­ 2m2E allows us to ultimately rewrite the expression into a form
Xdx '
Ii
that mimics a one­dimensional Schrodinger equation. Ifwe had defined the second derivative as
simply E, the connection to the one­dimensional Schrodinger equation would not have been as
obvious. (See also the next exercise.)
10.47. Defining
2
J[If
.
az
2
2
8­
10.49 . ­li2
­ ( ­8­ + ­8 ­ + ­
,
2m
2 By 2 2
ax
2
li
=--'
2m
[If( 8
y
n,,7TX',. n 1l)l • n z1TZ]
­­,sm­­·sm­­·sm­abc
a
.b
c
If"
(8
2
y7ZYJ . n Z 1TZ
X1lXJ
x1lX
z7lZ
.
ny7ZY
.
n
n
n
• n
- - ­­sm­­
·sm­­·sm­­+ ­­sm­­· ­ ­sm­­
·sm__
2
abc 8x
abc
abc
a
By2
b
c
2
•
If .
(8
2
n x1lX . ny7ZY ­­sm­­
• n z7lZJ]
+ ­­sm­­·sm­­
.
abc
a
b
&2
c
If
ti2 [ n x 2,. 2
=- 2m ­
a2
abc
•
sm-a-·sm-b-·sm-c- ~
­
ti2[
=­.­
[
.!!:..2m
,
2
8m
n z 2 ,. 2
c
2,.2
n
x
22
n
:
+ y2
a,
2
b
2
n x 1lX
2
•
If
n y2,.2
. n z7lZ
n y1l)l
.
n
J±
. n x 1lX . n y1l)l . n z7lZ
abc sm­a­·sm­b­·smc­
. n
y'~J
n 'TT\J
x1lX
z7lZ ]
­­sm­­·sm­­·sm­•
abc
abc
2'2] \f' =­h2[ nX22+--;­: + nZ22] \f', where the en~r"
+ nz :
c
8m
abc
2
/
eigenvalue is
);­,.1;
." /
(
'
nX2 + nY2 + nZ2 ]. This verifies that the wavefunction is an eigenfunction ofthe threea
'. b c
'
dimensional Schrodinger equation.
10.51. 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 (l ,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
10.53. For the two­dimensional box:
,
'¥
f£
ti ( ­82 + ­8­ 2J' the wave functions are
H~ =­­.
' 2m
2
ax
2
By
n 1l)l
.'.
h n
n '2]
= ­4.n1lX.
sm _X­sm­y_, and the quantized energies are ­ 2[ ~ 2 + ­­;­
.
ab
a
b
8m
.
a
b
10.55. We must evaluate the following integral:
(
(x)= II. V~(.x J-ms. -ms~
2'
8".1lX.7ZY. 7lZ
,2
V-;;Z;;;
abc
48
r
8'o1lX.ny.1TZ
f8"sm­.sm­.sm­'Lydydz
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:
(x') ~
RiSin:Jox'{ism :Jdx . . \
R~ .nYJ' ~.(
. J' (l·). J
. tryJ. ,/ ~/-·R
1lZ
­sm­
­sm­ ~. x.
­sm­
.,­.'­s.m. ­1lZ dz
b
b
b
b \ o. C
C
­ C ._ ._of­­.
The reason we do this is so
can now recognize that the second and third integrals represent
normalized one­dimensional wavefunctions, so the integrals are simply 1. Therefore, we get
x
o
(x 2 ) =
we
a{ {2sin 7lXJ'•x 2 , fI sin 7lX dx = ~
~l'1-;
a --
~a
a
jx
a
2
sin 2
o
7lX dx.
a
Using the integral table in the
appendix:
2[x
(x
=; 6- 4(nla)
=
3
2
1) . 27lX
X
27lX]a
'(>(nla)3 sm7- 4(nla)2 cos-;-
0
~[( a; -0-4:' )-(0-0- 4:' )] = a;
,
Finally, rather than repeat everything, by extension we can deduce that
3
3
and
...
2
(Z2) = C
(y2) = b
2
.
[2 sin 27lX dx.
'1(2
-;; sin 7lXJ'
a '1-;;
a
10.57. The first integralis set up as ai
~l
Using the appendix:
= Z[sin(nla-2nla)X +Sin(nla+2nla)X]a =2 [Sin(-n) + Sin(+)]=~oO
a
2(nla-2nla)
2(nla+2nla)
0
The reverse order of wavefunctions gives us
a -2nla
-2nla
ai f2 sin 27lXJ' [2 sin 7lX dx.
~l V;;. a V;; a
..
a
Solving:-
= ~[sin(2
I a - n I ah + sin(2n I a + n I a)x]a = 2 [Sin(n) + Sin(-n)] = 3.(0+ O)~
0
2(2nla+nla) 0 a 2nla
2nla
a
a 2(2nla-nla)
Therefore, order doesn't matter when showing that any two wavefunctions are orthogonal.
10.59. He-iE'11i"P(x) =
in)X(¥,~/lEi-:=e?
is the expression we need to show is satisfied. On the
,at
_
.
left side, we recognize that the Hamiltonian operator does not include time, so the exponential
function can be moved outside of the operator. On the right side; the spatial part of the
wavefunction, "P(x), does not depend on time, so it can be removed from the derivative. We get:
49
e-iEtlli . flq.t(X)
= i(i\fl(X)
=)1i: -
-iEllli)
e
.
Substituting:
e-iEtlli . flq.t(x)
o(
i: .
at
e-
iEll li
.
::l( -iEI/Ii)
-E
-tsu»
1
Evaluating the derivative: u e
=e
'­­
1i .
.
at
. 'P(x}.
The two exponential functions cancel. On the right side
v'".
of the equation, the!! terms cancel, as do the two i's with the negative sign. What is left is
qf¥(x) Q'¥(x), which.is.thejime­indepcndent Schrodinger equation. Since we accept this
. equation as a postulate, we affirm that the original equation does in fact satisfy the timedependent Schrodinger equation.
2
10.61. 1\fI(x,t)1 = (\fI(x,t»'\fI(x;t) =
= \fI(x)' . \fI(x) = 1\fI(x)1
2
•
[e-iEI/Ii . \fI(x»)' e-
iEll li
• \fI(x)
=
e+
iEl / li
. \fI(x)*
. e-iEtlli . \fI(x)
Thus, the square magnitudes of the time­dependent and the time-
independent wavefunctions are the same.
(
50
CHAPTER 11. QUANTUM MECHANICS: MODEL SYSTEMS AND THE
HYDROGEN ATOM
11.1. 3.558 mdynx
ldyn
xlOIOAx IN ==355.8 N.
A
1000 mdyn
1m
105 dyn
m
11.3. Objects in a gravitational field are experiencing a force that is directly proportional to their
height above .their equilibrium heights. Thus, they satisfy the fundamental mathematical
requirement for a Hooke's­law type of oscillator.
~: ~',
11.5. Starting with [ ­
1'12 d 2
[
­ ­­2
2m dx
[ ~:
221'1 2]
+a x
­ a'x'
2m
+ 21r'v'mx'
]'1' = E'¥ , substitute the definition to a to get
1'1 2
'P = E'Y. Dividing both sides ofthe equation By ­ ­'­
yields
2m
\
]'1' ~ ­ 2;:' '!'. Finally, bringing all terms to one side of the equation and
factoring out the 'P from two terms, we get
~
+ (2~E
_a 2x 2
)'P
=
0, which is ~ur
ultimately
result.
.
.11.7. Starting with a + 2an ­
.
an 2 an 1'1 2
­­+­­­ E
2m
m
2~E
1'1
= 0, first let us multiply all terms by ~ .
= 0, which rearranges to
.
definition of a: E
=
21Cvmli 2
2mli
+
.
21Cvmnli 2
mli
E
=­a1i
2
2m
2
2m
We get
anli 2
+ ­ ­ . Now we substitute for the
m
h
.
. Recalling that 1'1 =­ , we can substitute and cancel
.
21C
out the 2n terms. We also cancel mass out ofboth fractions. We are left with E
Factoring out the hv from both terms, we are left with E
= viz + n viz .
2
=( n + ~ )h v , which is the required
equation.
!ill = (6.626xlO­34 J·s)(1.00 s') = 6.63xlO·34 J.
11.9. (a) !ill = hv
(b) A =~
8
(c) This is in the (very­long­wavelength) radio
3.00xl0 1mls = 3.00xl0 8 m
v
1.00s·
wave region of the electromagnetic spectrum. (d) Such a long, low­energy radio wave would
have been undetectable by early­20th-century technology (and may still be undetectable today!).
51
11.11. Assuming that the energy change between each level is the same, the overall energy
change for the process will be four times the energy interval between adjacent levels, or tiE =
4hv. Now we need to convert our vibrational frequency to a true frequency.
17 = 3650cm·1
this light: v = ~
=..!­ . A. = 2.74xl0­
A.
4
cm=2.74xl0'6 m
Now, we calculate the frequency of
s
= 3.00x 10­ 6m1S =1.09 x 1014 S'I.
Now we can calculate the energy:
2.74x 10­ m
tiE:;::: 4(6.626xlO·34 J·s)(1.09xl0 14 s·l) = 2.89xlO­ 19 1. For one photon to have that amount of
A.
9
2.89 xl 0:: J = 4.36 x l 0 14 S·I (which is four times the
6.626xW' J·s "C.!......
•
original frequency). The wavelength of a photon having that frequency is .
energy, it must have a frequency of
A. = ~ = 3.00x lOS mls ,.= 6.88x 10­7 m.
v 4.36 x 1014 S·I
-/
(
11.13. Using equation 11.19, we find that '1'1 has the form:
Substituting into the Schrodinger equation: ­
a
2m ax
2
2
!:.­­2
)1/4(1)1/2
2
(2aI/2x)e-ax2/2.
N(2 a l/ 2x)e-ro:
2
/2
l'
2
+ _kx 2N(2a l / 2x)e-ro: /2
·2
(where we are using N to represent the collection of constants at the beginning of the
wavefunction, for clarity). First, let us evaluate the second derivative of the wavefunction.
. The first derivative is:
~(2al/
x)e- ax2/2 = 2a 1/2xe- ax2/2 . (-ax) + 2a1/2e-ax2/2
ax
The second derivative is:
.~ ~
2a 3/2x2e- ax2/2 + 2a 1/2e-ax2/2 ] = _4a 3/2xe-lU2 i2 _ 2a 3/2x 2e- ax2/2 (-ax) + 2a 1/2e-ar2/2 (-ax),
which simplifies to 2aS/2x3e-ux2/2 _4a 3l2xe-ax2/ 2 _2a3/2xe-ax2/2. Recalling the original
definition of'l'I, we can rewrite this as (if we include the N) a 2X 2'1'1­ 3d¥1' Substituting back
into the Schrodinger equation:
2
_!:.-(a 2x2'1'1 -3d¥/ )+!kx 2 '1't
2
~
2
.
.
Now we can substitute for the definitions of a and k:
.
2
1
2y2 m)x 2'1'
­h.­ (2mm)2
­ ­ x 2'1' +h.- -6nvm
- ' P +_(4n
2m
h.
1
2m h.
I
2
I
.
= _n 2y 2 mx2'¥.1 +3h.mJ¥I +n 2y 2 mx 2 'I'1
The first and last terms cancel, leaving 3h.mJPl' Writing in terms of h/2n:
3hny '1'1' which simplifies to ~ h y . '1'1 Thus, we have verified that we generate an eigenvalue of
2n
2
3/2 hv when we substitute 'PI into the Schrodinger equation.
+00
ar2
N 2 Jx 2e- dx = 1 Now we apply one of the integrals
­00
in the integral table from the appendix:
52
N
wavefunction is 'PI =2
:
1/4( )1/2
~
( )
Z
3J1I2
= 2[ :
N
= 2 112
[
3J1/4
Therefore, the complete
:
,
1/2 : 3J1/4 xe-crx2 / z , which can be shown to
be algebraically equivalent to
[
(2al/2x)e-ax2/2, which is the form of the wavefunction obtained when using
equation 11.19.
11.17. The classical turning point is the value of x at which the total energy equals the potential
energy. We have expressions for both ofthose energies, so we simply equate them and solve for
x:
hv(n + Yz) = Y2kx?
x
2
(2n+l)hv
= 2hv(n+1/2) = --'----'---
24
k
k
x=±
(2n+ l)hv
k
22
11.19. (2.435 x 10 kg)(2.995 X 10 kg) = 2.959x1022 kg.
2.435 x 10 24 kg + 2.995 x 10 22 kg
11.21. 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:
g. '
l•i d
c. 11'
. ­.
v = 1/#
ea s us to th e 10
owmg ratio:
' r * = ­ . Now we nee 'd
vfJ
fJ'
v
1/ v fJ
fJ
'
only calculate the reduced masses of the O­H and O­D 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
= (1)(16)/(1+16) = 0.941, while ~ ( O D )
= (2)(16)/(2+16) = 1.778. 'Therefore,
in grams: ~(OH)
v oc C1 an d vv" oc
we get
3650~m-1
~
= ~1.78
, v
2K
11.23. For 'P3 : N 2
Solve for v": v·
= 1.374
J(e
i3
¢
o
re
i3
,
¢ drjJ
2K
=1
J(e-
N2
o
N 2[rjJ]~1
=1
r
,
'
f(e i13fl e
i13
¢ drjJ
=1
N 2(2;r _ 0) = 1
}i3fl drjJ
=1
N=_l_,
.J2;r
2;r
21Z'
i3fl
o
N2 =_1_
For 'P 13: N 2
= 2656cm­1 •
0.941
2K
N2
'
J(e­i13 }il3fl drjJ = 1
¢
o
N 2 = _1
2;r
N
= _1_
.J2;r
Thus, both wavefunctions have the same normalization constant.
53
. m ~ 5.69x 1036
11.25. (a) 600 kg. m 2fs = m(6.626x 10­ J. s)
2rc
36
34
(b) E = ( 5 .69 X 10 ) 2 (6.626 x 10­ J . S)2 = 113 J
Classically, the child would have an
(2rc) 2 (2)(25 kg)(8 m)"
L2 (600k ·m 2 f s ) 2 .
.
g
2
= 113 J, the same amount ­ but this shouldn't be surprising,
energy of ­ =
21
2(25 kg)(8 m)
.
since we calculated the 'quantum­mechanical' amount of energy using the same values but in a
more roundabout fashion.
34
11.27. (0) E(m+I)-E(m) =
(m+~'A
rtiii)
m~;'
­
(b) IfE(l) -E(O) = 20.7 em", we can use the formula from part a to detennine1, then calculate
E(2) - E(1) using that value of I.
= (2· 0 +1)(6.6262 x 10­
34
J'S)2
1= 2.686 X 10-70 J2 m 2 fcm,t (we are not reducing
(2rc) (2)1
the units, since we're simply using this quantity to recalculate an energy difference in em" units)
.
(2.l+l)(6.626xlO­34 J'S)2
Now, calculating E(2) - E(l): E(2) - E(l) =
2
70 2 2
t
= 62.1 cm' .
.
(2rc) (2)(2.686 x 10­ . J s fcm' )
This is not a great agreement with the experimental value of 41.4 em", suggesting that the 2­D
. rotational model may not be the best model for HCl.
20.7cm·1
11.29. Refer to Figure 11.12. The easiest coordinate to relate is the z coordinate. Projecting the
r vector into the z axis, we find that the r vector is the hypotenuse of a right triangle. Since the
definition of cosine of an angle is "adjacent side over hypotenuse", we have cosO =!.., which
.
r
rearranges to z = r cose. The other leg of that triangle, by the way, is r sine, which is drawn in
Figure 11.12 as the dotted­line projection of the r vector in the xy plane. This dotted­line
projection serves as the hypotenuse of another right triangle. This dotted line's projection into
the x axis, the x coordinate, can be related to ~ through the cosine function, which is "adjacent
x
.
leg over hypotenuse". Thus, cos; = .
. Therefore, we have x = r sine cose, The other leg
rsmO
of that triangle can be related to the y coordinate through the sine function, as "opposite leg over.
hypotenuse. Therefore, we have sin; =
~
rSlnB
, or y = r sine sin~.
11.31. Because a potential energy is a multiplicative operator, ifit is simply a constant K, we
.have:
a
­­.:..­(a
--+cotO-+
li
2
2
.
21 00 2
2
2
ao
1
sin 2 0
a J\f'+I0¥ =E¥.
2
a¢/
2
J
a
a
1
a \f'
­Ii­ ( --+cotO-+
2
21 eo:
ao sin B a;2
= E¥ -K.'P = (E-K)\f'. =E
54
\f'.
new
So we could in fact use the same form of the Schrodinger equation if the potential energy were
some constant value.
2
11.33. (a)
i 2\}J3 2 =_n 2( B 2 +cotB~-1_2
,­
. BB
BB
2
B 2 ](.J105sin 2 BCOSBe-2i¢J
sin B BfjJ
4
Let us determine each derivative singly, then combine the three terms and simplify. The first
tenn:
.J105
. 3 BJ
.J105
. B cos 2 B­ sm
. 3 BrL -2'16
=­B\}J3 2 =(.JI05.
