sign in sign up
<<
Home , Mole fraction, Partial pressure

CHMY 361 HANDOUT#6 October 28, 2012 HOMEWORK #4 Key

Was due Friday, Oct. 26

1. Using only data from Table A5, what is the of water deep in a mine that is so far below sea level that the atmospheric is 1.17 atm?

0 ΔH vap = +44.02 kJ/mol

H20(l) --> H2O(g) Q= PH2O /XH2O = K, at ⎛ P2 ⎞ ⎛ K 2 ⎞ ΔH vap ⎛ 1 1 ⎞ ln⎜ ⎟ = ln⎜ ⎟ − ⎜ − ⎟ equilibrium, i.e., the Pressure ⎝ P1 ⎠ ⎝ K1 ⎠ R ⎝ T2 T1 ⎠ for the pure . ⎛1.17 ⎞ 44,020 ⎛ 1 1 ⎞ ln⎜ ⎟ = − ⎜ − ⎟ = ⎝ 1 ⎠ 8.3145 ⎝ T2 373 ⎠ ⎡1.17⎤ − 8.3145ln 1 ⎢ 1 ⎥ 1 = ⎣ ⎦ + = .002651 T2 44,020 373

T2 = 377

2. From table A5, calculate the Henry’s Law constant (i.e., ) for dissolving of NH3(g) in water at 298 K and 340 K. It should have units of Matm-1;What would it be in atm per fraction, as in Table 5.1 at 298 K?

o For NH3(g) ----> NH3(aq) ΔG = -26.5 - (-16.45) = -10.05 kJ/mol

ΔG0 − [NH (aq)] K = e RT = 0.0173 = 3 This is Henry’s Law. It says that at equilibrium the ratio of dissolved P NH3

NH3 to the of NH3 in contact with the liquid is a constant = 0.0173 (Henry’s Law Constant). This also says [NH3(aq)] =0.0173PNH3 or -1 PNH3 = 0.0173 [NH3(aq)] = 57.8 atm/M x [NH3(aq)]

The latter form is like Table 5.1 except it has NH3 in M instead of XNH3. Need the convert [NH3(aq)] to XNH3(aq)

Mole fraction of a 1 M in water is 1/(55 +1) = 1/56 = 0.018 moles NH3 / total moles. -1 -1 So, the conversion factor is: XNH3 = molL NH3 x 0.018 mol fraction/molL

-1 -1 57.8 atmM /0.018 M (mol fraction)M = 3200 atm (per mol fraction dissolved NH3 ) Note: this is similar to that for CO2 in Table 5.1, another fairly soluble gas.

Page 1 of 6

3. (a) How many moles of N2 gas are dissolved in the blood of a diver before entering the water. Assume 2.0 L of blood. (Use Table 5.1) (b) How many liters of gas at 1 atm and 298 K does this number of moles equal? (c) 1 atmosphere = 760 mm of Hg. What is 1 atm in feet of water? (d) If a diver stays at a depth in the water that is the answer to (c) until equilibrium is reached, what of N2 gas would be released from the blood if came to the surface quickly? [Note, you don't have to find the answer to (c) in order to know the answer to (d) ] a. 2.0 L blood (assume it is pure water) has close to 2.0 kg / .018kg mol-1 = 111 mol water

PN 0.78 atm at 1 atm air and 298 K, X dissolved = 2 = = 9.18 ×10−6 N2 k 85000 atm N2 −6 mols N 2 dissolved = 9.18 ×10 x 111 = 0.00102 nRT 0.00102 x 0.0821 x 298 PV = nRT V = = = 0.025 L = 25 mL P 1

nRT 0.00102 x 0.0821 x 298 b. PV = nRT V = = = 0.025 L = 25 mL P 1 c. 0.76 m Hg x 3.28 ft/m = 2.49 ft Hg. Hg is 13.6 time more dense, so 2.49 ft Hg x 13.6 ft H2O/ft Hg =33.9 ft H20 = 1 atm d. At this depth the total pressure would be 2 atm: 1 atm from the atmosphere and 1atm from the water, meaning that exactly twice the amount of N2 would be dissolved than at the surface. Thus 25 ml of N2 gas would be released as bubbles into the blood. The bubbles clog capillaries, causing low delivery to tissue.

4. Consider a 1x10-4 M polymer solution inside a dialysis bag that is permeable to the ligand, A, but not to the polymer. The following measurements of the free ligand outside the bag and total ligand inside the bag are measured at 3 equilibrium values. The units are M. From a Scatchard plot find Kd for the binding of A to its binding sites on the polymer and the number of binding sites per polymer . (the notation 5E-07 means 5 x 10-7, etc.) [A]out [A]in tot [SA] [SA]/[A] 1.100E‐05 7.443E‐04 7.33E-04 6.67E+01 9.100E‐07 3.821E‐04 3.81E-04 4.19E+02 8.300E‐08 6.139E‐05 6.13E-05 7.39E+02

