HW8 Answer Chem340 Spring 2011 (Version 2 Updated 4/11/2011)
P6.15) Consider the equilibrium CO g H 2O g CO2 g H 2 g . At 1000 K,
the composition of the reaction mixture is
CO2 H2(g) CO(g) H2O(g) Substance (g)
27.1 27.1 22.9 22.9 Mole %
a. Calculate KP and Greaction
at 1000 K.
b. Given the answer to part (a), use the H f of the reaction species to calculate
Greaction
at 298.15 K. Assume that Hreaction is independent of temperature.
(a)
11 80.2)4.1ln(1000314.8ln
4.19.229.22
1.271.27
2
22
KJmolKJKmolKRTG
xx
xxK
poreaction
OHCO
HCOp
(b)
1-1-1-
oreactionf,1000
oreaction15.298
1oCf,
oHf,
oCf,
oHf,
oreactionf,
mol -29.7kJ)1000K
1-
298
1() mol kJ 2.41(15.298 ) mol kJ 80.2(
1000K
298.15K
)1000K
1-
298
1(ΔH ,G
1000K
298.15K ,
2.41)5.110()8.241(5.3930ΔHΔHΔHΔHΔH222
KK
KG
KJmol
KKoreaction
OOO
P6.20) Calculate the Gibbs energy change for the protein denaturation described in
Problem 5.45 at T = 310.K and T = 340.K.
From P5.34 we have:
-1-1den mol K J 4.1109 K 310ΔS
-1-1den mol K kJ 43.93 K 310ΔH
K 310ΔGden is then:
1-1-1-1-
dendenden
mol kJ 3.31 mol K J 4.1109K 982mol kJ 43.93
K 310ΔS TK 310ΔHK 310ΔG
At 340 K we obtain:
1-
1--1
reaction
mol kJ 87.47
K 310
1
K 340
1mol kJ 343.9
K 310
mol J 13300K 340K 340 ΔG
P6.21) For a protein denaturation at T = 310. K and P = 1.00 atm, the enthalpy change is
911 kJ mol–1 and the entropy change is 3.12 J K–1 mol–1. Calculate the Gibbs energy
change at T = 310. K and P = 1.00 atm. Calculate the Gibbs energy change at T = 310. K
and P = 1.00103 bar. Assume for the denaturation V=3.00 mL mol–1. State any
assumptions you make in the calculation.
The Gibbs energy change at T = 310. K and P = 1.00 atm is:
den den den
-1 -1 -1 -1
ΔG 310 K ΔH 310 K T ΔS 310 K
911.0 kJ mol 310 K 3.12 J K mol 910.0 kJ mol
The Gibbs energy change at T = 310. K and P = 1.00103 bar is:
3den den 0
-1 -6 3 -1 8 1
ΔG 310 K,1.00 10 bar ΔG 310 K,1.00 atm V p p
91.0 kJ mol 3.0 10 m mol 10 Pa 101325 Pa 910.3 kJ mol
We assumed that the volume change is independent over the pressure range of 1000 bar.
P6.26) In this problem, you calculate the error in assuming that Hreaction is
independent of T for a specific reaction. The following data are given at 25°C:
CuO(s) Cu(s) O2(g)
H f
kJ mol1 –157
G f
kJ mol1 –130
CP,m J K1 mol1 42.3 24.4 29.4
a. From Equation (6.71),
d ln KPK
PT
0
K P Tf
1
R
Hreaction
T2T0
Tf
dT
To a good approximation, we can assume that the heat capacities are independent of
temperature over a limited range in temperature, giving
Hreaction
T Hreaction T0 CP T T0 where CP viCP,m i i . By
integrating Equation (6.71), show that
ln KP T ln K P T0 Hreaction
T0 R
1
T
1
T0
T0 CP
R
1
T
1
T0
CP
Rln
T
T0
b. Using the result from part (a), calculate the equilibrium pressure of oxygen over
copper and CuO(s) at 1200 K. How is this value related to KP for the reaction
2CuO s 2Cu s O2 g ?
c. What value would you obtain if you assumed that Hreaction were constant at its
value for 298.15 K up to 1200 K?
