HW8 Answer Chem340 Spring 2011 (Version 2 Updated 4/11/2011)


P6.15) Consider the equilibrium CO g  H 2 O 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


at 298.15 K. Assume that Hreaction
is independent of temperature.
Greaction
(a)
Kp 
xCO2 x H 2
xCO x H 2O

27.1  27.1
 1 .4
22.9  22.9
o
G reaction
  RT ln K p  8.314 JKmol 1  1000 K  ln(1.4)  2.80 KJmol 1
(b)
ΔH of,reaction  ΔH of,H 2  ΔH of,CO2  ΔH of,H 2O  ΔH of,CO  0  393.5  (241.8)  (110.5)  41.2 KJmol 1
298.15K o
1
1
G reaction ,1000 K  ΔH of,reaction (
)
1000K
298 K 1000K
298.15K
1
1

 (2.80 kJ mol -1)  298.15 K  (41.2 kJ mol -1)(
)  -29.7kJ mol
1000K
298 K 1000K
o
, 298.15 K 
G reaction
-1
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:
ΔSden 310 K   1109.4 J K -1 mol -1
ΔH den 310 K   343.9 kJ K -1 mol -1
ΔG den 310 K  is then:
ΔG den 310 K   ΔH den 310 K   T ΔS den 310 K 




 343.9 kJ mol -1  298 K   1109.4 J K -1 mol -1  13.3 kJ mol -1
At 340 K we obtain:

 
 13300 J mol -1
 1
1 

ΔG reaction 340 K   340 K   
 343.9 kJ mol -1  

310 K 
 340 K  310 K  

 47.87 kJ mol -1

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 = 1.00103 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:
ΔG den  310 K   ΔH den  310 K   T ΔSden  310 K 
  911.0 kJ mol-1    310 K    3.12 J K -1 mol-1   910.0 kJ mol-1
The Gibbs energy change at T = 310. K and P = 1.00103 bar is:
ΔG den  310 K,1.00  103 bar   ΔG den  310 K,1.00 atm   V  p  p0 
  91.0 kJ mol-1    3.0 10-6 m3 mol-1   108 Pa  101325 Pa   910.3 kJ mol1
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 H reaction
is
independent of T for a specific reaction. The following data are given at 25°C:
CuO(s)


–157


–130
H f kJ mol1
Gf kJ mol1

C P,m J K 1 mol 1

42.3
Cu(s)
O2(g)
24.4
29.4
a. From Equation (6.71),
T
 

1  Hreaction
d
ln
K

dT
P

R
T2
KP Tf
f
 
KP T0
T0
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
integrating Equation (6.71), show that
 where CP   i viC P,m i . By

 
ln K P T  ln K P T0 

 

T0  1 1 
Hreaction
T  T 
R

0
T0  C P  1 1  CP T
 T  T   R ln T
R

0
0
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



2Cu s  O 2 g ?
2CuO s


c. What value would you obtain if you assumed that H reaction
were constant at its
value for 298.15 K up to 1200 K?
Tf
a)

T
Tf

H reaction
1
d ln K P  
dT
R T0
T2

f
H reaction
 CP T  T 
1
ln K P T f   ln K P T0   
dT
R T0
T2
T

H reaction
T0  f dT CP



=
R
T2
R
T0
T
=
Tf

T0
dT CPT

T
R
Tf

T0
dT
T2

H reaction
T0   1  1   CP ln T f  CPT0  1  1 




T
R
T
T
R
T
T
R
0
f
f
0
0




b) 2CuO(s) 
 2Cu(s) + O2(g)

H reaction
T0   2  157  103 J mol1  314  103 J mol1
CP  2CP ,m  Cu, s   CP , m  O 2 , g   2CP ,m  CuO, s 
  2  24.4  29.4  2  42.3 J K 1 mol1
 6.4 J K 1 mol1
2  130  103 J mol1
ln K P 1200 K   
8.314 J mol1 K 1  298.15 K
2  157  103 J mol1  1
1
6.4 J K 1 mol1
1200 K




ln

1
1 
1
1
8.314 J K mol  1200 K 298.15 K  8.314 J K mol
298.15 K

6.4 J K 1 mol1  298.15 K  1
1




1
1
8.314 J mol K
1200 K 298.15 K 
ln K P 1200 K   10.1818
K P 1200 K  
PO2  3.78  105
PO2
P
bar
 3.78  105
c) This is equivalent to setting CP = 0. Neglecting the last two terms in the calculation above gives ln KP
= –9.6884 and
PO2 = 6.20  10–5 bar.
P6.39) You place 2.00 mol of NOCl(g) in a reaction vessel. Equilibrium is established