­ ­ · 2 sm B(cos B) cos B ­ ­
­ sm
e­ 2'16 =­
­ (2 sin
aB'-
4
BB;2\}J3,_2 = .J~05
I
4
I
4
(2cos3B-4in~¢
The second term:
cotB aBB 'P3,­2 = cotB
.J~5
~sinBco 2
B­ sin 3 B~-2i¢
= .J~05
~COS3
B­ sin 2 B cosB ~-2i;
The third term:
2
B \TJ
sin 2 B BfjJ 2 3,­2
1
­­­­T
_
1
­­­
sin 2 B
(.Jl
05 .. 2 Bcos Be- 2i¢ . ( ­ 2 ')2 J_ 4 ­­cos
Be-2i¢
.J105
­­sm
l
4
--
4
Combining these terms:
-li'(
1:5
..1
{zeos' 0 ­ 4 sin' OoosO
~ ~sin'
OeosO + 2eos' 0 ­ sin' OeosO
~ 4eosO~
­",
J
This simplifies to
-Ii' (
.I~5
(4 cos' 0- 8sin' 0 cos 0 ­ 4 cos
-Ii' ( ..11 cos 0(4 cos' 0 ­ 8 sin' 0 ­ 4
4°5
ok"')
k"')
Factoring a cosine out of the first­two terms:
If we recall that sin'e + cos'O
~
I, then we can
substitute the 4 inside the parentheses with 4sirt2e + 4 cos 2e (but don't forget to distrb~e
minus sign:
-li'(
..11:5 cos 0(4 cos' 0 ­ 8sin' 0 ­4sin' 0 ­ 400s' 0
~-'"
J The 400s'e terms cancel, and we
can group the sin 2e terms together to get
-li'(5~J.
e080(­128in'
the
.
ok'" J= +121i'( .I~5
an
J
sin? Oeos/F';' = +121i''P,,_,
Therefore, this wavefunction is eigenfunction of the square angular momentum operator with
an eigenvalue of 121'1. 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 e: the exponential function. We.have:
55
-
ih~
'~a
'¥3
= _ih~(.J105
-Z
4
~a
J
¢J= -2h'¥3_Z.
sin" eCOS()e-Zi¢ = -ih( _20(­J105 sin" ecosee­Z1
4
Therefore, this wavefunction is an eigenfunction of the 2­D angular momentum operator with an
eigenvalue of ­ 2h ­ as expected from the eigenvalue formula.
(c) To determine the energy eigenvalue, technically we have to evaluate the entire three­part
~z
.
we can use the answer to
derivative in equation 11.43. However, if we recognize that fI =
2/
.'
2
part a because we have already evaluated the eigenvalue to the L operator. Thus, we have
~.
.1 ~
.
1
12hz
.
Z
z
lJ'P 3.-Z = 2/ L '¥3,_Z = 2/ (12h
as expected from the eigenvalue formula.
!:..­,
)=U ­
z
11.35. (a) E(£ + 1) _ E(£) = (£+ 1)(£ + 1+ l)h Z (£)(£ + l)h = (2£ + 2)h 2
2/
2/
2/
(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:
34
20.7cm·1 = (2·0+2)(6.626x10­
J.s)z
/=5.370x10­70 JZ .sZ /cm' (we are not reducing
(271l(2)/
.
the units, since we're simply using this quantity to recalculate an energy difference in em" units)
. .
(2·1 + 2)(6.626 x 10­34 J ·s)Z
.
Z·
­70 Z Z
- 1 = 41.4 em".
Now, calculatmg E(2) ­ E(1): E(2) ­ E(1) =
.
(27l") (2)(5.370 xl 0
J s Icm )
. This is exactly the same as the experimental value of 41.4 ern", suggesting that the 3­D
rotational model is a good model for HCI.
11.37. For the four £ values in exercise 11.36:
For £ = 5: L2 = £(£+I)h2/(2n)2 = 5(5+1)(6.626xl0­34 J.s)2/(2ni = 3.34xlO'67 J2·S2.
angular momentum is the square root of this: L = 5.78xlO­34 Jvs.
For £ = 6: L 2 = £(£+I)h2/(2n)2 = 6(6+1)(6.626xl0·34 J·sil(2n)2 = 4.67x10­67 J2·S2.
angular momentum is the square root ofthis: L = 6.83xl0·34 Jvs.
For £ = 7: L 2 = £(£+I)h2/(2n)2 = 7(7+1)(6.626xlO­34 J.si/(2n)2 = 6.23x10­67 i·s2.
angular momentum is the square root of this: L = 7.89x 10­34 Jvs.
For = 8: L 2 = £(£+ 1)h2/(2n)2 = 8(8+ 1)(6.626xl0­34 J's)2/(2ni ~ 8.01x1 0­67 i·s2.
angular momentum is the square root of this: L = 8.95xto­34 J·s.
e
Therefore, the
Therefore, the
Therefore, the
Therefore, the
11.39. Physically, the two wavefunctions are the same except they are reflections through the xy
.plane.. One wavefunction has a positive z component, and the other wavefunction has a negative
z component.
11.41. V
=:
_~
=
41l'80 r
19
(1.602 x 10­ C)Z
= ­4.36 X to­18 J
12
2
41l'(8.854 x 10­ C / J . m)(0.529x 10,10 m)
.
11.43. Starting with equation 11.56:
56
[­ :~ [:' : (r' :)+ r' s~no:
(sino
:0)+ s~, 0:;,J+ V]~ ~
r'
E¥
.If V = 0 and r is constant, we lose two things immediately: the potential energy operator and the
first derivative term in r (since r is constant, any derivative with respect to r is zero)" We have
e· a·)
2
a
2
1
8 ("
sm ­
+ 2 .1 2 ·
2 JUJ
T
2Jl r sin e' ae
ae r sm e a¢
tt
­ ­(
2.
I7'\TJ
=1jT
N ow,.d""b"·
h deri
" wi·th·
istri utmgte
erivative
.
respect to theta through the parentheses:
J+
. ae
2
2
tt
e­a + sin e a
1 ( cos
.
2Jl r sin e
(
­ ­
2
ae
­­2
J
2
1 2 a 2 \fl = E'Y If we factor out the? in the
r sin e 8¢
2
denominator and combine it with ~ in the denominator, we get the moment of inertia, /:
_~(lcose+in
2
2
2/ sin e
a
a
J+_1_­
ae 2 sin 2 e a¢ 2
ae
J\f = E'Y
Finally, if we distribute the sinS term
.
in the denominator through, we get
2
­ ­tt ( cot
2/
G
a + -a- 2 + 1 - - a
J
." ..
- 2 \f = E¥ , which IS equation 11.46.
e­ae
ae sin e a¢
.
Cal
2
2
­­2
,t~F
we must convert the wavelengths to
1 . x 10 nm =15 230cm­
'
1em .
,
656.5 nm
wav~umber
units:
1
x 10 urn = 20 560cm­1
486.3 nm
1em
'
7
7
1
x 10 urn == 24,370cm­'
1
x 10 nm = 23,030cm­'
434.2nm
lcm
41O.3nm
1cm
These transitions are from quantum levels 3, 4, 5, and 6, respectively. Therefore, we can
calculate R as follows:
­I
( ­
1 ­­
1)
15 230cm=R
,
2 2 32
1· l)
(2"2-4"2
(12"2-""521)
R(;, ­;,) R~
.20,560cm­' =R
.
23,030em­' =R
24,370cm­'
~
R =109,700cm­'
R = 109,700cm­'
R =109,700cm­t
I09,700cm­'
.
Thus, to four significant figures, all four lines give the same value ofthe Rydberg constant.
(b) Since Z = 2 for helium, there is an extra Z2 term in the numerator of the Rydberg constant for.
He+, so R would equal 2 2 x 109,700 cm­1 = 438,800 em­I. Therefore, we would expect the lines to
appear at:
438,800 em
-I(;2 - 3\ ) = 60,940 cm ­lor 164 nm
57
1(2\ ­ :2) == 82,280cm­1orl22nm
438,800cm­
= 92,150cm­ t orl09nm
438,800cm­t2\ ­ 5\ )
438,800cm­
t(2\
= 97,510cm­t or103nm
­ 6\ )
11.47. Figure 11.17 in the text gives such a diagram.
11.49. The wavefunction '¥4,4,0 does not exist because Rcannot be equal to n. Similarly, a 3[
subshell does not exist because n would equal 3 and Rwould equal 3 also, and the constraints on
the hydrogen atom wavefunctions are that Rmust be less than n.
11.51. The setup is:
p=
JJ(
1
n(0.529A)3 )
1/2
(
e-rIO.529A.
)112 e-rlo.529A .r 2 sinfJdrdfJd¢
1
n(0.529A)3.
This is a triple integral, where the limits on ~ are 0 to 2n, the limits on e are 0 to n, and the limits
on r are 0 to IA. Thus, we have
21r
x
= Jd¢x JSinlMBX
. ° 0·
P
the e and
P == 4n x
..
~
(1
3
)O.tA
Jr2e-2rlo.529Adr As shown in the text (see example 11.24),
°
n(0.529A)
integrals collectively equal 4n. Thus, we have
1
)O.lA
3 Jr2e-2rlo.529Adr Using the integral table in the appendix, we can solve.
°
( n(0.529A)
.
.
r
the integral and evaluate:
'4
P
==
1l(
1
1l(0.529A)3 ) x [ e
P = 41C(2.150221A­, )x
-2r10529A
r
2
( (­2/0.529A)
.
[e ­'(._')"'.5>9(
-e.(
P
0­0­
(O.lA)' ~.529
2r
2
(­2/0.529A)2 + (­2/0.529A)3 . °
A) _ 2( O.lA)~
.529A) , _
2(0.5~9
A)' )
2(0.5~9A)' J
= 4n(2.150221A3)x[O.68518(­O.002645A3 ­ 0.01399A 3 ­ 0.03701A3) ] + 0.03701A 3
P == 0.0068, or 0.68%.
11.53. (a) For '1'28' there will be one radial node and no angular nodes, for a total of one node.
(b) For '1'38' there will be two radial nodes and no angular nodes, for a total of two nodes. (c)
For '¥3p, there will be one radial node and one angular node, for a total of two nodes. (d). For
'1'4fi there will be no radial nodes and three angular nodes, for a total of three nodes. (e) For'1'6g,
58
there will be no radial nodes and five angular nodes, for a total of five nodes. (f) For \f'7s, there
will be six radial nodes and no angular nodes, for a total of six nodes.
11.55.
a = 4m,/i
flez
z
= 4Jr(8.854 x 10­IZ C Z/J . m)(6.626x 10,34 J . s)z= 5.29585 x 10·1l m
(2Jr)Z (9.104x 1O­3! kg)(1.602xlO,19 C)
11.57. According to equation 11.67, \f'zp,
= ....;2
~ (\f'z P.I + \f'zP_I ).
Using the forms of the
wavefunctions from Table 11.4, we have
3
HI
T
_
Zp,
3
1 [1 (2Z JIIZ Zr -ZrlZa sm
. ee+i<p +-1.(2Z
Zr -z-rz» sm
. ee­i<PJ
- -JI/2 --e
a
.J2 ­8 .n:a­ ­3 --e
a
8 n:a 3
­­
~!(2Z:J/
i
Zr e-zrlzasine(e+i¢+e-i;) NowweusethefactthatcoS(J=ei¢+e- ¢ and
....;28 sa
a
2 .
3
1r:::: (2Z
JIIZ Zre -ZrIZa sin
. e COSy.
d.
­­3
substi
stitute: HI
T Zp =
s:
4....;2 sa
a
=
Also according to equation 11.67, \f'z
PY
~ (\f'z
=­ ....;2
P.I
­ \f'z . ). Using the forms of the
P­I
wavefunctions from Table 11.4, we have
HI
_
n~-i¢J
3Jl/Z z r
3JIIZ z r
i (1(2Z
-ZrlZa S
'fJe+i¢
1(2Z
---e
ill-- ---e -ZrIZa'
smDt:'
a
8 n:a 3 .
.J2 8 n:a 3
a·
:'
T Z
--p,
'
Zr l
_ I/ Z
•
l :«
_.). .
..
e''<P'­ e"'¢
i 1 2Z 3 J
r:::: - ( ­ ­ 3
- e Zr Za sm e~
I¢ - e I¢
Now we use the fact that sm (J ==
'.
....;2 8 sa
a
2z .
lIZ
.
substitute: \f'zp = 1r::::(2Z:J Zre-Zrlzasinesin(J. Note how the two iterms and the
4....;2 sa
a
Y
negative sign canceled.
=­
11.59. The combinations are:
1
.
'¥3d, =­('¥
.J2 3d.1 + '¥3d., )
.1 (
'¥3d) =.J2
- '¥3d+2 +'¥3d.2 )
and
i (\f' ­. '¥ )
\f'3dz ­­ ­.J2
­
3d.!
ss.,
. HI
T
3d.
= -i- ('¥
)
.J2' 3d+z ­'¥" : ' ,
\f'3ds = 'I'3do
The numerical labels on these composite wavefunctions are arbitrary; they are actually given
Cartesian coordinate labels that reflect their distribution in three­dimensional space. Note that
one of the wavefunctions is a pure eigenfunction, rather than a combination oftwo
eigenfunctions.
59
CHAPTER 12. ATOMS AND MOLECULES
12.1. Silver atoms have a single unpaired electron in their valence shell. Thus; there is an
unbalanced overall spin to the atoms, making them susceptible to magnetic fields.
12.3. The total mass being converted to energy is the mass of two elctrony~x9.10I-3
kg
= 1.822xl0­3o kg. Using Einstein's equation:
E=
~
(1.822xlO­
3o
kg'2.97~xl08
. 7'
~ ; li~e ~ 1 ~= :~ : ,number.
. ~.
9.86xlO Jim. 0
m/si = 1.638XIO­1@. 6.02xIO~ f
rl-Y'~
.Jf:FL-.,
/tvVl.
rep sent~:)
z e_~oLth
total.spin.angular momentum:. For el ~trons,
It c~.?e
either
+Yz or ­ ~. jb) If a particle has.~r":=<Q~!pen
m, can only ~qual
O. If a particle has S =;:~!)then
ms
can equal­3/2, ­1/2, 1/2, or 312.
can be­2, ­1, 0, 1, or 2. tfJ:lmltllclehass =-3.n,Jl!~i nt~
n2(2
2 2) 3e2 ­ . 3e2 ­ 3ez + e' + e 2 + -e'- 12.7. H = - - V e#, + V e # 2 + V e# 3 ­
2f.J
4ffBor,
4ffB or2
4ffBor3
4ffB Or12
4ffB o'i3
4ffBor23
The last three terms are what make this Hamiltonian non­separable.
A
12.9. 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
not­the correct energy eigenvalue for the atom.
12.11.
u', with o n ~ ,
would have the same wavefunction as the helium atom.
Therefore we simply ~;timcon:
'1'(1,2)
~
Jz
[(Is,a)fu"Q) t:V­ (!
Is,q
1 Is 2a
12.13. (a) '¥(Be) == r;:;;;
.
",24 Is 3a
Is 4a
ISla
Is 2a
1
'¥(B) == -J12O Is3a
120 Is 4a
Is lP
. Is 2P
Is3P
Is4[3
Is sa
IssP
Isl'P
Is2P
Is3P
Is4P
2sla
2s 2a
2s3a
2s4a
2sla 2s l [3
2s za2sz[3
2s3a 2s3[3
2s4a 2s 4[3
I
i";
.--1
(ls~),§].
.
-~
J.!.. I
If
Y7
y'L
2s l [3 2Px,Ia
2s 2P 2Px,2a
2s3P 2Px3a. Keep in mind that, because the sixp orbitals
2s 4[3 2PX:4(X
\
2s sa 2ssP
2px,sa
are degenerate, the last column could have been labeled 2px l}, 2pya, 2P.~I},·
2pza, .~ ro
(b) There are six possible determinants for the C atom that have the:tw9.­::UJlpairet!Jt­electrons
gni~ah
th.e sam~
sp~n.
'There are also six possible detrmi~
fot 1l(~ m dep n i g
on
WhICh spin­orbital is notused (analogous'to the wavefunctIon for Be III par<~
above).
12.15. Atoms that don't follow the aufbau principle exactly include Cr, Cu, Nb, Mo, Ru, Rh, Pb,
Ag, La, Ce, Gc, Pt, Au" Ac, Th,Pa, U, Np, and Cm. Note that none of these atoms are main
60
group elements; they are transition metals or inner transition (i.e. lanthanide or actinide)
elements.
12.17. (a) Li will have 6 possibilities for its unfilled valence shell: 2px1a, 2p/~,
2p/a, 2P/~,
2p/a, and 2pzl~.
(b) C will have six possibilities for its unfilled valence shell: 2px1a 2py1a,
2pyl~,
2p/~
2p/~,
and 2P/~
2Pzl~.
(c) K will have six
2p+la 2p z1a, 2p/a 2pz1a, 2pxl~
1a,
possibilities for its unfilled valence shell: 4p/a, 4pxl~,
4py 4P/~,
4p z1a, and4pzl~.
(d) Be
will only have one possibility for its excited state: since both electrons have­been excited to an s
orbital: 3S1Q '~2S3
(e) UwillhaVe6 possibilities for itsunfilled vaienc-e s~cr,7pxl~,
1a,
la,
7py 7pyl~,
7pz and tpz l~.
Note the siI1Jj~.
i~J'@!.or
Li"K;'\and U.
'---
.
""":.
12.19. The ground­state harmonic oscillator wavefunction is found in Chapter II.
*[(a)1I4'
e
_ax
K
(a)1I4 e -ax2/2dx_- c(a)t/2+
I'"xe -ax2dx .
2/2]·
4
·cx·
­
4'
K
~
-'"
According to the table in the Appendix, there is an appropriate form to use if the integral limits
are 0 to co. Since the function in the integral is even, we can split the interval and multiply the
value ofthe integral by 2. We get:
­
=:
2 . c( :
)
1/2( 2 1:a3 2:
)( )112
3
The two s:uare root terms cancel. Simplifying:
here c IS th e anh armonicity
.. constant.
= ­3c
­ 2 ' were
4a
­12.21. To determine a3, follow the same procedure used in Example 12.10 but use '£'1 and '£'3 for
the particle­in­a­box:
a,
J'¥; ·h·'¥,dx xd~niSl·~niSlf
~,
E, ­ E,
xd~niS xj~
U::', -::::)
=
=
)
~J
(-
P
l:
Again, we use the trigonometric identity sinax­sinex ='Y2[cos(a-b)x -cos(a+b)x] and stibstitute:
1 *~
a . = -2kma
-2- . 3
.
2
h
kma [a
= - - 2h
2
- c~s-+
2
K
-17
"This expression can be. integrated:
2
21lX" Itq' . 21lX
a
41lX ax. 41lX]a E I '
h I' .