0
0
0 0 0
2
0 2
02 2
0
0
1ln
1
1 1l
f f
f
f f f
T T
reactionP
T T
T
reaction PP f P
T
T T T
reaction P P
T T T
reaction P
f
Ha) d K dT
R T
H C T T ln K T ln K T dT
R T
H T C C TdT dT dT =
R T R T R T
H T C =
R T T R
0
0 0
1 1n f P
f
T C T
T R T T
b) 2CuO(s) 2Cu(s) + O2(g)
3 1 3 10
, , 2 ,
1 1
1 1
2 157 10 J mol 314 10 J mol
2 Cu, O , 2 CuO,
2 24.4 29.4 2 42.3 J K mol
6.4 J K mol
reaction
P P m P m P m
H T
C C s C g C s
3 1
1 1
3 1 1 1
1 1 1 1
1 1
2 130 10 J molln 1200 K
8.314 J mol K 298.15 K
2 157 10 J mol 1 1 6.4 J K mol 1200 K ln
8.314 J K mol 1200 K 298.15 K 8.314 J K mol 298.15 K
6.4 J K mol 298.15 K
8
PK
2
2
1 1
5
5
1 1
.314 J mol K 1200 K 298.15 K
ln 1200 K 10.1818
1200 K 3.78 10
3.78 10 bar
P
OP
O
K
PK
P
P
c) This is equivalent to setting CP = 0. Neglecting the last two terms in the calculation above gives ln KP
= –9.6884 and 2OP = 6.20 10–5 bar.
P6.39) You place 2.00 mol of NOCl(g) in a reaction vessel. Equilibrium is established
with respect to the decomposition reaction NOCl g NO g 1 2 Cl2 g .
a. Derive an expression for KP in terms of the extent of reaction .
b. Simplify your expression for part (a) in the limit that is very small.
c. Calculate and the degree of dissociation of NOCl in the limit that is very small at
375 K and a pressure of 0.500 bar.
d. Solve the expression derived in part (a) using a numerical equation solver for the
conditions stated in part (c). What is the relative error in made using the
approximation of part (b)?
a) Obtain an expression for KP in terms of the degree of advancement . NOCl(g) → NO(g) + 1/2 Cl2(g) Initial number of moles 2.00 0 0
Moles present at equilibrium 2.00 – 1/2 Mole fraction present
at equilibrium 2.00
12
2
12
2
1
21
22
Partial pressure at
equilibrium, i iP x P 2.00
12
2
P
1
22
P
1
21
22
P
We next express KP in terms of and P.
2
1
2
1
2
1
21 1
2 22 2
2.001
22
eqeqClNO
P eqNOCl
P PPP P P
P PK T
PPPP
b)
1
2
1
2
1
21
32
2
1
21 1
2 2 2 2 22 22.00
2.001
22
12 4 if 2
4
P
P PPP PP
K T
P
P
P PPP P
P PP
c) Calculate and the degree of dissociation of NOCl in the above limit at 375 K and a pressure of 0.500 bar.
3 1 3 1
3 1
NO, NOCl, 87.6 10 J mol 66.1 10 J mol
21.5 10 J mol
reaction f fG G g G g
3 1 3 1
3 1
NO, NOCl, 91.3 10 J mol 51.71 10 J mol
= 39.6 10 J mol
reaction f fH H g H g
3 1 3 1
1 1 1 1
298.15 K 1 1ln
298.15 K 298.15 K
21.50 10 J mol 39.6 10 J mol 1 1ln 375 K
8.314 J K mol 298.15 K 8.314 J K mol 375 K 298.15 K
ln 375 K 5.399
375 K 4.
reaction reactionP f
f
P
P
P
G HK T
R R T
K
K
K
352 10
13
22
221 3
3 32 2
1
4
4 4.52 104 8.68 10
0.500
P
P
PK
P
PK
P
The degree of dissociation is 0.045.2
d) Solve the expression derived in part a) using a numerical equation solver for the conditions stated in the previous part. What is the relative error in made using the approximation of part b)? Solving the expression of part a) without approximations gives = 8.4910–2.
The relative error is 2 2
2
8.68 10 8.49 102.2%.
