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Unformatted text preview: Chapter 6/ Chemical Equilibrium Chapter 6/ Chemical Equilibrium
AH ° _1_ 1 reaction R :rf 7 298.15K 02ag1+AH31(Hzag)'AH}(H20=1)AH}(C(18)
11 + 0 + 285.8x103 Jmol" +110.5x103 mm1 = 2.80x103 Jmor‘ 61‘ V :ra—TfF T T2 T2 _?2_ (aA/T 1 6A A s A_#A+TS_ U
% L AH}(C 6A/T aA/T 6T 8T V 2 aA/T
61/T V 6T V a1/T V 61/?" 6T V p 1'
62’ V P(298.15 K)
At constant V, 750X18: 29.342x10‘3 K—1
T2 T2 ‘ 3.32><1
id E 3 J‘ AU d .1.
T. T 71 T
Maw ) AA( T ) 1 1 he reaction is endothermic, K P increases with temperature. As T —> 00,
2 ~ 1 + AU T ———*— *
1” 298 15 K) 7
P6.2 7) A gas mixture with 3.00 mol of Ar, x moles of Ne, and y moles of Xe is prepared at a pressure. P( . 1
o 3 _
1 bar and a temperature of 298 K. The total number of moles in the mixture is four times that of Ar. AHMCW — 2'80 X10 J mol =112957 , __._._________..___.—_—————
" Rx298.15 K 8.314 J K" moil x298.15 K Write an expression for AG have, mixing in terms of x. At what value of x does the magnitude of AG mixing its minimum value? Calculate AG for this value of x. mixmg suming that AH ; is constant in the interval 275 K — 600. K, calculate AG° for the process If the number of moles of Ne is x, the number of moles of Xe is y = 8 — x. AG 2 nRTZ x, In x, mixing =nRT ——1ln4+—x—h1—+ x 9—x 9~x 1n __ o
4 12 12 12 12 .,, __ "(T;)+AH}~(71) 51:71:
dAGmg 1 x x 12 1 9—x 9—x 12 7i 2 1
———=nRT ——1n——+—————ln + — ~1) =0 3 .1 1 1
dx 12 12 12 x 12 12 12 9—x 219.9X10 Jmol # 3 1'1x _
241.8x10 Jmo ———K 29815K
1 x 1 9—x nRT x 298.15 K 495 '
:nRT —1n——+l——ln ~1 =—ln =0 3 1 *
12 12 12 12 12 9~x 219.9)(10 Jmol‘
x 9 —219.9x103 .lmol'1+228.6><103 JmorI _ 0
:1; x:— ‘ _______________._—————————————‘3.8A)
996 2 awe Change ‘5 —228.6x103 Jmol“
AG”’ix’"g = ”RT —~::1n4+§ln§ +31“; alculate the degree of dissociation of N204 in the reaction N204(g) —> 2N02(g) at 250. K and a L ‘ ' t’ t 'ncrease or decrease as the
:12 mol X 8314 J mol'1 K4 X 298'” K X 6054 = ”322 X103 J ssure of 0.500 bar. Do you expect the degree of dissocia 1011 o 1 P6.28) 1n Example Problem 6.9, K p for the reaction C0(g) + H200) ——> C02(g) + H2(g) was calculated
to be 3.32 x 103 at 298.15 K. At what temperature does KP = 7.50 x 103? What is the highest value that ature is increased to 550. K? Assume that AHoream(m is independent of temperature. KP can have by changing the temperature? Assume that AH ,3 is independent of temperature. 146 147 h ' lE uilibrium
Chapter 6/ Chemical Equilibrium Chapter 6/ C emlca q We set up the following table ‘ w ‘_ (298.15 K) # AHSWH _1__ 1
N204(g) <——> 2N02(g) , _ RX298.15 K R 7}. 298.15 K
Initial number of moles no 0
Moles present at _ :‘Jmol’l 55.3)(103 J in 01—} X 1 _ 1
equilibrium 1106 25 01'1x298.15 K ‘ 8.314 J K‘lmol'l 250. K 298.15 K
Mole fraction present —— 95 426
_ ‘ 3
at equilibrium no 5 25 '40X10
”0 + 5 ”o + 5 y
_ . () 10"3 _ —2
Partial pressure at q 4 : 4 x 6366 — 4'68 X10
Equilibrium, B = x,P no ﬂ); P i P A > 0, or increases as Tincreases.