. -' .-sm~
_ ·__ . ­=­­.­cos­­­sm­
va uating at t e units:
2
',-fl-<,./' 1(, / .a
16na 0 tz _4~·.
•
.
2
2
.
2
-a
a
-./
Jt2.. (1) ~ -;-(01- 100-2 t1~
kma [a
a3 =
0
21lX
41lX) dx
xcos­­xcos­
a
a
­.'
Co
.. \
.
0
]
4K (0)- [0 + 0 - 0 - 0]
15kma
= - 16~
3
.
.
From the Example, we found that k= 4-UX 7 kg-m/s', a = 1.15x10- 10 m, so*,e can evaluate this
expression explicitly for an' electron: a3 = -0.02998.
'{
/_~ .
12.23. The integral we need to evaluate is
61
!(~)/2
j 8 sa'
r,B,;
!-e- r/2a COSB.eErCOSB.(_1_)t/2 «:': .r 2sinBdrdBd¢
a
n:a 3 .
.
(Note that we have the
Zp, orbital as one wavefunction and the Is orbital as the other wavefunction.) Everything is
multiplicative, so we can separate this integral into three integrals as follows:
=
.J2e~
2Jd¢' jcos 2 B sin ()dB . jr4e-3r/2adr.
The integral over
8n:a 0
o.
0
over is easy to evaluate as a power function of cosine:
~
is simply 2n. The integral
e
I
Tr
1
3
1·
2
BsmBdB = ­­cos B = ­­(­1­1) =3
0
3
3
The integral involving r can also be evaluated using a formula from the appendix:
s
s
= 24·32a = 256a
jr4e-3r/2adr=' 41
(3/2a)s
243
81
o .
Combining the three integral solutions with the collection of constants, we get that;
jcos
o
Tr
2'
. 2:r . 2 . 256a 5
( E ) = .J2eE
4
t
8n:a
3
81
= 128.J2eaE
243
j(Ca,t'l't +Ca,2'1'2)*H(ca,l'l'. +c a,2'1'2)dr
..
. 'we distribute the operator and
j(Ca,t'l'l +Ca,2'1'2) (Ca,t'l'l +Ca,2'1'2)dT
.
..
.
multiply through to get the individual integrals:
12.25. Starting with
_ j(ca,t'l't +Ca,2'1'2)*(Bca,t'l't + BCa,2\2)dT
­
J(c•.,'I',.+.c•.,'I',)'(C•.,'I', +.c:.,'I',JdT
.~
••
.: j(ca,t'l't) HCa,t'l't + (Ca,2'1'2) HCa,l'l't + (Ca,l'l't) HC a,2'1'2 + (C a,2'1'2) HC a,'2'f2 d T
­
j
.
(Ca,l'l'J ca,.'I't + (C
..
.
.
\
a,2'1'2) Ca,t'l't + (ca,t'l't) Ca,2'1'2 + (C a,2'1'2) c a,2'1'2 d T\
The integral sign can be distributed through to all ofthe terms:
_ j(ca,t'l't)*BCa,t'l't dT + j(Ca,2~)2'l
BCa,t'l't dT + j(Ca,t'l't)* BCa,2 'l'2 dT + j(Ca,2 '1'2)* BCa,2 'l'2d T
­
j(Ca,t'l't)* Ca,t'l't dT + j (C a,2 '1'2)* Ca,t'l'ldT + j(Ca,l'l't)*Ca,2 'l'2 dT + j(Ca,2 '1'2)* Ca,2 'l'2 dT
Now the constants can be brought outside the integral signs:
C a,12
J~dT+C"'
...tCa,2 J'P2·ir¥tdT+Ca,lCa,2 J'Pt·ir¥2dT+Ca,/1'P2*ir¥2dT
Ca/ J'Pt·'PtdT+Ca:ICa,2 J'P2"'PtdT+Ca,t Ca,2 J'Pt·'P2dT+Ca,/ J'P2·'P2dT
At this point the definitions from equation 12.28 can be applied, and the expression becomes
2
2
.
2
2
Ca,t H 11 + Ca,.Ca,2 HI2 + Ca,t Ca,2 H12 + Ca,2 H 22 Ca,t H II + 2C a,t Ca,2 Ht2 + Ca,2 H 22
.
=
2
2
= -'2.:;~
­Ca,t 8 u + Ca,t Ca,2 8t2 + Ca,t Ca,2 8t2 + Ca,2 8 22
Ca,t 8 11 + 2Ca,t Ca,2 8t2 + Ca,2 8 22
This is equation 12.29.
=
12.27. An "effective nuclear charge" approach to determine appropriate wavefurtctions for H
atoms is unnecessary for two reasons. First, since we can solve the Schrodinger equation
analytically for the H atom; we have no need of approximation methods. Second ­ and perhaps
62
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.
12.29. (a) Using equation 12.31, the non­trivial solution is found by evaluating the determinant
-15-E·l -2.5-E·0
.
. == O. Because the baSIS wavefunctions are orthonormal, we mow that
-2.5-E·0 -4-E·1
-15-E -2.5
== O. We
'-2.5
-4-E
+19E + 53.75 = O. Using the quadratic
8 11 = 822 = 1 and 8 12 = 821 = O. We can simplify the determinant to
expand the determinant to get the quadratic equation K
formula, we can determine the two roots for E:
E==)57.3 ()1 4- 93~± 1-
E == ­3.46 or ­15.54
2
(b) The energies of the real system are still rather close to the ideal energies, but are farther.
2:fM? arid H21• As those values get
away than in Example 12.12. The reason iS~p'es
larger, the energies ofthe re~l
sy teir ~ <U@li ~m
those of the ideal system.
12.31. Given that any trial wavefunction ~ can be written as a linear combination of ideal
wavefunctions \fIj: rp == LC/f';, where fry; == E;\fIj. We need to show that JfiT¢xh ~ E) ,
where E 1 is the ground­state energy of the system. We start by substituting the linear
r
combination into the integral: J(Lcj\flj iT(Ic j\fIj}iT ~ E) . We CarI distribute the
f
.
J{2:
Hamiltonian operator through the second summation:
C j \fI;
r(L
C j i r Pj
~T
~
E1
•
Since the
\fIj's are eigenfunctions of the Hamiltonian operator, we can substitute Ej\flj in the second
summation: J{2:c;\fIJ{2:c jE j\fIj}iT ~ E) . Now we rearrange the summation arId'~teg l
signs, bringing the constants outside of the integral: LLcj*cjE j J'f'/\fIjdT ~ E) . Since the
j
j
ideal wavefunctions are orthogonal, the integral is exactly zero except for the terms where i = ).
This has the ultimate effect of eliminating one of the sums and making all of the subscript labels
the same. Thus, we get, ultimately, that rp* iT¢xh =L C j*C jEj ~ E) as the variation theorem we
J
.
j
are trying to prove.
The coefficients themselves may be positive or negative, but in either case, multiplying
the coefficient by itself will always yield a positive value. That is, clCj is always greater than
zero. Since we know that Ej ~ EI (that is, any arbitrary energy value is either the ground­state
energy or higher than the ground state energy), we can multiply Ej and E 1 by Cj*Cj, a positi ve
number, arid still have the same sign on the relationship: Cj*cjE) ~ Cj*cjEI. Furthermore, if we
sum over all)' s, the summation on the left will always be greater than or equal to the summation
on the right because the left sum is the sum of larger individual terms:
Lc/cjE j ~ Lc/c jEl' 0r Lc/cjE j ~E1Lc/Cj
j
j
j
j
. Thus, we can substitute this last
expression into our last integral equation from the last paragraph to get
Jrp*H¢xiT ==LC j *Cj~
j
~ E1LC j* c j .
j
63
Finally, if the trial wavefunction is normalized (which it should be), we have the
following relationships: 1
1 = L~,>i*Cj
=Jj?'ddt' =(~ '¥J(~ '¥ Jar ,
c;
cj
j
which can be rearranged as
J\f';*'Pjdr. Again, the original \f' wavefunctions are orthonormal, which means
;
j
that every integral is zero except for when i =j, in which case the integral equals 1. This again
has the effect of eliminating one of the summations, and we are left with 1 =
L
Cj * Cj •
j
Substituting this into the right side of the last inequality in the previous paragraph yields
J¢ f:I¢xl
*
t: :2: E 1 ' which is the variation theorem: any trial wavefunction will have an average
energy equal to or greater than then true ground­state energy of the system.
12.33. Mathematically, the Born­Oppenheimer approximation is given by equation 12.36:
'Pnuc . 'Pel. In words, the Born­Oppenheimer approximation is the assumption that
\}'molecule
electronic motion can be modeled as ifthe nuclei were not moving, because electrons more so .
much faster than the much­heavier nuclei.
12.35. Using the expressions in equation 12.45:
I1E = E 2
-
E1
= H 22 ­
H 12
1­ 8 12
-
H ll + H t 2
1 + 812
•
By combining the denominators, cross­multiplying and
collecting terms, you can easily show that this expression is equivalent to
+
. A 17 =. H 22 -2H12 ..,..HII + (2H22 + H ll )8 t 2 •
. tsi:
For H2 .Hi, = H 22 , so we can simplify further:
1­812
. I1E
= 2H1I812 ­
1­812
2H12
2
12.37. The bond order for the first excited state ofHt is ­112. Since this state is known to be a
dissociative state (that is, in this electronic state the two nuclei move apart spontaneously to
achieve a lower energy), we can deduce that any diatomic molecule that has a negative bond
order is inherently unstable, spontaneously increasing its internuclear distance to achieve a lower
(and more stable) energy.
12.39. Very simply, the first wavefunction in equation 12.43 is the bonding orbital because its
corresponding energy is lower than the second wavefunction in equation 12.43.
12.41. ol+: (cr1si (cr*1Si (cr2si (cr*2si (cr2pzi (n2px,y)4
O2­: (crlsi (cr*lsi (cr2S)2 (cr*2si (cr2pzi (n2px.y)4 (n*2px,y/
ol-: (crlsi (cr*lsi (cr2si (cr*2si (cr2pzi (n2pxJlt (n*2px,yt
12.43. According to Figure 12.24, NO has a bond order of 5/2.
64
CHAPTER 13. INTRODUCTION TO SYMMETRY IN QUANTUM MECHANICS
13.1. The more symmetry elements an object has, the simpler it is to mathematically describe
the three­dimensional shape of an object. Thus, the phrase "higher symmetry" ultimately
translates into mathematical simplicity.
13.3. Consult Figure 13.14 for the flow chart on determining the symmetry of an object. (a) A
blank: sheet of paper would have D 2h symmetry. (b) A three­hole sheet of paper would have C2v
symmetry. (c) A baseball (with stitching) should have C2v symmetry. (d) A round pencil
would have C",vsymmetry. (e) The Eiffel Tower would have C4v symmetry. (f) A book
(without printing) would have C2v symmetry. (g) A human body (without considering the gross
anatomical features) would have C, symmetry, since the front and back are different. (h) A
starfish would have DSh symmetry. (i) an unpainted stop sign would have D 8h symmetry.
13.5. (a) Yes, these two elements can constitute a complete group. (b). Yes, these two
elements can constitute a complete group. (c) These elements are not a complete group, since
the identity element E is missing. (d) These elements are not a complete group since the inverse
of C3 , C3 or C3 2 , is missing.
13.7. (a) Since S; involves C» followed by reflection in a plane perpendicular to the axis, the
matrix would be the same as C; but with a ­1 in the lower right comer:
cos O
­ sin 0
. S n::::
[
o
­1 0 0]
[
Therefore, i = 0
.
0]
0 . (b) Inversion involves changing the sign on every coordinate.
-1.
.
sinO
cos 0
0
0
-1
0
0
­1
.
13.9. (a) The following shows a multiplication table for the C3v point group (which you can
verify by using a molecule like NH 3 as an example):
.
2
cry"
E
C3
C3
C3
C3
E
C3
cl
E
cl
E
C3
E
cry'
cry
cry
cry"
C3
cry
cry"
crv'
crv'
.
.
cry"
cry"
cry'
C32
cry
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: cr v(EC3) = crv C3 = cry'.
Also, (crvE) C3 = crv C3 = cry'. Thus, the associative law is satisfied.
65
13.11. (a) In C2v, C20'Y = cs;', (b) In C2h, iC2 = O'h· (c) In D 6h, C60'h = 8 6 . (d) In D u , C2C/
84 • (e) In Oh, i84 = cl.
.
=
13.13. Porphine has D 2h symmetry as a neutral molecule. However, if a metal ion is substituted
into the center of the porphine ring, with the simultaneous loss ofthe two H atoms on the ring
nitrogens, the symmetry becomes D 4h .
13.15. The tetrahedron has Td symmetry: E, 8C3, 3C2, 684, and 60'd. the cube and the
octahedron have O, symmetry: E, 8C3, 3C2, 6C4 , 6C2', i, 886 , 30'h, 684, and 60'd.
13.17. Using the scheme in Figure 13.14: (a) H 202 has C2 symmetry. (b) Allene has D u
symmetry. (c) d­Glycine has C/ symmetry. (d) z­Glycine also has C/ symmetry. (e) cisDichloroethylene has e2v symmetry. (f) trans-Dichloroethylene has C2h symmetry. (g)
Toluene has C/ symmetry. (h) 1,3­Cyclohexadiene has C2v symmetry.
13.19. (a) The wavefunctions of deuterium oxide have C2v symmetry. (b) The wavefunctions
of boron trichloride have D 3h symmetry. (c) The wavefunctions of methylene chloride have C2v
symmetry.
13.21. (a) The possible formulas would be C4~,
CgHg, and C20H20. There wouldn't be a
hydrocarbon equivalent to the octahedron or icosahedron. (b) You should be able to verify that
C~
has all the symmetry elements of the TdPOint group, while CgHg has all of the symmetry
elements of the Oi; point group.
13.23. Of the listed species, the following will not have a permanent dipole moment:
phosphorus pentachloride, boron trifluoride, diborane, methane, carbon tetrachloride, 2,2dimethylpropane, and cubane.
66
13.25. E
XN
1 0
YN
0
0 0
0 0
0
1 0
0 0
0
ZN
0
0
x Hi
0 0
0
YHI
0 0
0
ZHI
0
0 0
xN
0 0 0 0 0 0
YN
0
0 0 0 0 0 0 0 ZN
1 0 0 0 0 0 0 0 0 XJ:II
0 1 0 0 0 0 0 0 0 YHI
0 0 1 0 0 0 0 0 0 ZHI
0 0
0
0
YH2
0
0
0
0 0
zH2
O. 0
0
0
x ID
0 0
0
0 0 0
1 0 0 0 zH2
0 0 0 1 0 0 x ID
Ym
0 0
0
0 0
0
0 0 0 0
zm
0
0 0
0 0
0
x H2
a
=
1 0
0 0 0 0 0
xN
1 0
YN
0 0
1 0 0 0 0 0 x H2
0 1 0 0 0 0 YH2
0
0 0
0 0
1 0 YH3
0 0 0 0 0 1 zm·
-
0 0
0
0
0
0
0
0
0
0
0
1 0 0
0
0
0
0
0
0
0
0
YN
ZN
0
0
1 0
0
0
0
0
0
0
0
0
ZN
x Hi
0
0
0
1 0
0
0
0
0
0
0
0 XH1
YHI
0
0
0
0
1 0
0
0
0
0
0
0 YHI
0 ZHI
0 0
0 0
0
1
0
0
0
0
0
0 0
0 0
0
0
0
0
0 ­1
0
YlI2
0
0
0 0
0 0
0
0
0
0
zm
0
0
0 0
0
0
0
0
0
x ID
0
0 0
0
0 ­1 ·0
0
0
0
YID
0
0
0 0
0 0
0
­1
0
0
0
zm
0
0
0 0
0 0
0
0
1
0
0
ZHI
x H2
=
0
0
0
xN
0 x H2
­1 0 YH2
1 zH2
0 xm
0 YH3
0 ZID
13~27.
By multiplying the corresponding characters together, one can show that the following
relationships exist for the D2 point group:
A1xA1=Al
A1XBl=Bl
A.1xB2=B2
BIXBl=A1
B 1xB2=B3
B 1xB3=B2
B 2xB3=Bl
B3xB3=A 1
Therefore, the closure requirement is satisfied.
A 1xB3=B3
B2XB2=Al
13.29. The specific solution depends on which pair of irreducible representations you select.
The following illustrates one pair for each point group.
(a) For C2 : AxB = (1.1·1 + 1·~)
=0
(b) For C2v : AlxB 1 = (1·1·1 + 1·1·­1 + 1·1·1 + 1·1·­1) = 0
(c) For D2h: AgxA u = (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 Os: TluxT2u = (1·3·3 + 8·0·0 + 3·­1·­1 + 6·1·­1 + 6·­1·1 + 1·­3·­3 +8;0·0
67
+ 3·1·1 + 6·­1·1 + 6·1·­1) = (9+0+3­6­6+9+0+3­6­6) = 0
ExTl = (1·2·3 + 8·­1·0 + 3·2·­1 + 6·0·1 + 6·0·­1) = 6+0­6+0+0 = 0
(e) For Td :
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.
13.31. Irreducible representations from different point groups will have different symmetry
classes and orders. Therefore, the GOT can't be applied properly and the concept of
orthogonality or normality is irrelevant.
