The units are determined by the standard states used in the Table: NH 3 (g) NH 3 (aq), means Q =[CO 2]/ p CO2 = M/bar.

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1 CHMY 36 Nov 5, 26 6 problems on 2 pages HOMEWOR #4: SOLUIONS. (a) From able 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 25 o C. What units does this equilibrium constant have? (b) Do the same as (a) at 37 o C (c) Convert the Henry s Law constants you obtained in (a) and (b) to the units of able 6. (bars/mol fraction) and compare with the numbers found in able 6. G f from A.5 H f kj/mol kj/mol CO 2(aq) CO 2(g) aq gas diff (a) 298=exp(-G /R) = exp(-85/(8.345*298) =.322 moll - bar - he units are determined by the standard states used in the able: NH 3 (g) NH 3 (aq), means Q =[CO 2]/ p CO2 = M/bar. (b)lechatlier says exothermic reactions shift left with increasing (i.e., get smaller, i.e., the solubility of all gases is smaller at higher temperatures). herefore H is absolutely required to find the at 3. (G changes as changes, so that just changing only from 298 to 3 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): 3 H R exp(.37).235m/bar 3 Note: a simple sign error will result in a higher value than 298.You should recognize that this is wrong by LeChatlier s principle, and find the problem. (c) First turn the upside down because the reaction is reversed. for evaporation = Because L of liquid water contains g/8 = mol, a solution has a mole fraction = /.322 =3.2 bar/m. XCO2 = [CO2] / ([CO2]+55.55). hus,.322 moll - becomes.322/( ) =.5793 mol fraction. 298 = 3.2 bar/(m *.5793 mol fract/.322m) = 725 bar/molefraction, which is not too far from 59 bar/molfraction in able 6.. Page of 5

2 At 3, mole fraction changes to.235/( ) =.4229 and 3 = /.235 = bar/m. his becomes *.235/.4229 = 2369 bar/mol fraction compared to 23 in the table. 2. A 4 lb person contains about 5 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 = bar (assuming that the solubility is the same as in pure water? (Use able 6. for 37 o C.) (b) How many liters of gas at bar and 3 does this number of mols equal? (c) atmosphere = 76 mm of Hg, and the density of Hg is 3.6 times that of water. What is bar in units of meters of water, assuming the density of water is. kg/l? (d) If a diver stays at a depth of 4 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) he partial pressure of N2 is.78 bar for bar air pressure assumed at sea level. he amount dissolved is /(98 x 3 ) molefraction/bar x.78 bar = mol fraction = 7.96 x -6. he moles N2 dissolved = 7.96 x -6 mol N2/(mol N2 + mol water) = very closely 7.96 x -6 * 5 L water x 55.5 mol water/l water) =.22 mol N2 gas dissolved. (b) = pv = nr V=nR/p =.22 x.8345 x 3/ =.57 L N2 gas. (c) bar = bar /(.3 bar/atm) x.76 mhg/atm x 3.6 m H2O/m Hg =.2 m H2O (d) 4 m H2O / (.2 m H2O bar - ) = 3.92 bar of additional air pressure. his means 3.92 *.78 = 3.6 additional pn2 which equates to 3.6 x.57 =.74 L of bubbles that would be produced upon coming to the surface. 3. Consider a x -5 M polymer solution inside a dialysis bag that is permeable to the ligand, A, but not to the polymer. he following measurements of the free ligand outside the bag and total ligand inside the bag are measured at equilibrium. he units are M. From a Scatchard plot find d for the binding of A to its binding sites on the polymer and the number of binding sites per polymer molecule. (the notation 5E-7 means 5 x -7, etc.) [A]out [A]in total [A bound]=sa SA/A 5.E-7 5.5E-5 5.E-5.E+2.E-7 3.E-5 3.E-5 3.E+2 2.E-8.E E E+2 Plot of [SA] vs [A] has slope = -E7 = -/d and intercept = = [otal Sites]/d = Stot/d [otal Sites] = * E-7 = 6 x -5 M [Polymer] = e-5 M Sites/polymer = 6 x -5 / x -5 = 6 Plot of [SA] vs [A] 6.E+2 4.E+2 2.E+2 y = -E+7x E+.E+ 2.E-5 4.E-5 6.E-5 Page 2 of 5

