Name ANSWER KEY Chemistry 25 (Spring term 2017) Midterm Examination Distributed Thursday, May 4, 2017 Due Thursday, May 11, 2017 by 1 pm in class or by 12:45 pm in 362 Broad a drop box will be left outside 362 Broad to return midterms during the week ***late penalties enforced*** Conditions Open this examination when you are ready to take it. This is a 3 hour examination that must be taken in one continuous stretch. You may use the Ch25 online lecture notes, problem sets and solutions, the course web site and a calculator. You may use handwritten notes you have made from thermodynamics texts, including EC and KKW. You may not use any books (including EC and KKW), exams and problem sets from previous years of Ch25 (unless they are posted on the course website), other web sites, non-calculator applications of Mathematica and related programs, discuss the exam with others, etc. This exam should have 20 pages total. Show your work! All work must be completed in the provided space. Getting the right answer is not enough the intermediate steps are needed for credit. useful relationships 1 µm = 10-6 m; 1 nm = 10-9 m; 1 Å = 10-8 cm = 10-10 m; 1 µ = 10-6 m = 10 3 nm. N A = 6.022 x 10 23 molecules mol -1 = Avogadro's number force/energy 1 J = 1 N m = 1 C V; 1 pn = 10-12 N; 1 pn nm = 10-21 J = 0.602 kj mol -1 ; 4.184 J = 1 cal. pressure units: 1 pascal (Pa) = 1 N m -2 = 1 J m -3 = 10-5 bar 1atm = 101325 Pa = 0.1013 pn nm -2 = 760 mm Hg = 14.696 lb/sq. in. = 0.1013 J cm -3 = 101.325 kj m -3 electrostatics F = 96.487 kj mol -1 V -1 = 96,487 C mol -1 ; q = 1.602 x 10-19 Coulombs = F /N A, 1 ev = 1.602 x 10-19 J, 4πε o =1.1126 x 10-10 C V -1 m -1 gas constant R 8.3144 J mol -1 K -1 = 1.986 cal mol -1 K -1 = 0.08206 liter atm mol -1 K -1 1.38066 10-23 J K -1 (= R/N A = k B, the Boltzmann constant) unless otherwise stated: you may assume T = 298 K, P = 1 atm RT = 2.48 kj mol -1 and k B T = 4.11 x 10-21 J at T = 298 K. for water, molecular weight = 0.018 kg mol -1, liquid density = 1000 kg m -3, ε = 80 liquid heat capacity = 4.18 J gm -1 K -1 = 75.4 J mol -1 K -1 ; H fusion = 6.01 kj mol -1 ; H vap = 40.66 kj mol -1 1
Name problem points 1a 5 1b 5 1c 5 1d 5 2a 2 2b 9 2c 9 3 15 4 12 5a 2 5b 9 6a 6 6b 2 6c 10 7a 2 7b 1 7c 1 total 100 2
Problem 1 (20 pts) A thermodynamics smorgasbord 1a (5 pts) We calculated in Problem Set 1 that an E. coli cell contains ~3 x 10 6 proteins in a 1 µ 3 volume. Assuming that this is all one type of protein (which, of course, it is not), what is the corresponding concentration of the protein in M (moles per liter)? molar concentration = moles protein / volume in liters = 3 10 6 6.02 10 23 10 18 m 3 1000 liters = 0.005M m 3 3
1b (5 pts) At the triple point of water (273.16 K, 611.7 Pa), the ice, liquid and vapor phases of water are in equilibrium. Under these conditions, H vap = 45.05 kj mol -1, S vap = 164.92 J mol -1 K -1 for the conversion of liquid to vapor H fus = 6.009 kj mol -1, S fus = 22.00 J mol -1 K -1 for the melting of solid to liquid From this data, calculate H sub and G sub (in kj mol -1 ) and S sub (in J mol -1 K -1 ) for the solid-to-vapor transition (sublimation) at the triple point of water. solid! vapor = solid! liquid + liquid! vapor H sub = H fus + H vap = 6.01 + 45.05 = 51.06 kj mol -1 S sub = S fus + S vap = 22.00 + 164.92 = 186.92 J mol -1 K -1 G sub = H sub T S sub = 0 kj mol -1 (as it must since the system is at equilibrium) 4
1c (5 pts) The average atmospheric pressure on the surface of Mars is 600 Pa. The radius of the planet is 3400 km and the acceleration due to gravity is 3.7 m s -2. From this data, compute the total mass M in kg of the Martian atmosphere. Hint: recall that atmospheric pressure reflects the force per unit area on the surface of a planet due to the weight of the atmosphere. Let M = mass of Martian atmosphere P = weight of atmosphere surface area of planet = Mg 4π R 2 M = 4π R2 P g = 4π ( 3.4 106 m) 2 600Pa = 2.4 10 16 kg 3.7ms 2 5