-4 Plot of [SA]/[A] vs. [SA]. The x intercept is 8 x 10 M = Stotal ; y intercept is 800 = Stotal/Kd. -4 -4 Therefore number of binding sites per polymer is Stotal/ [polymer] = 8 x 10 / 1 x 10 = 8 sites -4 -6 The value of Kd is Stotal /800 = 8 x 10 / 800 = 1 x 10 M

8.00E+02

6.00E+02

4.00E+02 Series1 2.00E+02

0.00E+00 0.00E+00 2.00E‐04 4.00E‐04 6.00E‐04 8.00E‐04

Page 2 of 6

5. If the for solute B = [Bmethanol]/ [Bwater] = 0.010, and the of B in water is 0.10 M at 25o C, answer the following: (a)What is ΔGo for the process B(methanol) --> B(aq)? (b) What is ΔGo for the process B(methanol) --> B(s)? (c) What is ΔGo for the process B(aq) --> B(s)? a. ΔGo for B(aq) --> B(methanol) = -RTln K = -8.3145 x 298 x ln( 0.01) = 11,400 J/mol (could have been written by inspection as 2 x 5,700).

Therefore ΔGo for B(methanol) --> B(aq) = -11,400 J/mol

o c. K for B(s) --> B(aq) is given as 0.1 M , so by inspection ΔG for this 5,700 J/mol Therefore ΔGo for B(aq) --> B(s) = -5,700 J/mol b. Adding the processes of a. and c. gives B(methanol) --> B(s). Therefore adding the ΔGo values of a. and c. gives the answer = -11,400 -5,700 = -22,100 J/mol

6. In certain mitochondria, the oxidation of glucose products pumps protons into the intermembrane space between the double membrane to reach a pH of 4.8. In the interior of the mitochondrian, the pH is 7.2. In addition, the intermembrane space has an electric potential difference of +80 mV relative to the interior. (“membrane potential” is -80 mV) This problem is very much like Prob. 5 of Chap. 5, with solution given in the Appendix.

(a) ATP is synthesized by the combined free energy change of the protons flowing down the concentration gradient from the intermembrane space to the interior and the change of electrical free energy coming from the voltage difference. Calculate the ΔG per mole of protons transferred.

⎛ a + ()outside ⎞ ⎜ H ⎟ Δμ = RT ln + ZF()φout − φin , where in means interior ⎜ a + inside ⎟ ⎝ H ()⎠ J ⎛1×10−7.2 ⎞ ⎛ C ⎞⎛ J J ⎞ = 8.314 ()298K ln⎜ ⎟ + ()1 ⎜96485 ⎟⎜0 − 0.080 ⎟ ⎜ −4.8 ⎟ molK ⎝1×10 ⎠ ⎝ mol ⎠⎝ C C ⎠ kJ = −13700 − 7700 = −21400 mol

(b) How many of ATP from ADP and P are obtained per proton if [ATP]/[ADP] = 3 and [P] 0.003 M?

⎛ ATP ⎞ ⎛⎛ ATP ⎞⎛ 1 ⎞⎞ o [] o ⎜ [ ] ⎟ ΔG = ΔG + RT ln⎜ ⎟ = ΔG + RT ln⎜⎜ ⎟⎜ ⎟⎟ = ⎝ [][]ADP P ⎠ ⎝⎝ []ADP ⎠⎝ []P ⎠⎠ kJ J ⎛ ⎛ 1 ⎞⎞ kJ 31.0 + 8.314 ()()298K ln⎜ 3 ⎜ ⎟⎟ = 48.1 mol molK ⎝ ⎝ 0.003 ⎠⎠ mol kJ 21.4 mol = 0.5 ; Therefore it requires about 2 protons to make 1 ATP kJ 48.1 mol

Page 3 of 6

+ + + + 7. Consider Fig. 5.22 but with [Na ]out/[Na ]in = 20.0 and [K ]out/[K ]in = 0.040 and φout - φin = 60 mV.

(a) What is Δμ for Na+(in) --> Na+(out)?

(b) What is Δμ for K+(out) --> K+(in) ?

(c) What is ΔG for the process: 3Na+ (in) + 2K+ (out)--> 3Na+ (out) 2K+(in)?

(d) If [ATP] = 0.08 [ADP] and [P] = 0.02 M will the process in (c) be spontaneous if the hydrolysis of ATP is coupled perfectly to it?

a.

a + ()outside Na = 20.0 aNa+ ()inside C ⎛ 1V ⎞ Z = 1 F = 96485 ΔΦ = 60mV ⎜ ⎟ = 0.06V mol ⎝1000mV ⎠