NO g  1 2 Cl 2 g .
with respect to the decomposition reaction NOCl 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)
Initial number of moles
2.00
→
NO(g)
0
+
1/2 Cl2(g)
0
Moles present at
equilibrium
2.00 – 

1/2 
Mole fraction present
2.00  
1
2 
2
at equilibrium
1

2
1
2 
2

1
2 
2
Partial pressure at

 2.00  

1
 2 
2

equilibrium, Pi  xi P


P



 

1
2 
2



P


 1
 2

1
2 
2



P


We next express KP in terms of  and P.
eq
eq
 PNO
  PCl2
    
 P  P
K P T  
eq
 PNOCl

  
 P 
1
2



  1
   P  2


1   
1
 2    P   2  
2  
2




 2.00    P

1  
 2  P

2 
1
 2
 P
 
 P 
 
b)

  1
   P  2


1   
1
 2    P   2  
2  
2

K P T  


 2.00    P

1  
 2  P
2 

1
 2
 P
 
 P 
 
1


P 2


2  2   2  P 


2.00  
1
 P   P 2
1
3


1 2  P 2
2 P  4 P 

     if   2
P
4 P 

P
c) Calculate  and the degree of dissociation of NOCl in the above limit at 375 K
and a pressure of 0.500 bar.

Greaction
 G f  NO, g   G f  NOCl, g   87.6  103 J mol1  66.1  103J mol1
 21.5  103 J mol1

H reaction
 H f  NO, g   H f  NOCl, g   91.3  103 J mol1  51.71  103 J mol1
= 39.6  103 J mol1
ln K P  T f   


 1

Greaction
 298.15 K   H reaction
1
 

R  298.15 K
R
 T f 298.15 K 
21.50 103J mol 1
39.6 103J mol 1  1
1





1
1
1
1
8.314 J K mol  298.15 K 8.314 J K mol
375
K
298.15
K


ln K P  375 K   5.399
ln K P  375 K   
K P  375 K   4.52 103
3
1
1  P 2
KP   2   
4 P 


  4KP


2
2
3
 4  4.52  103  3
 P 
2

 

  8.68  10

P
0.500 
 


1
2
The degree of dissociation is

2
 0.045.
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  = 8.49  10–2.
The relative error is
8.68  102  8.49  102
 2.2%.
8.49  102
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, SH
SH
2 O ,l
 70.0 J mol1 K1 and SH
2O, g
1
2 O,s
 48.0 J mol
1
K ,
 188.8 J mol1 K 1.
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?
  Gm  S m T (P  0)
  Gm  ( S Ho 2O ,l  S Ho 2O , s )(T )  (70.0 JK 1 mol 1  48.0 JK 1 mol 1 )  (5.00 K )
 110 Jmol 1
b) By what amount does the chemical potential of water exceed that of steam at
105.00ºC?
  Gm  ( S Ho 2O ,l  S Ho 2O , g )(T )  (70.0 JK 1 mol 1  188.8 JK 1 mol 1 )  (5.00 K )
 594 Jmol 1
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.
H mvaporization  1 1 
  
Pi
R
 T f Ti 
H mvaporization
Tf 
P 
 H mvaporization
 ln f 
R
RTi
Pi 

ln
Pf

At the normal boiling point, P = 101325 Pa.
Tb ,normal 
24.94 103 J mol-1
 270.0 K
101325 
24.94 103 J mol-1
-1 -1 
 ln
8.314 J mol K  

-1 -1
2232 
 8.314 J mol K  201K
At the standard boiling point, P = 105 Pa.
Tb , s tan dard 
24.94 x103 J mol-1
 269.7 K
24.94 103 J mol-1
100000 
–1 –1 
 ln
8.314 J mol K  

–1 -1
2232 
 8.314 J mol K  201K
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.
H mvaporization  1 1 
ln

