Page 1 of 7 Please leave the exam pages stapled together. The formulas are on a separate sheet. This exam has 5 questions. You must answer at least 4 of the questions. You may answer all 5 questions if you wish. Answering 5 questions can be an advantage if you are unsure of some of your answers (this will distribute the risk ). Answering 4 questions is advantageous if you are very sure of your answers. Each page is worth 0 points. The total exam grade will be normalized so that the maximum number of course points for this exam will be 0. For example, getting 80 points on 4 questions equals 100 points on 5 questions equals 0 points toward the final grade. Getting 80 points on 5 questions would be worth 80% of the maximum grade. If you leave a page blank, it will not be included in the grading. If you work on a page and then decide that you do not want it to be graded, be sure to mark the DO NOT GRADE THIS PAGE box at the bottom of the page. If you work on the page and fail to mark the box, the page will be graded. Work at least 4 problems (of your choosing) or all 5, as you prefer. Answers without explanations (where indicated) are not complete.
Page of 7 1. Ethanol (CH 3 CH OH) has an enthalpy of vaporization of ΔH vap = 38.6 kj/mol. The boiling point temperature for pure ethanol at sea level is 78.4 C. a) What is the vapor pressure of pure ethanol at T = 5 C? The boiling pressure is 1 atm (the sea level pressure). Using P 1 = 1 atm, T 1 = 78.4 C = 35 and T = 98, and plugging into the Clausius-Clapyron equation leaves 38,600 J P 1 1 ln = mol =.39. Solving for P gives P = 0.09 atm. 1atm 8.3145 J 98 35 mol b) Ethanol gas is fed into a distillation apparatus at a pressure of 0.3 atm and T = 100 C. No other gases are present in the apparatus. As the ethanol moves through the apparatus, its temperature drops. At what temperature will the ethanol condense out of the vapor phase? Boiling and condensation are the same thing looked at in different directions. This question is asking what the BP of ethanol is at P = 0.3 atm. Clausius-Clapyron gives the answer: J 0.3atm 38,600 = mol 1 1 ln. Solving for T gives T = 33 = 50 C. This is 1atm 8.3145 J T 35 mol the BP of ethanol at an atmospheric pressure of 0.3 atm. c) Suppose the apparatus in part b) was charged with nitrogen at atm pressure before the ethanol was introduced. Will this have an effect on the condensation temperature calculated in part b? (No credit unless you explain your answer.) Yes. The atmospheric pressure in the apparatus is now.3 atm. This must be the boiling pressure of the ethanol, which is the same as the condensation pressure. Because the P value in Clausius-Clapyron increases, the T value will also. d) Suppose the last person to use the distillation apparatus before you left some ethanol-soluble residue on the apparatus. Will the presence of this material have an effect on the condensation temperature calculated in part b)? (No credit unless you explain your answer.) Yes. According to Raoult s law, the vapor pressure of a solvent is reduced in the presence of a solute. The residue will go into solution in the condensate, thus reducing the mole fraction of ethanol. This will reduce the vapor pressure, meaning that the ethanol temperature must be higher to raise the vapor pressure to 0.3 atm. Thus, the condensation (boiling) temperature will rise. This may also be approached by considering the effect of a solute on BP using the BP elevation formula. In either case, the BP increases and therefore the condensation T will be higher than in part b).