"0 + 4: ”0 + 5 reaction n reacts with solid glycylglycine C4F18NZO3 to form urea CH 4N20 , carbon dioxide, and
We next express KP in terms of no , 5, and P. _ _ 6‘1 2 2 P 2
P— i a H N 04s)—>CH4Nzo<s>+3coz(g)+2H20<1>~
KT_P°_no+§P_ 452 P 452 P _482~ _ ‘
P ( ) — (5;, “ ”o _ (5 P * (no + 6) (no _ 5) P— — _(n )2 _ 52 T ‘ , and 100 atm solid glycylglycine has the following thermodynamic properties:
4.3. 0 7  v ‘ _
P° no + 5 F ”' 4 , 49—4915 Id moi“,AH} — 7460 kJ mol",S° — 1900 J K" “101 1
We convert this eXpression for Kp to one in terms of 0:. L“ '1: L _ at T = 298 K and at T = 310. K. State any assumptions that you make.
4982 P 4052 P
KP (T) : 2 2 o = _ 2 o
(no) 5 P 1 a P
P 2
K)(T)+4 a =KP(T) P,
(T)
KP (T) + 4 KP P
PO AG:eacti0n : ZAG; (N02 3g) _ AG} (N2047g)
'=2x51.3x103 Jmor1 —99.8><103 Jmor1 ~2.8x103 Jmor‘ AH:eacri0n : 2M; (NOZ’g) _AH; (N2049g)
=2x33.2x103 Jmor‘ —11.1x103 JmorI 55.3xlO3 Jmol'l 148 149 ' 1E 'l'brium
Chapter 6/ Chemical Equilibrium Chapter 6/ Chemlca qu1 1 A0,. (298.15 K) = AG; (urea, s) + 3A6; (C02, g) + 2A6; (H20, Z) bs;He1n1holtZ equation,
—AG; (glycylglycine, s) ‘
= —197.15 M mol—1+3x(—394.4 kJ mol“ )+ 2 x (2371 k] mol‘1)— 491.5 kJ mol’1 = —1363 kJ mol'1 1 1
725 K 298.15 K » —l
257%“)3 ””1 ~283.0x103 Jmor‘x AHR 098'” K) 2 AH; (urea, S) + 3”; (COZ’ g ) + 2M; (H20’ 1) ontaining 1.75 mol of an ideal gas at 325 K is expanded from an initial volume of p16 0 ﬂAH‘; (glycylglycine, S) 21311 volume of 75.0 L. Calculate AG and AA for this process for (a) an isothermal
= $372 kJ mol‘1 + 3 X (—3935 kJ mol" ) + 2 X (—2858 Id mol’1)— 746 kJ mol—1 ‘ath and (b) an isothermal expansion against a constant external pressure of 0.750 bar.
: _1 343 kJ mol“1 AG and AA do or do not differ from one another.
othermal reversible path
AG 298.15 K “1
1n KP (298.15 K) = ——4£(————)— = W = 549.879 ‘ 1),. 14
RT 8.314Jmol K X298.15 K P. nRTln—F=nRTln—V—
KP (298.15 K): 6.44x10238 , f L
_ 15.5 3
AG :r . 14 J mol'lK'1X325 Kx 1n————=—7.46><10 J
AGR(T2)=T2 __£_(_’_)+AHR i__1_ X83 75.0L
T1 T2 T1
AG}? (310‘0 K) 2 PdV = —nRTlnV‘f—
310.0Kx M—1343kJmoi‘x ——1—————1—— 1 _1 75.0L 103 J
298.15 K 310.0 K 298.15 K 5 molx8.3l4 Jmol K x325 lenﬁ=—7.46X
= —1364. kJ mol“1 the systems g0 {ause A and G are state functions, the answers are the same as to part a) because We have assumed that the enthal of reaction is inde endent of tem erature, which is a ood
py p P g hesame initial and ﬁnal states, T,Vi —+ T,Vf. AA = AH — AU = A(P V) = A(nRT). Therefore, AG = AA for an ideal gas if T is constant. approximation over the small temperature range. P632) Calculate AG° for the reaction CO(g) + 1/2O2(g) —> CO2(g) at 298.15 K. Calculate AG; a reaction cu have containers of pure O2 and N2 at 298 K and 1 atm pressure. Calculate AGmixmg relative 725 K assuming that AH 0 reaction is constant in the temperature interval of interest. iXed gases of AGjmm (298.15 K) = AG} (C02, g) — AG}(CO,g)—1AG} (02,g) _
2 xture of 15 mol of O2 and 15 mol of N2 =—394.4x103 Jmor1 +137.2x103 Jmor‘ ~~0
='—257.2><103 Jmor1 ate AGmixmg if 12 mol of pure N2 are added to the mixture of 15 mol of O2 and 15 mole of N2. ~ nRT(x1 ln x1 + X2 1n x2) .. . . 1 ..
