1

Problem Set #10

Assigned November 8, 2013 – Due Friday, November 15, 2013

Please show all work for credit

To Hand in

1.

2

2.

A least squares fit of ln P versus 1/T gives the result vaporizationH = 25.28 kJ mol–1.

3.

Assuming constant pressure and temperature, and that the surface area of the protein is

reduced by 25% due to the hydrophobic interaction:

molkJmmolmNr

mr

molggmLNMVr

r

A

A

A

/865)1052.2()4(/1002.625.0/0720.04N25.0 G

1052.2

)1002.6(/60000/73.03

4

4N25.0 G

29232

9

23

22

3

2

We think this is a reasonable approach, but the value seems high

Convert to per mole, ↓determine size per molecule

3

4. The vapor pressure of an unknown solid is approximately given by ln(P/Torr) = 22.413 –

2035(K/T), and the vapor pressure of the liquid phase of the same substance is

approximately given by ln(P/Torr) = 18.352 – 1736(K/T).

a. Calculate Hvaporization and Hsublimation.

b. Calculate Hfusion.

c. Calculate the triple point temperature and pressure.

a) Calculate vaporizationH and .sublimationH

2

2

3 –1

From Equation (8.16)

ln

ln ln ln

1 1

For this specific case

2035 16.92 10 J mol

sublimation

sublimation

sublimationsublimation

Hd P

dT RT

Hd P d P dT d PT

dT dT Rd d

T T

HH

R

3 –1

Following the same proedure as above,

1736 14.43 10 J molvaporization

vaporization

HH

R

b. Calculate .fusionH

3 –1 3 –1

3 –1

16.92 10 J mol 14.43 10 J mol

2.49 10 J mol

fusion sublimation vaporizationH H H

c. Calculate the triple point temperature and pressure. At the triple point, the vapor pressures

of the solid and liquid are equal. Therefore,

3

K K22.413 2035 18.352 1736

K4.061 299

73.62 K

2035ln 22.413 5.22895

Torr 73.62

5.36 10 Torr

tp tp

tp

tp

tp

tp

T T

T

T

P

P

4

5. The UV absorbance of a solution of a double-stranded DNA is monitored at 260 nm as a

function of temperature. Data appear in the following table. From the data determine

the melting temperature.

Temperature (K) 343 348 353 355 357 359 361 365 370

Absorbance (260 nm) 0.30 0.35 0.50 0.75 1.22 1.40 1.43 1.45 1.47

Plotting the relative absorbance versus temperature yields:

The plot indicates a melting temperature of approximately 355.9°C.

6. For the formation of a self-complementary duplex DNA from single strands H° = –177.2 kJ

mol–1, and Tm = 311 K for strand concentrations of 1.00 10–4M. Calculate the equilibrium

constant and Gibbs energy change for duplex formation at T = 335 K. Assume the enthalpy

change for duplex formation is constant between T = 311 K and T = 335 K.

first calculate equilibrium constant at 311 K, the melting

temperature, where f = 0.5:

f)-2(1

fK

2 C ,

10000

M101.0 0.5-1 2

5.0 K311K

4-2

Then we can calculate the equilibrium constant at 335 K:

2

1 2 1

K H 1 1ln

K R T T

, ,

( )

(

)

And finally we can calculate the Gibbs energy at 335 K:

( ) (

( )

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

340 350 360 370

T [degree Celsius]

Rela

tive A

bsorb

ance

Tm ≈ 355.9°C

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

340 350 360 370

T [degree Celsius]

Rela

tive A

bsorb

ance

Tm ≈ 355.9°C

𝐶𝑑𝑠

( −4 𝐶𝑑𝑠)

𝐶𝑑𝑠 2 ∗ −4𝑀 (𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 𝑓 )

𝐶𝑑𝑠

( −4 𝐶𝑑𝑠)

𝐶𝑑𝑠 ∗ −5𝑀 (𝑤ℎ𝑖𝑐ℎ 𝑚𝑒𝑎𝑛𝑠 𝑓 ≪ )

5

7. At –47°C, the vapor pressure of ethyl bromide is 10.0Torr 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 then 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?

(a) what is the overall composition of the system?

∗

∗

∗ ( )

2

b) We use the lever rule.

