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.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- name unit 11 review gas laws and intermolecular forces
- additel 760 automatic handheld pressure calibrators
- version 001 hw03 more ideal gas law practice sparks
- 48 practice problems for ch 17 chem 1c joseph
- jsp 760 tri service regulations for leave and other types
- understanding vacuum and vacuum measurement
- problem set 10 assigned november 8 2013 due friday
- answers to phase diagram worksheet livingston public schools
- chapter 10 answers to assigned problems
- energy freedman wappingers central school district
Related searches
- problem set 7
- icd 10 for anemia due to bleeding
- icd 10 sepsis due to uti
- icd 10 pain due to neuropathy
- icd 10 back injury due to fall
- icd 10 bacteremia due to escherichia coli
- icd 10 diarrhea due to medication
- 8 step problem solving template
- army 8 step problem solving process
- encephalopathy due to alcohol icd 10 code
- icd 10 dehydration due to alcoholism
- 8 steps problem solving model