13.33. In the exercise, the character for identity is given as 7. For the additional characters, we
can follow Example 13.10 and use the formulas in the Rh(3) character table, using thefact that}
must equal 3. We note thatforbitals have different phases when we reflect through the origin, so
we should use the "u" set of formulas (like the p orbitals in Example 13.10). We have:
x(E) = 2j + 1 = 7
X(C3) = 1 + 2cos(1200) + 2cos(2400) + 2 cos(3600) = 1
X(C2) = 1 + 2cos(1800) + 2cos(3600) + 2 cos(5400) =­1
X(C4) = 1 + 2cos(900) + 2cos(1800) + 2 cos(2700) =­1
X(C2 ) = 1 + 2cos(1800) + 2cos(3600) + 2 cos(5400) =­1
X(O = -(2j + 1) = ­7
X(S6) = ­1 + 2cos(600) ­ 2cos(1200) + 2 cos(1800) =­1
x(ah) = _(_1)3 = 1
X(S4) = ­1 + 2cos(900) ­ 2cos(1800) + 2 cos(2700) = 1
x(ad) =_(_1)3 = 1
Therefore, we have the following character set:
r
E
7
8 C3
1
3 C2
­1
6 C4
­1
6 C2'
­1
i
­7
. ·886
-1
3 ah
1
684
1
6 ad
1
We can reduce this into its irreducible representations by using the GOT:
. .
1
a(A1g ) = 48 (7 ·1·1 + 1·1·8+(­1) ·1· 3+(­1) ·1·6+(­1) ·1·6+(­7) ·1·1 + (­1) ·1·8+1·1·3
+ 1.1.
6 + 1·1 . 6) = 0 A Ig
.
68
a( Az g )
1
= 48
(7 . 1.1+ 1.1. 8 + (­1) . 1. 3 + (­1) . (­1) . 6 + (­1) . (­1) . 6 + (- 7) . 1.1+ (­1) .1. 8 + 1.1. 3
+ 1· (­1)·6 + 1· (­1).6) = 0 Az g
1
a(E ) =­(7.2.1 + 1· (~1)·
g
48
8 + (­1)·2·3 + (­1)·0·6 + (­1)·0·6 + (­7)·2·1 + (­1)· (­1)·8 + 1· 2·3
'
+1.0.6+1.0.6)=OE g
1
a(J;g) = 48 (7·3 ·1 + 1· 0 . 8 + (­1) . (­1)·3 + (­1) ·1· 6 + (­1) . (­1)·6 + (­7) . 3·1 + (­1) ·0·8 + 1· (­1) ·3
+ 1. 1. 6 + 1. (­1) . 6) = 0 I;g
1
a(T2g ) = 48 (7·3·1 + 1· 0·8 + (­1)· (­1)·3 + (­1)· (­1)·6 +(­1) ·1· 6 + (­7)·3·1 + (­1) ·0· 8 + 1· (­1)·3
a(Alu )
+ 1. (-1) . 6 + 1. 1. 6) = 0 T2g
1
.
= 48 (7 ·1·1 + 1·1· 8 + (­1) ·1· 3 + (­1) ·1· 6 + (­1)·1,6 + (­7)· (­1)·1 + (­1). (­1)·8 + 1· (­1)·3
+ 1. (­1) . 6 + 1. (­1) . 6) = 0 Alu
1
a(A zu ) = 48 (7 . 1·1 + 1·1· 8 + (­1) ·1· 3 + (­1) . (­1)·6 + (­1) . (­1) . 6 + (­7) ­(­1) ·1 + (­1) . (­1)·8
+ 1. (­1) . 3 + 1. 1. 6 + 1. 1. 6) = 1 A 2u
a(E u ) = :8 (7·2·1 + 1· (­1)·8 + (­1)·2·3+ (­1)·0·6 + (­1)·0·6 + (­7)· (­2)·1 + (­1) ·1­ 8 + 1· (­2)·3
+ 1·0·6 + 1· 0 . 6) = 0 E u
1
a(I;u)
= 48 (7 . 3 ·1 + 1· 0 ·8+ (­1) . (­1)·3 + (­1) ·1· 6 + (­1) . (­1) ·6+ (­7) . (­3) ·1 + (­1) . 0 . 8 + 1·1· 3
a(T2u )
+ 1·1· 6 + 1· (­1) . 6) = 1Tlu
1.
= 48 (7 . 3 . 1+ 1. 0 . 8 + (­1) . (­1) . 3 + (­1) . (­1) . 6 + (­1) . 1. 6 + (­7) . (­3) . 1+ (­1) . 0 . 8 + 1. 1. 3
+ 1·1· 6 + 1· (­1).6)= 1 T2 u
Thus, the irreducible representation that describes the seven f orbitals in octahedral symmetry is
A 2u + Tlu + T 2u.
13.35. In the case where the position of symmetry is the point x = n/2, the symmetry elements
are E, Cz (the x = n/2 axis), and 2 a's (the xy and the yz planes).
13.37. According to the scheme in Figure 13.14, ethylene (C214) has D 2h point group symmetry.
With the center of inversion being in the middle of the molecule, we can see that the inversion
operation changes the sign on the wavefunction; therefore, the character of i should be ­1. ..
According to how the axes are defined, the xz plane is the plane of the page; and since the
orbitals are being reflected onto themselves, the character of a(xz) should be +1. Finally, the yz
plane cuts through the plane of the page between the two orbitals. Reflecting the orbitals
through that plane reflects them onto an orbital of the same phase, so the character of a(yz)
should also be +1. TIlls is enough information to indicate that these orbitals have the B3u
irreducible representation in the D 2h point group.
69
1
13.39. (a) a(A) = 2 (5·1.1+1·1.1)=3A
1
'
a(B) = ­(5 .1.1+1.1.(­1))= 2B
2
Therefore, in this case, r = 3A + 2B.
(b) a(A)) =
~(6'1+023)=lA
1
a(A 2 ) ="6(6 ­1·1+0· 2·1+ 0·3· (­1))= 1A2
a(E) ,= !(6
6 ·1· 2 + O· 2·(­1)+ 0.3.0)= 2E
Therefore, in this case, r = Al + A z + 2E.
1
(c) a(A)) = ­(6 ·1·1 + (­2)·2·1 + 2 ·1: 1+2·2·1+­4· 2.1)= OA)
8
,
1
'
a(A 2 ) = 8(6 ·1·1+ (­2)·2'1 + 2 ·1·1 + 2· 2· (­1) + ­4· 2· (­1))= 1A2
a(B)
= !(6
·1·1+ (~2)·
2· (­1) + 2·1·1 + 2·2·1 +­4· 2.(­1))= 3B)
8 '
)
a(B2 ) = i(6.1.1 +(­2)·2·(­1) + 2·1·1 + 2·2 ·(­1) +­4· 2 '1)= OB2
,
1
'
,a(E) == ­(6·1­ 2+ (­2)­ 2·0+ 2 ·1­(­2) + 2·2·0+ ­4· 2· 0)= IE
Ther fO~,
in
~s
casEii0. .
(d) a(A)) = 24 (7 ·1·1 + (­2)·8·1 +3·3 ·1+1 ~6'·1+
(­1).6.1)= OA)
a(A2 ) :;;; ~(7
·1:1+(­2)·8·1 + 3·3·1 + 1:6·(­1)+ (­1)· 6 .(­1))= OA2
'"M
',"
1
a(E) = ­" (7 ·1· 2 + (­2)·8· (­1) + 3,3·2 + 1· 6·0 + (­1)·6 ­0) = 2E
24
a(~)
1
, " ,
= 24 (7 ·1·3+ (­2) ·8·0+ 3·3 ·(­1)+ 1·6·1 + (­1)· 6· (­1))= 1~
1
a(T2 ) = 24 (7 ·1·3 +(­2) ­8·0+ 3 ·3 ·(­1)+ 1·6 ·(­1)+ (­1).6.1)= OT2
V
.. ..
Therefore, in this c a E2E +
13.41. In order for a product of functions to be non­zero, the product of the irreducible
representations must contain the all­symmetric irreducible representation, usually labeled A I (or
A or A' or A I' or A Ig). If the product of irreducible representations does not contain A I, then the
integral must be exactly zero.
70
(a) In C3v , the product A lxA 2 yields the character set (1, 1, ­1), which is the A z irreducible
representation. Thus, A I xA z = A z. Thus, an integral having these irreducible representations
must equal zero.
(b) In C6v , the product EIXE2 yields the character set (4, ­1, 1, ­4, 0, 0). Using the GOT to
reduce this representation, we find that it equals BI + B2 + E 1• Thus, an integral having these
irreducible representations must equal zero.
(c) In D3h' the product Az'xA1"xE' yields the character set (2, ­1, 0, 2, ­1, 0), which is the E'
irreducible representation. Thus, an integral having these irreducible representations must equal
zero.
(d) In D 6h, the product B2gxB2u yields the character set (1, 1, 1, 1, 1, 1, ­1, ­1, ­1, ­1, ­1, ­1),
which is theA 1u irreducible representation. Thus, an integral having these irreducible
representations must equal zero.
(e) InD6h' the product BlgxBlg yields the character set (1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), which is
the A}g irreducible representation. Thus, an integral having these irreducible representations may
be non­zero.
(f) In Td' the product ExT} yields the character set (6, 0, ­2, 0, 0). Using the GOT to reduce this
representation, we find that it equals T1 + Tz. Thus, an integral having these irreducible
representations must equal zero.
(g) In Td, the product TzxTz yields the character set (9,0, 1, 1, 1). Using the GOT to reduce this
representation, we find that it equals Al + E + TI + Tz. Thus, an integral having these irreducible
representations may be non­zero.
. 13.43. Because s orbitals are spherically symmetric, they have the Dg(O) irreducible
.representation in the Rh(3) point group. If the operator has the irreducible representation D«(I),
the integral of interest has an overall symmetry given by the product Dg(O)x Du(l)x Dg(O). The
product of these three irreducible representations is equal to D« (I). Since this is not the allsymmetric irreducible representation (represented in this point group by D g(00, the integral is
exactly zero.
13.45. HzS has C2v symmetry. To determine the symmetry­adapted linear combination for the
molecular orbitals ofHzS, we follow the scheme from Section 13.9 but using only the valence
orbitals for H and S:
ISH!
Ism
3ss
3 xs
3 zS
IsH1
Ism
3ss
3px,s
s;~p3
E
Cz
ISH2
Ism
3ss
-3px,s
3pz,s
c
ISHI
Ism
3ss
3px,s
3pz,s
cr'
Ism
ISH}
3ss
-3px,s
3py,s.
3pz,s
The combinations for the A} wavefunctions are determined by determined by multiplying each
row of functions by the character of the row's symmetry element, adding each column, and.
dividing by the order of the group (4, in this case). We get:
1
1
q.t = ­(ISH) + Is H2 + ISH1 + Is H2 ) = ­(ISH) + Is H2 )
4
.
2
q.t
= :
(ls H2 + ISH) + Is H2 + ISH1) =
~
(ISH1 + Is H2 )
71
1
'P =­(3s s +3s s +3s s +3s s)=3ss
4
'P = .!.. (3P x,S
4
'P
­
3P x,S + 3P x,S
.
­
3P x.S ) = 0
= : (3p y,s -3p y,s -3p y,s +3pY,J= 0
'P =.!..(3pzs
4 ' +3pzs
, +3pzs +3Pzs)=3pzs
r
,
,
The only unique wavefunctions we get are'P = Y2(lSHI + ISm), 'P =3ss, and 'P = 3pz,s. For the
A 2 set of wave functions:
1
'P =­(ls HI +ls m ­ls HI -ls m)=O .
4
1
'P ::: -(Ism + ls Hl -Ism ­ls HI ) = 0
4
1
'P = ­(3s s + 3ss ­ 3s s ­ 3ss ) = 0
4
1
.
'P = 4 (3P xs ­ 3P x,S ­ 3P x,s + 3P x.S ) = 0
'P = : (3P y,S ­ 3P y,S + 3P y,S ­ 3P y,S ) = 0
. 1
..
.
'P = -(3pz,s + 3pz,s ­ 3pz,s ­ 3pz,s) = 0
4
Thus, there are no SALC wavefunctions that belong to the A 2 irreducible representation. For the
RI set of wavefunctions:
'P
= .!..(ls HI -Ism +ls HI
'P
= 4 (3]Jy,S + 3p y,s
-Ism) = 0
4
1 .
.
1
'P = -(ls Hl ­ls HI + ISm ­ls H1 ) = -(Ism ­lsHI )
4
2
1
.
'P = ­(3s s ­3ss +3ss -3s s )=0
4
. 1 '
'P = 4 (3p x.s + 3p x,s + 3p xs + 3p x,s ) = 3p x,s
1
.
­
3P y,S
­
~
p y,S )
=0
'P = : (3pz,s -3pz,s + 3pz,s -3pz,s)= 0
Thus, there are two SALC wavefunctions that have B 1 symmetry: 'P = ~(lsm
3px,s" For the B2 set of wavefunctions:
1
'P = -(lsHI -Ism -ls Hl + ISm) = o·
4
'P
1
=-(ls HI -Ism -lsHI + ISm) = 0
4
72
­ ISm) and'P =
1
'l'
= -(3s s -3s s -3s s +3s s )=0
'l'
=:
'l'
=: (3P y,S + 3P y,S + 3P y,S + 3P Y,S ) = 3P y,S
4
(3p 1(,S + 3p x.S
'l' = : (3P z,s
-
­
3P x.s ­ 3P 1(,S ) == 0
3 Pz,s - 3P Z,s + 3P Z,s ) = 0
Thus, there is one B2 wavefunction: 'l' = 3py,s, This gives us a full set of six molecular orbitals
having the proper irreducible representations for this molecule.
13.47. For many molecules, a reasonably good approximation for a molecular orbital may in fact
be an atomic orbital, especially if that particular atomic orbital is not involved in bonding. Many
core orbitals ­ orbitals not part of the valence shell­ do not participate in bonding, so by
themselves they may be good representatives ofmolecular orbitals.
13.49. We will be using four Is orbitals from the hydrogen atoms in CH 4 , along with the Is, 2s,
2p ,x, 2p.y , and 2p .z orbitals of the carbon atom. This gives us a total of9 atomic orbitals to
construct the SALCs, and ultimately we should get 9 independent combinations for the
molecular orbitals.
13.51, As indicated in Equation 13.12, the "first excited state" of'H, is actually a set of three
wavefunctions that have the same spatial part but different spin parts. Although we assume that
the spin part of a wavefunction will not affect its energy, in reality the different spin functions
will have a tiny effect on the energy of the overall wavefunction. A detailed enough spectrum
will show that the "first excited state" will have three separate lines, not one.
.
13.53.
(.1(s+ P.)} .1(s-p.)d< ~ [jssd<- jsP.dH jP;sd<- jp;p,d<]
The individual atomic orbitals are themselves normalized and orthogonal to each other.
Therefore, the first and last integrals are 1 and the second and third integrals are O. Subsituting:
= .!.(l­ 0 + 0 ­1) = O. Thus, the two hybrid orbitals are orthogonal.
2
13.55. If the CH3+ ion is roughly trigonal planar, then the carbon atom should have Sp2 hybrid.
orbitals making bonds to the hydrogens, as this is the only hybridization scheme that yields the
appropriate directions ofbonds.13.57. The character for E is 5, as all five orbitals operate onto themselves. The character for C3
would be 2. The character for C2 would be 1, for crh would be 3, for 83 would be 0, and for cry
2dhybrid
would be 3. Therefore, the set of characters for the five sp
orbitals would be (5, 2, 1,3,
0, 3). Using the GOT, this representation reduces to 2A i' + A 2" + E'.
-
73
13.59. The two orbitals are not orthogonal because they are on different atoms. Hence, there is
no requirement that they be orthogonal, (In fact, if they were, no bond would be fonned!)
13.61. The nitrogen atom not only must accommodate three bonds to hydrogen atoms, but also a
lone electron pair. Thus, the nitrogen atom needs four hybrid orbitals: sp3 orbitals.
74
CHAPTER 14. ROTATIONAL AND VIBRATIONAL SPECTROSCOPY
14.1. For a linear molecule, rotation about the molecular axis isn't recognized as a true rotation.
Even if it were, the moment of inertia about that axis is almost identically zero, suggesting that
only a negligible amount of energy is needed to promote rotation about that axis. A useful
spectrum couldn't ever be measured.
14.3. Using the equation c = AV:
v = 3.00xl08 S­I
(a) 2.9979xl08 mls = (1.00 m)v
8
5
(b) 2.9979xl0 mls = (4.77xlO· m)v
v = 6.28xl0 12 S·I
(c) 7894 A = 7.894xlO­ 7 m
2.9979xl08 mls = (7.894xlO·7 m)v
v = 3.798x10 14 S­I
(d) 2.9979xl08 mls = (1.903xl03 m)v
v = 1.575xl05 S·I
14.5. i.9979xl0 8 mls = A(8.04lxl012 S­I)
A = 3.728xlO­ 5 m
34
I2
E= (6.626x 10. J. s)(8.041xlO s")
E = 5.328xlO­21 J
E = hv
The speed of the photon is the same as any other photon's speed (in vacuum): 2.9979xl08 mls.
14.7. Since there are 1,000,000 micrometers in 1 meter and 100 m­I in 1 em", you can show
that for a photon of light having x em" and Y urn, xxy = 10,000.
14.9. (a) prolate symmetric top (b) spherical top (c) spherical top (d) asymmetric top (e),
.asymmetric top (f) asymmetric top (g) oblate symmetric top (h) linear (i) linear.
14.11. Both SF6 and UF6 are octahedral, spherical­top molecules. Therefore, their moments of
inertia are independent o~axis
of rotation. Therefore, we can choose a simple­to­envision
rotation and...u.s.e that t<y6aIc" ate the moments of inertia, and from that, the value of B. Let us
select rota~ e
thole~
e along one of the F-~
~r F­U­Faxes. That way, th~
three ato~s
alo
e axIS'"dO.p.ot contri te to the moment of inertia because they are not rotatmg. What IS
mg are four F\ttoms a tlrticular distance away from the rotational axis. The moments of
. rtia can be cal~nted
simply, by considering it as four atoms rotating about an axis. For SF6:
2
"4x(O.0190 kg! .022x 1023)(1.564x10­10
= 3.087xl0­45 kg· m
For ".". &;,••J= 4')('O.012,Q, kg/6.022x 1023)(1.996x 10­10
= 5.028x 10­45 kg' m 2
Now m ~ a 1 u e s
ofB:
.
