Student Solutions Manual for Physical Chemistry

Prévia do material em texto

i 
!,. 
1 
j 
1 
j 
! 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 100;~: Pa = 1.Q~~~~)Os Pa 
(d)	 1233mmHgx ltorr x 1atm x 1.01325bar =1.644 bar
 
1mmHg 760 torr 1atm
 
l cm"
(e) 125mLx-- =125cm3 
'lmL 
(f) 4.2 K-273.15 =-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. 
~ For this sample orgas under these conditions, F(p) = _ 0.0250L. =!J)4xl0-4 ~• 
....~. (33.0+273.l5)K
 
(Note the conversion to. .:' .he temperature.) tfthe volume is going to be 66.9 cnr' =
 
0.0669 L: 0.0669 L //' lOC -4 ~:~'~43 K . . \.0 4,-x II) ~ L .
 
. . .04 x 10 hL/ { {c::...
 
~, .>'f7;/ '.
 
1.9. There are many possi 0 versions. Using the fact that 1 cal = 4.184 J~' 
8.314 J x 1cal = 1.987 cal
 
mol.K 4.184J mol.K ~. i \; II_
 
,- . -" ..--~'_ .._....._---­
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 ofpressure, atm for STP and bar for SATP. 
1 
lb 1.13. The partial pressure ofNz = 0.80x 14:71b =11.8.
m. m. 
. I f 0 0 14.7 lb 2 9 lb The partia pressure 0 . z = .20x. =. -.. 
. m. m. 
1.15. (a) This is an equation for a straight line with slope = 5, so at x = 5 and x = 10, the slope is 
simply 5. 
d .. 
(b). The slope of this function is given by its first derivative: - (3x 2 - 5x + 2) = 6x - 5. At x = 
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: ~ (:) = -(:2 ). At x =5, the 
1 · 7 he slone i 7s ope IS - - ; at x = 10, t e s ope IS - - . 
25 100 
1.17. (a) (BVJ =~(nRTJ =- nR (b) (BV) =~(nRTJ = RT (c)
2TBp Bp p pan. r,» an p pT,n 
(BT) =~(PV)=L (d) (Bp) =~(nRT)=nR (e) .(.Bp) =~(nRT)=RTBV BV nR nR BT r» er v van T,V an v vp,n 
1.19. (~(dTJ J or (!!-(dT) J . . dV· dp v p dp dV p v 
2.. Using r, =..!!.-, and using data from Table 1.6, we have: 
/ bR/ .tJ
. 3.592 ea~ . .. . 
for CO2: Tn = mol. = 1026 K 
(0.04267 Llmol{0.08205 Latm) 
\. molK 
1.360 I!a~
 
T
B 
= mol = 521K
 
(0.03183L1mol{0.08205 Latm)
 
\. molK
 
2 
1.390ea~ 
forNz: Tn = . mol = 433 K 
(0,03913 Llmo/0.08205 Latm)
 
'\ molK
 
1.23. The C term is ;2 .In order for the term to be unitless, C should have units of (volumej', 
or e. 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, Nz, Oz, AI, CI-Lt, 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 X 104 cnr'. Therefore, we 
have for hydrogen: 
pV =1+ B = 1+ 15 cm3:mo~ =1.00067 , which is a,0.067% increase in the
 
RT V 2.2412xl0 em Imol ,--~
 
compressibility. For HzO, we have: (O,~ J~. .t. 0 0 ";;; 
3/mol·
pV 1 B ·1'· -1126cm 09497 hi h' 50°/ d . th - = + -= = + 4 3 ='. , w C IS a . 10 ecrease In e 
RT V 2.2412xl0 em Imol
 
compressibility with respect to an ideal gas.
 
~Y comparingth~ tw~ expressions from the~\ C 
~+(0 --!!:-) 1 +f( b) + ~ .. and Z = 1+'~ C + ..,\ 
f RT V V 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. 
1.31. In terms ofp, V, and T, we can also write the following two expressions using the cyclic 
rule: 
(av) = C~fl and (avJ = (:1 There are other constructions possible that would er p ( ap ) ap (aT)
av 
T
avoT 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 
compressl I ity ibili ISis de me fi d as - -1 (av)- , so K. WI'II have units its of0 1 . volume = I , 
V EJp T volume pressure pressure 
or atm-I or bar'] , 
1.35. For an ideal gas, K =_-.!-(av) =_.!-~(nRT) =...!.. nRT , Since nRT =V, this last 
T. V ap VEJp P V p2 P 
ion b 1 V I c. id 1 Th ,T. 1 dexpressIOn ecomes -- = - lor an I ea gas, e-"expresston -a IS eva uate asVp p P 
TTl (av) T a (nRT) T nR , ,-a =-- - =-- -- =--, For an Ideal gas, the Ideal gas law can be p pV er pVet p pV Pp 
' nR Vb' th hi I ..rearranged to give - =-, so we su stitute to get at t IS ast expression IS 
p T
:v ~ ,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 = M" = pM . The derivative of this expression with respect to RTf p RT 
temperature is (ad) = - P~ .Using the definition of V,. this can be rewritten as 
aT RT
(:a. =-~ 
p.« 
4
 