Page 3 of 7. The enzyme formyltetrahydrofolate synthetase employs ATP hydrolysis to form a carbonnitrogen bond between tetrahydrofolate and formate. The enzyme has four identical subunits, each with a binding sites for ATP. a) If the four ATP binding sites are identical with A = 1 10 4 and do not interact with each other, at what concentration of ATP will half of the enzyme sites be occupied by ATP? Half of the # of sites are two. The binding formula for number of sites bound per molecule is N ν =. When υ = and N = 4, 0.5 =, or A [ = 1. Thus, [ = 1/ A = 1 1+ 1+ 10-4 1 M = 0.1 mm. (The units come from the definition of activity.) b) If the four ATP binding sites are positively cooperative with a Hill coefficient of.5, at what concentration of ATP will half of the enzyme sites be occupied by ATP? The fraction of sites occupied according to the Hill analysis is c) On the graph at right, sketch the plot of f/[atp] vs. f for cases a) and b). For any straight line that you draw, indicate the value of the slope and the intercepts on both axes. Case a) gives a straight line with slope equal to - A and vertical axis intercept of (1) A. The f-axis intercept is at f = 1 (maximum binding). Case b) gives a curved Scatchard plot, the curvature being concave downward because the Hill coefficient,.5, is larger than 1, indicating positive cooperativity. n A = where n =.5. n 1+ Setting f = 0.5 shows that A [.5 = 1. Thus, [ = (1/ A ) 0.4 = (1 10-4 ) 0.4. 1 M =.51 10 - M = 5.1 mm. f\/atp] int = A d) Now suppose that three of the four ATP binding subunits in the protein are mutated so that ATP can no longer bind to them. Will this mutation change the answers to parts a) and b)? You don t need to solve for numbers but you do need to explain your answer. f slope = - A If three sites don t bind ATP, then N = 1 and υ = 0.5 in part a). Those values will lead to the same answer, 0.1 mm. The reason is that the sites are independent and equivalent. a) [ b) int = 1 The answer to part b) will most likely change if three sites no longer bind ATP. In most cases, cooperativity requires subunit-subunit interactions that induce ligand affinity changes in neighboring sites when one site is bound to a ligand. Here, there is only one site, and ligand binding there can t alter affinity on the other three subunits (because they no longer bind ATP). Most likely, the affinity of the single site would be similar to the affinity for the independ- ent, equivalent site result. f
Page 4 of 7 3. Three beakers contain three different solvents, A, B and C. A sample of the same compound (a nonvolatile solid) is placed in each beaker. All three solid samples partially dissolve into the solvents until the solutions become saturated. The concentration in solvent A is 15 mm, that in the solvent B is 60 mm and that in solvent C is 100 mm. All beakers are at T = 98 and 1 atm pressure. a) What is the value of Δμ, the standard free energy of transfer, for the transfer of the compound from the solid phase into the solution in solvent B? Δμ = μ solution - μ solid. The chemical potentials of the solid and solution are equal at equilibrium (when the solution becomes saturated). Thus, μ solution = μ solid or μ solution + RT ln a solution = μ solid. We see from this that μ solution - μ solid = -RT ln a solution = -(8.3145 J/ mol)(98 ) ln {(60 10-3 M)/(1 M)} = 6.97 10 3 J/mol = 6.97 kj/mol. b) The standard state chemical potential μ is highest in which of the solvents? X Solvent A. Solvent B. Solvent C. Explain: The chemical potentials of the all of the solutions are equal (because they all equal the standard state chemical potential of the solid). Thus, μ A + RT ln a A = μ solid or μ A = μ solid - RT ln a A. The solvent that has the lowest activity of dissolved solid is the one with the highest value of μ for the compound. c) The vapor pressure for pure solvent B is 0.078 atm at 98. One mole of this pure solvent occupies 00 ml at this temperature. What is the vapor pressure of solution B at this temperature? From Raoult s law, P solution = X solvent P* solvent. We are given P* solvent = 0.078 atm. We need to compute the mole fraction of solvent. There are 5 mol of solvent in 1 liter of solution. From the concentration above, there are 60 10-3 moles of solute in 1 L. Thus, the mole fraction of solvent is X solvent = (5 mol)/(5.06 mol) = 0.988. Plugging this value into Raoult s law gives the solution vapor pressure: P solution = (0.988)(0.078 atm) = 0.077 atm. d) Solvent C is water. What is the osmotic pressure of the aqueous solution above (relative to pure water)? Π = crt = (100 10-3 mol/l)(0.0805 L atm/mol )(98 ) =.45 atm =.48 10 5 Pa =.48 10 5 N/m. The osmotic pressure in the formula is the pressure relative to pure solvent.