1 1

1 1

tot tot

liq B B vap B B

B B A A A A

B B A A A A

n Z x n y Z

Z x Z x x Z

y z y z Z y

Therefore,

5

3

tot

liq EC EC

tot

vap EC EC

n y Z

n Z x

we know that xEC = 0.50 and yEC = 0.80

50.80 0.50

3

0.613

1 0.387

EC EC

EC

EB EC

Z Z

Z

Z Z

6

8. 1.053 g of beef heart myoglobin dissolved in 50.0 ml of water at T = 298 K generates sufficient osmotic pressure to support a column of solution of height d. If the molar mass of myoglobin is 16.9 kg per mole, calculate d.

We set the osmotic pressure equal to the pressure of the column:

h g V

RTnπ solutes

We can now solve for the height, h:

solutes

solutessolutes

-1

-1

3 -3 -2

m RT

Mn RTh

V g V g

1.053 g8.314472 J K mol 298 K

16900 g mol0.315 m

0.00005 m 1000 kg m 9.81 m s

9. Calculate the change in the freezing point of water if 0.0053 g of a protein with molecular

weight 10083 g mol–1 is dissolved in 100. mL of water.

-1

protein protein -1

f f solute f

solute

6

0.0053 g / 10083 g molm / MΔT K m K 1860 K g mol

m 100 g

9.78 10 K

10.

7

11.

12.

8

13.

( )

∗

( ∗ − ) (2 )

2

Assume we have 5g cholesterol dissolved in 95g Dipalmitoyl phosphatidylcholine.

(

)

2

(2 2 ) 2 ( 22 )

9

Extra practice for exam, do not hand in

Phase equilibrium

1. Use the vapor pressures for C2N2 given in the following table to estimate the

temperature and pressure of the triple point and also the enthalpies of fusion,

vaporization, and sublimation.

Phase T (°C) P (Torr)

Solid –62.7 40.0

Solid –51.8 100.

Liquid –33.0 400.

Liquid –21.0 760.

–1 –12

–11

1 2

–1 –1

100 Torrln 8.314 J mol K ln

40.0 Torr32.6 kJ mol

1 11 1

210.5 K 221.4 K

760 Torr8.314 J mol K ln

400 Torr26

1 1

240.2 K 252.2 K

sublimation

vaporization

PR

PH

T T

H

–1

–1 –1

.9 kJ mol

32.6 26.9 kJ mol 5.6 kJ molfusionH

To calculate the triple point temperature, take

, ,

, ,

221.4 K 100 Torr

240.2 K 400 Torr

solid ref solid ref

liquid ref liquid ref

T P

T P

3 –1 3 –1

3 -1

–1 –1

–1

3

1 1 32.6 10 J mol 26.9 10 J mol

32.6 26.9 10 J mol 221.4 K 240.2 K

100 Torr + 8.314 J mol K ln

400 Torr

10.00416 K , 240.3 K

32.6 10 J 100 Torr exp

tp

tp

tp

tp

T

TT

P

–1

–1 –1

mol 1 1402 Torr

8.314 J mol K 221.4 K 240.3 K

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2. Use the vapor pressures of Cl2 given in the following table to calculate the enthalpy of

vaporization using a graphical method or a least-squares fitting routine.

T (K) P (atm) T (K) P (atm)

227.6 0.585 283.15 4.934

238.7 0.982 294.3 6.807

249.8 1.566 305.4 9.173

260.9 2.388 316.5 12.105

272.0 3.483 327.6 15.676

A least squares fit of ln P versus 1/T gives the result vaporizationH = 20.32 kJ mol–1

.

3. It has been suggested that the surface melting of ice plays a role in enabling speed

skaters to achieve peak performance. Carry out the following calculation to test this

Hfusion = 6010 J mol–1, the density

of ice is 920 kg m–3, and the density of liquid water is 997 kg m–3.

a. What pressure is required to lower the melting temperature by 5.0°C?

b. Assume that the width of the skate in contact with the ice has been reduced by

sharpening to 25 10–3 cm, and that the length of the contact area is 15 cm. If a skater of

mass 85 kg is balanced on one skate, what pressure is exerted at the interface of the skate

and the ice?