AHW, (298.15K) = AHf (C02,g)—AHf (CO,g)——2—AHf (opg)
=~393.5x103 Jmol" +110.5x103 mm“ —0
=—283.0x103 Jmor1
AG° (T)=T erAH (T) i__1_
reaction 2 2 I; reaction 1 TZ Tl 150 151 Chapter 6/ Chemical Equilibrium
Chapter 6/ Chemical Equilibrium 1' o
1 ’ AH, ..
__ Icaclmn dT
dln K ) ”— —— 2
a) AG,,,,.,.,.g = 30 molx8.314 J K'1 morI x298 KX[%ln2—+§ln2] : 500 M 1 ’ R 7] T
5 5 5 11? AH‘T _ +AC},(T~7;)
K (T ) __ I Ieaclion dT
nK)(Tf)‘ln P 0 RT T2
_ l l I 1 1 1 _ 0 f
b) AGmimg —30 molx8.3l4 JK mol x298 KX(—2—ln§+§ln§j—~51.5 kJ dT AC. / 2,2— ACPT ﬂ
T2 R To T R [(1 T2
C) AGmL'xing : AG/nixing (separate gases) i_1— + ACP 1113+ ACI’YE) _1_ 1
—AGmixmg (15 mol A +15 mol B) Tf T0 R T0 R T, To
:42 molx 8.314 J K'1 mol'l x298 Kx[1—54ln%+%ln%]+51.5 . (3110(3) (—5 2Cu(s) + 02(g)
 3 ‘ = 4 103 J or1
=—67.8kJ+51.5kJ=—16.3k.l _rwcmm(T0)_2x157><10 Jmol 31 X m ,m (CH, S) + CP,m (029 g) _ 2C1)” (C110, S) P6.35) In this problem, you calculate the error in assuming that AH; is independent of T for the 1 4
(2 x 24.4 + 29.4 — 2 x 42.3) J K' mol reaction 2CuO (s) : 2Cu(s) + 02(g). 6 4 J K’1 mol'l The following data are given at 25° C:
Compound CuO(s) Cu(s)
AH;(1<J mol")  _ 2><129.7x103 Jmor‘
KP (1350' K): #8314 J mol'1 K'1x298.15 K —1 >1
2x157x103Jmor‘ 1 _ 1 j_ 6.4JK lmol ‘11n1350.K
8.314JK"mol‘ 1350K 298.15K 8.314JK' mol 298.15K 6.4 J K" mol'1x298.15 K 1 _ 1 ]
8.314 J mol" K" 1350. K 298.15 K in K,. (1350. K) = —6.32361 AG; (kJ mol“) Cf.” (J K‘I morl) 19(0) T, _ P
. 1 AH . , . 02 _ ~3
a. From equation 6.77 I d In K P = 71 I—Y’fTam‘idT. To a good approxrmation, we can assume that K1>(1200 K) 3 Po — 179X10
K1’(To) To .79 x 10'3 bar
the heat capacities are independent of temperature over a limited range in temperature, giving This is equivalent to setting ACp = 0. Neglecting the last two terms in the calculation above gives 1n
AH:eaclion (T) = Momma” (72)) + AC , (T — To) where AC}. = Z VICKI” (i) . By integrating equation 6.77, 3
" 55.7608 and P2 = 3.15 x10' bar. . O 3 ~1
show that 36) Consider the equilibrium in the reaction 302(g) :1 203(8)» Wlth AH R = 2854 X 10 J 11101 at 111KP(T)=1nKP(73)[email protected]_1+§x_A§£1 1 ACP T R T 71) R T To R TO 98K. Assume that AH; is independent of temperature. ’  .   ‘ ‘ ' ' ‘ 'll shift toward reactants or
b. Using the result from part (a), calculate the equilibrium pressure of oxygen over copper and CuO(s) at WlthOUt domg a calculatlon, predict whether the equillbrium posmon W1 products as the pressure is increased. (Without doing a calculation, predict whether the equilibrium position will shift toward reactants or 1350. K. How is this value related to KP for the reaction 2Cu0(s) 7: 2Cu(s) + 02(g)? c. What value would you obtain if you assumed that AHjeamon were constant at its value for 298.15 K up Products as the temperature is increased.