CHAPTER 2. THE 'FIRST LAW OF THERMODYNAMICS 
2.1. work =F· s=IFIlsl cosO 
(a) work =30 N . 30 m· cos 0° =900 Nrn =900 J. 
(b) work = 30 N· 30 m· cos 45° =900·0.7071 Nrn = 640 J. 
.	 . 1L .' 101.32 J 
2.3. W = -Pext8V =--c(2.33 atm)(450.mL - 50.mL)x = 0.932 L· atm x . =94.4 1. 
1000mL	 lL·atm· 
2.5. (a) The work would be less because the external pressure is less.	 I' 
(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\.!',/l'N-~I-" It-I lifz T-(,/·t~	 lk't;l#z.u- . 
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, similar-mass molecules. 
, I '?!l1'l,./.q~ 5-1 ~~~ ";~~PlI"1 ""'\11-{-1 °t~-""l1~fW:;'~?1 LVl~'/1~.<."1 01 i,.&t\;. -u-1;:.nIi. OZf· r« =t--L- ~'Vbcot. "'i/},~ ;_1f~LVr . 
2.9. First, calculate the energy needed to warm the water: y,~t~~l-l1~ ?f!AL 5"r~ 
q = m·c·I:1T= (1.00xl05 g)(4.l8 J/g·K)(L.Q.Q.K.1, ~~ l8xl 05 J
 
~~~~~~
 
Now, determine how many drops of a 20.0 kg weight falling 2.00 meters in gravity will yield
 
that much energy. The amount of energy in one drop is
 
mgh = (20.0 kg)(9.8l m/s2)(2.00 m) = 392.4 J. Therefore,
 
4.18xl05 J
# drops = =1070 drops. 
392.4 J/drop 
2.11. The verification of equation 2.8 follows from Boyle's law, which says PYi =P[Vf . This 
can be rearranged to give	 Vf = Pi . Substituting this into equation 2.7:
 
Vi Pf
 
W =-nRT~ Vf becomes W = -nRTln Pi (which is what we are supposed to verify); rev	 rev 
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 
2.15.	 W= -nRTln---:L = -(0.245mol)(8.3l4J/mol·K)(95.0+ 273.15 K)ln-'- ­
Vi !.OOOL
 
W= 5l80J 
5 
~.~t;. 
2.17. If any change in a system is isothermal, then the change in U must.bezero. 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 ofgas 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 = 572molN2 gas
 
Pi __ PI
(a) The final pressure can be determined using Charles' law: 
1'; Tf 
, PiTf (172 atm)(140.0+273.15 K)
 
r, =T= (20.0+ 273.l5K) 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.
 
VI' " J "010L ' 
+: .23. w=-nRTln-=~(0.505mol)(8.314 )(5.0+273.15K)1h '=+2690J~ 17; . mol-K 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. " 
. " (au) (au)0!j' In termsofpressure andvolume: dU = Up ,dp + av dV . 
p 
For enthalpy: , dH = (aH)- , (aH) dV.dp + ­
ap av pII 
~ In order for each term to have units ofJ/mol·K for each term, the first term has units 
\1bR!I-K; the second term has units J/mol·K2; and the third term has units J·K/mol. 
6 
Q(OHJ ==(O(U + PV») ==~ +(OPV) ==O+(OPV) ==(OPV) ==(ORT)
~ op T Op T(~ )T OPT OPT OPT OpT 
. v~~· 
which is zero at constant temperature~Therefore, (~H) is zero . . op T 
2 .3 1 This derivation is given explicitly in the text in section 2.7 . 
. ForRe:~2a ,. 2(0.03508 e .arm/mol") 36 I K ( d 40 K fr' )"T == - == ==. compare to om text 
Rb (0~08205 L· atm/mol- K)(0.0237 Llmol) 
ForH2: 
2a 2(0.244 L2 • arm/mol") . . '. 
T == - == == 224 K (compared to 202 K from text) 
Rb (0.08205 L· atm/mol- K)(0.0266 Llmol) , 
. ~~,)~bl.. . 
2.~ : Because the pressur~ change isn.'t!90 dras~ic, our ~swer to exercise 2.34 isprobably 
hin a few degrees ofbemg correcr--tf a truly isenthalpic process can be arranged. ~ 
" "P'(i6\ , . -: . . .. ~.41?1 b~, 1M;, :t\~" . 
2.37. Because strictly speakmg, heat capacities are extensiveproperties; they depend on the 
am t ofmatter in the system.. Thus, the form in equation 2.37 is the most general expression ~ 
/ t re "" the two ~s. . . 
2 . First, calculate L~al pressure and assume that this is the exte ~co stant throughout ~~:~ression (a good approximation; since t
 
== nRT == (0.145mol)(0.08205L·atmlmol·K)(273.15K) == 0.650 
~-~
 
P V 5.00L
 
Now we can calculate work as -Pext~V: 
. . 101321·
 
w == -(0.650 atm)(3.92L -5.00 L)x == +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 5.00L _ 3.92L T ==214K Thismeansthat~T==(214-273.l5)==-59'K. 
f1'; Tf ,273.15K Tr '> _____ 
Using q = n·c·~T: q =(O.l41\rr.~r;z.u.J.2.#m~gJ(-59 K) == -178 J. 
~u = q + w == -178 J + 71.1 J = -107 
.) 
2.41. Starting with - R In V (I v; == Cv lnT (I T,: both logarithm terms can be evaluated similarly. 
The logarithm on the left is evaluated as -R(ln(Vr) -In (Vi»' Since In(a) -In(b) = In (alb), this 
7 
2.4 
at 01, 
simplifies to -R·ln(VtlVi), which is the left side of equation 2.44. The right side gets evaluated 
Performing a similar substitution as in 
C C -R--R -R 
p	 - y 2 2 2 2/2 2 f n id 1di .r = = ' =-5- = -",-, =- or an 1 ea iatomic gas. 
c, ~R -R 5/2 5 
2 2 
.	 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 ofboth water and ice 
C: 
1mL ' 
for water: 1gram x=1.00016 mL 
,	 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 + (1atm)(-0.090,mL), m1. . = 333.5 - 0.0090 J ~ 333.5 J 
.. ,	 1000 1Lvatm 
This shows that fl.U and Mf can be very close, ifnot v y e same, for many condensed-
phase processes. ~ . 
2.47. Steam bums hurt m~water bums because steam gives up a considerable amount 
ofheat as the heat ofvaporization. 
' h ~1\\.. h fr . b ~~d inpart) th2.49. The eat of fusioh given up by t e eezmg water can e transferre (at least m part to e 
citrus fruit, keeping them warmer and (hopefully) keeping the fruit itself from freezing. 
and simplified similarly. 
R 
7 5 2 
aRe03(s) -7 Na (s) + Yz H2 (g) 
Arx"H = LfI. [H(prods)-LfI. [H(rcts) =(2mol)(26.5kJ/mol)- 0- 0 = 53.0kJ 
+ C (s) + 3/202 (g)] -2xfl.rH= +1901.62 kJ 
2 Na (s) + C (8) + 3/202 (g) -7 Na2C03 (s) fl.rH= -1130.77 kJ 
C (s) + 02 (g) -7 C02 (g) fl.rH= -393.51 kJ 
H2 (g) + Yz O2 (g) -7 H20 (1) ArB = -285.83 kJ 
.Ir--: 
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 constant-volume, qv = AU = -31,723 J. 
. ---- > 
8
 
w ~ 0 since the process is in a constant-volume calorimeter. To determine MI, we need to know 
the balanced chemical equation for the combustion ofbe~d: 
.CJI,COOH(s) +9 ~+ 3H,OOj 
For everymole ofbenzoic acid combust~ere isa Ch~~.sofq - 15/2) ~es ofg~ 
1.20 grams ofbenzoic acid are ~gx (~~ mo~ofbenz01c aCI . Therefore, the ( ~--'- ~ !-~~.Q~ .~, ..............­
:a~~:~~;n rmmbe:F-clinol~s~)X(~0.5) .= .-0,0~492Il1;()10f gas.__ :qsing the 
J
. Mi =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.
 l 
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/g-K)(-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 verall t:.rxnH is the sum of these three parts. Converting all energy values to kJ:
 t:. =-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 
melting point is DOC. (c) Spontaneous, because potassium compounds are generally so~~le in 
water. (d) Not spontaneous; an unplugged refrigerator should warm up. (e) Spontaneous, 
because ofthe effect ofgravity on the leaf. (f) Spontaneous, because both Li (s) and F2 (g) are 
rat~erlr~~~re. elements. (~) Not spontaneous, because water does not break apart into hydrogen 
and oxygen without some mput of energy. 
3.3.	 e=- wcyc/e =_(-334-115+72+150)1 =_ -227 ~0.267=26.7%
 
q\ 850J 850
 
· = l _ ~ow35.. e 0.440 = 1------'=-­ 0.560= T/ow T.low = 237K=-36°C 
Thigh (150 + 273.15) K 423K 
--:,,>~: Superhea\el steam has the advantage ofa higher temperature, so (hopefully) there will he a 
higher effici~~cyfor a heat engine using superheated steam. 
i ne definition of a p~~~~ m~tion machine is a machine that creates more energy than it 
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). 
~ll definiti~~s o~ e~~~~y are appl~~JP real-g~~ as well ~ ideal gases. 
~~~::r~l~~~~~n IS m{{ependent oft]}elype of~~tl involved m a p~~ss. 
~ A better statement of the secondlaw of thermodynamics includes the conditions under 
__~.~:ch the second .law is ~W7t~ly apyVf,able: for an isolated system, a ~<t~¥1eous change is 
~~~acc9mpanled by anincrease m entropy. 2, 1 
.....~Iv. 
units are standard, so let us simply substitute into the proper expression 
3.17. The number ofmoles of air being breathed in depends, of course, on the temperature of the 
air. Let's assume a normal room temperature o~ 295 K. (You may assume a slightly 
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 
Pf '	 758 torr -4
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 change-in-pressure part and a change-in-volume part: 
M. = -(1mol)(8.314J/mol·K)In 1atm·,...~5.2JIK 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 
The overall entropYfQaIige is-A.S'.-==::d5.2 + 42:8 ~ ore, for ~real gas we can 
~~n~~ffltulg~~~~~!... .4 J . - '-.,., 
~P-.R::i§Jlt.~~t.~::lPllcity of an.ideal gas under conditions of co stant pressure. However, 
the process in Example 3.3 is not a constant-pressure J2rocess. Therefore, while the value ofthe 
heat capacitygIven may be lliecoiieaexperi~ntal valu; for an ideal gas it probably wouldn"t 
~~.' ~ " 
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 ~h~f:um.J~w, 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. 
In this case; the hotcopperwill lose energy/and the cooler water will gain energr:_.1n te~§ of 
the second law;t1}is_~()Il.t~.!!~changewill 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 ·(T -1';):: -(5.33g)(0.385-)(Tf -372.85 K) = -2.052Tf + 765.107
 
~ f g·K
 
Heat gained =
 
J .'
 
+ m- c- /),T = m- c· (T - 1';) = (99.53 g)(4.l8-)(Tf - 295.75 K) :: 416.03T -123,042.470f g·K	 f 
Equate the two quantities and solve for Tf 
-2.052Tf +765.107= 416.03Tf -123,042.470 
. " 
123,807.577 = 418,082T/ Tr 296.13 K = 23.0°C.
 
- T- 29613K .
 
(c) M::mcln-L=(5.33g)(0.385J/g·K)ln .' =-0.473JIK=entropylossofCu 
~ 7;-'	 372.85 K 
I 
(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=--e-to-tal-=--entropy change for the system is M =-0.473 + 0.534 =0.061 JIK. 
(f) Because the over~n.entropychange of the isolated system is positive, so we would expect 
that the process - ~ equalization oftemperatures - would be spontaneous. -,". 
~~	 . . 
3.27. Since (for an isolated system) the first law of thermodynamics prohi~ts the crJalk,n of 
~energy,the concept of'vyou can't win" maybe used to cW\veXY- ifili.a~curately-that fact. 
Since the second law ofthermodynamics requires an efficiency ofless than 100%, yon will 
always get less energy out ofa 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. IfS = 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 ofa system is zero. However, numbers greater than 1 have positive 
logarithms, and if k-is a positive co~~~(which it is), then the absolute amount ofentropy S can 
never be negative. (This does not preclude that changes in S might have negative values.) 
12 
l.Q; l-{~,\ 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. ~~indeed a liquid at absolute zero, it would not have ~ropy. That's 
because according to the third law oftbennodvnamics, only a perfect ryst (that is, a solid) can 
have zero elW:opy at absolute zero 
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) ~ Ah03 (s) +' 2 Fe (s) 
!lS = [50.92 + 2 27.3)] - [2(28.30) + 87. 
.~ 0
'i39 For the formation ofH20 (1), L\S= 9~~ ...-uQ,6~ (205.14)= -163.29 JIK. For the 
ation ofH20 (g), !lS = 188.83 - 130.. - 1 2 (205. ~ -44.42 JIK. The difference is that 
e formation of the gas has a /).S that is hi y 118.87 JIK. The reason for this difference is 
the different phase ofthe product H20. 
3.41.	 L\S = nCln Tf = (800 Ib)(2200 gll~)(0.45 JIg· K)ln 923 K = 9.09 X 105 JIK 
1'; 293K 
3.43. At 37°C = 310 K, the number ofmoles ofgas is 
n = p V = (1 atm)(1 L) = 0.0393 moles. The change in entropy is then
 
RT (0.08205 L· atm/mol- K)(31 OK) .
 
!lS = -(0.0393mol)(8.314J/mol·K)ln 590mmHg = 0.0827 JIK
 
. 760mmHg
 
--'--' --.....	 ~ 
--.,.,-........<, e I,
 
'f~1-~ J-;
 
.... ~r" __ •. ~. 
-" --",,-"-,•..... 
-: 
~b ,~~,:- -t '2.('21{~)J - ~ (2- S 1,?6 ) + 8'7 I tI- ] 
_ ~- _	 ~~""--.- --. .. ---._----.-...-_.__.'. ~'.'--' .__ .... ..... _. __.--_.._"........__._._--,.
 
........... ~ -.
 
G~ ,q , ( j. ") \~ (1- ) ......---) ~l-.....o ('.P)\---~ +- ;"'1.-" 'l. 
-) I--k..-o t')"') ( ~ e.9§j 
131 'ho .. b ~ lO5\'(~ 
@ 
--~ -nl 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 -r-' 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 
ofconstant Tand V, the second and third terms are zero, so this spontaneity condition simplifies 
to 
(dA)T,v::; O. 
4.7. For internal energy, dU= 0 under conditions ofconstant volume and entropy.
 
For enthalpy,dH = 0 under conditions ofconstant pressure and entropy.
 
For Helmholtz energy, dA = 0 under conditions of constant volume and temperature.
 
4.9. The reaction can do up to ~3 kJ ofwork for every mole ofH20 formed. 
U= 1lH= 0 (for an iso~al process). To determine the pressure at the bottom ofthe 
(vl,llll/tU, if the water pressure--iiicreases by 1 atrn for every to.55 meters ofdepth and the depth is 
30 meters, then the pressure increase is 10,430 mx 1atm = 989 atrn. If this is the pressure 
10)~ 
