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Unformatted text preview: Prom/m 36k dJF23 ’SOW‘HOAS A\ 1' %r 0.010 mom was: 1: Aiksﬂooﬁ + (—1)7‘(3x0.01)]
1‘ 0.0(o
B) 1 #3“ 0.010 Mo\a\ Abbas; an? 0,005 mmL “£1904
3 £651” 001) +éa)z(3mo4> + @1)L(Zxo.005)+675(°'°0‘) 1: 0.46;
v— 4/3) , 4 3 4/1" , o 1‘ 0.0035 mo\a\
(1) W310,» \>.>m ’ 0'0; 1
95 (3+: Eimi == LO.80Q§ [14 43.4.0 $0.3m mm a E\ \0340Yt : ,O.$1O‘lt+\\%1I4/2_
I = 42—16010 +6010} g— 5 am .'. ﬁt? ' W3 \s 5%» 9:3}er 3mm Mun mwé) Value, ﬂ) 0.804, wMidq
\maw. 5105b JinCL WQL’HUM worn— beﬁ— in “MM 09 \0M) Rom! = XAB: + yap total . — 0.900 b 1— 0.550
130 : Bola! yB‘lDtolal = arx ( ) = bar
xA 0.450
__ J’AR;
xA _ x a a
P. +(e —P,, J» 0650 = w P; +0.450(P; —P;)
P; 0.650><(1—O.450) P; 0.450 ><(1— 0.650)
P; = 1.414 bar = 2.27 P95) A and B form an ideal solution at 298 K, with 294 = 0.600, P; = 105 Torr and P;
63.5 Torr.
a) Calculate the partial pressures of A and B in the gas phase. 13) A portion of the gas phase is removed and condensed in a separate container.
Calculate the partial pressures of A and B in equilibrium with this liquid sample at 298 K. a) Calculate the partial pressures of A and B in the gas phase.
PA = xAP; = 0.600x105 Torr= 63.0 Torr P5 = (1 — mp; = 0.400x63.5 Torr= 25.4 Torr b) A portion of the gas phase is removed and condensed in a separate container.
Calculate the partial pressures of A and B in equilibrium with this liquid sample at 298 K. The composition of the initial gas is given by PA _ 63.0 Torr
PA + PE 88.4 Torr X4: For the portion removed, the new x A and x E values are the previous y A and y R values.
PA = LIP/1‘: 0.713X105 Torr = 74.9 Torr PB =(1—xA)PB' = 0.287X63.5 Torr: 18.2 Torr PRU Assume that lbromobutane and lchlorobutane form an ideal solution. At 273 K,
P6210”, = 3790 Torr and Pb:m = 1394 Torr. When only a trace of liquid is present at 273 K: ychloro = a) Calculate the total pressure above the solution. b) Calculate the mole fraction of lchlorobutane in the solution.
c) What value would Zchlm have in order for there to be 4.86 mol of liquid and 3.21 mol of gas at a total presSure equal to that in part (21)? [Notes This compositionis different
from that of part (a).] a) Calculate the total pressure above the solution.
P ‘Ptatal — 'PCI‘IIOI‘OIDerMO __ chloro ychlnra _ a #
‘Ptoral (‘Pchlara — Pbromb) 3790 Pamel — 3790 Paxl394 Pa
PM, ><(3790 Pet—1394 Pa) Plum! = 3790 Paxl394 Pa : 2651Pa
3790 Pa  0.75x(3790 Pa — 1394 Pa) 0.75 = b) Calculate the mole fraction of l—chlorobutane in the solution. at n = X P + _ xchlora ) P’bromo chlora chlara P 431;,” _2651Pa—1394Pa total W“ Z P‘ —P;m _ 3790 Pa—l394 Pa chlara P total = 0.525 x c) What value would Zamora have in order that there are 4.86 moles of liquid and 3.21 moles of gas at a total pressure equal to that in part (a)? (This composition is different
than that in part (a).)