to 1350. K? Calculate KP at 750. K. = Calculate Kx at 750. K and 1.50 bar. 152 153 Chapter 6/ Chemical Equilibrium Ch t 6/Ch ‘ 1E 'l'b' , _
aper emlca quilrium forTz33gK,AGR=AHR“TASR’0 31 J K“ mor‘ = 780.8 J mot} a) The number of moles of products is fewer than the number of moles of reactants. Therefore, the I .
ure and pressure change IS g1ven by 338 K x 2 eQuilibrium position will shift towards products as the pressure is increased. nthalpy as both the temperat b) ,_R‘(338 K)+(P.1.—P338K)AV+ACP(T—338 K)
i  K = 0
Because AHZZacrion > 0: the CCl‘lﬂibfium Position will shift towards products as the temperature is = AHR (338 K) TAS R (338 i 338 K)
increased 5+(PT a 13338K)AV +AC,. (T ~338 K) = TASR(
C) ‘_ K)+(PT ~P338K)AV—ACP (338 K) — C.
Go 0 2 3 —l ASK (338 K) A I _1 6 3 1']
reaction :ZAGf (03:8): X163.2X10 Jmol 1r1+101325X999barX105 Pabar X3.10X10 111 m0
0 . o _ o _ 3 1 L * :
AHrmm” ~ 2AHf (03,g) —— 2 x 142.7 x10 Jmol _ K841 moi” x338 K 2.31 J K" mor‘ —7.98 J K“1 mol—1 an (T ) — — AG:eaCIiO” (298'15 K) __ AH:eaction 1 __ 1
1 f Rx298.15 K R Tf 298.15 K ‘ 3/2H ( ) 2%“) + NH3( g) comes to equilibrium at a total pressure of
2 x 163.2x 1031 mor1 _ 2 x 142.7x103 Jmor1 he ream” Fem” 2 g anP (750. K)=_ = 2.165 and 1.083, respectively, if 8.314 JK" mol‘l X29815 K 8.314 JK" morI 1 1
750. K _ 298.15 K In KP (750. K) = —62.3103
KP (750. K) = 8.69x10‘7‘8 X nt in the gas phase and Fe2N(s) was in excess. g) was initially prese ulate Kp at 700. and 800. K. 0 AS" at 700 K and 800 K and AH; assuming that it is independent of temperature.
111316 reaction ' ‘ d) Calculate KX at 750 K and 1.50 bar. . culate AGfmmn for this reactiOn at 298.15 K.
_AV +1 0 ' ’ 2Fe S + NH3(8)
KX = KP ’11; = 8.69x10‘28 x 115: bar =1.3Ox10_27 N28.) + 3/2 H2(g) H ( )
ar P6.3 7) For a protein denaturation the entropy change is 2.31 J K'1 mol'1 at P = 1.00 atm and at th
melting temperature T = 338 K. Calculate the melting temperature at a pressure of P = 1.00 x 103 atmi the heat capacity change ACp,m : 7.98 J K'1 mol'1 and if AV = 3.10 mL mol'l. State any assumptions yo make in the calculation.
likbar = 2.165 PHZ + PHz = 3.165 PH2 3 ,5 = 0.316 bar, PM, = 0.684 bar 0.684
(0.316) =3.85 3/2 ‘K I, (700. K) = :08800.K
lbar: 1.083 PHZ +PH2 =2.083 PH. P H 2 = 0.480 bar, PM); = 0.520 bar 154 155 Chapter 6/ Chemical Equilibrium 0.520
(0.480)32 K,, (800. K) = = 1.56 b) Assume that Aijcm is independent of temperature K]. (800. K) _ —AH° 1 1 reaction
W KP (700. K) 5 R 800. K _ 700. K KP (800. K)
—Rln
.. * KP (700. K)
1
800. K _ 700. K ln = ~42.0 kJ mol'1 reaction — 1 Aojmm (700. K) = ~RTln KP (700. K) = ~7.85 kJ mol“
AGjmm (800. K) = —RTln KP (800. K) = —2.97 kJ mol‘1 0 ASSWH (700. K) zﬂw_ ~ —48.9 J mol'] K‘1
700. K
ASSWH (800 K) = —48.9 J mol‘1 K“
AH° . 1 1
C 111K 29815 K :ln K 700. K __ reaction _
) P( ) P( ) R 298.15K 700.K
3 l
=ln3.85+ 42““) “ml x 1 =1” 8.314 I moi1 K“ 298.15 K C 700. K
G:W (29815 K) = —RTIn KP (298.15 K) =—8.314 J mol" K" X298.15 Kx11.1.=—27.4 kJ mol‘1 P6.39) Assume the internal energy of an elastic ﬁ dU=TdS~PdV—Fd€. nonexpansion work obtainable when a collagen ﬁber contracts from t = and T. Assume other properties as described in Problem 6.16. dU=TdS~PdV~7dZ dG =d(U+PV—TS)=TdS—PdV—ydl+PdV+VdP—TdS—SdT
=—7dZ+VdP—SdT 8_G = —y : kl
or I.)
