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Unformatted text preview: r‘l Li” Chapter 8 Thermodynamics of Nonelectrolytic Solutions 291 iturated The osmotic pressure for an ideal solution is given by Eq. 8.93: XB _ I mem (ton) 
~ highlv . ., _. ‘ ' 7 l . e 0.08200 L atm mol K l 300 K)(—0.0013014)
lesatits 7, W Rf “ X’* a l (0” L l] I) r 0.000 0.000
r .‘ _. r . mo
1:1?“ i l Fm 0.050 1.000
“d” = 9837Lat12 ‘2) 0.100 2.200
in Eq. 2, in“) is the partial molar volume of pure solvent. Since 5 moles occu 0.150 3.500
I 0.2 L, l/RU) —’ (0.25:5) L mol l. W
t air 0.250 7.000
 s
a satu' ' 3,23 Solute A is dissolved in solvent B at 320 K. The table at the right gives the mea 0300 8500
k» sured equilibrium partial pressures of B over the solution as a function of its 0.350 10.000
mole fraction in the solution, X 5. Is the solution ideal? Plot Para.01) V5. X E at 320 K. 0400 13.000
and on the same graph plot the idealsolution result. What can we conclude O 450 15 200
mime: about the nature of the A—B intermolecular forces relative to the pure compo ' ‘
nent intermolecular forces? 0.500 18.600
S I t. 0.550 21.800
0 U IOTT
_ _ 0.600 25.500
The accompanying figure shows a plot of PM”. vs. X E and the correspondim:l 0 650 28 900
(1) ideal—SUlution result where the table shows that 17‘}; : 51.858 torr at 320 K. ' '
Examination of the plot shows the solution to be nonideal. The negative devia— 0.700 33.300
tions from Raoult’s law suggest that the 24—8 interactions are stronger than 0.750 36.900
those between the pure components. 0 800 40 600
(2) ' '
Vapor Pressure vs. X5 0.850 43.300
iEq. 2, _ l _' 7 _" T " .l 0.900 46.100
.‘0 7
l ,5 — “08' 1 0.950 49.240
" ° N‘mldeal i 1 000 51 858
(3) 4o ; f ’ '
l' ——__——_—
i 1
E 30 —:
(4) E r. j
3 __
0 I: j
I 1
10 i —
(5) i ]
E _
o — ‘ 3
1.0
3 total
Em is ......._.......___—
' solu ‘
501“ .\ 8:24’“ The equilibrium partial vapor pressures of compound B at 320 K above a binary
‘ " solution of compound A in solvent B are measured and found to be those listed in
the table in Problem 8.23. Assume that these measured pressures can be repre
sented by the expression
‘pI‘L‘H r=~l
PBtsolni : [Tl where P}; is the pure component equilibrium vapor pressure over liquid B at
320 K, If the solution were ideal, we would have C! : l and C2 : C3 = C4 I 0.
For the measured data, determine the best values of the C, ti = 1, 2, 3, 4') by
using a linear least—squares fitting procedure. Prepare a plot showing the cal
culated fit and the data points on the same graph.,Use the analytical fit to the (1) vapor pressure data to determine the Henry's law constant for component B. :on—
ons,
this
the
' the it is
1)wl
sing (2) . L1.kuu.u uuLt — b u.  “muma..  uu _v b u. . 13.: u. u .LuntuJuteLu. u; 1.4;. .“,. (A) How many phases are present? Are any of the solutions saturated? How
do you know? (B) The chemist lowers the temperature of the mixture to 373.15 K. How
many phases are now present? If there are two phases, what masses of H20
and nbutanol are present in each phase? If there is only one phase, at what tem
perature will two phases appear. (See the phase diagram shown in Figure 8.21.
