Problem Set #10 Assigned November 8, 013 Due Friday, November 15, 013 Please show all work or credit To Hand in 1. 1
. A least squares it o ln P versus 1/T gives the result 3. Hvaporization = 5.8 kj mol 1. Assuming constant pressure and temperature, and that the surace area o the protein is reduced by 5% due to the hydrophobic interaction: G 4 3 r V M N 3 9 r.5 10 m G 0.5 N 4 A A r 0.73mL/ g 60000g / mol (6.0 10 3 9 0.5 N 4 r 0.070N / m 0.5 6.0 10 / mol (4 ) (.5 10 m) 865kJ / mol A Convert to per mole, determine size per molecule We think this is a reasonable approach, but the value seems high 3 )
4. The vapor pressure o an unknown solid is approximately given by ln(p/torr) =.413 035(K/T), and the vapor pressure o the liquid phase o the same substance is approximately given by ln(p/torr) = 18.35 1736(K/T). a. Calculate H vaporization and H sublimation. b. Calculate H usion. c. Calculate the triple point temperature and pressure. a) Calculate Hvaporization and H sublimation. From Equation (8.16) dln P H sublimation dt RT d ln P d ln P dt d ln P H T 1 dt 1 dt R d d T T For this speciic case H R sublimation sublimation 035 H 16.910 J mol sublimation Following the same proedure as above, H vaporization R 3 1 1736 H 14.43 10 J mol b. Calculate H usion. vaporization 3 1 H usion H sublimation Hvaporization 16.9 10 J mol 14.43 10 J mol.49 10 J mol 3 1 3 1 3 1 c. Calculate the triple point temperature and pressure. At the triple point, the vapor pressures o the solid and liquid are equal. Thereore, K K.413 035 18.35 1736 T T K 4.061 99 T T tp 73.6 K tp tp Ptp 035 ln.413 5.895 Torr 73.6 3 P 5.36 10 Torr tp tp 3
,, 5. The UV absorbance o a solution o a double-stranded DNA is monitored at 60 nm as a unction o temperature. Data appear in the ollowing table. From the data determine the melting temperature. Temperature (K) 343 348 353 355 357 359 361 365 370 Absorbance (60 nm) 0.30 0.35 0.50 0.75 1. 1.40 1.43 1.45 1.47 Plotting the relative absorbance versus temperature yields: Relative Absorbance 1.6 1.4 1. 1.0 0.8 0.6 0.4 T m 355.9 C 0. 0.0 340 350 360 370 T [degree Celsius] The plot indicates a melting temperature o approximately 355.9 C. 6. For the ormation o a sel-complementary duplex DNA rom single strands H = 177. kj mol 1, and T m = 311 K or strand concentrations o 1.00 10 4 M. Calculate the equilibrium constant and Gibbs energy change or duplex ormation at T = 335 K. Assume the enthalpy change or duplex ormation is constant between T = 311 K and T = 335 K. irst calculate equilibrium constant at 311 K, the melting temperature, where = 0.5: K, 0.5 K311 K 10000 (1- ) -4 C 1-0.5 1.0 10 M Then we can calculate the equilibrium constant at 335 K: K H 1 1 ln K R T T 1 1 ( ) ( ) C ds ( 4 C ds ) C ds 4 M (which means ) C ds ( 4 C ds ) C ds 5 M (which means ) And inally we can calculate the Gibbs energy at 335 K: ( ) ( ( ) 4
7. At 47 C, the vapor pressure o ethyl bromide is 10.0Torr and that o ethyl chloride is 40.0 Torr. Assume that the solution is ideal. Assume there is only a trace o liquid present and the mole raction o ethyl chloride in the vapor is 0.80 and then answer these questions: a. What is the total pressure and the mole raction o ethyl chloride in the liquid? b. I there are 5.00 mol o liquid and 3.00 mol o vapor present at the same pressure as in part (a), what is the overall composition o the system? (a) what is the overall composition o the system? ( ) b) We use the lever rule. tot 1 1 1 1 n Z x n y Z tot liq B B vap B B Z x Z x x Z B B A A A A y z y z Z y B B A A A A Thereore, n y Z n Z x tot liq EC EC tot vap EC EC 5 3 we know that x EC = 0.50 and y EC = 0.80 0.80 Z Z 0.50 Z Z EC EB 0.613 EC 5 3 1 Z 0.387 EC EC 5