0 10 0.10m 2 2
' '” 0.20 — 0.10
AG=wW = ikldl= lk12 =10Nm1x( m) i m) =—0.151
0.20111 2 0.20m 2 156 her under tension (see Problem 6.16) is given '1) Obtain an expression for (80/6010 T and calculate the maxim __, 20.0 to 10.0 cm at constan‘ Chapter 6/ Chemical Equilibrium  . . ‘ rm lactic acid
U def anaerobic conditions, glucose is broken down in muscle US$116 to f0
) n gerand lactic acid are given below. AH; (kJ moi“) C; (J K" moi") 8" (J K" moi“)
1273.1 219.2 209.2
673.6 127.6 192.1 (298.15 K) = ZAH; (lactic acid) — AH; (glucose)
i‘(—‘673.6 kJ mol‘1)+1273.l kJ mol“
kl mol’1 98.15 K) = 2S° (lactic acid) — S° (glucose)
92.1 J K” mol‘1 —209.2 J K51 mol‘l ‘K‘l meal”1 298.15 K): AHR(298.15 K)—TASR(298.15 K) = —]26 kl mor‘ 10. K) = AHR (298.15 K)+ACPAT .kJ mor‘ +(2xl27.6 1 1c1 mor‘ 219.2 J K"‘ mol"1)x(310. K—298.15 K)
kJ mol‘1 i_0‘K)=AHR(310. K)—310.K><ASR(310.K) 310.K (31100410. Kx ASR(298.15 K)+ACP1nm 310K _ A 7] _______
“moi"L310Kx 17SJK’1 mol 1+36.01K ‘ moi X1“ 298.15 K k1 moi“1 —54.7 kJ mol" : —128 kJ moi" 157 Chapter 7/ The Properties of Real Gases Chapter 7: The Properties Of Real Gases ‘ Can you conclude about the ratio of fugacity to pressure for N2, Hz and NH3 at 500 bar using and H2, and f < P for NH3 0 of fugacity to pressure greater or les Problem numbers in italics indicate that the solution is included in the Student ,5 SOIutiOnS Manual s than one if the attractive part of the interaction Questions on Concepts etween gas molecules dominates? e interactions dominate P < Pideal, SOf/P < 1 Q7.1) Using the concept of the intermolecular potential, explain why two gases in corresponding state gas is slightly above its Boyle temperature. Do you expect 2 to increase or decrease as P can be expected to have the same value for 2. Two different gases will have different values for the depth of the intermolecular potential and for the
62 d1stance at wh1ch the potential becomes p051t1ve. By normalizing P, T, and Vto the1r cr1t1cal values, th e Boyle temperature (5P ﬂ—T] > 0 , so 2 increases as P increases. differences in the intermolecular potential are to a signiﬁcant extent also normalized. xplain why the oscillations in the two—phase coexistence region using the Redlich—Kwong and s equations of state (see Figure 7. 4) do not correspond to reality
se. No real gas exhibits this behavior Q7 .2) Consider the comparison made between accurate results and those based on calculations using th‘ van der Waals and Redlich—Kwong equations of state in Figures 7.1 and 7.5. Is it clear that one of thes' Baal lations predict that as Vincreases, P will 1ncrea equations of state is better than the other under all conditions?
rmal reversible expansion under conditions such that z The Redlich—Kwong gives more accurate results for almost all of the values of pressure shown in thes van dﬂf Waals gas undergoes an isothe he work done more or less than if the gas followed the ideal gas law? ﬁgures. However, it is not better under all conditions.