8.49 10
P7.1) In this problem, you will calculate the differences in the chemical potentials of ice
and supercooled water, and of steam and superheated water all at 1 bar pressure shown
schematically in Figure 7.1. For this problem, SH2O,s 48.0 J mol1 K1,
SH2O,l 70.0 J mol1 K1 and
SH2O,g 188.8 J mol1 K1.
a. By what amount does the chemical potential of water exceed that of ice at –5.00°C?
b. By what amount does the chemical potential of water exceed that of steam at
105.00°C?
a) By what amount does the chemical potential of water exceed that of ice at –5.00ºC?
1
1111,,
110
)00.5()0.480.70())((
)0(
22
Jmol
KmolJKmolJKTSSG
PTSGo
sOHo
lOHm
mm
b) By what amount does the chemical potential of water exceed that of steam at 105.00ºC?
1
1111,,
594
)00.5()8.1880.70())((22
Jmol
KmolJKmolJKTSSG ogOH
olOHm
P7.3) Within what range can you restrict the values of P and T if the following
information is known about CO2? Use Figure 7.8 to answer this problem.
a. As the temperature is increased, the solid is first converted to the liquid and
subsequently to the gaseous state.
b. As the pressure on a cylinder containing pure CO2 is increased from 65 to 80 atm,
no interface delineating liquid and gaseous phases is observed.
c. Solid, liquid, and gas phases coexist at equilibrium.
d. An increase in pressure from 10 to 50 atm converts the liquid to the solid.
e. An increase in temperature from –80° to 20°C converts a solid to a gas with no
intermediate liquid phase.
a) The temperature and pressure are greater than the values for the triple point, –56.6ºC
and 5.11 atm.
b) The temperature is greater than the critical temperature, 31.0ºC.
c) The system is at the triple point, –56.6ºC and 5.11 atm.
d) The temperature is slightly greater than the triple point value of –56.6ºC.
e) The pressure is below the triple point pressure value of 5.11 atm.
P7.5) The vapor pressure of liquid SO2 is 2232 Pa at 201 K, and Hvaporization = 24.94 kJ
mol–1. Calculate the normal and standard boiling points. Does your result for the normal
boiling point agree with that in Table 7.2? If not, suggest a possible cause.
3 -1
, 3 -1-1 -1
-1
1 1ln
ln
At the normal boiling point, = 101325 Pa.
24.94 10 J mol
24.94 10 J mol8.314 J mol K
8.314 J mol
vaporizationf m
i f i
vaporizationm
f vaporizationfm
i i
b normal
P H
P R T T
HT
PHR
RT P
P
T
-1
5
3 -1
, tan 3 -1–1 –1
–1 -1
270.0 K101325
lnK 201K 2232
At the standard boiling point, = 10 Pa.
24.94 x10 J mol269.7 K
24.94 10 J mol 1000008.314 J mol K ln
8.314 J mol K 201K 2232
b s dard
P
T
The result for the normal boiling point is ~7 K higher than the value tabulated in Table
7.2. The most probable reason for this difference is that the calculation above has
assumed that Hvaporization is independent of T.
P7.6) For water, Hvaporization is 40.65 kJ mol–1, and the normal boiling point is 373.15 K.
Calculate the boiling point for water on the top of a mountain of height 5500 m, where
the normal barometric pressure is 380 Torr.
3 -1
, 3 -1-1 -1
-1
1 1ln
ln
At the normal boiling point, = 760 Pa.
40.656 10 J mol
40.656 10 J mol8.314 J mol K
8.314 J mol K
vaporizationf m
i f i
vaporizationm
f vaporizationfm
i i
b normal
P H
P R T T
HT
PHR
RT P
P
T
-1
354.4 K380 Torr
ln373.15K 760 Torr
P7.7) Use the values for Gf
(ethanol, l) and Gf (ethanol, g) from Appendix B to
calculate the vapor pressure of ethanol at 298.15 K.