  
Pi
R
 T f Ti 
H mvaporization
Tf 
P 
 H mvaporization
R
 ln f 
RTi
Pi 

Pf
At the normal boiling point, P = 760 Pa.
Tb ,normal
40.656 103 J mol-1
 354.4 K

380 Torr 
40.656 103 J mol-1
-1 -1 
 ln
8.314 J mol K  

-1 -1
760 Torr 
 8.314 J mol K  373.15K


P7.7) Use the values for G f (ethanol, l) and G f (ethanol, g) from Appendix B to
calculate the vapor pressure of ethanol at 298.15 K.
For the transformation C2H5OH (l) → C2H5OH (g)
ln K P 
ln PC2 H5OH ( g )

G f  C2 H 5OH, g   G f  C2 H 5OH, l 
P
RT
3
-1
3
-1
167.9 10 J mol + 174.8 10 J mol

 2.785
8.314 J mol –1K -1  298 K
PC H OH ( g )
KP  2 5
 0.0617 bar = 6.17  103 Pa
1 bar
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. CalculateHfusion.
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.
H mvaporization  1 1 
  
Pi
R
 T f Ti 
P
R ln f
Pi
H mvaporization  
 1 1
  
 T f Ti 
ln
Pf

H mvaporization  
74518 Pa
10539 Pa = 32.1103 J mol-1
1 
 1



 340 K 290 K 
8.314 J mol-1K -1  ln
H msublimation  1 1 
ln

  
Pi
R
 T f Ti 
P
R ln f
Pi
H msublimation  
 1 1
  
 T f Ti 
Pf
H msublimation  
1092 Pa
270 Pa = 37.4 103 J mol-1
1
1





250
K
232
K


8.314 J mol-1K -1  ln
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.
H mvaporization  1 1 
ln

  
Pi
R
 T f Ti 
H mvaporization
Tf 
P 
 H mvaporization
 ln f 
R
RTi
Pi 

Pf
At the normal boiling point, P = 101325 Pa.
Tb ,normal 
32.1103 J mol-1
 349.5 K
101325 Pa 
32.1103 J mol-1
–1 -1 
8.314 J mol K  
 ln

–1 -1
74518 Pa 
 8.314 J mol K  340 K
Smvaporization 
H mvaporization 32.1103 J mol-1
= 91.8 J mol –1K -1

Tvaporization
349.5 K
d) Calculate the triple point pressure and temperature. From Example Problem 7.2,
Ttp 
 H
vaporization
m
 H msublimation 
 P liquid
P solid H msublimation H mvaporization 
R  ln i   ln i  


solid
P
P
RT
RTi liquid 
i

 32.110 J mol
3
Ttp 
-1
 37.4 103 J mol-1 
 10539 Pa

270 Pa
37.4 103 J mol-1
ln
ln




–1
-1
5
5
10 Pa
10 Pa 8.314 J K mol  232 K 
-1
-1 
8.314 J K mol 


32.1103 J mol-1
+

–1
-1
 8.314 J K mol  290 K

 264 K
H mvaporization  1 1 
  
Pi
R
 Ttp Ti 
Ptp
32.1103 J mol-1  1
1 
ln



  1.3112
–1
-1
10539 Pa
8.314 J K mol
 264 K 290 K 
Ptp
 0.269
10539 Pa
Ptp  2.84  103 Pa
ln
Ptp

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:
K 315 K  
f D 1 - f N  1 - 0.965


 0.036
fN
fN
0.965
K 345 K  
f D 1 - f N  1 - 0.015


 65.67
fN
fN
0.015
The enthalpy of denaturation can now be calculated using equation from example
problem 7.8:
 K  345 K  
  65.67  
-1
-1
ln 
ln 
  R 
  8.314472 J mol K 
K
315
K
0.036




H  
 
 226.2 kJ mol-1
 1 1


1
1

  


 T2 T1 
  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
 P
3.6791  103
ln    23.593 
T
 Pa 
 31.317
K
a. Calculate the standard boiling temperature.
b. Calculate Hvaporization at 298 K and at the standard boiling temperature.
a)
3.6791  103
 P 
ln    23.593 
 ln105  11.5129
T
 Pa 
b
 31.317
K
T

3.6791  103   23.593  11.5129   b  31.317 
K

Tb
3.6791 103

 31.317  335.9
K  23.593  11.5129 
b)
H vaporization
1
–1
d ln P 8.314 J mol K  3679.1  298 K 
 RT