3
4
­24
(6.626 x 10­ J. S)2
B(SF6 ) = (21l")22(3.087 X 10­45 kg.m2) = 1.80 x 10 J
~
(6.626 X 10­34 J . S)2
­24
6
45
B(UF ) = (21l")22(5.028 x 10­ kg­rn") = 1.l0x 10 J
mi
mi
When you consider that the central atoms are so different in mass, these values of B are not that
much different from each other!
14.13. 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 K 2 • Therefore, for most energy levels, the degeneracy is 2x(2J
+ 1). When K = 0, there's not another value of K that yields the same energy, since -0 = O.
75
Therefore, the degeneracy of those energy levels is dictated solely by the various M J values ­ 2J
+ I of them.
14.15. Ethane is a prolate symmetric top with the two lowe~
same:
(6.626 x lO­34 J . S)2 .
23
.jJ
A = (2nY 2(1.075 x 10­46 kg' m") =_5.173 x 10­ J
2
(6.626 x lO­34 J . s)
.
­23 TI;:(.~B = C = (2nY2(4.200 x 10­46 kg­rn") = 1.324 x 10 7~
c~nseig
the
The lowest rotational energy level ~as
E .=...0. The second energy level co
value of the low of the two rotational constantsg
. E=BJ(J+1)
1+1).~-z3
J =2.648 x 10.23 J
The third ener
vel co:rrtesfrom the J = 2 value ofthe lower of the two rotational constants:
E=BJ(J+l)
+1)·1.324xJO­z3 J=7.944x10'23 J
T~e
evel corn:e;:::uom t J-~'
,.' alue of the higher of the two constants:
~=AJ( +1)
+1).5 J]~0-3
J= . 35xl0·22 J4
Fmally, the fift e ergy level willes from the J = 3 value ofthe lower of the two constants:
E=BJ(J+l) 3 +1)·1.324.?S,lO­23 J=1.589xl0·22 J
c...
III
!
14)7. (a) allowed (b) not allowed (c) not allowed (d) not allowed (e) not allowed (f)
allowed (g) not allowed (h) allowed
19.~fthe
..
s aced .......W4cm­ 1 this value represents
­I
~
/
2 so = .
alculating the reduced mass of!:
C
0 \-/'1
23'
­26
\(127)(35)
1 kg
P = (127 + 35) g x 1000 g + 6.02~ix
10 = 4556 x lO kg.­1
.
.'
rota!'
sp
Using the wavenumber\f~ s
B
=
olt.he equation34for the WFtional constant:
h
2
81r (prZ
81l'2
Solving for 1­, then for r:
,
i
'V l .6.626 x'
10­ J:­8
.
4.556 x 10­26 k r Z
6.626 x 10
~4
~
.
.9979 x lOlO cm / ~
­:
l
~
J·.s
=.D.1l52cm·1
.­­­­.
.
2
­19
r ­
­ 1 078 x 10m
_ ­ (81r 2)(4.556xl0- z6 kg)(0.O,57 cm­I)(2.9979 x 10 10 em/s) ­ "
r=3.~_'t
.
14.21. The easiest way to solve this question is to convert the B value into GHz units, then use
the fact that the first four absorptions will appear at 2B, 4B, 6B, and8B. If the rotational ..
spectrum consists oflines spaced by 15.026 em", this value represents 2B, so B = 7.513 em".
Converting this into gigahertz units:
1
1
A = ­;::­ = .
= 0.1331 em = 0.001331 m
v 7513cill'l
C = AV
2.9979 X 10 8 m/ s = (0.001331 m)v
0: (~j.it·_
/I~
Calculating the frequency:
v = 2.252 x 101I S·I
/~
te
~'!iJf
t:/~
If =~I<h )'! 7
\­
C
.'
-s.r ~
~ , /1/
I;:·
C__ :r
0,;
I~/
/0:.;1
~
=
225.2 GHz
Therefore, in GHz units, B = 225.2 GHz. Therefore, the first four lines of the rotational
spectrum should appear at 450.4 GHz, 900.8 GHz, 1351.2 GHz, and 1801.6 GHz.
14.23. Calculating the reduced mass ofHS:
1 kg .
.
( 32)(1)
­(­ 6.022 X10 23 = 1610 X10­ 27 k
=
x
J1 (32 + 1) g 1000 g
.
g
This lets us calculate B for the diatomic molecule:
­22
(6.626 x 10­ 34 J . S)2
10
27
B = (2nY 2(1.610 x 10­ kg)(1.40 x 10­ m)" = 1.762 x 10 J
Knowing that Jrnax is 8, we can estimate the sample temperature:
__ (1.381X10­23 J IK) II2
8~
Solve for T: T R: 1600 K
2TJ
'
2(1.762x10'2J)
.
.
14.25. From Table 14.2, we fmd that B (HCI) = to.59 ern". Smce B
2
2
=­h =­h ­
21
2J1r2'
we can
determine B (DCI) by determining the change in the reduced masses (everything else remains
the same):
.
J1
(H CI) = (1)(35.5) = 0.9726 ­
(l + 35.5) g
g
J1
(DCI) = (2)(35.5) =1.8933
(2 + 35.5) g
g
(Here we are simply using the atomic weight of CI, rather than a particular isotopic weight.)
'.Since the reduced mass is in the denominator of the expression for B, we can set up the
following ratio:
.
= J.l (HCI)
B (HCI) 1/J.l (HCI)
J.l (DCI)
B (DCI)
0.9726 g
_....0..­­­­­:.....,.... =__­­=1.8933 g
10.59 em"
B (DCI) = 1/ J1 (DCI)
Substituting and solving for B (DCI):
B (DCI) = 5.44 em'I.
14.27. First, at such high values of rotational quantum number, we expect that centrifugal
distortions will make the energy levels (and hence the energy difference) deviate quite a bit
from those of a rigidrotor, Second, as a nonpolar molecule, diatomic iodine wouldn't 'show a
pure rotational spectrum.
14.29. Using equation 14.27:'
3
D
R:
J
4B
l7 2
= 4(60.80cm'I)3 =
(4320em·I) 2
5
3
8.990x10 cm­
1.8662x10 7 em"
= 0.04817 cm'
This compares to 4.64x10'2 em", which is not far off (about 0.2%).
14.31. The total number ofnormal modes equals the number of vibrational degrees of freedom.
(IN e should differentiate the total number of normal modes from the total number of distinct
vibrational frequencies, which maybe different due to degeneracies.) Therefore, the total
number of normal modes for the molecules in exercise 14.30 are (a) 1; (b) 3; (c) 174; (d) 63;
(e) 48; (f) 7; and (g) 6.
77
. WIith th e equation
.. 1y t h e fiirst fracti
ri h t b y
. ­1 = ­ I
14.33. Startmg
+ ­I
, weI
mu tip
action on theeng
J1 . m l m 2
mim2 and the second fraction on the right by m-fmi:
! = ~+
J1
m1m 2
m1m2
= ml + m 2
m\m2
•
Now we can take the reciprocal of both sides of the equation to
get an equivalent expression for the reduced mass: . J1
=
m
jm 2
», +m 2
.
14.35. First, we need the reduced masses of the C=O and C=S bonds:
(12)(18)
(12)(32)
For C = 0: J1 =
g = 7.2 g
For C = S: J1 =
.
g = 8.73 g
12+18
12+32
Now we use the square root of the ratio of the two reduced masses to predict the vibrational
frequency ofthe C=S bond:
~
.2 g
v(C=S)
0.9082 =
.
(C = S) = 1215 em"
1338 cm'
8.73 g .
Although we predict 1215 em" for the frequency of the C=S vibration, it actually appears at 859
em". Thus, assuming that S is an "isotope" of 0 is not a good assumption.
v
::=
14.37. Nitrogen is used because, as a nonpolar diatomic molecule, its single vibrational motion
is Ik­inactive. Oxygen gas and any noble gas (He, Ne, AI, Kr) could also be used ­ but they are
more expensive. Thus, using nitrogen gas as a purge gas means that the infrared spectrum is
measuring the vibrations of the sample, not any IR­absorbing contaminant from the atmosphere.
v = I transitions. Overtone vibrations are the v
14.39. Fundamental vibrations are the v = 0 ~.
= 0 ~ v = n, where n is any number other than I. Hot bands are the v == n ~ v = n + I, where n
is any number other than I (that is, the vibrational transition starts in an excited vibrational
state).
14.41. Starting with the expression V =
~&V
.
respect to x twice: & = Joe
&2
~
kx" , we take the derivative ofthis expression with
.
::=
k , thus verifying the relationship. With regard to Units,
according to the original expression of Hooke's law:
(joules) = (N/m)(mi
J = N' m
J= J
Thus, the units are algebraically correct.
14.43. We need to go through a three­step process to determine a: first, use the data in Table
14.4 to calculate the "pure" anharmonicity constant X e ; second, use equation 14.40 to determine
De; third, use equation 14.37 along with data given in the exercise to determine a.
ForHF: x e = xev e
ve
=
90.07cm,1 =0.02176
4138.52 cm'
78
Rearranging equation 14.40:
=~4138.52cm'l
D
e
4xe
4(0.02176)
=47550cm­1
,
And finally, converting the wavenumber value of De into joules:
1
1
)1/2
k )1/2 ( 965.1 N/m
1
0cm!s·6.626xI0­
34 J . s
(
= 2(47,550cm,I)·2.9979xI0
a= 2D e
a = 2.26xlO 10 m,l = 2.26
ForHBr: x =
e
D =~
e
4xe
XeV e
ve
A-I.
1
= 45.2lcm· =0.01706
2649.67 cm'
Rearranging equation 14.40:
= 2649.67 cm,l = 38830 cm'
4(0.01706)
,
And finally, converting the wavenumber value of De into joules:
k
a = ( 2De
)112 (
411.5 N/m
1·
1
)112
34
locm!s
= 2(38,830 em ­1) . 2.9979 x 10
. 6.626 x 10­ J . s
a = 1.63x10 10 m,l = 1.63
A,I.
Comparing these values with 1.87 A-I for HCI, we find a trend that a decreases as the size ofthe
halogen atom increases.
14.45. Using the fact that De = Do + ~ hv, we must first convert the given Do values into cm­I
'units:
I
x
s
= 30 270cm,I
. 362kJ/molx 1000J x 1mol x
1kJ
6.02 x l 0 23 6.626 X 10­ 34 J. s 2.9979 x 10 10 em
'
.
for HBr, and for CO:
107tkJ/mo1x1000Jx 1mol x
1
x
s
=89560cm,l.
1
0
23
34
1kJ
6.02 x 10
6.626 x 10­ J .s 2.9979 X 10 em
'
The respective values of De are thus
f1- 7.
30,270+ ~ (2649.67)cm,1 = 31,590cm'1 forRBr
89,560 + !:.(2170.21) cm'' = 90,650cm,1 for CO
2
Now to determine Xe and VeXe. For RBr, using equation 14.40:
,
~
x = 2649.67 cm = 0.02097 . :. vex e = (2649.67 cm,I)(0.02097) = 55.6 em"
e
4(31,590 em")
For CO, using equation 14.40:
1
:. vex e =(2170.c~'I)6583ml
x = 2170.21cm· =0.65~
e
4(89,560 em")
. The predicted VeXe value for HBr is offby about 20%, while that for CO is rather close.
14.47. Starting with equation 14.39,£
~ hV.(v+ )~ ­hV.X.(V+ ~
J,
we can construct a
general expression for the change in energy between adjacent energy levels:
·79
AE=(V+l)-~h,
~)-hv,x(+H
+v,(v+ +v(,xVh-)~
n~
Multiplying out the binomials and collecting terms (which won't be shown here), we get
till = h v, ­ 2h vexe{v + 1). Ifwe want to express this change of energy in frequency units
directly (either in S­I or em"), we divide all terms by Planck's constant:
v = v e ­ 2v exe(v + 1). For the fundamental absorption, v= 0 and this becomes
v = v e ­ 2v exe. We know (see example 14.12) that we can use the ratio ofreduced masses as a
weighting factor to predict the shift of a vibrational frequency upon isotopic substitution. The
frequency of the heavier isotope,
.
v,', is shifted by a factor
. .
~
Jl. . Defining this factor as p, we
~
have that v, * == p·Ye • This variable also appears in the anharmonicity term as well. However,
considering equation 14.40, we also find that the anharmonicity constant X e is also defined in
terms of Y e, so X e also reduces by the same factor of p. We can thus state that X e* = p­x;
Assuming that, for the isotopically­substituted molecule, v· = v; ­ ~v;X
, we can substitute to
get:
v· = v; ­ 2v;x;
.
= pv, -
2(pve)(PXe) = pVe ­ 2p2vexe' which is the expression from exercise
14.46.
14.49. The first vibration has a g+ symmetry and is lR­inactive. The second vibration has a u+
symmetry and is lR­active.
14.51. (a) E and C3 are proper rotations; the avs are improper rotations. (b) The E, C3 s, and
C2s are proper rotations; the 8 4s and adS 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 83S, the 86S, and
all planes of symmetry. (d) The proper rotations are E and C2 , while the two 84s are improper.
(e) The proper rotations are E, the C(~), and the infinite C2S. The improper rotations are i, 8( ~),
and the infinite avs.. (f) All of the five classes of symmetry elements in 0 are considered proper
rotations.
14.53. (a) H 202 has C2 symmetry. A check of the C2 character table shows that both
irreducible representations, A and B, will be lR­active. Therefore,.H202 will have 6lR­active
vibrations. (b) Oxalic acid, (COOH)2, has Cs symmetry. A check ofthe Cs character table
shows that both irreducible representations, A' and A ", will be lR­active. Therefore, (COOH)2
will have 18 IR­active vibrations. (c) Sulfur trioxide, S03, has D 3h symmetry, and not all ofthe
irreducible representations are lR­active. Therefore, we must go through the steps to determine
the character ofthe vibrations:
80
E
2 C3
283
3 C2
3 cry
crh
2
4
1
4
1
2
Nstationarv
0
0
0
0
0
120
180
60
1800
0
180
8
.,.1
3
0
­1
­12
1+2cos8
.,2
12
0
­2
2
4
±N(l+2cos8)
3
­1
0
­1
­1
2
Xn (1+2cos8)
­2
3
0
­1
1
1
'X.tn ±(l +2cos8)
­2
6
0
4
2
0
Xvib
Using the GOT, this reduces to Al '+2E'+A2". According to the character table, E' and A 2"
modes are IR­active. Therefore, we expect S03 to have three IR­active vibrations.
(d) Formaldehyde is a C2y molecule. Not all of the irreducible representations are IR­active.
Thereitore, we have t0 go thr ou gh th esteps t0 d et ermine
. th ech arac ters of the vibrations:
,E
C2
cry
cry'
4
2
4
2
Nstationarv
0
0
0
180
180
0
1800
8
­1
­1
­1
3
1+2cos8
'.
­2
4
12
2
±N(1+2cos8)
­1
­1
­1
3
Xr, (1+2cos8)
­1
1
1
3
'lIn ±(1+2cos8)
2
0
4
6
Xvib
Using the GOT, this reduces to 3A I+2B I+B2, all of which are IR­active. Therefore, CH 20 will
have 6 IR­active vibrations. (e) Acetone is a C2y molecule. Notall of the irreducible
representations are IR­active. Therefore, we have to go through the steps to determine.the
characters of the vibrations:
E
C2
cry
cry'
,10
2
6
2
Nstationarv­'
0
0
0
'0
1800
180
180
8
­1
­1
­1
3
1+2cos8
­2
6
2
30
±N(l+2cos8)
­1
­1
3
­1
'X.r, (l +2cos8)
1
­1
I'
3
'X.tr, ±(l +2cos8)
2
6
24
0
Xvib
Using the GOT, this reduces to 8Al+4A2+7BI+5B2. Of these, the zl}, BI, and Bfvibrations are'
IR­active, so acetone will have 20 IR­active vibrations.
81
14.55. IfKrF4 were ever synthesized, one could tell ifit were square planar or tetrahedral by
determining the number ofIR­active (and Raman­active) vibrations the molecule has. lithe
molecule were planar, it would have 3 IR­active vibrations, while if it were tetrahedral, it would
have 4 IR­active vibrations.
14.57.
6Cz' i
886
3ab 684 6ad
4
0
0
0
0
16
0
0
0
8
Nstationarv
180° 90°
180° 180° 120°
180° 90° 180°
0°
120°
e
­1
­1
­1
­1
0
1
­1
3
0
1
1+2cose
48
0
0
0
0
0
0
0
0
4
±N(l +2cose)
­1
­1
­1
1
0
­1
1
3
0
­1
'l.r, (l +2cose)
­1
1
1
­1
­1
1
1
3
0
0
'X.tr, ±(l +2cose)
·0
0
­2
2
0
0
42
0
2
8
'X.vib
Smce only T1u IS IR­actIve, all we need to do IS determine how many T1us there are m this
reducible representation. Using the GOT, we find that there are 3Tl u s as part of the irreducible
representation. Therefore, there are three (triply­degenerate) JR.­active vibrations. This
represents 9 of the 42 possible vibrations of CgRg.
E
3Cz
8C3
6C4
14.59. Combination bands are typically (but not always) weak because they are not usually
allowed by the quantum­mechanical selection rules.
14.61. Both dioctyl sulfide and hexadecane are largely composed of long hydrocarbon chains,
the only difference being that diocty1 sulfide has a sulfur atom in the middle. Thus, their
vibrational spectra should be very similar.
14.63. SiRt is a spherical top whose rotational energy levels are given by the expression
.
J(J + 1)1i 2
21i 2 1i 2
­­­'­­"'""'­­­ and whose spacing between rotational levels is 2B = ­ ­ =­ . Since 2B = 16.72
21
21
1
em", we convert this to joule units: 2B = 3.32x10­z2 J. Solving for 1:
34
3.32 X 10­ 22 J = (6.626 x 10­ J 'S)2
41£21
1 = 3.35 X 10­ 47 kg­m? =
_
"­­""
.