1d (5 pts) One way to fix N 2 is through reduction with water to form O 2 and NH 3 : 2N 2 (g)+ 6H 2 O(g)! 4NH 3 (g)+ 3O 2 (g) 1d.i Calculate G in kj mol -1 for this reaction as written from the free energies of formation of H 2 O (g) and NH 3 (g) of -228.6 and -16.6 kj mol -1, respectively. 1d.ii What is the equilibrium constant at 298 K? 1d.iii Will the application of pressure shift the equilibrium composition towards the reactants or products? 1d.iv Based on the thermodynamics, do you think that this reaction could be competitive with the Haber-Bosch reaction dissected in PS3? Explain briefly (1 sentence). 1di G = 4G NH 3 + 3G O2 6G H2 O 2G N2 = 4( 16.6)+ 0 6( 228.6) 0 = +1305 kj mol -1 1dii G = RT ln K G RT K = e ( ) =1.8 10 229 idiii Since there are 8 moles reactants and 7 moles products, increasing pressure will favor products since the total volume is smaller than for the reactants. idiv No - the Haber-Bosch is much more energetically favorable at room temperature (K = ~6x10 5 ) and even at high temperature the equilibrium is still ~10-4 to 10-5 and can be shifted to favor products by the application of reasonable amounts of pressure. 6
Problem 2 (20 pts) On the edge of stability The wildtype form of a particular protein heat denatures at a melting temperature T m = 332 K with an enthalpy of unfolding, H m = 372 kj mole -1 at T m. 2a (2 pts) What is the entropy of unfolding, S m, in kj mol -1 K -1, at T m? ΔS m = ΔH m = 372 =1.12 kj/mole/deg T m 332 2b (9 pts) A variant form of this protein is generated that has a native state 6 kj mol -1 more stable than the wild type form over the temperature range near T m. Furthermore, the thermodynamic properties of the unfolded state are unaffected by this mutation. If S m calculated in prob. 2a is identical for both the variant and wildtype proteins, at what temperature will the variant protein unfold? Assume that the entropy of unfolding is independent of temperature over this temperature range. G = µ N,variant - µ N,wt = -6 kj mol -1 G T = S m T = G S m = 6 1.12 = +5.4 T = 332 + 5.4 = 337.4K = 64.3 C 7
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2c (9 pts) A different protein from that in problem 2a/b contains N R = 150 residues and has a G of unfolding = +75 kj mol -1 at the temperature of maximum stability, T* = 300 K. Assume that the entropy change for unfolding at this temperature, S(T*) depends on only two terms: S(T*) = S(hydrophobic effect) + S(conformational) where S(hydrophobic effect) is the entropy change associated with the hydrophobic effect (that stabilizes the native N state) and S(conformational) is the conformational entropy change associated with the large numbers of degrees of freedom in the unfolded D state. Furthermore, assume at this temperature T* that the contribution of the hydrophobic effect to protein stability is exclusively entropic, with S(hydrophobic effect) = -3000 J mol -1 K -1, and that the conformational entropy is given by the Boltzmann expression, S (conformational) = RlnΩ N R where Ω is the number of conformational states per residue available to D, relative to N. Under these conditions, what are the values for S(T*), H(T*) and Ω? at T*, d G/dT = - S(T*) = 0 S(T*) = S(hydrophobic effect) + S(conformational) = 0 S(hydrophobic effect) = - S(conformational) S(hydrophobic) = -3000 J mol -1 K -1 = - RlnΩ N R = -RN R ln Ω = -8.3 (150) ln Ω Ω=e (3000/(8.3 150)) = e 2.4 = 11, so that each residue has approximately 11 times as many conformational states available to it in the D state, relative to the N state. H(T*) = G(T*) + T* S(T*) = 75 + 0 = +75 J mol -1 9