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c. What is the melting point of ice under this pressure?

d. If the temperature of the ice is –5.0°C, do you expect melting of the ice at the ice–skate

interface to occur?

a)

2 2

–1 –1

–3 –3

–3 –3, ,

7 –1 –1

22.0 J mol K

18.02 10 kg 18.02 10 kg

997 kg m 920 kg m

1.44 10 Pa K 144 bar K

fusion fusion

m m

fusion

fusion m

H O l H O l

S SdP

M MdT V

The pressure must be increased by 720 bar to lower the melting point by 5.0ºC.

b) –2

7 2

–2 –5

85 kg 9.81 ms = 2.2 10 Pa = 2.2 10 bar

15 10 m 25 10 m

FP

A

c) 21 C

2.20 10 bar 1.5 C144 barfusion

dTT P

dP

; Tm = – 1.5ºC

d) No, because the lowering of the melting temperature is less than the temperature of the ice.

4. Consider the transition between two forms of solid tin, Sn(s, gray) Sn(s, white).

The two phases are in equilibrium at 1 bar and 18°C. The densities for gray and white tin

are 5750 and 7280 kg m–3 Htransition = 8.8 J K–1 mol–1. Calculate the

temperature at which the two phases are in equilibrium at 200 bar.

In going from 1 atm, 18 C to 200 atm, and the unknown temperature

At equilibrium

0

gray gray gray

m

white white white

m

gray white gray white gray white

m m

gray white

m m

gr

T

G V P S T

G V P S T

G G V V P S S T

V V PT

S

–3 –1 5

–3 –3

–1 –1

1 1

1 1118.71 10 kg mol 199 x10 Pa

5750 kg m 7280 kg m9.8 C

8.8 J K mol

8.2 C

Sn

gray white

ay whitetransition

f

M P

SS

T

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5. 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:

0.0360.965

0.965-1

f

f-1

f

fK 315K

N

N

N

D

67.560.015

0.015-1

f

f-1

f

fK 345K

N

N

N

D

-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 1

T 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.

6. 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

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.

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We make the table of concentrations:

Cinitial Cequilibrium

AB 0 f C

A = B C/2 C/2 – 2f C

The equilibrium constant at the melting temperature with f = 0.5 is given by:

C

2

C f 22

C

C f

CC

CK

2

BA

AB

And for C = 1.00 10–4

M:

20000

M101

2

C

2K

4

Ideal and Real Solutions

1. Predict the ideal solubility of lead in bismuth at 280°C given that its melting point is

327°C and its enthalpy of fusion is 5.2 kJ mol−1.

14

2. The vapour pressure of 2-propanol is 50.00 kPa at 338.8°C, but it fell to 49.62 kPa when

8.69 g of an involatile organic compound was dissolved in 250 g of 2-propanol. Calculate

the molar mass of the compound.

3. The addition of 5.00 g of a compound to 250 g of naphthalene lowered the freezing

point of the solvent by 0.780 K. Calculate the molar mass of the compound.

4. The osmotic pressure of an aqueous solution at 288 K is 99.0 kPa. Calculate the freezing

point of the solution.

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5. The molar mass of an enzyme was determined by dissolving it in water, measuring the

osmotic pressure at 20°C, and extrapolating the data to zero concentration. The

following data were obtained:

c/(mg cm−3) 3.221 4.618 5.112 6.722

h/cm 5.746 8.238 9.119 11.990

Calculate the molar mass of the enzyme.

6. a

16

7. a

17

8. a

18

9. A and B form an ideal solution. At a total pressure of 0.900 bar, yA = 0.450 and xA =

0.650. Using this information, calculate the vapor pressure of pure A and of pure B.

*

*

*

* * *

*

* * *

*

*

*

0.900 bar 1 0.550 0.623 bar

0.650

0.4500.650

0.450

0.650 1 0.450 2.27

0.450 1 0.650

1.414 bar

total A a B total

xtotal B totala

A

A BA

a B A A

B

A B A

B

A

B

P x P y P

P y PP

x

y Px

P P P y

P

P P P

P

P

P

10. The heat of fusion of water is 6.008 103 J mol–1 at its normal melting point of 273.15 K.

Calculate the freezing point depression constant Kf.

22 –1 –1 –3 –1

3 –1

–1

8.314 J mol K 18.02 10 kg mol 273.15 K

6.008 10 J mol

1.86 K kg mol

solvent fusion

f

fusion

f

RM TK

H

K

11.

19

12.