The calculations need be done only as accurately as permitted by this diagram.) Solution (A) Figure 8.21 shows that a 20%—bymass mixture of nbutanol in 80% H20 at
410 K corresponds to a point that lies above the solubility curve of nbutanol in
water. Therefore, only a single phase is present at 410 K. The solution is not sat—
urated. If it were, the point would have to lie on the solubility curve shown in
the figure, and it does not. (B) When the temperature of the mixture is lowered to 373.15 K, the point
now lies below the solubility curve in the figure. Therefore, we are in the two
phase region, where we have a water solution saturated with nbutanol and
an nbutanol solution saturated with water. The tie line at 373.15 K in the phase
diagram intersects the solubility curve at approximately 9% nwb'utanol by mass
in the water solution and 33% water in the nbutanol solution. Let to, be the
amount of water present in the watersaturated nbutanol solution, and let w2 be the mass of nbutanol in this solution. The phase diagram tells us that we
must have w] — = —. 1
w; 67 ( ) By mass balance, the masses of water and nbutanol in the nbutanol—saturated
water solution must be mwm, = 80 H w, and mkbmml = 20 — :02. The phase
diagram shows that, in the water solution, we have 80  w
20 _ w: = 291. (2)
Using Eq. 1, we get
33w2
:01 = 67 . (3)
From Eq 2, by cross multiplication, we obtain
9(80 — wl) = 91(20 — w,). (4)
Substituting Eq. 3 into Eq. 4 produces
9[so — 3:02] = 91(20  102) = 1,320 — 91:02. (5)
Rearranging terms in Eq. 5 yields
—4.433 w: + 91302 = 86.57102 = 1,820 — 720 = 1,100. (6)
Solving for wz, we obtain
1,100
w2 = 8657 g = %0f 11 — butanol. (7)
Using this result in Eq. 3 yields
(33)(12.71)
w1=—E7—g=EZigofHZO. (8) Thus, the watersaturated nbutanol solution contains 12.71 g of nbutanol and
6.26 g of water, so that its total mass is 18.97 g, which gives 67% by mass of
butanol and 33% by mass of water, as required. Mass balance now tells us that
in the nbutanol—saturated water solution, we have 73.74 g of water and 7.29 g
of nbutanol, for a total mass of 81.03 g. This solution is, therefore,
7374/8103 X 100 = 91% water, as indicated by the phase diagram. m— C 3 3"” Chapter 8 Thermodynamics of Nonelectrolytic Solutions 307
‘J ' .j “‘3” Solution n I ' .  . . .
ml The phase diagram shows that all points along a horizontal line at 935 K he in n the the sin glephase region, where we. have a homogeneous liquid solution.
Consequently, the chemist will observe no phase changes in this solution, no n the matter how much Mg he or she may add. When the mole percent of Mg is less If so, than 1,3, the solution components are Man; and Zn When the H1019 Per‘l‘nt t_ the Mg exceeds '1 3, the solution will contain N’Ian2 and Mg. 'f the The lilsmolespercent Mg solution at 773 K is initially in the smgle—phase
region above the solubility curve, Therefore, the solution begins as a singlephase,
homogeneous liquid solution whose components are NIan; and Zn. As Mg is
added, we move horizontally across the phase diagram along the 773K tempera d ture line. When we reach a composition of about 18.5 mole percent of Mg, we Ems encounter the equilibrium line between a homogeneous solution of Zn saturated ng with Man: and solid Mang. At this point, there is a single phase: a homoge tat it neous Zh solution saturated with Man 2. The phase diagram then tells us that if X is the number of moles of Mg present in the solution, we must have tera X  K29. 09 + X 0.185, (1) \e a since the number of moles of Zn is fixed at 0.90 by the conditions of the prob Iuld lem. Thus, ml" (0.185)(0.9) , _ “‘1‘ X : 7 _ moles ot Mg = 0.204 mole ot Mg. (2) am 1  018:: the Since we began with 0.1 mole of Mg, we reach this point after the addition of 11‘“ 0.104: additional mole of Mg. L As soon as we add more Mg, we move into the twophase region, where
we have solid Mg2112 and a Mmesaturated Zn solution. As more Mg is
added, more Man2 precipitates from the solution. Finally, we reach the point
at which we have 1,93 mole percent of Mg on the vertical line through Point C.