8. 1.053 g o bee heart myoglobin dissolved in 50.0 ml o water at T = 98 K generates suicient osmotic pressure to support a column o solution o height d. I the molar mass o myoglobin is 16.9 kg per mole, calculate d. We set the osmotic pressure equal to the pressure o the column: n solutesrt π g h V We can now solve or the height, h: msolutes RT n M solutesrt solutes h V g V g 1.053 g -1 8.31447 J K mol 98 K -1 16900 g mol 3-3 - 0.00005 m 1000 kg m 9.81 m s 0.315 m 9. Calculate the change in the reezing point o water i 0.0053 g o a protein with molecular weight 10083 g mol 1 is dissolved in 100. ml o water. 6 9.78 10 K m protein / Mprotein -1 ΔT K msolute K 1860 K g mol m solute -1 0.0053 g / 10083 g mol 100 g 10. 6
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13. ( ) ( ) ( ) Assume we have 5g cholesterol dissolved in 95g Dipalmitoyl phosphatidylcholine. ( ) ( ) ( ) 8
Extra practice or exam, do not hand in Phase equilibrium 1. Use the vapor pressures or C N given in the ollowing table to estimate the temperature and pressure o the triple point and also the enthalpies o usion, vaporization, and sublimation. Phase T ( C) P (Torr) Solid 6.7 40.0 Solid 51.8 100. Liquid 33.0 400. Liquid 1.0 760. P 1 1 100 Torr R ln 8.314 J mol K ln P 1 40.0 Torr H sublimation 3.6 kj mol 1 1 1 1 T T 10.5 K 1.4 K 1 760 Torr 1 1 8.314 J mol K ln 400 Torr 6 Hvaporization 1 1 40. K 5. K 3.6 6.9 kj mol 5.6 kj mol H usion 1 1.9 kj mol 1 1 To calculate the triple point temperature, take T T 1.4 K P 100 Torr solid, re solid, re 40. K P 400 Torr liquid, re liquid, re 1 1 3.6 10 J mol 6.9 10 J mol 3-1 Ttp 3.6 6.9 10 J mol 1.4 K 40. K 1 T P tp tp 1 1 + 8.314 J mol K ln 400 Torr 1 0.00416 K, Ttp 40.3 K 100 Torr 3 1 3 1 3 1 3.6 10 J mol 1 1 100 Torr exp 40 Torr 1 1 8.314 J mol K 1.4 K 40.3 K 9
. Use the vapor pressures o Cl given in the ollowing table to calculate the enthalpy o vaporization using a graphical method or a least-squares itting routine. T (K) P (atm) T (K) P (atm) 7.6 0.585 83.15 4.934 38.7 0.98 94.3 6.807 49.8 1.566 305.4 9.173 60.9.388 316.5 1.105 7.0 3.483 37.6 15.676 A least squares it o ln P versus 1/T gives the result Hvaporization = 0.3 kj mol 1. 3. It has been suggested that the surace melting o ice plays a role in enabling speed skaters to achieve peak perormance. Carry out the ollowing calculation to test this H usion = 6010 J mol 1, the density o ice is 90 kg m 3, and the density o 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 o the skate in contact with the ice has been reduced by sharpening to 5 10 3 cm, and that the length o the contact area is 15 cm. I a skater o mass 85 kg is balanced on one skate, what pressure is exerted at the interace o the skate and the ice? 10
c. What is the melting point o ice under this pressure? a) d. I the temperature o the ice is 5.0 C, do you expect melting o the ice at the ice skate interace to occur? usion usion 1 1 dp Sm Sm.0 J mol K usion dt M M usion Vm 997 kg m 90 kg m HO, l HO, l 1.4410 Pa K 144 bar K 7 1 1 3 18.010 kg 3 18.010 kg 3 3 The pressure must be increased by 70 bar to lower the melting point by 5.0ºC. b) F 85 kg 9.81 ms P A 5 1510 m 510 m 7 =. 10 Pa =. 10 bar c) dt 1C T P dp 144 bar usion.0 10 bar 1.5 C ; T m = 1.5ºC d) No, because the lowering o the melting temperature is less than the temperature o the ice. 4. Consider the transition between two orms o solid tin, Sn(s, gray) Sn(s, white). The two phases are in equilibrium at 1 bar and 18 C. The densities or gray and white tin are 5750 and 780 kg m 3 H transition = 8.8 J K 1 mol 1. Calculate the temperature at which the two phases are in equilibrium at 00 bar. In going rom 1 atm, 18 C to 00 atm, and the unknown temperature T G V P S T gray gray gray m G V P S T white white white m At equilibrium gray white m m T gray white S S m m gray white gray white gray white G G 0 V V P S S T M V V P Sn 1 1 gray S transition 3 1 1 1 5 118.7110 kg mol 199 x10 Pa 3 3 5750 kg m 780 kg m 9.8 C 1 1 8.8 J K mol T 8. C white P 11