Problem 3 (15 pts) Having the vapors In lecture 7 (April 25), we calculated that the vapor pressure of water in equilibrium with the liquid phase at 25 C is 23.6 mm Hg, from an analysis of the temperature dependence of the chemical potentials for the liquid and vapor phases of water. Since air is saturated with water vapor under these conditions, the vapor pressure of 23.6 mm Hg corresponds to 100 % relative humidity at 25 C. What is the relative humidity corresponding to a water vapor pressure of 23.6 mm Hg at 37 C? (ie what is the ratio of 23.6 mm Hg to the equilibrium vapor pressure of water at 37 C?) For this calculation, the following thermodynamic information at 298 K may be useful and you may assume that the values of H and S are temperature independent. phase G (kj mol -1 ) H (kj mol -1 ) S (J mol -1 K -1 ) liquid -237.2-285.8 70.0 vapor -228.6-241.8 189.0 (as an irrelevant aside to working this problem, the vapor pressure of water at 37 C sets the upper limit to the altitude survivable by humans in an unpressurized environment, since water will boil when the atmospheric pressure equals this value at ~18 km or 60,000 feet; the Armstrong limit ) solution 1 10
There goal is to determine the value of q that corresponds to µ between liquid water and vapor at 310 K. There are multiple ways of deriving this value but they reflect the following relationships and the similarity of triangles ABC and ADE p + q = 8.6 kj mol -1 p = S or p = S T = 0.119 12 = 1.4 kj mol-1 T q = 8.6 p = 7.2 kj mol -1 µ = 7.2 kj mol -1 = RT ln P P! = (0.0083144 * 310)ln P P! P = P! e 7.2/(0.0083144*310) = P! e 2.79 = 0.061P! = 46.5 mm Hg = 0.061 atm = 6200 Pa and the relative humidity = 23.6/46.5 = 50.8% solution 2 - G = H-T S µ = µ liquid ( T ) µ gas T ( ) = ΔH T S = ( 285.8 + 241.8) T (.070.189) = 44.+ 0.119T µ ( 310) = 7.1 kj mol -1 = RT ln P P = (0.0083144 * 310)ln P! P! P = P! e 7.1/(0.0083144*310) = P! e 2.76 = 0.063P! 23.8/48.2=55% = 48.2 mm Hg = 0.063 atm = 6422 Pa solution 3 integrated Clausius Clapeyron equation dg = VdP - SdT = 0 assume V = ideal gas volume = RT / P dp P = SdT RT ln P 2 P 1 = H R P 2 = P 1 e 23.6 46.9 = 50% = HdT RT 2 " 1 1 % $ ' # T 2 T 1 & H " 1 1 % $ R # T 2 T 1 & ' = 23.6 e 44 " 1.0083144 310 1 % $ # 298 & ' = 46.9mmHg These numbers are all derived assuming ideal gas behavior, and that H and S are independent of temperature. The relative humidity does depend slightly on the details of these approximations, but the calculated results are close to the experimentally observed values of 23.8 and 47.1 mm Hg, giving a relative humidity = 50.5% 11
Problem 4 (12 pts) Melting under pressure Phospholipids in a bilayer arrangement undergo a variety of phase transitions. An important example is the conversion between the ordered Lβ gel phase which melts to the fluid Lα phase. The Lα phase is biologically relevant, since it allows diffusion of the various membrane components in the plane of the bilayer. Not surprisingly, the Lα phase is more disordered (which is why it is favored at higher temperatures) than the Lβ, and, since the packing is less efficient, there is an increase in volume, V, upon this melting transition. Lβ Lα gel phase - molecules pack together more tightly together. Alkyl chains are more highly ordered - larger bilayer thickness Temp liquid crystalline - represents bulk of lipids in biological membranes - considerable disorder in alkyl tails (Gennis, pg. 40) Because the Lα phase has the greater volume, addition of pressure stabilizes the smaller Lβ phase, with the consequence that higher melting temperatures are observed when bilayers are placed under pressure. For the melting of the phospholipid DPPC, the pressure dependence of the melting temperature is illustrated below. For DPPC at P = 1 atm, T m = 41.3 C and H m = 25 kj mol -1. Estimate dt m /dp from this figure, assuming it is constant under these range of conditions. From this information and the Clausius-Clapeyron equation, calculate the change in volume in m 3 mol -1 and Å 3 molecule -1 for DPPC for the Lβ to Lα transition. 12
