CHMY 361 Nov 5, 2016 6 problems on 2 pages HOMEWORK #4: SOLUTIONS
1. (a) From Table A.5 (back of text), calculate the Henry’s Law constant (i.e., equilibrium constant) for dissolving CO2(g) in water (i.e., solubility) at 25oC. What units does this equilibrium constant have? (b) Do the same as (a) at 37oC (c) Convert the Henry’s Law constants you obtained in (a) and (b) to the units of Table 6.1 (bars/mol fraction) and compare with the numbers found in Table 6.1 0 G f 0 from A.5 H f kJ/mol kJ/mol
CO2(aq) -413.80 - 385.85
CO2(g) -393.51 -394.36 aq –gas diff. -20.29 8.5 1
0 -1 -1 (a) K298=exp(-G /RT) = exp(-8510/(8.3145*298) = 0.0322 molL bar
The units are determined by the standard states used in the Table: NH3 (g) NH3 (aq), means Q =[CO2]/ pCO2 = M/bar.
(b)LeChatlier says exothermic reactions shift left with increasing T (i.e., K get smaller, i.e., the solubility of all gases is smaller at higher temperatures). Therefore H0 is absolutely required to find the K at 310K. (G0 changes as T changes, so that just changing T only from 298 to 310 in the above equation is wrong, and worth no credit.)
Use Van’t Hoff equation for temperature dependence of all equilibrium constants (e.g., vapor pressure, Henry Law constants, solubilities, partition coefficients, etc): K H0 1 1 K - 20290 1 1 ln 310 ln 310 0.317
K298 R 310 298 0.0322 8.3145 310 298
K310 0.0322exp(0.317) 0.0235 M/bar
Note: a simple sign error will result in a higher value than K298 .You should recognize that this is wrong by LeChatlier’s principle, and find the problem.
(c) First turn the K upside down because the reaction is reversed. K for evaporation = Because 1L of liquid water contains 1000 g/18 =55.555 mol, a solution has a mole fraction = 1/0.0322 =31.02 bar/M.
-1 XCO2 = [CO2] / ([CO2]+55.55). Thus, 0.0322 molL becomes 0.0322/(0.0322 + 55.55) = 0.0005793 mol fraction. K298 = 31.02 bar/(M * 0.0005793 mol fract/ 0.0322M) = 1725 bar/molefraction, which is not too far from 1590 bar/molfraction in Table 6.1.
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At 310, mole fraction changes to 0.0235/(.0235+55.55) = 0.0004229 and K310 = 1/.0235 = 42.64 bar/M. This becomes 42.64 * .0235/0.0004229 = 2369 bar/mol fraction compared to 2130 in the table.
2. A 140 lb person contains about 50 L of water (about 4/5 of which is intracellular). Assuming all of this water is saturated with dissolved N2 gas, answer the following questions: (a) How many mols of N2 gas are dissolved in a person when the air pressure = 1 bar (assuming that the solubility is the same as in pure water? (Use Table 6.1 for 37o C.) (b) How many liters of gas at 1 bar and 310 K does this number of mols equal? (c) 1 atmosphere = 760 mm of Hg, and the density of Hg is 13.6 times that of water. What is 1 bar in units of meters of water, assuming the density of water is 1.00 kg/L? (d) If a diver stays at a depth of 40 m in water until equilibrium is reached, how many liters of N2 gas would potentially be released in the form of microbubbles if the diver came to the surface very quickly? (Remember, Henry’s Law is just saying that the amount of a particular gas dissolved is directly proportional to the partial pressure of that gas in equilibrium with the liquid.)
(a) The partial pressure of N2 is 0.78 bar for 1 bar air pressure assumed at sea level. The amount dissolved is 1/(98 x 103) molefraction/bar x 0.78 bar = mol fraction = 7.96 x 10-6 . -6 -6 The moles N2 dissolved = 7.96 x 10 mol N2/(mol N2 + mol water) = very closely 7.96 x 10 * 50 L water x 55.5 mol water/L water) = 0.0221 mol N2 gas dissolved.
(b) = pV = nRT V=nRT/p = 0.0221 x 0.083145 x 310/1 = 0.570 L N2 gas.
(c) 1bar = 1 bar /(1.013 bar/atm) x 0.760 mHg/atm x 13.6 m H2O/m Hg = 10.20 m H2O
-1 (d) 40 m H2O / (10.20 m H2O bar ) = 3.92 bar of additional air pressure. This means 3.92 * 0.78 = 3.06 additional pN2 which equates to 3.06 x 0.570 = 1.74 L of bubbles that would be produced upon coming to the surface.
3. Consider a 1x10-5 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 equilibrium. 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]in [A]out [A bound]=SA SA/A total Plot of [SA] vs [A] 5.00E-07 5.05E-05 5.00E-05 1.00E+02 6.00E+02 1.00E-07 3.01E-05 3.00E-05 3.00E+02 2.00E-08 1.00E-05 9.98E-06 4.99E+02 4.00E+02 2.00E+02 y = -1E+07x + 598.7 0.00E+00 0.00E+00 2.00E-05 4.00E-05 6.00E-05
Plot of [SA] vs [A] has slope = -1E7 = -1/Kd and intercept = 598.7 = [Total Sites]/Kd = Stot/Kd [Total Sites] = 598.7 * 1E-7 = 6 x 10-5 M [Polymer] = 1e-5 M Sites/polymer = 6 x 10-5/1 x 10-5 = 6
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4. The partition coefficient of solute B between water and ethanol cannot be directly measured because water and ethanol are completely miscible. (a) Calculate this partition coefficient ([B(aq)]/ [B(ethanol)]) if the solubility of B in water = 0.10 M and the solubility of B in ethanol = 5.0 M at 25o C. From the data in (a), calculate the following: (b) What is ΔGo for the process ([B(aq)]/ [B(ethanol)]? (c) What is ΔGo for the process B(aq) --> B(s)?