"inCreaSe, then the pressure at that depth must be ~?"_::'=j.~-fitrn. Therefore: .
 
~_.----......._----------- ·990atIn.
 
w~ d'R~~=--:<'.:"~1~8!!4 Jlmol· K)(~2: K) In '1";. -15;1001 . 
Since thisis areversible J?Jo.ce~~l~s ~qu~~:jls~elL:Fiii!lllY, for!':.S: .
 
.. - i . 990 atrn
 .~;
M ~R In p', =~ mol)(8.314 J/mol· K) In 1atm =+57.3 J/K.. . . 
~~ ~np-~~~ . . ... 
4.13. For NaHC03 (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 ofdata is not given for that phase
 
under that condition. However, since we know that O°C is the normal melting point ofH20
 
under standard pressure, we expect that the value of I1G would be exactly O.
 
14 
/4.1 
\ 0 
-, .~ 
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 ) .dU = dA - aT y dT for our final answer. We don't use the chan~~_f1: wi~!espect to( 
volume. 
4.21. To show that (au) := -p has consistent units on both sides of the equation, let us look at av s . 
.the units distribution through the equation. U has units ofenergy (J), V has units ofL, and p has 
units of atm. Thus, we get .!- =atm, which does not appear to make much sense. But recall that 
L 
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. eo (~( :J)=(~( 8(X; y»)) = (~ (1))= 0 
( ; ( :)) = ( ; ( 8(X; 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) (~(:))=(~( 8(X;;'»)) = (~ (nx"y'-' l) = n'x'-'Y'-' 
1-7" . -r)~" ~i 
o V~ 
.z<. 'S"" h/!-"'''1: 
(;(:))= (;(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'-' 
( ; ( :)) = ( ; ( B(X;:'))}= ( ; (mx--' y' l) = mnx"?'y'-'. Since the two derivatives are
 
equal, they are exact differentials.
 
e) (;(:))
= (;(B(YS:(XY)))) = (; (y' eos(xyl)= 2yeos(xy)- y'xsin(xy). 
~tCoH'i.'i)
( (~( :))= (~ (B(YS:(XY)))) = (~ (sin(xy) +xyeos(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) as T as . 1 dS .
- =T- +V =V(1 +--) All we need to do IS show that a.T = ---. Using the ( Op T Op V ap V dp 
definition of a.:
 
"a = ..!..(av) =..!..(_ as)_ ,where we have used a Maxwell relation as a substitution.
 
V aT p V Op ~ .
 
Multiplying by T: aT = T (-- as) ,which is the desired relationship. Therefore,
 
V Op T
C:l =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 
. 
o demonstrate the cyclic rule ofpartial derivatives, let us evaluate the derivatives listed in 
e 1.11 using the ideal gas law: 
=~(nRT) =nR (av) =~(nRT) = nR 
aT v V aT p "aT p p 
v) a(nRT) nRT .(aap T =Op P =-7 and substitute: 
16 
part of II 
" 
_ap_ ) _ (aVlaT)p _nR_=_ (nRlp) whichreducesto-nR_=+p orpV=nRT,
( et v - (aV lap)T V -(nRTl p2) V T 
YerJ~}g the cyclic rule. V,.l ~ 
-L [: s-)r v ,/ ,Y, 
-:.. ,t (~-;r'/ 
(JP .(av) . Using a Maxwell relati?nship, substitute for the fin,al partial 
as)T JY: ~ r: ~ (f!-) 
I ap T' 01 <l r" <lJ7. '( 
e, and rewrite the denominator as its reciprocal in the numerator (note that the 1 Vs , 
( av) (ajJ) (aT . Inspection ofthe three partial derivatives shows that the three variables aT p av T Op V 
are organize . such a way as to give the cyclic rule ofpartial differentiation, so that the product 
d~deriVativesis--l: +1)~1. 
',')/,// ~-' ',,0- or ' to a Maxwell relationship, (fJp) =_(aT) For an ideal gas, 
iJ'(( (/ as av 
( \, \ (', / V) =
< 
__ P;;-_ 'or a van der ~aals gas: p : 
er fl p nI!J ~ ~ -- b) . fY' ,,\,\ 1=. ~ 
, , Therefore: _(aT) =-__1 [(- 2~n2 J(w -nb)+ (p + an: J]. Here, 
av p '11l{! " V V 
o take the derivative of the two V-containing terms using the chain rule. 
4.35 ..The derivation of a Gibbs-Helmholtz 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: ~(A) =_Uor
,aT T p T2' 
:r(~), =--~U-2 . 
6.37 " =nRTln A, RT i' (0.988'mol)(8.314 J/mol.K)(350 K)ln 25.0 L = "':'967 J . fu 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. We rewrite this as dA = -pdV. Integrating both sides: M = - JpdV . oV T 
17 
4.31. Subs ituting for the definitions ofq and K: 
(IIV) ev 
~ = 
K as T (l/V) ~V 
__________/' 
\ \"," 
"<:<~:.: ~' '-]<v 
V 
2 
nR 
Substituting for p from the ideal gas law: M =- Jn~T dV, which integrates to 
. V
 
M =-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 single-phase, single-component 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 
the 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. bar equals 0.987 atm. Therefore, going from 1.00 atm to 0.987 atm: ~ 
\,,~fiL -#;niriat =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. 
4.47. (a) .1 mol ofH20 (g). (b) 10.0 g ofFe 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 
~etermining which reagent is the limiting reagent: . 
100 g Zn x 1molZn =1.53 mol Zn O.l500Lx 2.25 molHCI = 0.3375mo1HC1 .. 
65.4g . L 
The reaction requires a 1:2 molar ratio ofZn to HC1, but we have a 1:0.221 mole ratio, so HCI 
, will be the~Therefore, according to the definition of~:~ = 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 heme-O, + CO ~~ heme-Cf) + 02 is 
2.3x 1023/9.2x 1018 = 2.5x104, indicating a strong preference by heme for CO - explaining its 
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\Y' 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 
InK= -68,000J/mo1 InK=27.432 K=8.2x101l • (d) Thereaction 
(-8.314J/mol· K)(298.15 K) 
would move toward the direction ofmore products. 
5.13.	 For example, consider the following gas-phase reaction:
 
2 H2 (g) + O2 (g) ~~ 2 H20 (g)
 
2 
Its equilibrium constant expression is K = P~20 • Now, divide all coefficients by 2. We get 
PH2 • P02 
H2 (g) + Y2 O2 (g) ~~ H20 (g) 
19 
PH
For this reaction, the equilibrium constant expression is K = PH20 1/2 • The exponents on all 
2 
• P0
2 
ofthe partial pressures are halfofwhat 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 I1Gequals 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 lnp == -6.6161 P = 1.34 X 10-3 atm or bar for the
 
(8.314)(298.15) .
 
reaction to reverse. 
5.17. Addition ofan 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. I1rxnOO = [-41.9 -73.94] - [2(51.30~ 228.61] = 10.2 kJ. 
10,200 J/mol =-(8.314 J/mol· K)(298.15 K) InK InK =_ 10,200 J/mol 
(8.314 J/mo1· K)(298.l5K) 
lnK=~4.1l48... K= 1.63xl0·2. 
~ Using the 11(? from the previous problem: I1Go = -RTln a dia , and substituting from 9 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 =5.34x 10-3 L for the molar volume of graphite.
2.25 g 1000 em 
3 
12.01g x 1cm x 1L 3 =3.42 x 10-3 L for the molar volume of diamond. Substituting: 