~
~r-.
r?
_
I
Now we need to determine a formula for the moment of inertia. If we assume that the molecule
is spinning about an Si­R bond, then there are three equivalent H's making a triangle whose
angle is determined by the tetrahedral bond angle, 109.45°. The angle that the bond makes from
the rotational axis is half that, or 54.725°. The component of this bond that is rotating is related
164. Finally, there are three hydrogens spinning. Thus, we have:
to the sine of this angle, or
3.35xlO­47 kg­m?
3(1.00g / 1kg
1000 g
/~
10
C 1.00xlO­ m =.lQ0 A. _
1
2~rY
.lQ J) ).~
~
Solving for r:
.
This information could n01 be determined from a pure rotational spectrum because SilLi does
not have a permanent dipole moment, so does not have a pure rotational spectrum.
.82
14.65. If centrifugal distortion were negligible, there would be no DJ tenus in either equation
14.41 or 14.42. They would simplify to:
M = hv - 2xev e + (BI + Bo)(.lrower + 1) + (BI ­ Bo)(.lrower + Ii
. and
M = hv ­ 2xeve ­ (B I + Bo).lrower + (B I - Bo)J2lower
14.67. The spectra are different in that they will appear in different regions of the
electromagnetic spectrum. The Raman spectrum generated using the He­Ne laser as excitation
source will appear in the red part of the visible spectrum, while the Raman spectrum generated
using the green light from the Kr+ laser will appear in the green region of the visible spectrum.
However,
the
pattern of new absorptions and their relative intensities
should be virtually the
"..
.
.
same.
14.69. If a point group contains i, a center of inversion, then vibrations that are infrared­active
are not Raman­active, and vibrations that are Raman­active are not infrared­active,
83
CHAPTER 15. INTRODUCTION TO ELECTRONIC SPECTROSCOPY AND
STRUCTURE
15.1. Although you can check all of the irreducible representations in the D6h character table
individually, it is probably easiest to recognize that the only way the transition moment integral
will be nonzero is if the allowed excited states also had E 1u symmetry. That's the only way that
. the product ofthethree irreducibie representations will contain A 19.
15.3. First, we need to determine the reduced masses ofD and T. For D:
= (9.109x10­31 kg)(3.344 x 10­27 kg) = 9.1065 x 10­31k
g
f.l 9.109xlO­31 kg+3.344x10· 27 kg
= (9.10~xIO-3
ForT:
f.l
kg)(5.008xlO'27 kg) = 9.1073 x 10­31k
g
9.109xlO­31 kg+5.008xlO­ 27 kg
Thus, the new Rydberg constants are:
31
(1.602 X 10­,19 C)4(9.1065 X 10­ kg)
= 2.1783x 10­18 J = 109660
3
4
1
2
4
8(8.854x10­ C /J ­m)? (6.626 x 10­ J 'S)2
' .
19
31
= 2.1785 X 10­18 J = 109 670
R(T) =
(1.602 x 10­ C)4(9. 1073 x 10­ kg)
34
1
2
4
8(8.854xlO­ C /J ·m)2(6.626x10­ J ·S)2.
'
R(T)'=
cm'
cm'
These are very small­ but detectable ­ changes.
. 15.5. The drawing is left to the student.
15.7. (a) The possible values of L are.Z..J,.-...and. o. M L goes from -L to Land so can be at most
­2, ­1, 0; 1, and 2 (depending on the value of L). S can be 1 or 0, with M s going from ­S to S
(and so can be ­1, 0, or 1, depending on S). J depends on the values of L and S, but can range
'from 3 to O. M J will depend on the value of J, but will range from .-J to J, so might be ­3, ­2, ­1,
0,1,2, or 3~ (b) The possible values of L are 6,5,4,3, 2, 1~ and O.ML goes from -L to L and so
can be at most ­6 through 6 (depending on the value of L). S can be 1 or 0, with Ms going from
-S toS (and so can be ­1,0, or 1, depending on S). J depends on the values of L and S, but can
range from 7 to O. MJ will depend on the value of J, but will range from -J to J, so might be ­7
through 7. (c) The possible values of L are 3, 2, 1, and o. M L goes from -L to L and so can be at
most ­3, ­2, ­1, 0, 1, 2, and 3 (depending on the value of L). S can be 1 or 0, with Ms going from
­S to S (and so can be ­1, 0, or 1, depending on S). J depends on the values of L and S, but can
range from 4 to o. MJ will depend on the value of J, but will range from -J to J, so might be ­4,
­3, ­2,.­1, 0, 1, 2,3, or 4.
,.
15.9. The aluminum atom has a single unpaired p electron in its valence shell. The spin ofthat
single electron can show hyperfine coupling ­ that is, the spin can interact with the spin of the
aluminum nucleus (which, by the way, has spin of 5/2) to exhibit closely­spaced but slightly
different energy levels.
'
15.11. We need to convert the wavelengths into energy values and evaluate the difference:
5890Ax 1m
.. 1010 A
= 5.890x 10­7 m
8
v = 2.9979 X 10 mls = 5.090x1014 S·1
5.89 X 10,7 m
84
E= (6.626xlO­ 34 J·s)(5.090xlO I4 S"I) = 3.373xlO­19 J
5896Ax 1m =5.896xlO­ 7m
v=2.9979xl08m/s=5.085xlO,4s"
10 10 A
5.896xlO· 7 m
34
I4
E = (6.626xl 0. J.s)(5.085xlo S·I) = 3.369xlO­19 J .
The difference between these two energies is 4xl0·22 J, which corresponds to ab ut 240 J/mol.
.
~tV
15.13. 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 m; values ­ is zero, as they all cancel. Thus, half­filled subshells­will always have their
hi est multiplicity for an S term symbol.
015.15. An h electron has .e = 5; therefore, possible L values for two h electrons are 0, 1,2,3,4,
@ 5,6,7,8,9, and 10. These lead to S, P, D, F, G, H, I, J, K, and Lterms, For two electrons, the S
v es can be either 0 or 1, leading to the combinations 3S, IS, 3p , Ip , 3D, ID , 3F, IF, 3G, IG, 3R,
H, 31, \ 3J, IJ, 3K, IK, 3L, and IL. Ifwe were to include Jvalues, the term symbols would be
3S0,1, IS0, 3p2,1,0, l p I, 3D3,2,1> ID2, 3F 4,3,2, IF3, 3G5,4,3, IG4, 3H6,5,4, IH5, 317,6,5, II6, 3J· S,7,6, 13
J7, K9,S,7,
IKs, 3LlO,9,S, and IL9. (Only one Jvalue would be present for each individual term; the Jvalues
for the triplet states are grouped together for reasons of clarity.) We are not excluding any
possible term symbol, but some ofthe above may be forbidden by Pauli principle arguments.
According to Hund's rules, the higher multiplicity is preferred for the ground state (3, here); the
higher value of L is preferred (L, here), and for a less­than­half­filled subshell, the lower value of
J is preferred (8, here). So the ground state term symbol should be 3Lg.
15.17. Since /)S = 0, any allowed excited state will have S = 1. M = 0 or ±1, so allowed excited
states can be either D, F, or G terms, Finally, since M= 0 or ±1, allowed excited states can have
a value for J of 1,2, or 3. Combining these
e items... elbis~.4t
terin symbols are ~2,
3
3 J
3 3 33
"
­.­._­. .. .. . . . . . '
p~dy.1Q,;theJ:laOnsm
between
D3, FI, F2, F 3 G
.
e
L, ,and J: on&h
3~2 .;~.
v_~ . : "~ bOIS.·
.
15.19. A diatomic molecule is, in a first approximation, a rigid rotor. Thus, its behavior can be
modeled to some degree of accuracy by the 3­dimensional rigid rotor ideal system, which
ultimately predicts that the angular momentum of the system is quantized.
15.21. If Liz has two electrons in 1t u orbitals, then the term symbols. derive from the product
2
TIuxil u in the D h point group. This gives us the set of characters (4, 4cos24>, 0, 4, 4cos 4>, 0). By
analogy to the discussion of the term symbols of diatomic molecules in the text, we rationalize
the irreducible representations that contribute to this character set, rather than use the GOT
directly. Using the same arguments, we should get L g+ + L g­ + Llg •
15.23. The acetylide ion, Cl", is isoelectronic with O2, so it should have the same molecular
it should have the same ground­state term symbol, 3Lg­...
orbital diagram. Th~s,
85
15.25. (a) No unpaired electrons, so N03­ should be colorless. (b) Mn0 4­ has unpaired
electrons in the Mn's d subshell, so it is highly likely that the ion will absorb visible light and
have some color. (It is, in fact, strongly purple.) (c) There are no unpaired electrons, so NH4+
should be colorless. (d) Cr 2 0 l 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, yellow­orange.) (e)
There are no unpaired electrons in ol-, so we expect it to be colorless. (f) Acetylide ions have
unpaired electrons in its molecular orbitals (see the answer to exercise 15.23)~
so we expect it
might have a color. (It's grayish­black in color.)
15.27. According to the point group character table in the Appendix, the electric dipole operator
has the irreducible representations Al, B}, or B2 (depending on which three­dimensional axis is
involved). In order to guarantee that the A, irreduci,ble representation is part of the product
'P upperf.t¥lower' an excited state must also have the irreducible representations A}, BI, or B 2 . Thus,
possible excited states will have irreducible representations lA}, IB}, or IB 2 .
_
15.29. Nothing would change in the Huckel approximation of ethylene if deuterium atoms were
substituted for hydrogen atoms in the molecule. Deuterium has the same electronic structure as
hydrogen (with the exception of a small change in the reduced mass of the system), so
substitution shouldn't change the electronic approximation.
15.31. Cyclopentadiene easily accepts an electron because when it does, it has six
and is aromatic.
7t
electrons
­15.33. (a) Ideally, neutral cyclopolyenes will be aromatic if they have 4n+2 carbon atoms with
7t electrons in the ring. (b) Ideally, cyclopolyenes having a single positive charge if they have
4n+3 carbon atoms with 7t electrons in the ring. (c) Ideally, cyclopolyenes having a double
positive charge if they have 4n+4 carbon atoms with 7t electrons in the ring.
15.35. Heating a potentially laser­active substance excites systems (atoms and molecules) into'
excited states in a manner that ultimately mimics thermal equilibrium. Thermal equilibrium is
not a population inversion; the higher an energy level is, the less populated it is. Thus, it would
be extremely difficult (ifnot impossible) to construct a population inversion by purdy thermal
means; some other method would be necessary.
15.37. Light.from fireflies would be a (chemically­induced) example of phosphorescence.
15.39.10.6,urnx
v=~ 2.9 7 xI08m1s=2.83XI0 3s-1
1m =1.06xl0­ sm
10 6,urn
34
13
A
1.06 x 1O-~
m
2o
E = hv = (6.626xl0­ J.s)(2.83xl0 s") = 1.87xl0­ J per photon
300,000 J
1photon
Therefore,
x
20
= 1.60 x 1025 photons per second.
s
1.87 x 10­ J
15.41. Power is defined as energy per unit time, so for a 300 mJ pulse in 2.50 ns:
_
Is
IJ
_
2.50nsx 9
=2.50xlO 9 s
300mJx
=3.00xlO 1J
1000mJ
10 ns
86
3.00x10· 1 J
8
power =
. 9 = 1.20 x 10 J/s == 120 MW (megawatts)
2.50x10· s
.
87 .
CHAPTER 16. INTRODUCTION TO MAGNETIC SPECTROSCOPY
16.1.. A magnetic field vector is the magnetic phenomenon formed when a current I flows
through a straight wire, forming a circular magnetic field. A magnetic dipole vector is a linear
magnetic effect formed by a current I flowing in a circle.
16.3. (a) If the electron had a positive charge, the value ofthe magnetic dipole wouldn't change
because its value is based on the magnitude ofthe charge, not its positivity or negativity.
However, because the charge is opposite, the direction of the magnetic dipole vector will be
opposite its original direction. (b) For positronium, we can argue that the magnetic dipole will
be zero overall. That's because in positronium, we have two particles "orbiting" each other that
have the same mass, and rather than having the tiny electron moving around a massive,
.unmoving nucleus, we have two equal­mass particles in orbit about a mutual center. The orbits
will be, on average, equal­ producing equalmagnetic dipoles ­ but opposite ­ implying that the
two magnetic dipole vectors will cancel each other out.
16.5. According to equation 16.6, the Bohr magneton JlB equals
= (1.602x10­
constants:
PB
19
34
C)(6.626 x 10­ J ·s)
2n­.2(9.l09x10­3 J kg)
~.
Zm,
Substituting for these
= 9.273x10­ 24 C.J ­s/k
g
Within truncation error, this is the same value as given in the text. However, the units are
unusual.­ But if we recall that a tesla is defined as a kg/Cis, we note that the units of JlB are
­ equivalent to Jrr. Thus, we verify both the value and units of the Bohr magneton.
16.7. (a) .Of the three states in the IS ­7 Ip 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: I1E = J.!B·MeB:
8£= (9.274x10­24 Jrr)(±1)(2.35 T) = ±2.18x10­23 J. This corresponds to 1.10 em".
.
p
(b) For the I ­7 In transition in the presence of a magnetic field, there will be 15 possible
transitions, as the I p state as three possible M L values while the In 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 I1E as for the IS ­7 Ip transition: I1E = ±2.18x10­23 J.
16.9. We need the g, value for the 2 P 3/2 term symbol. Using equation 16.13:
=1+ (3/2)(3/2+ 1)+ (1/2)(1/2 + 1) ­(1)(1 + 1) =1.3333 ...
gJ
2(3/2)(3/2+ 1)
.
Now we use equation 16.11 and the fact that the maximum M J 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):
24
1lE = (1.3333)(9.274xlO­ Jrr)(3)(0.00006 T) = 2.23x10­27 1. This is equivalent to 1.12xlO­4
em", a difficuit energy difference to differentiate. 16.11. Using equation 16.12:
88
gJ
= 1+ [(4)(4 + 1) + (2)(2 + 1)­ (2)(2 + 1)](2.0023 ­1) = 1.50115
2(4)(4+ 1)
Using equation 16.13:
= 1+ [(4)(4 + 1) + (2)(2 + 1) ­ (2)(2 + 1)] = 1.5
gJ .
2(4)(4+1)
The difference is less than 0.08%.
16.13. 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:
.
.
34
24
9
E = (6.626x10­ J)(lOx10 s") = 7x10· J/photon.
Of course, the energy depends on the frequency of microwave radiation used. This can be compared
to 3xlO­19 ­ 5xlO'19 J for a single photon of visible light. .
16.15. For the amine radical (NHz·), we can consider Example 16.7b ofNH3·, the ammonia
radical. In this example, we had a maximum possible Af of5/2, leading to 2(5/2) + 1 = 6 ESR
signals. For the amine radical, we have one less hydrogen, so a maximum possible MI of2,
leading to 2(2) + 1 = 5 ESR signals.
16.17. Since the carbon atoms have 1= 0, they do not contribute to the hyperfine coupling ofthe
electron. However, the seven hydrogen atoms all have 1= Yz, and the possible combinations of
those spins can lead to MI values of +7/2, +5/2, +3/2, + 1/2, ­1/2, -3/2, -5/2, and 7"'7/2. Thus, we
would expect 8 ESR signals for the cycloheptatrieny1 radical.
16.19. M' = (2.0058)(9.27,4xlO,z4 J/T)(0.3476 T)
M'= 6.466x10,z4 J
Converting this to wavenumbers, we find that till = 0.3255 cm,l.
..
16.21. With six lines in the ESR spectrum, an atom in the radical can have a spin of up to 512
(assuming that there are other atoms in the radical that are contributing to the signal); ifit has a
spin higher than that, then that particular atom probably isn't in the molecule. Checkingthe table
of nuclear spins, we fmd that (a) 42K (I= 2), (b) 35C1 (I= 3/2), (c) 37CI (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 non­zero nuclear spin; in all other cases, other
atoms with non­zero I are necessary in order to produce the given ESR spectrum.
16.23. The NMR transitions should appear at the same wavelength. Equations 16.22 and 16.23,
which give expressions for the frequency of resonance, do not have a dependence on the MI
values (since!!.MI = ±l).
16.25 .. Since the working frequency is given in MHz, we will use the hertz version of the
resonance condition, equation 16.23. Nuclear gN values are found in Table 16.1 and the
appendix:
For 2H : 330x106
27
S·1
= (0.85744)(5.051 X 10­ J/T)B
.
6.626x10­34 J·s
89 .
]j =50.49T
For 19F: 330x10 6
S·l
For 31p: 330x 106
S·l
27
= (5.2567)(5.051x10­ J/T)B
6.626x 10­ 34 J. s
27
= (2.2634)(5.051 X 10­ J/T)B
6.626 x 10­34 J; s
B
= 8.235T
B=19.13T
27
For 55Mn: 330x10 6 s' = (1.387)(5.05lx10­ J/T)B
6.626x10- 34 J·s
B = 31.20T
16.27. (a) An acceptable molecule that yields this NMR spectrum is CH3COOCH3, or methyl
acetate. (b) An acceptable molecule that yields this NMR spectrum is 1­bromo­2­ _
methylpropane, or CH3CH(CH3)CH2Br.
16.29. The differences between energy levels is the same, since there is a unit change in the MI
quantum number for all ofthe allowed transitions. Using equation 16.23:
27
V= (0.5479)(5.051 X 10­ J/T)(3.45T) ~1.4 x 07
s" = 14.41 MHz
6.626 x 10.34 J . S
16.31. 2.45 x 109 s'
= (2.0823)(5.051 X 10­27 Jff)B
6.626 x 10­34 J . S
90
B
= 154.35T
CHAPTER 17. STATISTICAL MECHANICS: INTRODUCTION
17.1. 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 ofballs in each box, then there are 20 different ways of putting
the balls in the boxes.