We now ha \'e only solid Manz. When more Mg is added, we move to the right of the vertical line through
Point C into a region where we have solid Man2 and a Mangsaturated
Mg solution. When we reach a system composition that is about 56 mole per
cent of Mg, we encounter the equilibrium line along which we have a single
phase, which is a Man 1saturated Mg solution. If the number of moles of Mg
present at this point is Y, we have Y
Y + 09 — 0.56, (3)
so that
(0.9)(056)
Y = r—~_ﬂ moles of Mg = 1.145 moles of Mg. (4)
1  0.56
We have, therefore, added 1.045 moles of Mg to reach this point on the phase
diagram. The addition of more Mg moves us back into a singlephase region in ‘d which we have a solution of Mg in Man2 that is not saturated. As more Mg is ﬁt added, the solution becomes increasingly concentrated in Mg, until we reach 1‘ the intersection of the equilibrium curve for Mg and a Mgsaturated solution of 7“ Man 3, This takes place when the mole percent of Mg is about 82911.. If Z is the ‘1‘ number of moles of Mg present at that point, we have \t‘ 1s Z = 0.82. (5) Z t 0.9 308 Section 2 Solutions and Comments on Problems 8.39 Solving for Z, we obtain (O‘BUJUJBZ) _
7. : T 0 m moles of Mg : 4.10 moles of Mg. (6) We have, therefore, added 4.00 moles of Mg to reach this point.
The addition of more Mg moves us back into a twophase region, where
we have a Man: solution saturated with Mg and solid Mg The HlOeIleO4 phase diagram is shown in Figure 8.30 on the next page.
How many compounds of H20 and fIgSO4 can be formed? What are their
chemical formulae? Label the phases present in each region of the diagram. Solution The phase diagram shown in Figure 8.30 exhibits three distinct maxima.
Therefore, three compounds of water and sulfuric acid are formed. The phase
diagram shows that the first maximum has 20 mole percent sulfuric acid.
Consequently, its. molecular formula must be H1804  4HIO. The second maxi
mum occurs at 33.339}: mole percent sulfuric acid. This means onethird of the
moles in the compound are sulfuric acid. Therefore, the molecular formula of
this compound is 112504  21120. The final maximum has an equal molar mix—
ture, so the formula is H380,  H20. The contents of various regions of the phase diagram in Figure 8.30 are
given as follows: Region A: One phase, a homogenous solution of water and sulfuric acid. Region B: Two phases. One is solid H20. The other is an tho—14380, wifllO
solution that is saturated in H30. Region C: Two phases. One is solid crystals of sulfuric acid tetrahydrate
(H3504  4H20). The other is an HgO—l {ISOl ‘ 41130 solution that saturated in H2804 1H20. Region D: Two phases. Solid crystals of sulfuric acid tetrahydrate.
(HZSO4  4HzO) and an HZSO4  4H307HzSOi  2H30 solution that is
saturated with HZSO4  4H3O. Region E: One solid phase containing a mixture of solid H380, wit1:0 and
solid H3504  ZHZO. Region F: Two phases. Solid crystals of sulfuric acid dihyd rate (HESOI  EH30)
and an 113504  ~1H3(.)—f~12‘504  3120 solution that is saturated with
H2504 ' ZHIO. Region G: Two phases. Solid crystals of H;SO4  21130 and a solution of
H350,1  ZHZO and HQSOT  H20 that is saturated in HISU4  2H20. Region H: One solid phase containing a mixture of solid H350.1  2H;O and
solid also,  ZHJO. Region l: Two phases. Solid H250,1  H20 and a solution of H350,  H30 and
HitSO4  2H30 that is saturated with H250,1  H30. Region J: Two phases. Solid HESO4  H20 and a solution of HZSO4  H30 and
H350, that is saturated with HISOJ  H30. Region K: One solid phase containing a mixture of stwlid 1—12504 . HSO and
solid H2504. Region L: Two phases. Solid 113504 and a solution of H380, v H20 and H380,
that is saturated with H2504. Chapter 8 Thermodynamics of Nonelectrolytic Solutions 309 (6)
Jre E
E
ge.