__ IJChIoro‘Plaral _ ychloro * t :
Rotal (‘Pchlora — Barnum) _ 3790 Pa><2651 Pa — 3790 Paxl394 Pa = 0 750
2651 Pax(3790 Pa—1394 Pa) ' I I P chlnro bramo a x _ ychlora Pbromo
* ‘ chlora nu
Iachloro + (1317mm;  Iachlara ) ychlaro 0.750Xl394 Pa 
—_— ______———————— = 0.525
3790 Pa + (1394 Pa— 3790 Pa)X0.750 to! _ tot _
nliq (Zchloro _ xchlaro)  "vapor chlaro Zchlaro) m to! 321 m01x0,750 + 4.86 molX0.525 nvaparychlam + "liq xChlaro = _____________.—————— = Z :
chlora n10! + rig: m01+ m0]. P9.9) At —47°C, the vapor pressure of ethyl bromide is 10.0 Torr and that of ethyl
chloride is 40.0 Torr. Assume that the solution is ideal. Assume there is only a trace of
liquid present and the mole fraction of ethyl chloride in the vapor is 0.80 and answer
these questions: a) What is the total pressure and the mole fraction of ethyl chloride in the liquid? b) If there are 5.00 mol of liquid and 3.00 mol of vapor present at the same pressure as in
part (a), what is the overall composition of the system? b) We use the lever rule.
"it: (ZR “x3) 2 "if; (ya ‘23)
ZBxB =(1—ZA)_(1“XA)=xA_ZA # t .. PE _PEC _ _ _ __
a) BUM—m yB_ZB(1“y.4)_(1 ZA)_ZA yA _ 10.0 Torr—40 Torr Therefofe, ‘ 40.0 Torr+ (10 Torr_40 Torr)><0.80 ngj; yEC —ZEC _ 5 225.0 Torr n32; _ ZEC —xEC — 3
Bata1:xECP;c+(1xEC)PEB _080
x 2 13mm] _P£B = 250 TOH_1O_O Torr we know that xESC— 0.50 andync . EC ch —PE'B 40.0 Torr—10.0 Torr (0.8025C) 23(ZEC — 0.50)
=0'50 ZEC =0.613 253 = (1 — 25C) = 0.387 P9.I3) At 399°C, a solution of ethanol (x1 = 0.9006, P;=l30.4 Torr) and isooctane (P;= 43.9 Torr) forms a vapor phase with y1 = 0.6667 at a total pressure of 185.9 Torr. a) Calculate the activity and activity coefﬁcient of each component.
b) Calculate the total pressure that the solution would have if it were ideal. a) The activity and activity coefﬁcient for ethanol are given by
_ yIBm, _ 0.6667x185.9 Torr _ 0 9504
P; 130.4 Torr ' a1_0.9504__1055 x1 _ 0.9006 _ ' Similarly, the activity and activity coefﬁcient for isooctane are given by at 1__. a2 : (l—yJPwml = 0.3333X185.9 Torr :14” p; v 43.9 Torr
1.411 7/2 =i=_————:14.20
x2 1—0.9006 b) If the solution were ideal, Raoult’s law would apply.
PTolal : xl‘P; + x213;
= 0.9006 x1304 Torr + (l —— 0.9006) X 43.9 Torr =121.8 Torr P9.15) At 399°C, the vapor pressure of water is 55.03 Torr (component A) and that of
methanol (component B) is 255.6 Torr. Using data ﬂow the following table, calculate the
actiVity coefﬁments for both components using a Raoult’s law standard state. See the solution to Problem P915 for the method used. The results are shown below. xi P Torr)
0.0490 257.9
0.3120 211.3
0.4750 184.4
0.6535 156.0
0.7905 125.7 ul
.
' P9.16) The partial pressures of Brz above a solution containing CC14 as the solvent at
25°C are found to have the values listed in the following table as a function of the mole
fraction of Brz in the solution [G. N. Lewis and H. Storch, J. American Chemical Society
39 (1917), 2544]. Use these data and a graphical method to determine the Henry’s law constant for Brz in CC14 at 25°C. xBrz 0.005 0.01 0.015 0.02 0.025 0.03 XBFZ The best fit line in the plot is PB,z (Torr) = 413 xBrz — 0.063. Therefore, the Henry’s law constant in terms of mole fraction is 413 Torr. “an”. 7w“.. gum.» ,0 Vi P9.18) The partial molar volumes of ethanol in a solution With xH20 = 0.60 at 25°C are 17 and 57 cm3 mol‘l, respectively. Calculate the volume change upon mixing sufﬁcient
ethanol with 2 mol of water to give this concentration. The densities of water and ethanol
are 0.997 and 0.7893 g cm‘3, respectively, at this temperature. V = HHZOVHEO + 77E, V51 —V_H=o =17.0 cm3 mol'1 and E737 = 57.0 cm3 mol‘1 n
1730 =2.00 and x30 =———”£—=0.600
2 1 "1920+"51 . 2mm 20.600; nE,=1.333
2mol+nEl The total mixed volume is given by med = nHzofgzo + 7151757
2 2.00 molx17.0 cm3 mol‘1+1.333 molx57.0 cm3 rnol‘1
= 109.98 cm3 ..
