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Page 1 of 6 This Is Henry's Law. It Says That at Equilibrium the Ratio of Dissolved NH3 to the Partial Pressure of NH3 Gas In

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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 boiling point of water deep in a mine that is so far below sea level that the atmospheric pressure 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 Vapor Pressure ⎝ P1 ⎠ ⎝ K1 ⎠ R ⎝ T2 T1 ⎠ for the pure liquid. ⎛1.17 ⎞ 44,020 ⎛ 1 1 ⎞ ln⎜ ⎟ = − ⎜ − ⎟ = 1 8.3145 ⎜ T 373 ⎟ ⎝ ⎠ ⎝ 2 ⎠ ⎡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., equilibrium constant) 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 mole 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 partial pressure of NH3 gas 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 concentration in M instead of XNH3. Need the convert [NH3(aq)] to XNH3(aq) Mole fraction of a 1 M solution 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 volume of N2 gas would be released from the blood if the diver 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 oxygen 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 molecule. (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 partition coefficient for solute B = [Bmethanol]/ [Bwater] = 0.010, and the solubility 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 molecules 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 aqueous solution of a protein with molecular weight of 5 x 104 g/mol, what is the osmotic pressure in mm of liquid water at 298 K, using the relationship that reminds one of the ideal gas 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.
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  • Topic720 Composition: Mole Fraction: Molality: Concentration a Solution Comprises at Least Two Different Chemical Substances

    Topic720 Composition: Mole Fraction: Molality: Concentration a Solution Comprises at Least Two Different Chemical Substances

    Topic720 Composition: Mole Fraction: Molality: Concentration A solution comprises at least two different chemical substances where at least one substance is in vast molar excess. The term ‘solution’ is used to describe both solids and liquids. Nevertheless the term ‘solution’ in the absence of the word ‘solid’ refers to a liquid. Chemists are particularly expert at identifying the number and chemical formulae of chemical substances present in a given closed system. Here we explore how the chemical composition of a given system is expressed. We consider a simple system prepared using water()l and urea(s) at ambient temperature and pressure. We designate water as chemical substance 1 and urea as chemical substance j, so that the closed system contains an aqueous solution. The amounts of the two substances are given by n1 = = ()wM11 and nj ()wMjj where w1 and wj are masses; M1 and Mj are the molar masses of the two chemical substances. In these terms, n1 and nj are extensive variables. = ⋅ + ⋅ Mass of solution, w n1 M1 n j M j (a) = ⋅ Mass of solvent, w1 n1 M1 (b) = -1 For water, M1 0.018 kg mol . However in reviewing the properties of solutions, chemists prefer intensive composition variables. Mole Fraction The mole fractions of the two substances x1 and xj are given by the following two equations: =+ =+ xnnn111()j xnnnjj()1 j (c) += Here x1 x j 10. In general terms for a system comprising i - chemical substances, the mole fraction of substance k is given by equation (d). ji= = xnkk/ ∑ n j (d) j=1 ji= = Hence ∑ x j 10.