2
dT
T  31.37 
2
2
 38.19 kJ mol –1 at 298 K and 37.20 kJ mol-1 at 335.9 K
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:
K
CAB
f C/2
4


2
CA CB  C
C

 (1  f ) 
2

And for C = 1.00  10–4 M:
K 
4
4

 40000mol1
C 1104 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
ln
  20.801  2.6524  103
PT
Pa
T
 33.402
K
The density of CS2 at this temperature is 1255.5 kg m–3. By what factor does the vapor
pressure change?
P  298 K 
2.6524  103
 20.801 
Pa
298  33.402
P  298 K 
 4.79  104
Pa
ln
M
 P V
ln   
P
liquid
m
0
P  P0   liquid
RT
P  P0 
RT
76.14  10 –3 kg mol –1
 200  105 Pa  4.789  104 Pa
 P
1255 kg m –3
 0.488576
ln   
8.314 J mol –1K –1  298 K
 P0 


P  1.6299 P0  1.6299  4.789  104 Pa = 7.806  104 Pa
P7.40) Calculate the difference in pressure across the liquid–air interface for a water
droplet of radius 150 nm.
From Equation (7.24)
Pinner  Pouter
2 2  71.99 103 Nm -1


 9.60  105 Pa
–9
r
150  10 m
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.)
P T 
A  2
3.6971  103
 A 1 
 23.593 
 11.0043
T
Pa
325
31.317

 A  3
K
P  6.01  104 Pa
ln
P 
2 2  22.07  10 3 Nm 1

 4.41  10 4 Pa
4
r
10 m
Pinside = Pvapor + P = 1.041  105 Pa
2 2  22.07 10 –3 N m -1
P 

= 4.41106 Pa
–8
r
10 m
Pinside  Pvapor  P  6.01104 Pa + 4.41106 Pa = 4.47 106 Pa
For a very large droplet, P → 0, and the vapor pressure is P0 = 6.01  104 Pa . For the
small droplet, the vapor pressure is increased by the factor
32.04  10 3 kgmol 1
 ( 4.41  10 6 )Pa
( )
3
791.4kgm
ln(P / P0 )  M r 
 6.607  10  2
1 1
RT
8.314Jmol K  325K
2
P  P0 exp(6.607  10 )  1.068P0
 2
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:
 p
ln
 p0
 2 M
 
 rRT
 2γ M 
p  Exp 
 p0
r ρ R T


2  71.99 103 mN m -1   18.02 103 kg mol1 

   23.75 Torr 
 Exp
8
1
1
-3
 1.0 10 m   1000 kg m    8.314472 J mol K    293 K  
 26.4 Torr
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:
Γ

c2
ab
R T 2.303 1  b c 2 
0.01 M 
8.314472 J mol
1
0.0298 N m  19.6 L mol 
 291 K  2.303  1  19.6 L mol  0.01 M 
-1
K 1
-1
-1
 8.76435  10 -7 mol m -2
For the other concentrations in the lists:
c2 [M]
0.01
0.1
0.2
0.4
0.8
1.0
 [mol
8.76435×10-
3.54127×10-
4.26104×10-
4.74306×10-
5.02741×10-
5.08843×10-6
m-2]
7
6
6
6
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 mol 
1
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 nonpolar 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:
  water    ethanol alanine    16.3 kJ mol -1    19.4 kJ mol -1   3.1 kJ mol -1
  water    ethanol valine    12.3 kJ mol -1    19.4 kJ mol -1   7.1 kJ mol -1
  water    ethanol  phenylalanine    8.28 kJ mol -1    19.4 kJ mol -1   11.12 kJ mol -1
  water    ethanol  proline    8.61 kJ mol -1    19.4 kJ mol -1   10.79 kJ mol -1
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)
7
6
ln(p)
5
4
3
2
Y = 17.8663 - 3040.12181 x X
1
-3
4.0x10
-3
4.5x10
-3
5.0x10
-3
5.5x10
1/T
A least squares fit of ln P versus 1/T gives the result H vaporization = (-3040.12181)x
8.314=25.28 kJ mol–1.
ln
(b)
P Tb 
torr
 17.8663 
3.0401  103
Tb
K
3.0401  103
Tb 
 270.64 K
760torr 

 17.8663  ln

torr 
