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Problem Set #10 Assigned November 8, 2013 – Due Friday, November 15, 2013 Please Show All Work for Credit

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Problem Set #10 Assigned November 8, 2013 – Due Friday, November 15, 2013 Please Show All Work for Credit Problem Set #10 Assigned November 8, 2013 – Due Friday, November 15, 2013 Please show all work for credit To Hand in 1. 1 2. –1 A least squares fit of ln P versus 1/T gives the result Hvaporization = 25.28 kJ mol . 3. Assuming constant pressure and temperature, and that the surface area of the protein is reduced by 25% due to the hydrophobic interaction: 2 G 0.25 N A 4 r Convert to per mole, ↓determine size per molecule 4 r 3 V M N 0.73mL/ g 60000g / mol (6.02 1023) 3 2 2 A r 2.52 109 m 2 23 9 2 G 0.25 N A 4 r 0.0720N / m 0.25 6.02 10 / mol (4 ) (2.52 10 m) 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) dPln H sublimation dT RT 2 dln P d ln P dT d ln P H T 2 sublimation 11 dT dT R dd TT For this specific case H sublimation 2035 H 16.92 103 J mol –1 R sublimation Following the same proedure as above, H vaporization 1736 H 14.43 103 J mol –1 R vaporization b. Calculate H fusion. HHHfusion sublimation vaporization 16.92 103 J mol –1 14.43 10 3 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, KK 22.413 2035 18.352 1736 TTtp tp K 4.061 299 Ttp Ttp 73.62 K P 2035 lntp 22.413 5.22895 Torr 73.62 3 Ptp 5.36 10 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: 1.6 1.4 1.2 1.0 0.8 0.6 Relative Absorbance Relative 0.4 Tm ≈ 355.9°C 0.2 0.0 340 350 360 370 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 –1 –4 mol , and Tm = 311 K for strand concentrations of 1.00 10 M. 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 퐶푑푠 temperature, where f = 0.5: −4 ( 퐶푑푠) f 0.5 K , K311 K 10000 −4 2(1- f)2 C 2 1- 0.52 1.010-4 M 퐶푑푠 2 ∗ 푀 (푤ℎ푖푐ℎ 푚푒푎푛푠 푓 ) Then we can calculate the equilibrium constant at 335 K: K H 1 1 ln2 , KRTT1 2 1 퐶푑푠 ( −4 퐶 ) ( ) ( ) 푑푠 퐶 ∗ −5푀 (푤ℎ푖푐ℎ 푚푒푎푛푠 푓 ≪ ) 푑푠 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. tot tot nliq Z B x B n vap y B Z B ZBBAAAA x 11 Z x x Z yBBAAAA z 11 y z Z y Therefore, tot nliq yZEC EC 5 tot nvap Z EC x EC 3 we know that xEC = 0.50 and yEC = 0.80 5 0.80ZZ 0.50 EC3 EC ZEC 0.613 ZZEB1 EC 0.387 5 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: n RT π solutes g h V We can now solve for the height, h: m solutes RT n RT M h solutes solutes V g V g 1.053 g -1 -1 8.314472 J K mol 298 K 16900 g mol 0.315 m 3 -3 -2 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. m / M 0.0053 g / 10083 g mol-1 protein protein -1 ΔTf K f m solute K f 1860 K g mol msolute 100 g 9.78 106 K 10. 6 11. 12. 7 13. − ( ) ( ∗ ) (2 ) ∗ 2 Assume we have 5g cholesterol dissolved in 95g Dipalmitoyl phosphatidylcholine. ( ) 2 (2 2 ) 2 ( 22 ) 8 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. P2 –1 –1 100 Torr R ln 8.314 J mol K ln P 40.0 Torr H 1 32.6 kJ mol–1 sublimation 11 11 210.5 K 221.4 K TT12 –1 –1 760 Torr 8.314 J mol K ln 400 Torr H 26.9 kJ mol–1 vaporization 11 240.2 K 252.2 K –1 –1 H fusion 32.6 26.9 kJ mol 5.6 kJ mol To calculate the triple point temperature, take TPsolid,, ref221.4 K solid ref 100 Torr TPliquid,, ref240.2 K liquid ref 400 Torr 1 1 32.6 103 J mol –1 26.9 10 3 J mol –1 3 -1 Ttp 32.6 26.9 10 J mol 221.4 K 240.2 K –1 –1 100 Torr + 8.314 J mol K ln 400 Torr 1 –1 0.00416 K , Ttp 240.3 K Ttp 32.6 103 J mol–1 1 1 Ptp 100 Torr exp –1 –1 402 Torr 8.314 J mol K 221.4 K 240.3 K 9 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 Hvaporization A least squares fit of ln P versus 1/T gives the result = 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 –1 Hfusion = 6010 J mol , 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? 10 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) fusion fusion –1 –1 dP SSmm 22.0 J mol K fusion –3 –3 dT V MM18.02 10 kg 18.02 10 kg fusion m –3 –3 H22 O,, l H O l 997 kg m 920 kg m 1.44 107 Pa K –1 144 bar K –1 The pressure must be increased by 720 bar to lower the melting point by 5.0ºC. F 85 kg 9.81 ms–2 b) P = 2.2 1072 Pa = 2.2 10 bar A 15 10–2 m 25 10 –5 m dT 1C 2 c) TP 2.20 10 bar 1.5 C ; Tm = – 1.5ºC dP fusion 144 bar d) No, because the lowering of the melting temperature is less than the temperature of the ice.
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