—.P Because V > Kdeal, for z < 1, AV < AVm'ea], Therefore Q7 .3) Why is the standard state of fugacity, f, equal to the standard state of pressure, Po? reversible expansion P external If this were not the case, the fugacity would not become equal to the pressure in the limit of lo PdVl< H Pd Meal he value of the Boyle temperature increases with the strength of the attractive interactions es of the gases Ar, CH4, and C6116 in increasing order. pressures. (27.4) Explam the s1gn1ﬁcance of the Boyle temperature. molecules. Arrange the Boyle temperatur The Boyle temperature provrdes a way to classrfy the way in which 2 var1es wrth P at low values of tions increases with the number of electrons and atoms, Ar < " the strength of dispersive interac \ 6H6 :A system containing argon gas is at pressure P1 and temperature T1. for different gases. If T > TB, 2 increases with increasing P; if T < TB, 2 decreases with increasing P. A the Boyle temperature, the repuls1ve and attractive 1nteractions cancel one another out so that the gas How would you go about behaves ideall over a lar er ran e of ressure than at other tem eratures. . .
y g g p p ting the fugac1ty coefﬁcrent of the gas? 7.5 A van der Waals as under oes an isothermal rever ibl e an 10 ' ‘ at 2 >1 ‘ . . . 
Q ) g g S e xp S n under cond1t1ons SUCh th e crltical constants of the gas to determme the reduced pressure and temperature. W1th these a ‘ (7
1. Is the work done more or less than if the gas followed the ideal gas law. ts,restimate the fugacity coefﬁcient using Figure 7.11. For a 'reversrble expans1on, Paternal : P' Because V > Vida” for Z > 1’ AV > AVideal' Therefore, )5 By looking at the a and b values for the van der Waals equation of state, decide whether 1 mole w§=HPdVl >ledeea, Q7.6) For a given set of conditions, the fugacity of a gas is greater than the pressure. What does this tell
etween 02 molecules. Consequently, 02 will have a higher pressure. you about the interaction between the molecules of the gas?
5) Will the fugacity coefficient of a gas above the Boyle temperature be less . . . . . . . at low
If the fugacrty rs greater than the pressure, the repuls1ve part of the potential dominates the 1nteract10I1 than one between the molecules. 158 159 % C
hapter 7/ The Properties of Real Gases Pzzs) F01” a va
n der Waals
gas: Z:K11/(V _b)_ Ta 1 . . . .
y or series in the limit Vm >> I) to obtain 2 l (b
z + —a/RT)(l/V) f(x)=f(0)+ M H 1 h 1
nti
dx x=0 Scase,f(x)=‘andx=_
V 1”
Z:;/i.._ﬁa__ 1 a In
Iii—b _ _
RTV,n 13 RTVm
Vm
Because _1__z1+x+x2
l—x —27—+...
1 b
7ZI+_
1*“ Vin
b
Zs1+~_L_ 1
VI” RTVm “1+“[ *5?) 172 a R; l .
H I d the Relative Stability of Solids, Liquids, Chapter 8: Phase Diagrams an “roblem numbers in italics indicate that the solution is included in the Student ’5 Solutions Manual. Questions on Concepts
08.1) Why is it reasonable to show the y versus T segments for the three phases as straight lines as is 1? More realistic curves woul sus T plot? Explain your answer. d have some curvature. ls the curvature upward or one in Figure 8. ownward on a y ver c curves will curve downward, but not very s slowly with T, the realisti (9E) = —S . Because S increase
_ 6T p apidly.
ves in the solid, liquid, and gas regions of Figure 8.5 have Q82) Why do the temperature versus heat our different slopes?
f these curves are equal which has different values for the three The slopes o to the inverse of prm, as can be seen in Figure 2.5.
0 scale. What would be
e drawing were to scale phases
axis of the liquid + Q8.3) Figure 8.5 is not drawn t
solid, liquid, and liquid + gas segments for water if th of H20? For the liquid + solid segment, the relative lengths on the q p
and the system consisted Mo“ , for the liquid segment, the length is the length of the segment is AH,“ e length is AH;f"’""z“""" .Numerically, for water, using C "4“” AT , and for the liquid + gas segment, th
and 40650 J mol'l. P,m
Table 8.2, the relative lengths are 6010 J mol'1 7530 J rnol'1 ,
—> d a e ——> f of the P—V—T phase diagram of Q8.4) Showthepathsn—aoap—éqandaabac Figure 8.13 in the PaT phase diagram of Figure 8.4. 173 ' th Relative Stability of Solids, Liquids, and Gases
Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases Chapter 8/ Phase Diagrams and e . .  ‘7
h physical origin of the pressure difference across a curved liquid—gas interface.