For the transformation C2H5OH (l) → C2H5OH (g)
2 5
2 5
( ) 2 5 2 5
3 -1 3 -1
–1 -1
( ) 3
ln C H OH, C H OH,ln
167.9 10 J mol + 174.8 10 J mol2.785
8.314 J mol K 298 K
0.0617 bar = 6.17 10 Pa1 bar
C H OH g f fP
C H OH gP
P G g G lK
P RT
PK
P7.13) Carbon tetrachloride melts at 250 K. The vapor pressure of the liquid is 10,539
Pa at 290 K and 74,518 Pa at 340 K. The vapor pressure of the solid is 270 Pa at 232 K
and 1092 Pa at 250 K.
a. Calculate Hvaporization and Hsublimation.
b. CalculateHfusion.
c. Calculate the normal boiling point and Svaporization at the boiling point.
d. Calculate the triple point pressure and temperature.
a) Calculate� Hvaporization and Hsublimation.
-1 -1
3 -1
1 1ln
ln
1 1
74518 Pa8.314 J mol K ln
10539 Pa = 32.1 10 J mol1 1
340 K 290 K
vaporizationf m
i f i
f
vaporization im
f i
vaporizationm
P H
P R T T
PR
PH
T T
H
-1 -1
3 -1
1 1ln
ln
1 1
1092 Pa8.314 J mol K ln
270 Pa = 37.4 10 J mol1 1
250 K 232 K
sublimationf m
i f i
f
sublimation im
f i
sublimationm
P H
P R T T
PR
PH
T T
H
b) Calculate Hfusion.
Hfusion = Hsublimation – Hvaporization = 37.4 103 J mol-1 – 32.1 103 J mol-1
= 5.3 103 J mol-1.
c) Calculate the normal boiling point and Svaporization at the boiling point.
3 -1
, 3 -1–1 -1
–1 -
1 1ln
ln
At the normal boiling point, = 101325 Pa.
32.1 10 J mol
32.1 10 J mol8.314 J mol K
8.314 J mol K
vaporizationf m
i f i
vaporizationm
f vaporizationfm
i i
b normal
P H
P R T T
HT
PHR
RT P
P
T
1
3 -1–1 -1
349.5 K101325 Pa
ln340 K 74518 Pa
32.1 10 J mol = 91.8 J mol K
349.5 K
vaporizationvaporization mm
vaporization
HS
T
d) Calculate the triple point pressure and temperature. From Example Problem 7.2,
3 -1 3 -1
3
5 5-1 -1
ln ln
32.1 10 J mol 37.4 10 J mol
10539 Pa 270 Pa 37.4 10ln ln
10 Pa 10 Pa8.314 J K mol
vaporization sublimationm m
tp liquid solid sublimation vaporizationi i m m
solid liquidi i
tp
H HT
P P H HR
P P RT RT
T
-1
–1 -1
3 -1
–1 -1
3 -1
–1 -1
264 KJ mol
8.314 J K mol 232 K
32.1 10 J mol+
8.314 J K mol 290 K
1 1ln
32.1 10 J mol 1 1ln 1.3112
10539 8.314 J K mol 264 K 290 K
1053
vaporizationtp m
i tp i
tp
tp
P H
P R T T
P
Pa
P
3
0.2699 Pa
2.84 10 PatpP
P7.19) A protein has a melting temperature of Tm = 335 K. At T = 315 K, UV
absorbance determines that the fraction of native protein is fN = 0.965. At T = 345. K, fN =
0.015. Assuming a two-state model and assuming also that the enthalpy is constant
between T = 315 and 345 K, determine the enthalpy of denaturation. Also, determine the
entropy of denaturation at T = 335 K. By DSC, the enthalpy of denaturation was
determined to be 251 kJ mol–1. Is this denaturation accurately described by the two-state
model?
We first calculate the equilibrium constants at 315 K and 345 K:
0.036
0.965
0.965-1
f
f-1
f
fK 315K
N
N
N
D
67.56
0.015
0.015-1
f
f-1
f
fK 345K
N
N
N
D
The enthalpy of denaturation can now be calculated using equation from example
problem 7.8:
-1 -1
-1
2 1
K 345 K 65.67ln R ln 8.314472 J mol K
K 315 K 0.036H 226.2 kJ mol
1 1 1 1T T 345 K 315 K
This result deviates from the DSC result, indicating that the denaturation process is not
accurately described by a two-state model.