3.51 g 1000cm 
2900Jlmol=-[p.42x10-3 Umol)(p-1)-(5.34x10-3 Umol)(p-1)]1 \+:-H-­
. \ : 
p ~ 14,900 atm. 
..".-_.,. 
5.25. (a) ~Go=-(8.314J/mol·K)(298.15K)1n(1.2x10-Z) ~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.OIO-x. The amounts of sol- and 
W are therefore +x. We therefore set up the following expression: 
-K = 1.2x 10-z = o.O~~x_xand solve for x using the quadratic formula: x = -0.0185 or 0.00649. 
e reject the negative root: x = 0.00649.- Therefore;-the-fmal-cOilce~trations are [:H+] = [SOl-] 
.00649 m and [HS04-] = 0.010 - 0.00649 = 0.00351 m. 
5 .2 . Use equation 5.20 and letK2 = 2KI: 
I' -100,000J/mol( 1 1 J 'ln2- -100,000 J/mol ( 1 1 J S 1 .~ T' . = - - - - - 0 ve lor 2. 
'. K I 8.314 J/mol· K 298 K t; 8.314 J/mol· K 298 K 1; ~ 
. (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 K2 = 
lOKI and perform the same calculations: 
1n lOKI = -100,000J/mo1( 1 -~J' 1n10 = -100,000 J/mol (1 ~J Solve for T2: 
K I 8.314 J/mol· K 298 K t; 8.314J/mol·K 298K t; 
(ln10)(8.314) = _1__~ T = 282 K.
 
-100,000 298 t; 2
 
For a reaction whose Mi = -20 kJ, repeat both sets ofcalculations using this value of Mi. 
Letting K2 = 2KI : 
1n 2KI - - 20,000 J/mol ( 1 - ~J 1n2 - -20,000J/mol( 1 1 J S(jlv~ for T2: 
K I 8,.314J/mol·K 298K t; - 8.314 J/mol· K 298K - Tz 
(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: 
lOKI - 20,000J/mol (1 1 J In10 = - 20,000 J/mol (1 _1J Solve for T2: . 
n~= 8.314J/mol.K 298K - Tz 8.314J/mol·K 298K T2 
21 
I 
(ln10)(8.314) 1 1 
-'--~-'-- =- - - T2 = 232 K. 
- 20,000 298 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:
 
dInK =_(_1)(_ Ml)dIn; =_Ml and bring the -(l/r) term to the other side: 
-(lIT )dT R . .. dT T 2 R 
dInK MI hi h i '5 18or .dT = RT 2 ,w IC IS equation . . 
5.31. If 1.0 mol of glycine is made into 1.00 L ofsolution, let us assume that the original pH is 
7.00 and calculate the amount ofprotonated glycine is in the solution. According to equation 
5.21: 
34 K =[gly][H+] =10-2. • If [gly] = 1.00 M and [W] = 1.00xlO-7 M, we can calculate the
 
1 .. [glyH+] .
 
. f[ I +] -234 (1.00)(1.00 x 10-
7 
)concentration 0 g yH : 10 . = -'----"-~-_--:..
 
. . . ,_ [glyH+]
 
Similarly, according to equation 5.21: 
60 •..K =[gly-][H+] =10-9. Again, assuming [W] == 1.00xlO-7 M, we can calculate the resulting 
2 . [gIy] . 
[gly-]: 1O~9.60 == [gly-](1.00xlO-
7 
) [gly'] == 2.51xlO-3 M. 
(1.00) 
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 only-chemically stable, single-component materials that can be
 
made from iron aiiOCh~ .>:
 
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:e on the w~ter sUf~siently 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~ve because e!!~!.ID'..p.as.l0 go.into a solid in order to sublime it. (b)
 
Mf sh~ld be n~e 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.
 
'. AC' Q'+30,700J IK fairl 1 h ldb . edb6.13. LW = ~ =' = +86.9 J , Ian y c ose to w at wou e predict y
T (80.1+273.15)K . .
 
Trouton's rule.
 
q !!.	 H 
6.15. Using the defmition ofentropy for thisisothennal change: 6S = ~ = fw and 
T T 
tin' 1247JI 1 K 510,400J/molb t· su SItU	 g: . mo· =----- T = 4093 K or 3820°C. 
T 
6.17. The derivation ofequation 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 = 123.9cnr' 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
 
-0.0070 L. Using the Clapeyron equation:	 - ~ -= ~ r- = x-- ­
!!.T ·!!.V ~.5 K - 0.0070 L 101.32 J/,
 
23	 /
j<-: ~ 
I~ -. . 
/ 7!.!(~y,·,« j\·t~ 
1~1C=;::fi3-atm.--Th@f-efuI:e,-~he pressure-b¥_abou~tm, should.be 
en to make rhombic sulfur the stable fonn at 1OOo~ 
. ~.-')~ .. ~~~ 
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 monoclinic-to-rhombic phase change must have a Mf of -0.368 kJ/mol: 
!¥J = (Mf)ln Tf = ( -368 J/mol )In (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 ofchemical hot packs cannot be described using the Clapeyron equation 
or the Clausius-Clapeyron equation because supersaturated solutions are not equilibrium 
systems, nor does the process involve a phase change (it involves a solubility change). 
6.27. Ifwe use dp = Mf ·t as the form ofthe Clausius-Clapeyron equation, we can substitute 
. ~RT . 
.directly the values given in the problem: 
.dp = (71,400 J/mol) . (7.9 x 10-
5 
bar) = 7.8 x 10-6 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 ofenergy, and dA has units of'rrr', r must have 
units ofJ/m2 • (b) The derivative ofA, in terms of r; is dA = 87trdr and the derivative of V in 
terms of r is dV=41tr2 dr. Substituting into the right side of the equation: 
2 
2dV 2(4nr dr) 8 d d. A () B . 1 . th .--= = nr r = ri , C y sunp y rearrangmg e expression, we can get 
r r 
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 ofdG. 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!) =-8, the units of I! are J/mol and the unit on Tis K, so the 
aT p,n 
overall unit on the left side is Jzmol-K, 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, ifwe remember that the unit J can be written in terms ofLatm, 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 ofmolar volume. 
6.35, Figure 6.6 shows 12 phase boundaries. 
6.37. If there were such a thing as a single-axis phase diagram, equation 6. probably be 
rewritten as 'degrees of freedom = 2 - P', because there would be 0 1 s variable n eded to 
determine the exact state of the system. 6\:0 \"~ 
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 7 7 .. For water: n =- = x =. x rno 0 
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 sincep/ = 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 = • X2~2 • , which is the equation of interest. 
Pt +(pz - Pt )x2 
7.13. p(ethanol) = (0.0006)(115.5 torr) = 0.0693 torr = 0.07 torr (1 sig fig) 
• 
7.15. Equation 7.19 is Y. = • XI~' • • First, let us multiply through the XI in the 
P2 +(Pl - pz )xt . 
denominator: Y1 = • XtP~ •. Now, multiply the denominator over to the other side 
P2 +X,Pt - xtlh 
26 
\ 
• • 
• • 
• • • • 
of the equation: Yt (P2* + XtPt* - X1P2 .)= XtPt*, or YtP2* + YtXtPt' - YtXtP2 * = 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): 
• 
X t = • YtP'2 •. Now factor out the YIS from the last two terms in the denominator: 
Pt - YtPt + YtP2 . 
• 
Xl = • Yt~2 • , which is our ultimate expression. 
P, + (P2 - Pt )Yt . 
7.17. Consider the equation P tot = • P2 .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 