5 e smiicaé point «w s
1
f} . ‘ ro
* hesive forces in the liquid. Across an interface, the resultant force vector is not ze
he CO If th interface is curved the resultant force tends to minimize the surface area, leading to
. e , rence across the interface. . ‘ n .
nt refers to a point in a P4” phase diagram for which three phases are in equilibrium, s“ ﬁstsmite lepOi . . .
pints correspond to a gasliquid—solid equilibrium? _ several solid phases there can be a triple point corresponding to equilibrium between 3
re ’ K?
does the hquid—gas coexistence curve in a PwT phase diagram end at the critical point. ”temperature point the gas and liquid have the same density, and can no longer be distinguished. This Q8.5) At a given temperature, a liquid can coexist with its gas at a single value of the press all pomts on the liquid—gas coexistence curve extended beyond the critical point. However, you can sense the presence of H20(g) above the surface of a lake by the humidity, and _ can you get a P— T phase diagram from a P—V Tphase diagram? still there if the barometric pressure rises or falls at constant temperature. How is this poss1ble. Tphase diagram by projecting a P—V— T phase diagram on the P—T plane The statement that at a given temperature, a liquid can coex1st With its gas at a Single value of the y does the triple point in a PﬁT diagram become a triple line m a P V diagram. pressure holds for a system with only one substance. For the case described, the system consists of W ase because a gas and a liquid are in equilibrium for the range of values of volume from that and air. The chan e in barometric ressure is e uivalent to an external ressure exerted on a li uid_
g p q p q iquid to that of the pure gas at a given temperature. discussed in Section 8.5, this will change the vapor pressure only slightly. y does water have several different solid phases but only one liquid and one gaseous phase? (28.6) Why are the triple point temperature and the normal freezing point very close in temperatur ‘ aSeous phases are disordered, so that except for special cases such as superﬂuid and normal most substances? s no way to distinguish two different liquid or gaseous samples of the same pure substance This is the case because the freezing point changes only slightly with pressure. the pressure is increased at _4 50 C ice I is converted to ice 11 Which of these phases has the (28.7) Give a molecular level explanation as to why the surface tension of Hg(l) is not zero. 9
ty . I 
density phase is more stable as the pressure is increased according to LeChatelier s prinCiple. nei hbors than an atom in the li uid. The ex erience a net force e endicular to the s rface that 
g (1 y p p rp 11 Ce I is the less dense phase. sublimation furzon vaporization ?
is AHm — —AH + AH”, m lead to a pressure increase inside the droplet. H isa state function, AH has the same value for solid ——> gas or solid —> liquid —’ gas as long as larger than the critical volume, and c) slightly larger than the critical volume.
a) No, because the liquid and gas are not distinguishable above the critical density. b) No, because all the liquid will evaporate. . .   ' 1 l te the enthalpy of
c) Yes, because the denSity is less than the critical density the vapor pressures of C1F3 glven 1“ the followmg table to ca cu a Q83) Why are there no points in the phase diagram for sulfur in Figure 8.9 that show rhombic 11 using a graphical method or a least squares fitting routine. monoclinic solid phases in equilibrium with liquid and gaseous sulﬁir? According to the Gibbs phase rule, four phases of a pure substance cannot coexist. 174 175 Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases , dlnP _ 8.314 J mol‘1 K" x(298 K)2 x2.7738><103 (ion 298K :RT_____._ ’
_ ( ) dT (298—5308)2 11;} mol" at 298 K 8.314 1 4 , 2 . 3
(352.8 K) = ”2 LnP _w
0” (352.8—53.08) :31}?
4.25 0k] mol”1 at 335.9 K Section 8.8, it is Stated that the maximum height of a water column in which cavitation does 4.38 _I
_, _ 11
$83 8“?” 4x63 ccur is ~9.7 KL Show that this is the case at 298 K,
bining the two parts of equation 8.28, K5888
3.313
3.583 bles will form when Pom, has the value of the vapor pressure of water. 3.25
£38368 85.8843 {3.68852 {3.38533 {3.884% the equilibrium
15mg”) 0(1) +— 1120(8)
. . . 1 . va oriza ion __ '1
A least squares ﬁt of In P versus 1/ T in units of K gives the result AHmp ’  30.58 kJ mol . _ Gf (H20, g)  AG} (H20, D 2 —228.6 k] mol'1+ 237.1 M mol'1= 8.5 kl mol'l. P8.2) The vapor pressure of benzene (l) is given by 3
ln .1: = 20.767 — 2;”; X10 AGjmm, _
Pa K 4 53.08 RT ‘ H204
—8.5 xl03 J mol“ exp —“__—_Ti_'——i_—'_——__
8.314Jm0l K X298.15K a. Calculate the standard boiling temperature.
4 0.032 bar = 3242 Pa b. Calculate AHXPWWW at 298 K and at the standard boiling temperature. , uppose that Pimer = 1 atm = 101325 Pa a.