Vunmixed = "3,0 ﬂ; "57 ME,
‘ szo p5:
= 2.00 molx18.02 g molx I‘m—3
0.997 g
+1333 molX 46.07 g moi1 x—l‘lnj—
0.7873 g
236.15 cm3 + 78.00 cm3
2 114.15 cm3
AV = Vm — me 2109.98 cm3 114.15 cm3 = —4.2 cm3 P9.21) The dissolution of 5.25 g of a substance in 565 g of benzene at 298 K raises the
boiling point by 0.625°C. Note that Kf= 5.12 K kg mol‘l, Kb = 2.53 K kg rnol‘l, and the
density of benzene is 876.6 kg m4. Calculate the freezing point depression, the ratio of
the vapor pressure above the solution to that of the pure solvent, the osmotic pressure, it and the molecular weight of the solute. lime“ = 103 Torr at 298 K. _AT,, _ '0.625K AT =K ; —————————=0.247molk '1
b bmrolule msalule Kb mold g
M = = 37.6 gmolI
0.247 molkg X0565 kg ,
ATf = —Kfmm,m = —5,12 Kkg mol‘I x 0.247 molkg‘l ='—1.26 K
‘benzene = xbenzene = "benzene
Iabwene "benzene + nsolute
565 g
—1
= 565 78.11 gmol = 0.981
———g—_l + 0.247 molkg“ x0565 kg
78.11 gmol
—3
is3%9ﬁq—x8314 Jmor‘K'1 x298 K
7z_ nmlmRT _ 37.6X10 kgrnol _ 5 37X10'5Pa
“ V 565x10‘3kg _ ‘ 876.6 kg m“3 P925) A solution is made up of 184.2 g of ethanol and 108.1 g ofHZO. If the volume of
the solution is 333.4 cm3 and the partial molar volume of H20 is 17.0 c1113, What is the
partial molar volume of ethanol under these conditions? V : HHZOVHZO + "ethanol VethanDI ‘
V : nHZOVIIZO + net/1mm] Kahuna
.1
— 333.4 cm’ —JQL%x17.0 cm3mor‘
I7  ————V _ nHzoVHzO  1802 g mm  57 8 cm3mol'1
ethanol nethana] &_ .
46.04 gmol'1 P9.2 8) At a given temperature, a nonideal solution of the volatile components A and B
has a vapor pressure of 832 Torr. For this solution, 32,; = 0.404. In addition, xA = 0.285, P;= 5 91 Torr, and P3“ = 503 Torr. Calculate the activity and activity coefﬁcient of A and
B. PA = yAPwmI = 0.404X832 Torr = 336 Torr
PB 2 832 Torr — 336 Torr = 496 Torr
(111 = P11: 336 Torr 20.569
PA 591 Torr
7,. = 51¢ = m = 2.00
xA 0.285
as = 53— : ———4% To“ = 0.986
PE 503 Torr
x3 0.715 P929) Calculate the activity and'activity coefﬁcient for CSz at xCSz = 0.7220 using the data in Table 9.3 for both a Raoult’s law and a Henry’s law standard state. P
61‘“ — €52 —M=0.87z3 C51 _ PCS! ' 512.3 Torr _ c2552 0.8723 — — = = 1.208
7/552 xCS2 0.7220
a” = PCS! = w = 0 2223
“3 ka 2010Torr '
H
ygsz : aCS2 = 0.2223 : 0.3079 x“: 0.7220 ...
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This note was uploaded on 03/06/2011 for the course CHEM 340 taught by Professor Staff during the Spring '08 term at Ill. Chicago.
 Spring '08
 STAFF

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