17.3. According to Stirling's approximation, the natural logarithm of 1,000,000! is
In (1,000,000!) = (I,OOO,OOO)ln(I,OOO,OOO) ­ 1,000,000 = 12,816,000. Therefore, 1,000,0001 is
about e 12,816,OOO, or about 105,566,000.
17.5. The logarithm of the expression is
In(N ­1).=
In[e-
N
• NN-1f2 •
(271/ f 1 • (1 + _1_ +
1 + ...)] Using properties oflogarithms,
.
12N 288N 2
'.
this reduces to
In(N ­1)!=
-N (N
+
­!)lnN +!ln27l" +In(I+_I_+
1 ). To evaluate In(5000!), we set
12N 288N 2
2
. 2
N to 5001 and substitute:
In(50 1- ).= + ~ ln50 1 27 " ( +
2
1
+
1
12(5001) 288(5001)
2
2J
Evaluating each term:
In(5000)! = ­5001 + 42,591.22 + 0.9189 + 1.6663xl0­5 = 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.)
17.7. According to the description in the exercise, we need to solve the following expression:
12
TI-(7-xY +38}tx
°
Expanding and evaluating:
12
12
1[,,­ 49 + 14x - x
2
12
12
= ~'(7
2_-_1_1}tx_
+ 38}Ix = ...::..f_{1_4_X_-_X_
(1
2
= --:7_x__ -_
)12
3_-_1_1_X",--o::.....
3_X_
12
12
·144 ­ 1728 ­11·12 ­ 0) = 25. Thus, on average, there will be 25 insects per month.
3
.
12
17.9. For a grand canonical ensemble, equations 17.4 ­17.6 would be rewritten as
V= 2:Vj <i-)',
T='Tj
P=Pj
j
The last relationship is due to the fact that chemical potential is also an intensive property.
91
17.11. Ifwe refer to Figure 17.6, we see that the distribution labeled (1,0,1,1) appears six times
out of twelve, making it the most probable distribution (with a probability of 0.50). For this
system, the thermodynamic properties are probably not dictated by this one distribution, because
there are only twelve possible distributions and the other distributions will have some obvious
effect on the overall average. But for a system that has a larger number of possible distributions,
the most probable distribution has a larger and larger impact on the overall thermodynamic
properties ofthe system as a whole.
17.13. (a)
to
~1:
a~l
~ = Cl~1
a¢J = a(Cl~J
+
C2~
a~1
+
a(C2~)
+
C3~
is the complete equation. Taking the derivative with respect
a(C3~)
+
a~1
derivatives with respect to
~
and
= C1 + 0+ 0 = C1 •
.
a~1
~3.
Similar expressions exist for
(b) The general expression is
a~i
a¢ = c..
.
This general
expression Was applied three times, once for each summation in equation 17.14, to get the four
remaining terms (in one, the chain rule of derivation was applied) that resulted in equation 17.15.
17.15. Use equation 17.21: Nt = !Le-llElkT. Ifwe assume the degeneracies are the same, the
.
No go
glJgO term cancels. Converting 200 em" to J units:
8m/s=5.996xlO
.It=!=
=0.005cm=0.00005 m v=2.9979xl0
1
V 200cm­1
0.00005m .
E = hv = (6.626xfo­34 J·s)(5.996xlO I 2 S­I) = 3.973xl0­21 J
Now we can use equation 17.21: Nt =
e-(3.973xIO-
21
23
J)I(1.38h<10·
JIK)(298K)
I2s­1
= 0.38.
No
Thus, there ate almost three times as many atoms in the ground state as there are in the first
excited state.
92
1 == e~
­(5.38xIO­
22
J)/(1.38IxIO·
23
Solving for T: _0.51082 ... == _
JIK){T)
22
8.64 xl 0­
.
(1.381xl0- 23)T
3
T== 122.5 K
17.19. (a) The state function A will always have a slightly lower value (i.e. a larger magnitude
negative value) than G, based on the definitions in equations 17.44 and 17.45 . (b) It is difficult
to tell which is greater because the expressions are so different. One can imagine circumstances
in which either is greater than the other.
17.21. For equation 17.46, Pj == -kT In..!L, the restriction is that the amount of any other
Nj
"*
substance in 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.
17.23. (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.
1 "s. .e-O;lkT]
17.25. We need to show that lim k In(Lgje-o;lkT)+- ~
19, _ .lkT
== klng o ' The first term
T-+O [
T LJgje 0,
is easy. Consider the summation. The first term in the summation is go, because eo· in the
exponent is zero (as the ground state) and the limit of that term as T goes to zero is just go. For
all of the remaining terms, as T goes to zero, the exponentials become e'", which is zero .. So for
the first term, we get k In go as T goes to zero. The second term is more problematic, and we will
need to apply L'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:
e-I/kTIsjgjeo/
1 LSigie-OllkT
1 e-l/kTLsigieOI
.;:::::;:::=------,-,-
T Igje-&/lkT
=
. T e-1/kTL:gje&/
= ----,-,======---
Te-1/kTLgje&'
By factoring the e-lIkT out ofeach term, we are taking advantage of a property of exponentials.
-l/kT
All we need to do now is determine the limit of ;
-11
.le
kT as T approaches O. We can't simply
.
substitute T= 0 into the expression because we will have trouble with infinities and zeros.
However, we can use L'Hopital's rule and take the derivative of the numerator and denominator:
1
1
1 -IIkT
e
II kT)
kTi
D(e_
D(Te- I/kT) ­ e-IIkT Tk/I-e_l~T+
kT 2
==
kTi
==
1
. Evaluating this.
1
kT
2
2
_kT_+_l == kT • kT+1 == kT +T
kT
kTi
1+_1_
kT
expression at the limit of T = 0 is still problematic, but we can apply L'Hopital's rule again to get
93
D~l)
=
D(kT +T)
0
, which is an expression that can be evaluated at T = O. The fraction
2kT+l
equals 0 at T = 0, so the entire second term goes to zero as T goes to zero, and we are left with
the fact that lim S = k In go .
T-?O
17.27. If N = N A in the Sackur­Tetrode equation, the equation itself doesn't change, but the
value of S calculated using that value is now the molar entropy.
. 1/2
17.31. At 25 K: A =
(6.626 x 10­ 34 J 'S)2
= 1.745 x 10­ 10 m
2rc( 0.004kg )0.381X10­23 JIK)(25K)
6.02x10 23
.
. 1/2
(6.626 x 10­34 J 'S)2
At500K: A=
=3.903 x 10­11 m
2rc( 0.004kg )(1.381 X10­23 JIK)(500K)
6.02x10 23
17.33. (a)
23
)3/2
23
2rc(0.012kg/6.02x10 )(1.381x10­ JIK)(1000K)
{ (. .
(6.626x10·34J·s)2
S = (6.02x10 23)(1.381x10­23 JIK) In
[
23
3
X(1.381 X10­ JIK)(1000K) .1L·atm. 1m }+~]
.
(1.000 atm)
101.32 J 1000 L
Solving: S = 164.9 JIK
94
2
(b)
23
s =(6.02 X 10 23)(1.381 X 10­23 JIK)[ln{(27r(0.0558 kg/6.02 X 10 23 )(1.381
X 10­
JIK)(3500 K)J3/2
34
(6.626 x 10­ J'S)2
.
X
23
3
(1.381 X 10­ JIK)(3500K) .IL·atm. 1m }+~]
(1.000 atm)
101.32 J 1000 L
.
2
S=210.1JIK
23
23
(c) S = (6.02 x l 0 23)(1.381 X 10­23 JIK)[ln{(27r(0.2006 kg/6.02 x 10 )(1.381 X 10­ JIK)(298 K)J3/2
(6.626 x 10­34 J. S)2
X
23
3
(1.381 x 10­ JIK)(298 K) . 1L· atm. 1m } +~]
(1.000 atm)
101.32 J 1000 L
2
S = 174.9 JIK.
17.35. First, we can calculate the de Broglie wavelength of electrons going O.Olc (= 2.9979xl0 6
6.626 x 10~34
J .S
=2.43 x 10­10 m
6
31
(9.109xlO­ kg)(2.9979xl0 mls)
Next, we use this as the wavelength in the thermal de Broglie equation and solve for T:
­34
2
JI/2
2.43 X 10­ 10 m =
(6.626xl0 J ·s)
( 21Z"' (9.109 x 10­31 kg)(1.381 x 10­23 JIK)T
mls): A. =
T= 94,100 K
95
CHAPTER 18. MORE STATISTICAL MECHANICS
18.1. (a) 12C has a nuclear spin of zero, so there are 2(0) + 1 = 1 nuclear spin state. Therefore,
qnuc = 1. (b) 56Pehas a nuclear spin of zero, so there are 2(0) + 1 = 1 nuclear spin state.
Therefore, qnuc = 1. (c) IH has a nuclear spin ofyz, so there are 2(YZ) + 1 = 2 nuclear spin states.
Therefore, qnuc = 2. (d) 2H has a nuclear spin of 1, so there are 2( 1) + 1 = 3 nuclear spin states.
Therefore, qnuc = 3.
18.3. 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.
18.5. The minimum value of qelect is 1, because this is the minimum degeneracy ofany
electronic ground state.
18.7. According to example 18.3, the value of Do is
432kJ 1000J
1 mol
=7.18xlO­19 J
6.02x 10 23 molecules
kJ
mol
for one molecule of Hj. Using this value in the definition of the partition function:
23JIKX298K)
= 5.89x 10 75. This is somewhat lower than the original value
qeJect = 1·e7.18xlo­'9 J/(1.381xlO·
of about 2x 108°, but still a large number.
18.9. (a) Assuming that the 89.8 J is the value for De (to be accurate, we would need to include
the zero­point vibrational energy):
89.8 J..
= 1.49 x l 0­22 J Substituting into
1mol
23
mol 6.02 x 10 molecules
= 1. e1.49xlO­22 J/(1.381xlO·2J JIK.)(4.2K) = 13 1
q
elect
.
qelect:
•
(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.
H;
18.11. The ratio of qvib(H2) to Qvib(D2) can be written using equation 18.20:
q (H) (kT / h v H ) v D
vib
2 =
2
=_ 2 • Since v =­1 ­, where k is the force constant and J.1 is the
)
vH 2 .
21l IJ
.
.
qvtD (D 2 )
(kT / h v D2
. .
.
.. :
...
. (H) (1 / 21l) Ik / IJ
2
reduced mass, this ratio of partition functions becomes q vtD 2 =
V
D
1I2l)~k(
/ IJHt
qvtD (D 2 )
reduces to q.. (II,)
qvtD (D2)
=
,
which
t'H' .
We cau substitute the reduced masses for H2 and 0, into this
IJD 2
expression or, recalling that deuterium has twice the mass of hydrogen, recognize that deuterium
96
will have twice the reduced mass of hydrogen. Thus, the ratio simplifies and we get our final
answer: q V1b (H2) =
qVlb(D 2)
=3N_6(
18.13. qVlb
qvib
0.
fi
e-
O/ 2T
[[
1 - e -OIT
1
J
= (
e-1870/2.298
1 - e -1870/298
J3(
e-2180/2.298
1 - e -2180/298
= (0.04347i(0.0258Ii(0.0009149)(0.0007114i
=
J2(
e-4170/2'298
1 - e -4170/298
J(
e-4320/2-298
1- e
-4320/298
,
J3
1.802x10­20
18.15. For a gas­phase molecule, the minimum value of qiluc 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 18~3)
if either Twere very low (i.e.
oK) or the vibrational frequencies were very large. However, for any nonzero temperature arid
finite vibrational frequency, qvib is greater than zero.
18.17. Using the expanded form of equation 18.34 (in which the definition of O, is given
" 18.19. Acetylene with one deuterium substituted for a hydrogen should not show the same
intensity variations in its rovibrational spectrum. Because the molecule no longer has hcenter of
symmetry, the symmetry implications on the allowed rotational states no longer apply.
18.21. For NH 3 : ammonia is a symmetric top, so we use equation 18.41 and the three rotational
temperatures (two of them the same) from Table 18.3:
1/ 2
K)( K
(298
298 )112 = 74.8
3
13.6 K 8.92 K
Carbon tetrachloride, CCLt, is a spherical top, so we can use equation 18.38:'
«;
= 7r
1/ 2
q rot
(
7r
298 K )3/2
=­
=32182
12 0.0823 K
18.23. Cp will always be greater than Cv , due to the additional Nkterm in it. Note that thisis
consistent with the conclusion from phenomenological thermodynamics, which states that Cp and
C; differ by R.
18.25. Table 18.5 gives a useful summary of the contributions of each partition function to the
thermodynamic property. Let us simply use those terms for a diatomic molecule, looking up the
. values that we need from the various tables, and solve.
ForE:
3
E=-RT-D
+RT(B
_v + Bv IT ) +RT
e
2
2T e Ov l T _ l
97
3
.
E = ­(8.314 Jzmol K)(298 K) ­ 431.6 kl/mol­i (8.314 Jzmol­ K)(298 K) (~nK ­ ­ 2 '
2·298K
4227 K/298 K)
.
4227K1298K
+ (8.314 Jzmol K)(298 K)
e
'-1
E = 3.72 kJ/mol­ 431.6 kJ/mol + 17.57 kl/mol + 2.48 kl/mol = ­407.8 kl/mol
+
ForH:
5
(8 8
H =-RT-D e +RT+RT _ v + 8 ~T IT +1 ) +2RT
2
2T e v -1
7 '
(4227K
H = ­(8.314 Jzrnol­ K)(298 K) ­ 431.6 kl/mol + (8.314 Jzrnol­ K)(298 K) ­ ­ 2 ·
.
2·298K
4227 K/298 K
)
+ 1 + 2(8.314 Jzmol­ K)(298 K)
-1
e
E = 8.67 kllmo1­ 431.6 kl/mol + 20.05 kl/mol + 4.96 kl/mol = ­397.9 kl/mol
+ 4227K1298K
For S, the ground electronic state is a singlet, so the electronic participation is zero:
S
.S
=R[m{(2:,kT)'" '~v l~:;e[R+]}
m(l-e-,.IT)]+Rm;' +R
26kg)(1.381
= (8.314 Jzmol­ K)[1n( 2;r(6.06 x 10­
'.
x 10.
(6.626 X 10­34
+ (8.314 Jzmol­ K)[42:~98
.
23
J'S)2
3e5/2]
l/K)(298 K))3/2 0.02445m
.
6.02 X 10 23
K In(1_e-4227/298)1 +(8.314 J/mol.K)ln,:98 K + 8.314 Jzmol­K
­1
15.2K
e
S = 153.59J/mol·K + 8.73><10'5 Jzmol­K + 24.74 Jzmol­K + 8.314 Jzmol­K = 186.6 Jzmol­K
J
Finally, for G, it is obvious from Table 18.5 that G = H ­ TS; so,
G = ­397.9 ­ (298)(0.1866) kJ/mol = ­453.5 kl/mol
(but feel free to use the expressions in Table 18.5 to determine G; you should get the same
answer, within truncation error).
18.27. From Tables 18.1 and 18.3, we get Ov and Or for H2: 6215 and 85.4 K, respectively. By
calculating the reduced masses of Hr, lID, and D2, we can determine the following relationships
between them: Jl(H2)/Jl(HD) = 0.5/0.66666 ...= 0.75 and Jl(H2)/Jl(D2) = 0.5/1 = 0,5. From these
ratios and knowing the relationships between the Ov and Or and Jl, we get 5382 K and 64.0 K for
Ov and 9r oflID (respectively), and 4395 K and 42.7 for 9v and Or of D, (respectively).
In terms ofthe changes in H and S, the only contributions will he from rotational and
vibrational partition functions, since the translational and electronic contributions are the same,
while nuclear contributions are ignored. In addition, for MI, the rotational contributions are the
same as well, leaving only changes in vibrations. Thus, we have for MI:
98
sn = 2(8.314 Jzmol­ K)(298
K)( 5382 K + 5382 KI 298 K + 1)
.
2(298 K)
e5382K/298K -1
-[(8.314J/mol.K)(298K)( 6215K + 6215K/298K +1)
.
2(298 K)
e6215K/298K -1
.
+ (8.314J/mol·K)(298K)
( 4398 K
4398 KI 298 K
)]
+ 4398K/298K
+1
2(298K) e
-1
Ml = 49701 ­ 28313 ­ 20760 = 628 Jzmol = 0.628 kl/mol
ss = 2· (8.314 J/mol· K)(53~ ;8
-[(8.314 Jzmol­
K)(62:~8
+ (8.314 J/mol· K)(43~;5 8
In(l­ e­5382K1298K)1 + 2. (8.314 Jzmol­ K)(ln 298 K + 11
)
64.0K )
K
­1
e
e
K _ In(1_e­6215K1298K)1 + (8.314J/mol. K)(ln 298 K + 11
­1
)
85.4K )
e .
K .; In(l­ e­4395K1298K)1 + (8.314 J/mol. K)(ln 298 K + 11]
­1
')
42.7 K
)
/:is = 42.21 ­ 18.70 ­ 24.47 IzmolK = ­0.96 J/mo1·K
18.29. Using the expression E
q N)
= kT 2(aln
= NkT 2.! aq
: the expression for the
qaT
translational 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:
E. =NkT2
elect
1
D
~e'
er
a(gleD,lkT)=NkT2
aT
IkT
1
IkT
D
~e'
.
gl
._ De eD,lkT
kT2
.
.
The degeneracies cancel, the exponentials cancel, the kT terms cancel. The only terms that
don't cancel are N and -De, giving us Eel ect = -ND e. For (polyatomic) rotations, we have
2
E
rot
= NkT
1
~(
T3
°
A
(J'
= NkT 2
a(~
a
e
0'
)1/2J
aT
)Jl2
OBOe
.n
1
n l/2 (
3
T .
AOile
T3
)1/2
l/2
(J'
(
1
)1/2 . ~ T 1I2
AOBOe
2
°
0AOBOe
All of the terns (z, cr,8s) cancel, and what we have left is (after collecting all of our terms in 1):
I/2
E=~NkT2
T
=~NkT2!