becomes • P2 ~t • , which becomes P2 ~t = Pt· , as it should when the system contains 
Pt + P2 - Pt P2 
only component #1. 
7.19. ~mixG = (2 mol)(8.314 J/mol·K)(293.15 K)[(0.5)(ln 0.5) + (0.5)(1n 0.5)] = -3380 J
 
~mixS = -(2 mol)(8.314 Jzmol-K) [(0.5)(ln 0.5) + (0.5)(1n 0.5)] = +11.5 JIK.
 
7.21. If an initial composition ofXl = 0.1 were used, the tie lines that you would draw would
 
eventually lead you to the.minimum-boiling azeotrope, near the middle of the phase diagram.'
 
Thus, the azeotrope will be your ultimate product.
 
7.23. 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 hasbeen
 
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 of4.17xlO-s implies that there are 4.17xlO-s 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 ofwater is
 
27 
3 
. I 18.01g 1.00em 1 3 c: he molari f thi0.9999583 mo x x = 8.0 em , or 0.0180 L. Therefore t e mo anty 0 t IS
 
1mol 1.00 g .
 
smol 
· . 4.17xlO- 000232M T d .. h H ' I 1soIution IS = . . 0 etennme t e enry s aw constant, use atm = 
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 ofair 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 ofnitrogen. (b) If air is 20% oxygen and the mole fraction 
of air in water is 1.388x10.5, let us assume that oxygen would be 20% ofthat 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) 
=2.78x10·6• (c) If oxygen is 20% of 1 atm = 101,325 Pa, then the partial pressure ofoxygen is 
(0.20)(101,325 Pa) = 20,265 Pa. Therefore: 20,265 Pa = Kjx(2.78x10-6) x, =7.29xl09 Pa. 
This is the same answer we got in exercise 7.30, and it should be - all we did was multiply both 
numerical values by 0.20, so the calculated value ofIG. should be the same. However, Table 7.1 
shows that the actual value ofK, 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 ofphenol 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 1mL = 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 ofnaphthaiene in n-decane 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 128.0gCIOH g I mol 0.730g 39.8 g naphthalene 100/134 29.7g---x x x =	 x =----'' ­
0.689 mol mol 142.0gdecane 1mL 134mLdecane 100/134 100mL 
In terms ofmolarity, 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 10-3 ) = - 14,900 J/mol ( I _4.828314... = -1792( 1 T_1-)
8.314J/mol·K 298.15K T 298.15 
0.002694 = (0.003354- l/T) l/T= 0.006600.... T= 1515 K = 1242°C. 
7.41. We can tell by looking.at a phase diagram of the NaCI-H20 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 freezing-point depression. 
7.43. The drawing is left to the student. 
7.45.	 lux!t =- 2,600J/mol ( 1 - 1 J Inx=-0.30184.: 
soue 8.314 J/mol·K 273.15 K . (97.8 + 273.15) K 
x=0.739. 
7.47. Molarity includes the concept ofpartialmolar volume because it is defined using the 
numbenof liters ofsolution. Thus, even ifpartial molar volumes may change during a range of 
compositions, only the total volume is used to define molarity.' 
7.49. First, let us calculate Kf and Ks: 
K ::: MsolvRTM/ = (18.02 g/mol)(8.314J/mol· K)(273.15 K)2 = 1.86 Klmola1
 
f 1000AfusH (1000 glkg)(6009 J/mo1) ,
 
29 
K; = MsolvRTB/ = (18.02 glmol)(8.314 J/mol· K)(373.15 K)2 = 0.513 Klmolal 
1000Ll vapH (1000 glkg)(40,660 J/mo!) 
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 ofwater (assuming negligible volume change from the solute), 
that's 55.5 moles of H20 . 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 ofwater, or 0.01802 L. 
Using the van't Hoffequation, 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/ = (159.8g1mol)(8.314llmol·K)(273.15-7.2 K)2 = 8.89KJmolal
 
. / 1000LlfusH (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 
1000,6vapH (1000 glkg)(29,560 Jzmol) . 
n Xi n	 30 Pa 1bar
7.55.	 m =-.-~ =_. m = x-"------­
#kg L water RT (0.08314L· bar/mol·K)(310K) 100,000Pa 
m = 1.16x10-5molal. For 1 kg of solvent (water); 
. -5 185000 g
1.16xl0	 molalxlkgx ' =2.15gpolymer.
mol 
30
 
CHAPTER 8. ELECTROCHEMISTRY AND IONIC SOLUTIONS 
0.0225N= q1(1.00C) ql=2.50x10-8C 
4n-(8.854 xl 0.12 C2/J .m)(1 00.0 m)" 
2q:
8.3. (a)	 F= qtq2 1.55xlO-6N= 1 Solveforq:
41l"&or2 4n-(78)(8.854 xl O' 2 C IJ· m)(0.06075 m)? 
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 xl0-: N = 311 J/C .m , while the electric 
4.98x10' C 
field e oth . I . 1.55x10-{;N Ie on the 0	 er partie e IS 9 = 156 J C· m. ' 
9.96x10" C 
19	 19 
8.5. F = (+1.602 xl0- C)(-1.602 xl0. C) ,F = .8.24 X 10-8 N. This may not be 
4n-(8.854x10·12 C2/J .m)(5.29 x 10.11 m)" 
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 ofJ/C and not newtons. ' 
9Thetwo halfreactions are: 
Mn02 + 2 H20 -7 Mn04" + 4 H+ + 3e" E = -1.679 V . 
4 e' + O2 + 2 H20 -7 4 OH" 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 H20 and consolidating the water molecules on both sides, we 
get 
4 Mn02 + 3 02 + 2 H20 -7 4 Mn04' + 4 W_ E= -1.278 V 
To calculate LSG': --­
AGO = -nFE =-(12 mol)(96,485 C/mol)(-1278 V) = +1.480 x 106 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': 
(
 
Dfj,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:
 
Br2 + 2 e' -7 2 Br E= 1.087 V
 
2 P" -7 P2 + 2 e" E= -2.866 V
 
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: <;j
. HzOz + 2 W + 2 e- ~ 2 HzO . E = 1.776 V
 
2Cr~Clz+2e- E=-1.358V
 
Because both half reactions have two electrons, they can be combined directly without
 
multiplying through either reaction. The overall reaction is
 
2 HzOz + 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 104 J =
 
8.11. The two half reactions are:
 
Fez+ -7 Fe3+ +' e- E = -0.771 V
 
Fez+ + 2 e- ~ Fe E=-0,447 V
 
To balance the electrons, we multiply the first reaction by 2. For the overall reaction, we get
 
. 3Fez+~2Fe3++Fe 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) =2.350x10s 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 thevoltage 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 half-reactions typically occur in an.aqueous solvent, the interactions 
between the ionic species. and the solvent molecules has an impact on the overall energy change 
(in terms ofE and tJ.G) ofthe process. Although all alkali metal ions have a +1 charge,the 
.smaller, higher-charge-density lithium ion interacts more strongly with water molecules,
 
increasing the energy of the process. . .
 
8.17.	 E = EO - RT In [Zn2+] EO for this reaction is 1.1037 V, so we have
 