3 4. he 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 =. M and the solubility of B in ethanol = 5. M at 25 o C. From the data in (a), calculate the following: (b) What is ΔG o for the process ([B(aq)]/ [B(ethanol)]? (c) What is ΔG o for the process B(aq) --> B(s)? (a) Solubility = for B(s) --> B(aq) = [B(aq)]/X B(s) =. M/ for pure solid B Solubility = for B(s) --> B(ethanol) = [B(ethanol)]/X B(s) =5 M/ for pure solid B Partition Coef. [B(aq)]/ [B(ethanol)] = for B(ethanol) B(aq) = ([B(aq)]/X B(s))/ [B(ethanol)]/X B(s) )=./5 =.2 (b) ΔG o = -R = *298(.2) = 9693 (c) his is reverse of solubility so = /. = ΔG o = -R = *298() = -575 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.. In the interior of the mitochondrion, the ph is 7.. In addition, the intermembrane space has an electric potential difference of +75 mv relative to the interior. AP 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 AP that can be obtained from ADP and P at ph 7 when 4 moles of protons are transferred if [AP]/[ADP] = 2. and [P].5 M? (a) C R C in out zf ( 7. in out ,5coulombs ( (.75J / coulomb) J ) (b) ΔG = 4 x Δ =4 x (-8649) = -74,595 J Page 3 of 5

4 (c) AP reaction = ADP + Pi AP ΔG = +3, J ΔGproton-transfer = n protons Δproton-transfer ΔGAP = ΔG + RQ/Q = 3, + R(2./.5) = +45,845 J/mol (at ph 7 the Q cancels [H + ] in Q) Maximum will be when reversible, i.e., at equilibrium, i.e., when ΔGtotal = ΔGproton-transfer + ΔGAP =. moles protons x (-8649) ) = moles protons =45845/8649 = 2.46 on average 6. he osmotic pressure of sea water is known to be 25 bar at. Using this number: (a) what is the activity of water in sea water at? (b) what is the vapor pressure of sea water at, given that for pure water it is 62 Pa? (c) what is the melting point of pure water ice in seawater? (d) what is the point of this sea water, assuming the activity is the same as at.? (a) a - exp( V R m 25.bar.8L ) exp( ) (b) Using Rauolt s Law, ph2o(g) / ah2o(l) = vapor pressure of pure liquid = 62 Pa, (i.e., the equilibrium constant for H2O(l,a)H2O(g, psea water ) : psea water = 62 *.98 = 6. Pa (c) he reaction is H2O(s,pure) H2O(l,a) Q = a/. At, this is spontaneous to right because a <. (salt lowers the melting point) o 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 constant = a/. So, use Vant Hoff with H fusion = 67 J/mol m a H R fusion melting melting melting Page 4 of 5

5 (d) Similarly, for the point raising: he reaction is H2O(l,a) H2O(g, ph2ogas) Q = /a he point = with applied pressure = bar because the vapor pressure of pure water =. bar at this temperature. For sea water at, this is spontaneous to LEF because the applied pressure = at sea level, causing ph2o in any bubbles in the water to also =, which exceeds the vapor pressure of the sea water, which has vapor pressure less than because of the salt as shown in 6.(b). o reach equilibrium we must increase the temperature to make the process more spontaneous until the equilibrium constant = /a, i.e., so that the vapor pressure of the solution =. bar = the applied pressure. Again, use Vant Hoff with H vaporization = 4,657 J/mol bp a H vaporazation R , Page 5 of 5