P7.21) The vapor pressure of methanol(l) is given by
lnP
Pa
23.593
3.6791 103
T
K 31.317
a. Calculate the standard boiling temperature.
b. Calculate Hvaporization at 298 K and at the standard boiling temperature.
a)
35
3
3
3.6791 10ln 23.593 ln10 11.5129
Pa 31.317K
3.6791 10 23.593 11.5129 31.317K
3.6791 1031.317 335.9
K 23.593 11.5129
b
b
b
PT
T
T
b)
21 –12
2
–1 -1
8.314 J mol K 3679.1 298 Kln
31.37
38.19 kJ mol at 298 K and 37.20 kJ mol at 335.9 K
vaporization
d PH RT
dT T
P7.22) Suppose a DNA duplex is not self-complementary in the sense that the two
polynucleotide strands composing the double helix are not identical. Call these strands A
and B. Call the duplex AB. Consider the association equilibrium of A and B to form
duplex AB
A B AB
Assume the total strand concentration is C and, initially, A and B have equal
concentrations; that is, CA,0 = CB,0 = C/2. Obtain an expression for the equilibrium
constant at a point where the fraction of the total strand concentration C that is duplex is
defined as f. If the strand concentration is 1.00 10–4 M, calculate the equilibrium
constant at the melting temperature.
We make the table of concentrations:
Cinitial Cequilibrium
AB 0 f C/2
A = B C/2 C/2 (1-f)
The equilibrium constant at the melting temperature with f = 0.5 is given by:
AB2
A B
C f C/2 4K
C C CC(1 )
2f
And for C = 1.00 10–4 M:
1
4
4 4K 40000mol
C 1 10 M
P7.33) Calculate the vapor pressure of CS2 at 298 K if He is added to the gas phase at a
partial pressure of 200 bar. The vapor pressure of CS2 is given by the empirical equation
lnP T
Pa 20.801
2.6524 103
T
K 33.402
The density of CS2 at this temperature is 1255.5 kg m–3. By what factor does the vapor
pressure change?
3
4
298 K 2.6524 10ln 20.801
Pa 298 33.402298 K
4.79 10Pa
P
P
00
0
–3 –15 4
–3
–1 –10
4 40
ln
76.14 10 kg mol200 10 Pa 4.789 10 Pa
1255 kg mln 0.488576
8.314 J mol K 298 K
1.6299 1.6299 4.789 10 Pa = 7.806 10 Pa
liquidliquidm
MP
V PP
P RT RT
P
P
P P
PP
P7.40) Calculate the difference in pressure across the liquid–air interface for a water
droplet of radius 150 nm.
From Equation (7.24)
3 -15
–9
2 2 71.99 10 Nm9.60 10 Pa
150 10 minner outerP Pr
P7.41) Calculate the factor by which the vapor pressure of a droplet of methanol of
radius 1.00 10–4 m at 45.0°C in equilibrium with its vapor is increased with respect to a
very large droplet. Use the tabulated value of the density and the surface tension at 298 K
from Appendix B (table 7.3) & table 7.5 (p154) in Ch 7 for this problem. (Hint: You
need to calculate the vapor pressure of methanol at this temperature.)
3
4
2 3.6971 10ln 1 23.593 11.0043
Pa 325 31.3173K
6.01 10 Pa
P T AA
TA
P
Pam
Nm
rP 4
4
13
1041410
10072222
..
Pinside = Pvapor + P = 1.041 105 Pa
–3 -16
–8
4 6 6
2 2 22.07 10 N m = 4.41 10 Pa
10 m
6.01 10 Pa + 4.41 10 Pa = 4.47 10 Painside vapor
Pr
P P P
For a very large droplet, P → 0, and the vapor pressure is P0 = 46.01 10 Pa . For the
small droplet, the vapor pressure is increased by the factor
02
0
2
11
6
3
13
0
0681106076
1060763253148
104144791
1004322
PPP
KKJmol
Pakgm
kgmol
RTrMPP
.).exp(
..
).(.
.)(
)/ln(
P7.44) Calculate the vapor pressure for a mist of water droplets, where the droplets are
spherical with radius 1.00 10–8 m. Assume T = 293 K. The vapor pressure of water with
a flat interface is 23.75 Torr.