(a) Solubility = K for B(s) --> B(aq) = [B(aq)]/X B(s) =0.10 M/1 for pure solid B Solubility = K for B(s) --> B(ethanol) = [B(ethanol)]/X B(s) =5 M/1 for pure solid B Partition Coef. [B(aq)]/ [B(ethanol)] = K for B(ethanol) B(aq) = ([B(aq)]/X B(s))/ [B(ethanol)]/X B(s) )= 0.1/5 =0.02
(b) ΔGo = -RTlnK = -8.3145*298ln(0.02) = 9693
(c) This is reverse of solubility so K = 1/0.1 = 10 ΔGo = -RTlnK = -8.3145*298ln(10) = -5705 J/mol
5. Suppose in certain mitochondria, the oxidation of glucose products pumps protons into the intermembrane space between the double membrane to reach a pH of 5.0. In the interior of the mitochondrion, the pH is 7.0. In addition, the intermembrane space has an electric potential difference of +75 mV relative to the interior.
ATP is synthesized from the combined Gibbs energy change per mol of protons (Δ) flowing down the concentration gradient from the intermembrane space (out) to the interior (in) and from the change of electrical free energy coming from the voltage difference.
(a)What is Δ for H+(out) --> H+(in) for this mitochondrion?
(b) Calculate the ΔG if a total of 4 moles of protons are transferred.
(c) What is the maximum number of moles of ATP that can be obtained from ADP and P at pH 7 when 4 moles of protons are transferred if [ATP]/[ADP] = 2.0 and [P] 0.005 M?
(a)
0 Cin RT ln zF (in out) Cout 107.0 0 8.3145 298ln 1 96,500coulombs (0 (0.075J / coulomb) 105.0 -11412- 7236 18649 J
(b) ΔG = 4 x Δ =4 x (-18649) = -74,595 J
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0’ (c) ATP reaction = ADP + Pi ATP ΔG = +31,000 J
ΔGproton-transfer = nprotons Δproton-transfer
0’ 0’ ΔGATP = ΔG + RTlnQ/Q = 31,000 + RTln(2.0/0.005) = +45,845 J/mol (at pH 7 the Q0’ cancels [H+ ] in Q)
Maximum will be when reversible, i.e., at equilibrium, i.e., when ΔGtotal = ΔGproton-transfer + ΔGATP = 0. moles protons x (-18649) + 45845) = 0 moles protons =45845/18649 = 2.46 on average
6. The osmotic pressure of sea water is known to be 25 bar at 273 K. Using this number: (a) what is the activity of water in sea water at 273 K? (b) what is the vapor pressure of sea water at 273 K, given that for pure water it is 612 Pa? (c) what is the melting point of pure water ice in seawater? (d) what is the boiling point of this sea water, assuming the activity is the same as at 273 K.?
- V 25.0bar 0.018L (a) a exp( m ) exp( ) 0.980 RT 0.083145 273
(b) Using Rauolt’s Law, pH2O(g) / aH2O(l) = vapor pressure of pure liquid = 612 Pa, (i.e., the equilibrium constant for H2O(l,a)H2O(g, psea water ) : psea water = 612 * 0.98 = 600.0 Pa
(c) The “reaction” is H2O(s,pure) H2O(l,a) Q = a/1. At 273, this is spontaneous to right because a < 1. (salt lowers the melting point) To reach equilibrium we must decrease the temperature to the melting point of the sea water i.e., make the process less spontaneous until equilibrium is reached, i.e., until the equilibrium 0 constant = a/1. So, use Vant Hoff with H fusion = 6007 J/mol
K a H0 1 1 ln m ln fusion K273 1 R Tmelting 273
8.3145 0.98 1 1 ln 0.00368 6007 1 373 T melting Tmelting 270.9 K
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(d) Similarly, for the boiling point raising:
The “reaction” is H2O(l,a) H2O(g, pH2Ogas) Q = 1/a
The boiling point = 373 K with applied pressure = 1 bar because the vapor pressure of pure water =1.0 bar at this temperature.
For sea water at 373, this is spontaneous to LEFT because the applied pressure = 1 at sea level, causing pH2O in any bubbles in the water to also = 1, which exceeds the vapor pressure of the sea water, which has vapor pressure less than 1 because of the salt as shown in 6.(b). To reach equilibrium we must increase the temperature to make the process more spontaneous until the equilibrium constant = 1/a, i.e., so that the vapor pressure of the solution = 1.0 bar = the applied 0 pressure. Again, use Vant Hoff with H vaporization = 40,657 J/mol
K 1 H0 1 1 ln bp ln vaporazation K a R T 373 373 boiling
8.3145 1 1 1 ln 0.00268 40,657 0.98 373 T boiling T 373.576 K boiling Tboiling 0.576 K
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