Page 5 of 7 4. Hydrogen (H ), nitrogen (N ) and xenon (Xe) all act like ideal gases. The molar mass of each gas is, respectively, g/mol, 8 g/mol and 131 g/mol. a) Suppose 1 mole each of these gases is in a sealed Xe container. The temperature and pressure of each is 98 0.00 and 1 atm. A plot of the fraction of molecules at a 0.015 particular speed is shown at right. Indicate which curve is for which gas. The most probable speed (at the peak of each curve) is proportional to m -1/. Thus, the gas with the highest mass (Xe) has the smallest <u> mp and that with the lowest (H ) has the largest <u> mp. b) The sample of nitrogen gas is examined at 3 different temperatures, 100, 400 and 1000. A plot of the fraction of molecules at a particular speed is shown at right. Indicate which curve is for which temperature. The most probable speed (at the peak of each curve) is proportional to T 1/. Thus, the gas sample at the highest temperature (1000 ) has the largest <u> mp and that with the lowest T has the smallest <u> mp. F(u) F(u) 0.05 0.010 0.005 0.000 0.00 0.015 0.010 0.005 N H 0 1000 000 3000 4000 5000 u (m/sec) 100 400 0.000 0 500 1000 1500 000 u (m/sec) 1000 c) The collisional cross sectional diameter σ of H is.5 10-10 m. The collisional cross sectional diameter of N is 3.7 10-10 m. The collisional cross sectional diameter of Xe is 10.8 10-10 m. Which of these three gases has the longest mean free path at T = 300? Explain. 1 The mean free path is given by l =. The value of the number of molecules per unit N π σ V volume (N/V) is the same for each gas because they all are ideal. This can be shown by saying that n/v = P/RT. The term on the right is the same for all gases. Multiplying both sides by N A gives N/V. The longest mean free path (largest value of l) thus occurs for the smallest value of σ, i.e. for H. d) Which of these three gases has the highest average kinetic energy per molecule at T = 300? Explain. All three have the same average kinetic energy. For an ideal gas, the total kinetic energy of the gas is 3 U = N k T. Because each sample has the same number of molecules (Avogadro s number) and A B temperature, the average E per molecule is the same for H, N and Xe. Another way to look at this is that the E of a single molecule is proportional to the mass of the molecule and its velocity squared. Averaging over all molecules, the average E is thus proportional to m and <u > = (u rms). u rms is proportional to m -1/, so the m cancels, leaving a dependence on T only. Thus all gases have the same average E per molecule.
Page 6 of 7 5. A sample of the protein DNA ligase is purified and prepared for centrifugation. The molar mass of the protein is known to be 74,000 g/mol. The density of the buffer solution is 1.01 g/cm 3 and the viscosity of the buffer solution is 9.94 10 - Pa sec. (Pa sec is the unit of viscosity in the SI system.) The temperature for all parts of this problem is 0 C. a) 15 micrograms of the protein is added to 5 microliters of buffer. The volume of the protein solution is carefully measured to be 5.011 μl. What is the partial specific volume v of the protein in this buffer? Here we have that ΔV = 0.011 μl and w = 15 μg. Therefore, v = (0.011 μl)/(15 μg) = 7.33 10-4 L/g = 0.733 cm 3 /g. The latter is the more typical unit for partial specific volume. b) A DNA ligase sample is centrifuged in the same buffer as in a) at 56,000 rpm. The position of the protein boundary layer vs. time is measured, and plotted as at right. The slope of the plot is 3.4 10-4 min -1. What is the sedimentation coefficient for DNA ligase? (Hint: there are π radians in one revolution.).303 d log r1.303 The formula relating slope, s and ω is s = = ( slope). ω dt ω We need to put ω into SI units. ω = (5.6 10 4 rev min -1 )(π rev -1 )(1 min/60 sec) = time (min) 5.86 10 3 sec -1. The slope must be in SI units as well: slope = (3.4 10-4 min -1 )(1 min/60 sec) = 5.67 10-6 sec -1. Plugging these values into the above formula gives.303 5.67 10 6 5860sec sec 1 3.8 10 13 s = =. Using Svedberg units, this value is s = 3.8 S. 1 ( ) ( ) sec log(r 1/ ) 0.8 0.81 0.80 0.79 0.78 0.77 0 0 40 60 80 100 10 140 c) What is the value of the frictional coefficient for DNA ligase in this buffer? The formula relating s, ρ, m, v and f is ( vρ ) m s = 1. The value of m, the single-molecule mass, is m = f M/N A = (74,000 g/mol)/(6.03 10 3 mol -1 ) = 1.3 10-19 g = 1.3 10 - kg. Rearranging the 3 cm g 1 0.733 1.01 ( 1.3 10 kg) formula, plugging in known values, and solving for f gives ( 1 vρ ) m = = 8.41 10-11 kg/sec = 8.41 10-11 Pa m sec. d) If DNA ligase can be modeled as having a spherical shape, what is its radius? f = s 3 g cm 13 3.8 10 sec The formula relating f, viscosity and radius for a sphere moving through a viscous medium is f = 6πηa. Rearranging to solve for a leaves a = f/6πη. Plugging in the known values gives a = (8.41 10-11 Pa m sec)/{6(3.1416)(9.94 10 - Pa sec)} = 4.49 10-11 m. This would mean that the molecular diameter is about 0.9 Å, much too small to be reasonable. Clearly, the molecule is not a sphere.
Page 7 of 7 Scratch pad: For grading purposes: question 1 3 4 5 Tot score