experimentally, dt m /dp = (44.6 41.3)/(136-1) = 0.0244 K/atm 0.0244 K/atm / (1.013*10 5 Pa/atm) = 2.41 x 10-7 K/Pa from dt m /dp = T m V/ H m (Clausius Clapeyron) V = (dt m /dp) H m /T m = ((2.41x10-7 K/Pa)*(25000 J)/314) = 1.92x10-5 m 3 /mole 1.92x10-5 m 3 /mole x 10 30 Å 3 /m 3 x 1 mole/(6.02*10 23 molecules) = 32 Å 3 /molecule 13
Problem 5 (11 pts) Fueling up in the microbial world One of the fascinating relatively recent discoveries is the existence of life under extreme conditions (by our standards, not these organisms, of course) of temperature, pressure, ph, etc. Three reactions (in aqueous solution) that could serve as potential sources of metabolic energy for these organisms are listed below: A B C CH 3 CO 2 H (aq) CO 2 (aq) + CH 4 (aq) HCOOH (aq) CO 2 (aq) + H 2 (aq) 0.5N 2 (aq) + 1.5H 2 (aq) NH 3 (aq) The G values under standard conditions for these reactions provide estimates of their metabolic energy yields. The free energies of formation, G f, for the reactants and products, in aqueous solution at 25 C are tabulated below. G f (kj mol -1 ), compound P = 1 atm T = 25 C CH 3 CO 2 H (aq) -396.48 CH 4 (aq) -34.46 CO 2 (aq) -385.98 HCOOH (aq) -372.30 H 2 (aq) 17.72 N 2 (aq) 18.18 NH 3 (aq) -26.71 5a (2 pts) Why don t the G f entries = zero at T = 25 C for H 2 and N 2 in this Table? The standard state for gases like H 2 and N 2 are the gas at 1 atm, 298 K, not in aqueous solution; the G f for the latter will need to include the energetics of the gas-water interactions. 14
5b (9 pts) Calculate the standard free energy changes, G, for the three reactions, A, B and C, at T=25 C. Which reactions in principle could likely serve to drive the energy metabolism of an organism (identified as having G < 0)? Which reactions could in principle drive formation of ATP ( G < -40 kj mol -1 )? reaction A G f, reaction (kj/mole) B C reaction ΔG f = ΔG f,products ΔG f,reactants G f, reaction (kj/mole) i A -385.98-34.46 +396.48-23.96 B -385.98 +17.72 +372.30 +4.04 C -26.71-0.5*(18.18) -1.5*(17.72) -62.38 i reactions A and C in principle could drive metabolism, but only C could potentially drive ATP synthesis. 15
Problem 6 (18 pts) Thermodynamics of ethanol solutions The vapor pressure of ethanol (EtOH) in a mixture of ethanol and water at 25 C is tabulated below as a function of x 2, the mole fraction of ethanol. x 2 P(EtOH) P(H 2 O) mm Hg mm Hg 0.000 0 23.8 0.042 8.43 22.6 0.089 16.88 21.27 0.144 23.52 19.96 0.207 28.78 18.81 0.281 32.38 17.57 0.370 35.33 16.39 0.478 37.89 15.53 0.610 42.56 13.24 0.779 48.78 9.02 1.000 59.1 0 The vapor pressures of ethanol and water in equilibrium with a solution of indicated concentration are depicted with circles and squares, respectively. 6a (6 pts) Using the Raoult's law standard state for ethanol, determine from this data the non-ideal contribution to the free energy of transfer of ethanol from pure ethanol to dilute aqueous solution. This quantity equals RT ln γ 2 in the limit as x 2 0. (hint: follow the discussion in lecture 8 for n-propanol). Give your answer in kj mol -1. approximate the limiting value of γ 2 by the value calculated for x 2 = 0.042 using the Raoult's law standard state: γ 2 = P x 2 P = 8.43 0.042 59.1 ~ 3.4 RT lnγ 2 = 0.0083144 298 ln 3.4 = 3.0 kj mol -1 (the partial pressure of the hypothetical state extrapolated from dilute n-propanol to x2 = 1 is ~8.43/.041 = 201 mm Hg. Another way to estimate activity coefficient in dilute solution is from the ratio of this hypothetical Henry s law partial pressure to that of the pure solvent γ 2 = 201 59.1 ~ 3.4 16