To obtain the vapor pressure of the droplets we use:
T R r
M 2
p
pln
0
0
3 -1 3 1
8 -3 1 1
2 Mp Exp p
r ρ R T
2 71.99 10 mN m 18.02 10 kg molExp 23.75 Torr
1.0 10 m 1000 kg m 8.314472 J mol K 293 K
26.4 Torr
γ
P7.45) In Example Problem 7.8 we found that the surface tension of solutions of
aliphatic acids is given by 0 – = a log(1 + bc2) where c2 is the solute concentration and
0 is the surface tension of pure water. The constants a and b are determined at T = 291 K
for n-butanoic acid: a = 0.0298 N m–1 and b = 19.6 L mol–1. Calculate the surface
adsorption for c2 = 0.01M, 0.10M, 0.20M, 0.40M, 0.8M, and 1.0M. What is the
maximum surface adsorption that can be achieved in solutions of n-butanoic acid?
Example problem 7.7 indicated that the surface adsorption, , can be obtained for
aliphatic acids. For a concentration of 0.01 M we obtain:
2-7-
1-
1-1-
11
2
2
m mol 108.76435
M 0.01 mol L 19.612.303
mol L 19.6m N 0.0298
K 912K mol J .3144728
M 0.01
c b1 2.303
b a
T R
cΓ
For the other concentrations in the lists:
c2 [M] 0.01 0.1 0.2 0.4 0.8 1.0
[mol
m-2]
8.76435×10-
7
3.54127×10-
6
4.26104×10-
6
4.74306×10-
6
5.02741×10-
6
5.08843×10-6
P7.49) The table lists the chemical potential difference water ethanol
for transferring
various amino acids from water into ethanol. T = 298 K.
Amino Acid Glycine Alanine Valine Phenylalanine Proline
water ethanol
–19.4 –16.3 –12.3 –8.28 –8.61
kJ mol1
Source: Klotz, I. M. “Energy Changes in Biochemical Reactions.” New York, Academinc
Press, 1967.
All of the quantities in the table are negative because all amino acids consist of amino
and carboxyl groups, both of which are polar, in addition to side chains, which may be
polar or nonpolar. Assume the contributions to water ethanol
from the polar and
nonpolar groups are simply additive. Assume further that for glycine only polar groups
contribute to water ethanol
because a side chain is essentially absent. For each amino
acid, calculate the contribution to water ethanol
from the side chain alone. Assuming
ethanol is representative of the interior of a protein, explain how amino acids with non-
polar side chains contribute to folding. If in a protein a phenylalanine were replaced by a
glycine, would the stability of the folded state be increased or decreased? Explain.
The contribution to water ethanol
from the side chains alone in each of the amino
acids in the table are:
-1-1-1 mol kJ 1.3mol kJ 9.41mol kJ 3.61 alanineethanolwater
-1-1-1 mol kJ 1.7mol kJ 9.41mol kJ 3.21 valineethanolwater
-1-1-1 mol kJ 12.11mol kJ 9.41mol kJ 28.8 inephenylalanethanolwater
-1-1-1 mol kJ 79.10mol kJ 9.41mol kJ 61.8 prolineethanolwater
Hydrophobic sidechains in the protein contact each other in order to avoid contact with
hydrophilic groups and thereby contribute to the folding process. If the substitution of a
phenylalanine with a glycine, would increase or decrease depends on the total
composition of the protein.
Z: Please check the last statement.
Q1. (Modified from P7.11)
Use the vapor pressures of n-butane given in the following table. (a) Calculate the enthalpy of vaporization using a graphical method or a least-squares fitting routine. (b) How much is the standard boiling temperature of this compound? Use the equation you obtained from (a).
T (K) P (Torr) T (K) P (Torr)
187.45 5.00 220.35 60.00
195.35 10.00 228.95 100.00
204.25 20.00 241.95 200.0
214.05 40.00 256.85 400.0
(a)
4.0x10-3 4.5x10-3 5.0x10-3 5.5x10-31
2
3
4
5
6
7
Y = 17.8663 - 3040.12181 x X
ln(p
)
1/T
A least squares fit of ln P versus 1/T gives the result vaporizationH = (-3040.12181)x
8.314=25.28 kJ mol–1.