Problem Set #10 Assigned November 8, 2013 Due Friday ...

To Hand in 1.

Problem Set #10 Assigned November 8, 2013 ? Due Friday, November 15, 2013

Please show all work for credit

1

2.

A least squares fit of ln P versus 1/T gives the result H = vaporization 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:

G 0.25 N A 4 r 2 Convert to per mole, determine size per molecule

4 r3 3

V2

M2

NA

0.73mL /

g

60000g

/ mol

(6.02 1023)

r 2.52 109 m

G 0.25 N A 4 r 2 0.0720N / m 0.25 6.02 1023 / mol (4 ) (2.52 109 m)2 865kJ / mol

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

2

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 Hvaporization and Hsublimation.

From Equation (8.16)

d ln P dT

H sublimation RT 2

d ln P d ln P dT T 2 d ln P Hsublimation

d

1 T

dT

d

1 T

dT

R

For this specific case

H sublimation 2035 R

Hsublimation 16.92 103 J mol?1

Following the same proedure as above,

Hvaporization 1736 R

Hvaporization 14.43 103 J mol?1

b. Calculate H fusion.

H fusion H sublimation Hvaporization 16.92 103 J mol?1 14.43 103 J mol?1 2.49 103 J mol?1

c. Calculate the triple point temperature and pressure. At the triple point, the vapor pressures of the solid and liquid are equal. Therefore,

22.413 2035 K 18.352 1736 K

Ttp

Ttp

4.061 299 K Ttp

Ttp 73.62 K

ln Ptp 22.413 2035 5.22895

Torr

73.62

Ptp 5.36 103 Torr

3

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:

Relative Absorbance

1.6 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0

340

Tm 355.9?C

350

360

370

T [degree Celsius]

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:

K f 2(1- f)2 C

,

K311 K

2

1 -

0.52

0.5 1.0 10-4

M

10000

Then we can calculate the equilibrium constant at 335 K:

ln

K2 K1

H R

1 T2

1 T1

,

,

( )

(

)

(

-4 )

2 -4 (

)

(

-4 )

-5 ( )

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

( )

(

( )

4

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.

ntot liq

ZB xB

ntot vap

yB ZB

ZB xB 1 Z A 1 xA xA Z A

yB zB 1 yA 1 zA Z A yA

Therefore,

ntot liq

ntot vap

yEC ZEC ZEC xEC

5 3

we know that xEC = 0.50 and yEC = 0.80

0.80

Z EC

5 3

ZEC

0.50

ZEC 0.613

ZEB 1 ZEC 0.387

(

)

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download