eir
Jr} ' l l I
m‘ ‘ ll 2n 40 on an IOU
{50 IN.) . . H2504
191 " Mol percent sulfuric acrd
xi—
he A FIGURE 8.30
of The HZO—HZSO4 phase diagram. The letters mark the
iv various regions of the diagram. Note that all temperatures
are given in degrees Celsius.
ire
8.40 The phase diagram for two metals A and B contains one eutectic and no com"
pound between A and B forms. The freezing points of solutions of A and B are
found to vary linearly with the mole percent of B in the solution. The melting
points of pure A and B are 650 K and 600 K, respectively. The cooling curve for a
mixture containing ~10 mole percent B exhibits a single flat region at 400 K. Using
this information, sketch the A 7 S phase diagram and indicate the phases present
in each region,
Solution
Since the cooling, curve of the +0molepercent 8 mixture has only one break. at
400 K, this mixture must correspond to the eutectic composition. Therefore, the
equilibrium line showing the freezing points for metal A from [LB solutions
700 K
)i Liquid solution hill! K A l
:4 I,
'5
F5 Son K 
‘1 I _ . Solid 3 +
E ’ ‘SIUIKI T . saturated solution a
p. saturated‘solutton 01.5 .1" A V of A In 3
4E?“ }\ Solid A + solid B e. Mole percent H 9.2 Twenty—five grams of A1C13 are dissolved in 2 kg of H200). Assuming com
plete ionization and no ionpair formation, obtain an expression for the chemi
cal potentiai of the solute in terms of the standard chemical potentials, the
temperature, and the mean ionic activity coefficient. Solution The ionization reaction is AlCl3 + :c H200) ———> Ala}: + 3 C15“ Therefore,
u+ = 1 and v_ = 3, so that v = LL. + v_ = 4. The chemical potential of the
solute is given by Eq‘ 9.12: m : #5 + m Inivni(m/m°ll = [with + 3 #Ei'l + M Intuiymr‘nni. m 313 314 Section 2 Solutions and Comments on Problems The molality of the solution is moles of AlCl3 25 g X (1 moi/133.34 g) A
m ’ = = 0.093745 moi kg ’, (2)
kg of H10 2 kg so that 0.093745 mo] kg”
1 mol kg"1 m/m" = = 0.093745, (3) and we also have
pi = [vitv‘flm = W33?” = 27"“ = 2.2795. (4}
Substituting into Eq. 1 yields a, _ [am + 3,..31—1 + 4RT ln[(2.2795)7i(0.093745)1 mm + sail—1 + 4RTln[O.21369yil , (5) which is the required expression. “nu—unnu— Solutions and Comments on Problems 9.81 320 Section 2 m (moi kg") "y,
0.001 0.888
0.002 0.851
0.005 0.787
0.010 0.727
0.020 0.664
0.050 0.577
0.100 0.517
0.200 0.469
0.500 0.444
1.000 0.495
2.000 0.784
5.000 5.907 10.000 43.1 ——_——__ The Handbook of'Chemistry and Physics lists the data in the table to the left for the
mean ionic activity coefficients for aqueous solutions of CaCl1 at 298.15 K. (A) Compute the value of 34,. predicted by the Debye—Hiickei theory at each oi
these concentrations, and determine the percent error in the theory. (B) Use Eq. 9.68 to compute yi at each of the given concentrations, and deter
mine the percent error in the equation. (C) Prepare a plot showing all of the data on the same graph over the range oi
concentrations 0 S m S 2.0 molal. Solution (A) The ionization reaction is CaCl2 + x 1120 —> Cain. + 2Cl,lal,,; as a
result, we have V, = 1, i’_ z 2, z 2 2, and z : 1. The ionic strength of the
solution is, therefore, given by a '1 S : 0.5[m(2)‘ wt 2 mtil': = 3m, (1) where m is the morality of CaClz.