nF [Cu 2+]
 
32 
(8.314JIK)(298.l5K) I [Zn 2+] Th' 
01oOOOV = 11037V -	 n IS rearranges to 
(2 mo1)(96,485 C/mol) [Cu 2+] 
[Zn2+]	 [Zn2+] 
In 2 =8.07279... 2 =3.21x103 • Unfortunately, we can't mathematically
[Cu "l . [Cu +] ..
 
determine specific concentrations without more information; we can onlyspecify the ratio. 
8(a) 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 =0.0375 V. (d) The
 
. [Fe3+] 0.08 (3 mol)(96,485 C/mol) 0.08 . .
 
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) 144.96­
130.6S = 14.28 JIK. The number of electrons transferred is 2 so we have:
 
o~ -0.044 V +( 14.28 JIK )I1T Solving for I1T: I1T= 595 K. Therefore, ifwe
 
(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. . 
8.23. Since heat capacity (at constant pressure) is defined as (OH) ,we can take the derivative oT	 . . 
p	 '. 
..	 (O(MI)) (OEO [OEO 0 2 ED]Jof equation 8.30 WIth respect to temperature: =-nF - + - + T-.-.. -t .' 
.	 oT oT. oToT .p 
oE D o2E O] 
which simplifies to I1Cp =-nF 2-+T--2 •( .oT oT 
8.25.	 E = 0 V - (8.314JIK)(298.15 K) In 0:0077 =0.0194 V
 
(2 mol)(96;485 C/mol) 0.035
 
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 Solve for K : K = 1.74xl0·10 •
 
(l mol)(96,485 C/mol) sp sP sP
 
33 
8.29. First, we need to determine the overall reaction. Using Table 8.2, fmding the Mn04-lMn2+ 
half reaction, and combining it with the hydrogen electrode reaction, we find a 1a-electron 
overall reaction: 
2 Mn04- + 6 H+ + 5 H2 ~ 2 Mn2+ + 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 
7.88x1051 = (0.288)2 [If] = 4.6x10-9 Therefore, the pH = -log (4.6xlO·9) ::= 8.34. 
(0.034)2[H+ r (1)5 
8.31. TheKsp for Hg2Ch was determined in exercise 8.28 and is 1.29xlO·18 . If'x moles per liter
 
ofHg2Ch dissociates, one gets x M Hgzz + and 2x M cr. Therefore, we have:
 
1.29xlO-18 ::= (x)(2xi 1.29xlO-18 = 4x3 X = 6.86xl0·7 Since the equilibrium concentration
 
of'Cl' is twice this, [Cn = 1.38xlO-6 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 )= 0.0750m 
2 
(d) 1= ~ (0.0250m)(+3)2 +(0.0750m)(-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 ofproducts-minus-reactants to determine Ml, we need the 
heats of formation of the products (one ofwhich 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 non-zero I:1rH is that for r. Therefore, we 
have 
-11O.38kJ = (2 mol)(l:1rH[f]) I:1rH [f] =-55.19 kl/mol. 
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 3.5xlO-4 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 Ais worked out in the text. The studentneed simply verify 
that the numbers and units do reduce to 1.171 molarll2 . 
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) =0.154 molal. Therefore, the ionic
 
. 0.100kg
 
strength is 1= ~ (0.154 molal)(+1)2 + (0.154 molal)(-1)2)= 0.15~ 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:.:t912~. .
 
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 (Fe3 + 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 molafl12)(+3)2In ) =_ . (0.750 molalyl2 
". YFe • 1+ (2.32 X 109 m "molal'!" )(9.00 X 10-10 m)(O.750 molal)'? 
InyFe). == -3.250 Y (Fe3) = 0.0388 Therefore, the activity ofFe3 + is (0.0388)(0.100 m) 
=0.00388 m. 
Similarly, for the Cu2+: 
l12)(+2)z'(0.200
lny l. =_ (l.i71 molal molalyl2
 
Cu 1+ (2.32 X 109 m "molal?" )(6.00 x 10-10 m)(0.200 molalj'?
 
Inycul • = -1.291. Y (Cu
l 
) = 0,275 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 =0.379 + 0.0075 =0.386 V
 
(6 mol)(95,485 C/mol) (0.0138)3 . .
 
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. 
8.45. Equation 8.61 is I = e2 ·1 z 12 .( N j J.A .~. Equation 8.5 shows that E has units of 
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 ofkg/m-s. 
The numbers 6 and 1t have no units. The radius r has units ofm. Combining all of these units: 
35 
1 ) 2 N/C. C
2 -rn' -Nvm-s
C2 • - • m • This can be rearranged to get . One of the C ( m' (kg/m-sjun) C·m3 -rn-kg 
units cancels; as do the three m units in the numerator and three of the four munits in the . 
denominator. This results in C· N -s . Ifwe break down the newton unit into kg-m/s', we have 
kg-rn 
C ~ kg . m . s . The kg and m units cancel, as do one of the second units in the numerator and
 
s -kg-rn
 
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
 
:;o.",--, •• 
:-.J:"'.'.: 
li;.c. 
CHAPTER 9. PRE-QUANTUM 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 
d (aL J oL d . h Ii 11 . deri .- -- =-, we etermme teo owmg envatives: 
dt oi az 
(aL) 0 (1 . ) •ai. =oi 2 mz 2 - mgz =mz
 
!!-(OL) =!!-(mi) =mz
 
dt Oi dt 
2: =~ (~ mi - mgz) =-mg 
Combining the last two lines, we have for the Lagrangian equation ofmotion: 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 . 
.(OH) 1 (OH) 1 0(1 ;2 ) 1 ( . 0)' . d Th deri . f th--. = - -. =--.. -mz + mgz =- mz + =z, as reqUIre. e envative 0 e 
Ornz m Oi mOi 2 m .
 
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::) = ~ (~ mi2 + 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? = 11 00 2 = O. For the Lyman series: 
wavenumber =109,700 cm-tl~ - 0)= 109,700 cm·i is the series limit. For the Brackett series: 
wavenumber ~ 109,700 em-t;, -0) ~ 6856 cm' is the series limit. 
37 
9.11. First, we need to convert the wavelengths to wavenumbers:
 
1m 100 em -5 1 1
656.2nrnx 9 x =6.562xlO em -== 5 = 15,240cm'l 
10 run 1m A 6.562xlO- em 
486.1nmx 1m x100cm=4.861xlO-5cm .!..= 1 =20,S70cm·1
 
109 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" ~ R(;, -3 15,240 cm' = R(0.1388888. __l R= 109,730cm-' 
20,570cm·1 =R(~-~) 20,570cm·1 =R(0.1875) R =109,710cm·1 
.' .. 2 4 
23 040cm·1 = R(~ __I) 23,040cm·1 = R(0.21) R == 109,71Ocm·1 , 22 52 
. 109,730+109,710+109,710 109720 -IThe 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~1.612xI0·27+ 
~}..9'i~~1Q.'27) kg =Q ~~. The ratio of these two masses will tell us how-~~ybeta
 
27
 
. le . ak I h f 1 l' h . I 6.694 x l 0- :::: 7349 b eta partie es to I . partie es It t es to equa t e mass 0 a p apartie e: -31 . 
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. 
2 W -8 W 4) •9.15. For a flux of 1.00 W/m: 1.00-2 = (elj~Qs.~_~~~:'~1{4 )(T Solvmgfor T:T = 65 K . mm 
. 2 W W . 
Forafluxof10.00W/m: 10.00-2 = (5.6705xl0-
s 2 4)(T 4) SolvingforT:T=115K 
m mK 
2 W' -SW)4Fora fluxof 100.00W/m: 100.00-'-2 ::::(5.670SxlO 24 (T) Solving for T :T == 205K 
m mK 
