CO2 H 2 g) CO( ) H O( ) Substance · PDF fileΔG 310 K ΔH 310 K T ΔS 310 K...

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HW8 Answer Chem340 Spring 2011 (Version 2 Updated 4/11/2011) P6.15) Consider the equilibrium CO g H 2 O g CO 2 g H 2 g . At 1000 K, the composition of the reaction mixture is CO 2 H 2 (g) CO(g) H 2 O(g) Substance (g) 27.1 27.1 22.9 22.9 Mole % a. Calculate K P and G reaction at 1000 K. b. Given the answer to part (a), use the H f of the reaction species to calculate G reaction at 298.15 K. Assume that H reaction is independent of temperature. (a) 1 1 80 . 2 ) 4 . 1 ln( 1000 314 . 8 ln 4 . 1 9 . 22 9 . 22 1 . 27 1 . 27 2 2 2 KJmol K JKmol K RT G x x x x K p o reaction O H CO H CO p (b) 1 - 1 - 1 - o reaction f, 1000 o reaction 15 . 298 1 o C f, o H f, o C f, o H f, o reaction f, 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 . 393 0 ΔH ΔH ΔH ΔH ΔH 2 2 2 K K K G KJmol K K o reaction O O O

Transcript of CO2 H 2 g) CO( ) H O( ) Substance · PDF fileΔG 310 K ΔH 310 K T ΔS 310 K...

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

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.

(b)

3

3

3.0401 10ln 17.8663

torrK

3.0401 10270.64

76017.8663 ln

torr

b

b

b

P TT

T Ktorr