⎛ a + ()outside ⎞ J ⎛ C ⎞ Δμ = RT ln⎜ Na ⎟ + ZFΔΦ = 8.314 ()()()298K ln 20.0 + 1 ⎜96485 ⎟()0.06V ⎜ a + inside ⎟ molK mol ⎝ Na ()⎠ ⎝ ⎠ kJ Δμ = 13.2 mol

a + ()outside a + ()inside K = 0.04; K = 25. aK + ()inside aK + ()outside C ⎛ 1V ⎞ Z = 1 F = 96485 ΔΦ = −60mV ⎜ ⎟ = − 0.06V mol 1000mV b. ⎝ ⎠ ⎛ a + ()inside ⎞ J ⎛ C ⎞ Δμ = RT ln⎜ K ⎟ + ZFΔΦ = 8.314 ()()()298K ln 25. + 1 ⎜96485 ⎟()− 0.06V ⎜ a + outside ⎟ molK mol ⎝ K ()⎠ ⎝ ⎠ kJ Δμ = 7.98 − 5.79 = 2.19 mol c. 3Na + ()in + 2K + ()out → 3Na + ()out + 2K + ()in ⎛ kJ ⎞ kJ 3Na + ()in → 3Na + (out )3Δμ = 3⎜13.2 ⎟ = 39.6 ⎝ mol ⎠ mol ⎛ kJ ⎞ kJ 2K + ()out → 2K + ()in 2Δμ = 2⎜2.19 ⎟ = 4.38 ⎝ mol ⎠ mol kJ 3Na + ()in + 2K + ()out → 3Na + ()out + 2K + ()in ΔG = 44.0 mol

o ⎛ [][]ADP P ⎞ J J ⎛ (0.02)⎞ kJ d. Δμ = Δμ + RT ln⎜ ⎟ = −31000 + 8.314 ()298K ln⎜ ⎟ = −33.4 ⎝ []ATP ⎠ mol molK ⎝ 0.08 ⎠ mol

Page 4 of 6

8. For a 0.3 g/L of a protein with molecular of 5 x 104 g/mol, what is the in mm of liquid water at 298 K, using the relationship that reminds one of the law?

ΠV=nRT approximately, where V is volume of solution, and n = moles solute

⎛ g ⎞ ⎜ 0.3 ⎟ w ⎛ Latm ⎞ Π = RT = ⎜ L ⎟⎜0.08205 ⎟()298K =1.47 ×10−4 atm M ⎜ g ⎟⎝ Kmol ⎠ ⎜ 50000 ⎟ ⎝ mol ⎠

−4 ⎛ 760mmHg ⎞⎛13.6mmWater ⎞ 1.47 ×10 atm⎜ ⎟⎜ ⎟ =1.52 mm water ⎝ 1atm ⎠⎝ mmHg ⎠

9. (a) What is the activity of water in a solution whose melting point is 270 K? (b) What is the boiling point of this solution, assuming the same activity holds for the boiling point . (c) What is its (ignoring any vapor pressure from the solute)? (d) What is the osmotic pressure of this solution at 270 K? a. H2O(s,a=1)---> H2O(liq, a=?) K270 = a/1= a for equilibrium at the new melting point; K273 = 1 because pure water is at equilibrium at 273

⎛ K ⎞ − ΔH fus ⎛ 1 1 ⎞ − 6007J / mol ⎛ 1 1 ⎞ ln⎜ 270 ⎟ = ⎜ − ⎟ = ⎜ − ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1 ⎠ R ⎝ Tmp 273 ⎠ 8.3145 ⎝ 270 273 ⎠

⎛ aH O ⎞ ln⎜ 2 ⎟ = −0.0294 ⎜ ⎟ ⎝ 1 ⎠ a = e −0.0294 = 0.971 H2O

9. b. boiling: H2O(liq,a=0.971)---> H2O(g, PH2O=1) K= PH2O/a =1/a = 1/0.971 = 1.0298 Endothermic reaction, K is increasing from 1 to 1.0298, so T must be greater than 373 K.

⎛ Kbp ⎞ − ΔH vap ⎛ 1 1 ⎞ − 40.66J / mol ⎛ 1 1 ⎞ ln⎜ ⎟ = ⎜ − ⎟ = ⎜ − ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ K373 ⎠ R ⎝ Tbp 373 ⎠ 8.3145 ⎝ Tbp 373 ⎠

⎛1/ aH O ⎞ 1.0298 ⎜ 2 ⎟ ⎛ ⎞ ln⎜ ⎟ = ln⎜ ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠ ⎛1.0298 ⎞ 8.3145ln⎜ ⎟ 1 1 1 ⎝ ⎠ + = = 0.002675 − 40660 373 Tbp

Tbp = 373.84 c. H2O(liq,a=0.971)---> H2O(g, PH2O=1) K= PH2O/a

PH2O = aK where K is obviously the vapor pressure of pure liquid at 298 K PH2O = 0.971 x 0.031 = 0.030 atm

Page 5 of 6

−V Π a = e RT V Π = − ln a d. RT RT 0.0294 (0.08205 LatmK -1mol -1 )(298 K ) Π = − ln a = V 0.018 Lmol -1 = 40 .0atm If this seems large, recall that from problem 8., ΠV = nRT gives approximately Π = 25 atm for a 1M solution. The freezing point depression is approximately 0.86 x , which is about 1 m, so that would be a change of only -0.86 K. But in this problem the melting point was -3 K, suggesting a value considerably higher than 25 atm.

Page 6 of 6