.
2
T 3/2 2
.T 2
18.31. If the reaction deals only with isotopes, the translational arid 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
99
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 V1bq roJmixed 2
­ (qVlbqroJ'4 N2 (qVlbqroJl5 N2
this becomes
K -
K
=
(OrT °vT) mixed
Using the high­temperature limit expressions for each q,
2
The T variables cancel. On the assumption that the 9v and
C:" ;))2:,.;J'N'
ar 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.
18.33. 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 statistically­determined entropy S values are very
close to the experimentally­determined 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 GcaJ,mot be construed as the "total
energy" of the atoms.
100
CHAPTER 19. THE KINETIC THEORY OF GASES
19.1. A postulate is a statement that is assumed but not proven. Other statements can be based
on them and supported or not supported by them, but the postulate itself is only supported based
on what is derived from it.
19.3. The drawing is left to the student.
19.5. Let us rewrite equation 19.8 in such a way as to get the expression for kinetic energy all by
itself on one side. We get
.!.2 mv
2
= 3p V. Note that the right side of this expression has
m
.
pressure, volume, and amount ­ three of the four variables that the conditions ofa gas depend on.
The fourth variable is temperature, and by comparing the right side to the ideal gas law, we can
argue that 3PV is proportional to temperature, T. (We just don't know the proportionality
2N
constant yet.) Therefore, we substitute: .!.mv 2
2
IX
T, leading to the conclusion that the (average)
.
kinetic energy of a gas is dictated solely by the temperature of the gas.
19.7. Cesium atoms have a mass of 0.1329 kg/mole:
200 mls =
3 . (8.314 J/mol· K)T
0.1329 kg/mol
T=213K
400 mls =
3· (8.314 J/mol· K)T
0.1329 kg/mol
T=853K
600mls=
3·(8.314J/mol·K)T
0.1329 kg/mol
T=1920K
800 mls=
3 . (8.314 J/mol· K)T
0.1329 kg/mol
T=3410K
1000 mls = 3· (8.314 J/mol· K)T
T=5330K
0.1329 kg/mol
The temperature is related to the square ofthe speed, rather than directly to the speed.
+00
19.9. We need to solve the integral
JAe(1/2)KV dv = 1 for the proper value of A(this is equivalent
2
­00
to normalizing a wavefunction...). This is an even function centered about zero, so we can
rewrite this integral as
2jAe(1I2)KV dv = 1.
2
The integral is ofthe form
j.e-
o
our case, the constant b = -%K. Therefore, we have:
bX 2
dx =
2
1'~
~
5Afl>7'(.V
11
.!..(n)1I2.
2 b
0
>.
dv ::. \
hi
J
eo
2A e(1I2)Kv dv = 1
2
o
1( tr
[
2A ­
2 -K/2
)1/2]_-1
2tr)1I2
A( ­ ­
=1
K
_(
K
A- ­ -
2tr
)1/2
This final expression is equation 19.25.
1 g '(v)
19.11. Showing that K is also equal to
19.13.
vnm
= ..J3RT/ M =
VlOOstprob
..J2RT/ M
Y
and ­
1 g '(v )
Z
Z
follows the same steps as
v y g/v y)
V z gz(v z)
presented 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.
Y
(3 = 1.2247...
'\)"2
19.15. Rubidium atoms have a mass of 85.5 g/mol or 0.0855 kg/mol. Using the formula for the
most probable velocity:
T=5.14x1O- 7 K
O.Olm1s=· 2(8.314J/mol·K)T
0.0855 kg/mol
19.17. 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 19.13).
19.19. 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.
19.21. 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.
19.23. 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).
.
19.25. Is,I
= trp(4.0x1O- IO m)2~6(1.38xO-
JIK)(298K)r
2
(tr. 2.18x 10­ 25 kg)ltl
P = 6.42x10 l5/m3• That is, there need to be 6.42x10 l5 atoms ofXe per every cubic meter. Since
6.02~
1023
23,atoms,
1o
we need 42x 1015 3~/
:= 70J.~8?
m" = 9.38x10 L.
1 mole ofXe has 6.02x10
That's' the
~olurie
of a cube that
is\~4
m~A
(just under half a kilometer) on a side.
19.27. The total number of collisions is Z. V= 3.21x1Ol5 s·lm· 3 x 9.38x107 m3. = 3.01x1023
collisions per second.
19.29. What the question is asking is which quantity is larger, ZHe~Ar
or ZAr~He?
The value of z
depends on the density of the other gas. In this case, since the two gases have equal
concentrations, we expect that the two values of z should be the same.
19.31. Using equation 19.51:
rate= O. 10 mm 2 x
00014 romH gx
l atm
101,325Pa
(l m)?
x.
x­.....:...­_(l000mm)2
760mmHg
I atrn
x [ 2,,(0.2006 kg/6.02 x IOn ;(1381 x 10·n JIK)(295 K) )'"
r;:. ~
rate = 2.02 x 1014 S·1
14
That~jS'r"
.02x10 mercury atoms are diffusing through per second.
..
eneraIly, we can use equation 19.51 for this.. but again we need to watch our units.
e should be expressed in Pa units, so we convert the 0.100 torr:
p{~s
./
0.100torrx latrn xI01325Pa=13.33Pa
...
760 torr
1atrn
.
x 1...C;Iq.MM
If the diameter of the tube is 0.01625 inches, its radius is 0.008125 inches, or 0.206375 mm.
Thus, its area is nr 2 = n(0.206375 mmi = 0.1338 mnr' = 1.338x10·7 m 2 • Now substituting into
equation 19.51:
.
. rate = (1.338 x 10·' m' )(13.3 3 N/m' { 2,,(.0399 kg/6.02 x 10"
rate = 4.294 x 1016
_
1mol
SiX·
.
6.02xl0
23
39.9
~,p2-I6)<'_O.
~'l)o
I L
= 2.85 xlO
mol
.
X­­
)~1.3
81x \O­n JIK)(300 K) )'"
•
6
g/s = 2.85 mtcrograms per second .
.
19.35. Substituting only units into the expression for D I 2 in equation 19.54:
1
(llmol· K)(K)
2
kg/mol
m ·lIm 3
The other quantities in equation 19.54 don't have units, and recall that p is the particle density of
the gases, meaning that irs the number of particles per cubic meter. The Kelvinand mole units
in the square root term cancel, and we can decompose the joule unit to cancel the kilogram unit.
What's left is
~
1
m
1
V7' m ' ­L' m" =­;­'l/m =~.m=-s
m
m'
as the standard unit on D l2 . As mentioned in the' text, D I2 values are more often expressed in
cm 2Is units. .
.
I
.
19.37. Diffusion stops because the term del, the concentration gradient, is zero when a minor
dx
component is evenly distributed throughout the mixture.
19.39. If E = 3/2 kT, then the energy of an ammonia molecule at 295 K is
103
21
E = 3/2 (1.381 x 10­23 JIK)(295 K) = 6.11 x 10­ J
If this energy were kinetic energy, then we can use the equation ~
the molecule:
(
I
6.11xlO­2 1 J=~(
mV to determine the speed of
=
0.017kg )V 2
v 658m1s
2 6.02x10 23
Let us use the definition of root­mean­square speed to determine another velocity for ammonia
molecules using a different route:
= 3(8.314 J/mol· K)(295 K) =658 mls
=~3RT
v
.' 0.017 kg/mol. .' .
besurprising; as the two­expressions for speed are related via
This coin de c sIi~uldn't
derivation of the root­mean­square spe d."'·- , _ ~
nns
M
19.41. (a) If the ammonia were diffusing through helium, diffusion will likely be faster. That's
because the value of (rl + r2i will likely be smaller (since helium atoms are smaller than air
. molecules), and the reduced mass I­t ofNH3­He is smaller as well. Both of these terms are in the
denominator of the expression for D l 2 in equation 19.54, so D12 itself will be larger, implying
faster diffusion. (b) If ammonia were diffusing through SF 6, the arguments are reversed. The
value of (r1 + rzi will be larger, and so will u. Therefore, D12 will be smaller and diffusion
slower.
19.43. It may be naive to assume that ions in solution behave like individual gas particles,
especially since they are charged particles. Furthermore, ions don't travel as individual particles,
.as the kinetic theory assumes for gas particles. They move with groups of water molecules or
other ligands surrounding them, suggesting that the assumption that the travel of ions in solution
.
should not be based on the masses of the ions themselves.
104
CHAPTER 20. KINETICS _
20.1. For the reaction aA + bB
~
cC + dD, we can write several other forms of the rate:
d[C]
rate = _ d[A] = +~
d[D]
c dt
dt
d dt
or, in terms of the concentration ofB:
rate =_ d[A]
= +~
dt
rate = _ d[B]
= +!!- d[C]
dt
rate =_ d[B]
c dt
=.: d[D]
dt
d
dt
There are several other possibilities in terms of d[C]/dt and d[D]ldt as well.
+
5 .
20.3. If 1.00 mmol of'H are consumed, then there are ­(1.00 mmol)== 0.3125 mmol of Fe (s)
16
is consumed. Therefore, the rate with respect to Fe (s) is
3
0.3125xl0­
av lue'It was WIith respect to. Trt·
1­
­ ­ ­ ­mol
­ Fe
­ == 406
. x 10­7 mo 11s (Whi ch iIS th
e same
n ill
5
153.8s
exercise 20.3). If 1.00 mmol ofW are consumed, then there are
~(1.0
16
of Mn04­ is consumed. Therefore, the rate with respect to Mn04­ is
mmol) = 0.125 mmol
3
406 x 10­7 mo 1/s (Whi ch'IS th
'It was WIt
. h respect to B+'III
-1 0.125x10­ mol Fe ==.
e samei
va ue
2
153.8s
.
exercise 20.3). In fact, all of the rates for each species tum out to be4.06x10·7 molls.
20.5. The species could be a catalyst, or perhaps an inert gas or solvent that provides collisional
energy to the reactants or products.
20.7. First, let us convert the times into rates by dividing the number of moles of A reacted by
the seconds elapsed. We get
O.lOrnol == 2.72x 10­3 Mis 0.10mol == 4.00x10­3 M/s 0.10 mol == 1.00 x 10­2 M/s
36.8s
25.0s .
. 10.0s
respectively, for the three trials. Using the first and second trials:
3
2.72 x 10­ Mis = k(0.20t (OAO)bb ' were
.
hi h duces to.
0 68 =.
0 667 a. By inspection,
• • h ld b
It s ou
e
4.00 x 10' Mis k(0.20t (0.60)
.
clear that a = 1. Using the second and third trials:
-~3
3
4.00
10­­M/s
­
­x ­
­ = k(0.20t (0.60)b
2
hi h d uces to 0·4
. .It sh ou ld b e
were
. = 0 .4 a. B'
y inspection,
k(0.50t (0.60)b '
1.00 x 10. M/s
clear that b == 1 also. Using the first set of data to determine k:
2.72xl0·3 Mis ==k(0.20 M)l(OAO M)'
k> 3Axl0·2 M­ls· l
20.9. If a rate is given in units of Mzs, in order for that to be the ultimate unit, k would have to
have units ofM·3s· l.
20.11. 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 straight­line
105
equation has the fonny = mx + b, whereas this equation is of the fonny= eX, which would not
plot as a straight line.
noJ3)First, we need to determine how many grams of HgO needs to decompose in order to
~
.00 and 10.0 mL of 02 gas at STP. Using the fact that at STP there are 22.4 L of gas
volume:
1.00mLx
1L x 1mol x 2molHgO x 216.6gHgO =0.0193gHgO
1000mL 22.4L
l mol O,
molHgO
must decompose in order to form 1.00 mL of 02. Of course, for 10.0 rnl., ten times that amount,
or 0.193 g HgO, need to be decomposed. Ifwe 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:
Taking the logarithm of both sides:
(a) 0.9807 g = 1.00 g x e·(6.02xI0-4 s·1)t
In 0.9807 = In 1.00 ­ (6.02x1 0­4 s·l)t
t = 32.4 s
Taking the logarithm of both sides:
(b) 0.807 g = 1.00 g x e,(6.02 xI0-4 s·l)t
In 0.807 = In 1.00 ­ (6.02xlO­4 S·I)t
t == 356 s
20.15. Start with equation 20.19: ­
tl~]
= k -dt. Integrate both sides, 'with the left side's limits
being [A]o through [A]I and the right side's limits are 0 to some time t:
d[A]
1--=
Ik.dt
[A
(Al,
I
JAlo'
]2
The integral on the left side is simple l/[A], evaluated at the limits, while
0
the integral on the right is simply kt, integrated at its limits. We have:
1
[Al,
[A] (Al o
= ktl'
Evaluating at the limits, we get
0
1
1
---=kt
[A], [A]o
(which is equation 20.20).
20.17. (a) The rate law is ­ d[A] = k[A]3. We can rearrange this to get ­ d[~]
= k -dt , which
dt
.
[A]
·
lv
u1timate
y imtegrates to
1
2[A1
2 ­
1
2[A]o
Z
= kt .
,.
1 2 = kt +
1 2 ' we see that we would
(b) If we rearrange the integrated expression to
.
2[A],
2[A]o
have to plot 1/2[A]l versus time t to get a straight line having slope k and intercept 1/2[A]/.
20.19. For a zeroth order reaction, the integrated rate law is written as [A]o ­ [A], = kt, which
can be rewritten as [A]I = -kt + [A]o. This shows that if [A]l were plotted versus time, the slope
would be -k and the intercept would be [A]0.
106
20.21. For a zeroth­order reaction, (A]o ­ [A]I = kt, so we want to know the value of t when [A]I
=0:
[Aj, ­ 0 = kt
t = [A]olk
[A]o = kt
20.23. When a reaction uses H 20 as a solvent, the concentration of H 20 is typically so relatively
high that it can be considered as constant throughout the course ofthe reaction. This 'constant'
concentration of H20 can be incorporated into the value of the rate law constant for a pseudo
first order reaction, whose rate constant will have units of S·l.
20.25. Four experimentally­determined 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.
20.27. 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 zeroth­order reaction will continue at a
constant rate for two complete half lives.
[A]o (k r + kre-(kr+kr)I). If the reverse reaction is negligible,
(k r + kJ
that suggests that k; is negligible and can be ignored in the above expression. If so, equation
20.29. Equation 20.33 is [A], =
20.33 becomes [A]t =
[~]o
(kre-
kft),
which reduces to [Al = [A]oe- krl , which is the integrated
t
rate law for a normal first­order reaction.
20.31. The initial ratio is equal to the appropriate ratio of the rate constants, in this case ka
divided by k 1 (because the product A­B is made by the second reaction):
ratio
=~
= 3.95 X 10-4 S·1 = 8.98
k1
4.40x10­s
S·1
The ratio of A-BIB-C at equilibrium cannot be determined based on the information given.
because we don't know the rate constants for the reverse processes.
20.33. Starting with equation 20.38, [A], = [A]oe-(k 1+k2)t, we take the logarithm ofboth sides:
·In [A], = In [A]o ­ (k 1 + k2 )t
.
This equation is in the form y = mx + b, the form for a straight line. In this case, the slope would
be given by the expression -(k1 + k2) . Unfortunately, the slope is the sum of the two rate
constants. It is not possible from a plot such as this to determine the individual values ofthe two
rate constants; other experiments (and/or plots) are necessary.
20.35. With the understanding that straight­line plots of equations 20.41 and 20.42 would be .
difficult to define and/or interpret, the best way to determine k 1 and k2 may be by numerical trialand­error: try to fit the experimental data by assuming various values for k 1 and k2 • Analytic
solutions for.the two rate constants may not be forthcoming.
107
20.37. At t = 0, the amount of 21oBi will be at maximum; that will be the time before any of it
decays. Mathematically, the amount of 206Pb will be at maximum at t = 00.
20.39. (a) When k1 »k2, the second rate constant is negligible with respect to the first. The
equations reduce to
[A], = [A]oe- kl' (no change)
[B], = k~]o
(e-kll _ e-k21 ) = ­[A]o (~
I
0 ­ e-k21) =[A]o (e-k21)
,
[C], = [AJ,[1+ :,
(k,e ­',1 ­ k.e ­'1 l] ~ [AJ,[1+ :, (- 0­ k.e ­',1 l]
=[AJ,[I + :, (- k.e -~I
l] =[AJ, [I­e­·,I]
(b) When k 1 «k2, the first rate constant is negligible with respect to the second. The equations
reduce to
[A], = [A]oe- k" (no change)
[B], = k,[A]o (e-k,1 _e-k21)= k)[A]o (e-kl_~O)=
k2
k2
[C], = [AJ,[I+
= [A]o
[1­ e-
,~_
k,[A]o (e-kll)
k2 .
(V" -k,e-.l")]=[AJ,[I- :, (k,e-"'--O)]
kt
• ]
.
0.4 : This problem needs to be answered graphically, using the techniques illustrated in
ple 20.8. Your final answer will depend on the accuracy of the graph, but you should get a
pre­exponential factor around 4x10-1
°.
. 20.43. If295 K is the nominal base temperature, then a 10 degree increase would make the
second temperature 3<15 K. At 295 K, ifthe rate constant were k, then at 305 .the rate constant is
2k. Using equation 20.52:
.
.
,In :k
EA
=­ 8.314~0'
K
(29~K
30~K)
The k's cancel; solving for EA :
= 51,850 J/mol = 51.85 kJas the activation energy.
20.45. Using the Arrhenius equation: 9.9xlO­12 cm3/s =
Solving for A: A = 1.2xlO­ 1I cm3/s
A.e-1900J/mo1l(8.314 J/nx.'1.KX1l53K)
. 20.47.' In
k
= _ 20000J/mol ( 1 _
1 )
1.77xlO­6/M·s
8.314J/mol·K 373K 298K
Solve for k: k = 8.97xlO­ 6 /Ms
20.49. (a) To determine the energy of the photon (which represents the activation energy):
108