6b (2 pts) How does this value of RT ln γ 2 compare to that calculated for n-propanol in lecture 8 (+6.6 kj mol -1 )? Does this make sense in terms of our discussion of the hydrophobic effect? Explain briefly (1-2 sentences) n-propanol has an additional methylene group (carbon) so that it has more exposed non-polar surface than ethanol and a more unfavorable transfer from a nonpolar phase to water which works out to about 3 kj mol -1 per carbon. 17
6c (10 pts) From the vapor pressure data, calculate the G mixing in kj mol -1 when 1.00 mole of ethanol is mixed with 10.23 moles of water at 298 K. Compare this to the value expected for an ideal solution with the same composition. Hints: Use mole fraction concentration units. You may use the nearest entry in the vapor-pressure table with this composition for this calculation. the mole fraction of ethanol in the mixture is 1/(1+10.23) ~ 0.089 G mixing = G solution G pure solvents = G mix =1 ( µ ethanol in mixture µ pure ethanol )+10.23( µ water in mixture µ pure water )!!! =1 ( µ ethanol + RT lna ethanol in mixture µ ethanol RT ln( 1 1) )+10.23( µ water! + RT lna water in mixture µ water recall a = P where P is the appropriate reference vapor pressure (pure solvent for Raoult's law) P # =1 RT ln 16.99 & # % (+10.23 RT ln 21.27 & % ( $ 59.1 ' $ 23.8 ' = 5.96 kj mol -1 for the ideal solution =1( RT ln0.089)+10.23( RT ln 0.911) = 8.36 kj mol -1 not asked for, but as an observation: G for the ideal solution is more negative (favorable) than the real solution by -2.4 kj mol -1 due to the unfavorable free energy associated with the transfer of a hydrocarbon to water. RT ln( 1 18
Problem 7 (4 pts) Last call 7a (2 pts) Predict the residual entropy per mole at 0 K of a crystal containing the methane species CH 3 D, where one of the 4 hydrogens found in CH 4 is replaced with the hydrogen isotope deuterium (D). Assume a tetrahedral arrangement of H and D atoms around the central carbon atom, and that the Hs and D can freely substitute for one! another in the crystal lattice. Express your answer as Rln m $ " # n % &, where m and n are integers, and evaluate your answer numerically in J mol -1 K -1. Compare your residue with the experimental value of 11.5 J mol -1 K -1. 4 different ways to orient one CH 3 D molecule at one site in the crystal lattice, or W = 4 N A ways of orienting an Avogadro s number of CH 3D in a crystal lattice. Hence S = k B lnw = k B ln 4 N A = N A k B ln 4 = Rln 4 ~ 11.5 J mol-1 K -1 19
7b (1 pts) How many different types of branched amino acids are found in ribosomally synthesized proteins? Name 2 of them. discussed in class there are 4 branched amino acids leucine, isoleucine, valine and threonine 7c (1 pts) Briefly explain (1-2 sentences) the thermodynamically relevant contributions of one of the following two scientists: (i) Max Rubner (ii) Barclay Kamb discussed in class Max Rubner - German nutritionist/physiologist who experimentally demonstrated that the first law of thermodynamics applies to living systems and determined the modern values of the caloric content of different types of foods. Barclay Kamb - Caltech geophysicist/chemist who was an expert in the phases of ice; he was a Caltech undergraduate and graduate who was a student of Linus Pauling, married his daughter Linda and has been characterized as the "smartest man in the world". 20