5. A protein has a melting temperature o T m = 335 K. At T = 315 K, UV absorbance determines that the raction o native protein is N = 0.965. At T = 345. K, N = 0.015. Assuming a two-state model and assuming also that the enthalpy is constant between T = 315 and 345 K, determine the enthalpy o denaturation. Also, determine the entropy o denaturation at T = 335 K. By DSC, the enthalpy o denaturation was determined to be 51 kj mol 1. Is this denaturation accurately described by the two-state model? We irst calculate the equilibrium constants at 315 K and 345 K: K315 K K345 K D N D N 1-1- 0.965 N N 0.965 1-1- 0.015 N N 0.015 0.036 65.67 K 345 K 65.67-1 -1 ln R ln 8.31447 J mol K K 315 K 0.036 H 6. kj mol 1 1 1 1 T T 345 K 315 K 1-1 This result deviates rom the DSC result, indicating that the denaturation process is not accurately described by a two-state model. 6. Suppose a DNA duplex is not sel-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 o A and B to orm duplex AB Assume the total strand concentration is C and, initially, A and B have equal concentrations; that is, C A,0 = C B,0 = C/. Obtain an expression or the equilibrium constant at a point where the raction o the total strand concentration C that is duplex is deined as. I the strand concentration is 1.00 10 4 M, calculate the equilibrium constant at the melting temperature. 1
We make the table o concentrations: C initial C equilibrium AB 0 C A = B C/ C/ C The equilibrium constant at the melting temperature with = 0.5 is given by: CAB C C C C C K A B C And or C = 1.00 10 4 M: K 0000 4 C 110 M Ideal and Real Solutions 1. Predict the ideal solubility o lead in bismuth at 80 C given that its melting point is 37 C and its enthalpy o usion is 5. kj mol 1. 13
. The vapour pressure o -propanol is 50.00 kpa at 338.8 C, but it ell to 49.6 kpa when 8.69 g o an involatile organic compound was dissolved in 50 g o -propanol. Calculate the molar mass o the compound. 3. The addition o 5.00 g o a compound to 50 g o naphthalene lowered the reezing point o the solvent by 0.780 K. Calculate the molar mass o the compound. 4. The osmotic pressure o an aqueous solution at 88 K is 99.0 kpa. Calculate the reezing point o the solution. 14
5. The molar mass o an enzyme was determined by dissolving it in water, measuring the osmotic pressure at 0 C, and extrapolating the data to zero concentration. The ollowing data were obtained: c/(mg cm 3 ) 3.1 4.618 5.11 6.7 h/cm 5.746 8.38 9.119 11.990 Calculate the molar mass o the enzyme. 6. a 15
7. a 16
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9. A and B orm an ideal solution. At a total pressure o 0.900 bar, y A = 0.450 and x A = 0.650. Using this inormation, calculate the vapor pressure o pure A and o pure B. P x P y P * total A a B total P x * a A * B * A * B P 0.900 bar 1 0.550 total ybptotal x 0.63 bar x 0.650 * A B * * * a B A A 0.450P 0.650 P P P P P P A yp P P P y * B * * * A 0.450 B A 0.650 10.450 0.450 10.650 1.414 bar.7 10. The heat o usion o water is 6.008 10 3 J mol 1 at its normal melting point o 73.15 K. Calculate the reezing point depression constant K. K K RM T 1 1 3 1 solvent usion H usion 3 1 6.00810 J mol 1.86 K kg mol 1 8.314 J mol K 18.0 10 kg mol 73.15 K 11. 18
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