9.17. (a) power per unit area == (5.6705 x 10.8 W/m2K4)(5800K)4 == 6.42 x 107 W/m2 
2'(b) 6.42xl07W/m2 x 6.087xl012 m = 3.91x1020 W 
38 
365 d 24 hr 3600 s 
(c) 3.91x102oW=3.91xl02o J/s x --x--x =1.23xl028 Jperyear.
1year 1d 1hr 
9.19. (a) Using Wien's law: Ivmax·5800 K =-,89§ !JIW:~ Ivmax = 0.4997 urn = 4997A. 
(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~C2 (ehC/L-1 )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 2 
dE - 2wc ( )dA - 21ihc ( 1 )(- .s,d v) This expression simplifies into 
- AS e /»:J-1 - (clv)s ehc/(c/v)kT -1 v 2 
dE = - 2~ v, (ehV/~T _1)dV. Ifwe now substitute x =h v I kT so that dx = hi kT .dv , or 
dv = kT h- dx and v = xkTIh, we substitute again: 
4 3w(xkT I h)3 ( 1 ) kT·. 21leT x 
+--:-"'2--'-- -- -dx. Collecting terms: dE= 3 2 '--. Now we can 
. c eX -1 h h c e' - 1 
ate from x = 0 to x = (you should verify that when v = 0, x = 0, and when v = , x = : 
3 
co 2JTk4T4 x3 2JTk4T4 co x 21ik4T4 1f4 21fseT4 
l dx=JdE=J-: h3c2 'ex_ldx= h3c2 J x _ = h3c2· '15=- 15h3c2 which is eo 0 
the Stefan-Boltzmann law. (The negative sign indicates that energy is being givenoff.) 
9.23. Substituting the values of the various constants: 
s 23 4
 
21f (1.;81 x ~ 0- JIK) 34 3 5.6-7'6:'i;;;;;;;)which is almost exactly the value
 
15(2.9979xl0 mls) (6.626xl0- Js) ... .. ""..-"'" ..
 
ofthe Stefan-BQltzIllmm£2.lJ,§,~tierted in the text. 
19 
functi 16' I 16 V 1.602x10· J 3·46 1 -19 J9.25. A work ction of2. eV IS equa to 2. e x =. xO . 
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 ofmotion, ~ mv". 
8
 
550nrnx 1m =5.50xl0-7 m v == 2.9979xl0 mls = 5.45xl014 S·I
 
(a) 109 run 5.50xlO·7 m 
E=(6.626x 0-34 Js)(5.45xl014 S · 1) 
= 3.61xl0-19 
If3.46x 10' J are used in overcoming the work function, there is 0.15xl0'19 J l~:ft. Using the 
formula for kinetic energy and the mass of the electron: 
39 
B 
v = 2.9979x10 mls = 6.66x1014 S·I450 nm x 19m =4.50x10-7m 
(b) 10 run	 4.50xlO:7 m 
IE = (6.626 X10-34 Js)(6.66x1014S· ) 
E= 4.41x10-19 J.
 
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:
 
0.95x10-19 J =.!-(9.109X 10-31kg)v2 v=457,000mls.
2	 .
 
8
 
. 1m -'350 10- 7 _2.9979x10 rn/s_857 1014-1(c) 350 nm x 9 -. x m v - 7 -. x s
 
10 nm 3.50xlO' m
 
E = (6.626 x 10-34 Js)(8.57 x1014s·1) 
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 10-19 J =.!-(9.109x 10-31 kgjv" V = 698,000mls.
2
 
9.27. By combining the equations E = hv and e = AV, we get an expression that can be used 
..directly: E = he . For the wavelength of 10m: 
A 
8
(6.626xl0-34 j .s)(2.9979xI0 mls) 199 10- 26 J h . 602·2 10231 I 00120JI I hE =	 =. x per p oton, x. x mo == . mo p otons 
10m
 
For a wavelength of 10.0 em = 0.100 m:
 
8rn/s)
(6.626xl0-34 J.s)(2.9979xl0 199 10- 24J h 6022 1023/ I 120JI I h .E =	 ==. x per p oton, x o. x mo =. mo p otons 
0.100m
 
For a wavelength of 10 microns = 0.00001 m:
 
34 8m1s)
(6.626 x 10- J.s)(2.9979xl0 199 10- 20J h 6022 10231 1 12000JI 1 hE =	 =. x perp oton,x. x mo = mo p otons 
O.OOOOlm
 
For a wavelength of 550 DIn = 5.50x 10'7 m:
 
-~ 8 .
 
(6.626xl0 J·s)(2.9979xl0 mls) 361 10-19 J h 6022 10231 1 217500JI 1 h tE =	 7 == . x per p oton, x. x mo == , mo p 0 ons 
5.50xl0' m
 
For a wavelength of300 DIn = 3.00x10'7 m:
 
34 8
 
(6.626 x 10- J .s){2.9979xl0 mls) 662 10-19 J h 6022 10231 J 398700JI 1 h E ==	 7 ec • x per p oton, x. x mo == , mo p otons 
3.00x'10· m .
 
For a wavelength ofl A= 10-10 m:
 
0-34 8
(6
E - .626xl J·s)(2.9979xl0 rn/s);"199 10- 15J h 602210231 l-12020'JI I h - 10.10 m -. x perp oton,x. x mo -. x mo p otons 
9.29. Using equation 9.34, for n =4: 
40 
r= son
2
h
2 
=,i,8.854x10-I2g;kUn)(42)(~<i~~~s)2 =8.47xl0-IOm=8.47 A 
1lmee
2 
Jr(9~g)n~~)2	 . 
For n =	 5: 
2 2 12 2IJm)(52)(6.626xlO-34 
r = son	 h = (8.854xl0- C J 'S)2 =1.32x10-9 m=13.2A 
2mnee Jr(9.109x10-
31kg)(1.602x10·19 C)2 
For n =	 6: 
2 12 2IJm)(62)(6.626xlO-34
r = son	 h
2 
= (8.854x 10- C J 'S)2 = 1.91x10-9 m=19.1A 
2mn 
ee 
Jr(~.109x10-31 kg)(1.602x10·19 C)2 
9.31. Angular momenta are simply integral units ofh/2n: for n = 4,
 
L = 4·(6.626x 10-34 J.s)l2n = 4.22x10.34 J.s.
 
For n = 5: L = 5·(6;626x10.34 J·s)l2n = 5.27xlO-34 J·s.
 
For n = 6: L = 6.(6.626xlO'34 J.s)l2n = 6.33xlO·34 Jvs.
 
2 
9.33. (a)	 v = 
e
(b) For n = 1,
2sonh 
. -19 2 
(1.602 x 10 C)	 1 106 mI Thi . b 0 f th dv = 12 2	 34 = 2. 8 x S IS IS a out .73% 0 e spee 
2(8.854x10- C /Jm)(1)(6.626xl0' Js) 
oflight. (c) L = mvr = 9.109xl0·31kg· 2.19x106 m/s- 5.29xlO·11 m = 1.06x10.34 Ls, which is 
equal to h/2n. 
9.35. Starting with n'A = 27tr, substitute for wavelength using de Broglie's relation A = ~: 
mv 
n ~ = 21CY , which rearranges to nh = Znmvr . Finally, dividing both sides by 2n yields 
mv 
mvr =;; ,which is Bohr's postulate that the angular momentum, mvr, is a quantized multiple of 
sn« 
9.37.	 A = ~ = 1.00 x 10':'10 m = (6.626 x 1~:34 Js) v = 7.27 X 106 mls for an electron, 
mv (9.109x10' kg)(v) 
A=~=1.00x10-IOm= (6.626x1~7-34 Js) v=3.96x103 mlSfOraA.. 
mv ~ ~ 
41
 
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 single-valued. 
(h) No, the function is not acceptable because it is not continuous. (g) No, the function is not
 
acceptable because it is not single-valued.
 
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 - 7 - ~ . 
x 
10.7. (a) Px(4,5,6)=(-4,5,6) (b) PyPz(O,-4,-l)=(O,4,l) 
.(c) Px~(5,0,0) =(5,0,0) (the two operations cancel each other out) 
(d) Py~(n,n /2,0) =(-n~-n /2,0) 
(e) Yes, PXPy should equal PyP for any set ofcoordinates, because the x coordinate and the yX 
coordinate are independent ofeach other. . , 
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 ofclassical mechanics, K= 1/2. mv", or in terms oflinear momentum p =mv, 
K = p2 . Substituting for the definition of the linear momentum operato~~
 
2m ~
 
2 2 2 
-2 8 .K
A
=-1 ( -In'f:. -8)2 =-1 ( - rt*-2 8 ) =---21i ,whiICh'IS the one-d'tmenSlOnaI kinetic. energy
2m ox 2m ax 2m ax . 
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 
34 
&(1.12 x 10.30 kg. 10,000 m J:2: 6.626 X 10- J . S &:2: 4.7lx10-9 m. 
'S 4Jr 
10.17. First, we need to convert the 1.0Q em'! to energy; 
8 
1 ( .1 m J= 0.01 m ="t, v = ~ = 3.00x 10 mls = 3.00x 1010 S-l 
1.00cm' 100,cm' A. 0.01m 
Now calculating the equivalent M;; M; = h v =(6.626 X 19-34 J . s)(3.00 X 1010s·1 ) 
M; = 1.99x10.23 J Calculating the t:.t from the expression