. s Usin =998k '3
ln ~15— :20.767—2T7738X10 =111105 ath ‘gp gm
a 16—5308 W 100
998kgm3x981ms'2‘ ‘ m '4) Use the vapor pressures for PbS given in the following table to estimate the temperature and 3
27738)“) +5308 = 352.8 K
n, vaporization, and sublimation. T” : 20.767—1n(105) Ssure of the triple point and also the enthalpies of fusio 176 177 ‘ Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases Phase T (°C) P (Torr) 'olid ,ref Solid 1048 40.0
Solid 1108 100. _i
 T . , T
IqUid 1221 400.11qmd,rcf
1quid 1281 760. [in ‘PsolidJLf/
To estimate AH,;”"””’””"” , use the vapor pressures of the solid phase
+111 Bu 11id,ref’
In 3: = __ sublimation __1_ _ i : L‘ I ‘
[)1 R 112 T], i‘ i “ i yblimation AHmpm‘izaI/on
i — ”1 +Rln PM “1,2131 _
Rh’lﬁ 8.314 J m01_1 K X In 100 Torr lulu/(114 Id ./ llqzud,rcf
AHsublimat/‘on = ID] _ 400 TOIT
m 1 1 1 — 1
f7; (1048+27315) K (1108+273.15) K
= 231.7 kJ mol'1
To estimate AH,:;“P””Z“"”" , use the vapor pressures of the liquid phase  207 X103 J m 01.1
1494 K
P 0
Rlni 8.314 J mol‘1 K‘1 xln 1
AHvaporizalion = P1 400
”2 1 1 1 ~ 1
f‘i‘ 1221 +273.15 1281 +273.15
AH,:“"””Z“”"" : 206.5 kJ mol‘1 _
AH £15m” : AHriublimation __ AHfzaparization _ I 1 ' .' : V ‘  i from either the equation for vapor pressure of the 801i (1 .
= 231.7 kJ mol‘1 —206.5 kJ mol‘1 . . , ,
= 25.2 kJ mol‘1 At the triple point, the solid and liquid have the same vapor pressure. Increased from 65 to 80 atm no interface Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases Chapter 8/ Phase Diagrams and the Relative Stability of Solids, Liquids, and Gases T (°C) P (Torr)
1048 40.0 sublimation 1 1
AH!» _ _ M R Two/Id ,rcff' T Liquid 1221 400. T T,,q,,,d,rq; T
l
  ”'— + 1n Bro/idjef
To estimate AH 51”!” ””0““ , use the vapor pressures of the solid phase T
l 1 1 P
In E 2 __ AHsublimation _l____1__ T ——_ + n liquid,ref'
R R I; I; quu/dJeff
AH’21ublimulion AH,:;ap0/~izalmn
R Inf: 8.3 14 J IIIOI—1 Kll X In *100 TOIT (ion sublimation “”‘iw—__ — —]—_'——————————— + R111 [JAB/Mid _ R111 qull'dﬂ’f
AHsublimation 2 B = 400 TOIT ‘_ ' _ AH,” solia’Jef liquid ,ref
”1 1 1 1 — 1
f”f (1048+273.15) K (1108+273.15) K
= 231.7 k] mm“ Emu/1114' = 400 Torr
Tliquid ref 2 1494 K
' vaporization . . 7 i
To estlmate AH”, , use the vapor pressures of the liquid phase 232 X 103 J mol'1 207 X103 J mol‘1
>< ___—____ _ ____________
P x103 Jmol" 1381 K 1494K
18er2 8.314 J mol‘1 K'1 xln % ) 00 T
AH];aporizat/on I 1 z 1‘] K“! 1 1 . OIT
i_i 1 _ 1 “mo X n 400. Torr
Ti 712 l221+273.15 1281 +273.15
AH;W’Z""“" = 206.5 kJ mor‘
AH fusion :: AHsub/imalion _ AHvaporizaIion _ .
”7 231m7 k J 11 :06 5 k I 1—1 ‘e at the triple point can be obtained from either the equation for vapor pressure of the solid
2 . mo — . mo
2 25.2 kJ mol‘1
At the triple point, the solid and liquid have the same vapor pressure. AHiiilbhmmm 1 . ._ .1.
R TrolldJef T
232 x 103 J moi" 1 1 X __._...__
8.314Jmo1"‘K'1 1381K 1398K pressure on a cylinder containing pure C02 is increased from 65 to 80 atm, no interface 178 179 ...
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