The Debye—Hﬁckel activity coefficient is togm) : —0.50926(2_)(irainy'2 : —1.7641m"3. (2) Inserting these moialities yields the following table of values: In 111$th .73, (IDebyeéHiickei)“ % error
0001 0.888 0.879 — 1.01
0.002 0.851 0.834 M 2.00
0.005 0.787 0.750 ~4.70
0.010 0.727 0.666 — 8.39
0.020 0.664 0.563  15.21
0.050 0.577 0.403 30.16
0.100 0.517 0.277 —46.42
0.200 0.469 0.163 —65.24
0.500 0.444 0.057 —87.16
1.000 0.495 0.0172 —96.52
2.000 0.784 0.0032 —99.59
5.000 5.907 0.00011 10.000 43.1 0.0000026 ————__..____________ (B) The empirical equation given by Eq. 9.68 predicts that the activitv coefe
ficient is ‘ (3m)l'2 ‘
 .3 7 0.30i3m)
1 + (3m)l" I 103m) = 70.50926(2){1) 1.7 5 1"?
: —i.ii'1852l—i20 m —0.90rn ii +1.73205m” . (3}
Using Eq. 3, we compute the values shown in the table at the top of the next page.
Obviously, the empirical equation fits well for concentrations at or below 0.10
molai. Above this level, however, the results become very poor. Chapter 9 Thermodynamics of Electrolytic Solutions 321 1:31 the m 7:: (EXPL) 7: (Eq' 9‘68) 0/“ err0.1;:
nach Of (1001 0,888 0.887 v0.11
0.002 0.851 0.848 —0.35
we“ 0.005 0.787 0.782 *0.63
mac of 0.010 0,727 0.722 — 0.69
" 0.020 0.664 0.658 —0.90
0.050 0.577 0.577 —0.00
0.100 0.517 0.539 4.25
0:35ch 0.200 0.469 0.548 16.84
0.500 0.444 0,790 77.92
1.000 0.495 1.866 276.99
(1) 2.000 0.784 12.884 1,543.36
5,000 5.907 5,940.6
10.000 43.1 2.020 x 10"
(2) ____._.__———————
Activity Coefficrents or C'aC‘l1
'71 l I 1 T—r I l‘ ‘ l I 1 [—1— ‘
14 f7 —— Eq. 9,68 
E — — ' Dchye—liiickle Theory—J
1" E 0 Experiment
E 1 U :— —
E : Et]. 9.05 4
:5 2 11h E Experintﬂnl ‘1
E O h '
"C 11,1 ‘:' _
0.: i:— _'
E ..._____ DelayeeHiickle theory ‘
0.0 _ ————————————— _ _ .1
rillll‘illlthLJ. illllLLLJ'l'LLJLLLJJ.J_LlJ—Ld_l——l—Lv1—‘I
11.0 (1.2 0.4 11.0 0.8 1.0 1.2 1.4 1.0 1H 2.0
3.101211in [mo] kg"!
9.9 The Dehye—l—liickel theory predicts that the mean ionic activity coefficient is a
monotonically decreasing function of the ionic strength (i.e., as S increases, 9/:
oef— steadily decreases]. In contrast, the empirical relationship given in Eq. 9.68
‘ exhibits a minimum when yt is plotted against the molality of the solution. In
it many cases, the experimentally measured activity coefficients also exhibit such
I a minimum. (A) Show that the minimum in v1 predicted by the equation occurs at the
E point when the molality of the solute is
(3’ g 0.709959 ,.,
tn. : 1.10 1"1 (B) At what concentration would we obtain a minimum value of Vi for CaClg? ...
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This note was uploaded on 01/15/2012 for the course CHM 4411 taught by Professor Ohrn during the Spring '08 term at University of Florida.
 Spring '08
 OHRN
 Kinetics

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