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

Home , Melting point, Triple point, Vapor pressure

Problem Set #10 Assigned November 8, 2013 – Due Friday, November 15, 2013 Please show all work for credit

To Hand in

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

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 109 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

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 lnln P dT d P H    T 2 sublimation 11  dTdTR dd   TT   For this specific case H sublimation 203516.92 H 10 J mol   3 –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

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  , K311 K   10000 −4 2(1- f)2 C 2 1- 0.52 1.010-4 M 퐶푑푠 2 ∗ 푀 (푤ℎ푖푐ℎ 푚푒푎푛푠 푓 ) Then we can calculate the equilibrium constant at 335 K: K H  1 1  ln2      , KRTT1   2 1  퐶푑푠

( −4 퐶 ) ( ) ( ) 푑푠

퐶 ∗ −5푀 (푤ℎ푖푐ℎ 푚푒푎푛푠 푓 ≪ ) 푑푠 And finally we can calculate the Gibbs energy at 335 K:

( ) ( ( )

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 0.80ZZ   0.50 EC3 EC

ZEC  0.613

ZZEB1  EC   0.387

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   m100solute g   9.78 106 K

10.

11.

12.

13.

− ( ) ( ∗ ) (2 ) ∗ 2

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

( )

(2 2 ) 2 ( 22 )

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,, refsolid221.4 ref K 100 Torr

TPliquid,, refliquid240.2 ref K 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

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

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

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?

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 –1–1   144 bar K

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.

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 –3 –1 –1 are 5750 and 7280 kg m Htransition = 8.8 J K mol . 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 T gray gray gray GVPST m    white white white GVPST m    At equilibrium

gray white gray white gray white GGVVPSST   0  mm     

11 MP gray white Sn  VVPmm gray white T   gray white S  S  Stransition 11 118.71 10–3 kg mol –1    199 x10 5 Pa 5750 kg m–3 7280 kg m –3   9.8 C 8.8 J K–1 mol –1

Tf  8.2 C

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:

f 1- f  1- 0.965 K315 K  D  N   0.036 f N f N 0.965

f 1- f  1- 0.015 K345 K  D  N   65.67 f N f N 0.015

K 345 K   65.67  ln   R ln    8.314472 J mol-1 K -1  K 315 K  0.036 H       226.2 kJ mol-1 11  11      TT 21 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.

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:

CAB f C 2 K   2  CACB  C  C   2 f C  2  And for C = 1.00  10–4 M: 2 2 K    20000 C 1104 M 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.

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.

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

7. a

8. a

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.

* Ptotal x A P a y B P total

* Ptotal y B P total 0.900 barx  1 0.550 P0.623a  bar xA 0.650 yP* x  AB A *** Pa P B P A y A

0.450P* 0.650  B ** * PPPAB0.450 A   * PB 0.650 1 0.450 2.27*  PA 0.450 1 0.650 * PB  1.414 bar

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.

2 –1 –1 –3 –1 2 RMsolvent T fusion 8.314 J mol K 18.02  10 kg mol  273.15 K K f    3 –1 H fusion 6.008 10 J mol –1 K f 1.86 K kg mol

11.

12.