Chemistry 25 (Spring term 2015) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!!

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1 Name ANSWER KEY Chemistry 25 (Spring term 2015) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!! Distributed Thursday, June 4 Due by Thursday, June 11 at 1 pm to 362 Broad 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 your Ch25 lecture notes, problem sets and solutions, the textbook (Kuriyan, Konforti and Wemmer), the course web site and a calculator. You may not use any other books, notes, exams and problem sets from previous years of Ch25, Ch24(ab), other web sites, non-calculator applications of Mathematica and related programs, discuss the exam with others, etc. This exam should have 22 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 = m; 1 µ = 10-6 m = 10 3 nm. N A = x mol -1 = Avogadro's constant force/energy 1 J = 1 N m = 1 C V; 1 pn = N; 1 pn nm = J = kj mol -1 ; J = 1 cal. pressure units: 1 pascal (Pa) = 1 N m -2 = 1 J m -3 = 10-5 bar 1atm = Pa = pn nm -2 = 760 mm Hg = lb/sq. in. = J cm -3 = kj m -3 electrostatics F = kj mol -1 V -1 = 96,487 C mol -1 ; q = x Coulombs = F /N A 4πε o = x C V -1 m -1 gas constant R J mol -1 K -1 = cal mol -1 K -1 = liter atm mol -1 K 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 J at T = 298 K. for water, molecular weight = kg mol -1, liquid density = 1000 kg m -3, ε = 80 heat capacity = 75.4 J mol -1 K -1 ; H fusion = 6.0 kj mol -1 ; H vap = kj mol -1 1

2 Name problem points 1a 5 1b 5 1c 5 2a 5 2b 8 3a 5 3b 8 3c 7 4a 5 4b 8 4c a 5 6b 8 6c 5 7a 2 7b 5 2

3 1 Should this problem be 50% reduced or 100% oxidized? 1a (5 pts) An organism has been reported to grow on H 2 and CO 2 as an energy source, producing acetate (CH 3 CO - 2 ) as product. Write a balanced reaction for the reduction of 2 CO 2 to form 1 acetate, coupled to the oxidation of H 2 to form H +. reduction reaction: 2 CO 2 CH 3 CO 2 - the summed oxidation states of carbon in the reactant is 2 x (+4) = +8 and in the product is (-3) + (+3) = 0; hence this is an 8 e - reduction. To balance O need to add 2 H 2 O to product To balance H need to add 7 H + to reactant oxidation reduction 4 H 2 8H + + 8e - (as an 8 e- oxidation) total 2CO 2 + 4H 2 CH 3 CO H + + 2H 2 O 1b (5 pts) For an organism to survive with this metabolism, the reduction of CO 2 by H 2 to form acetate (the reaction balanced in problem 1a) must be energetically favorable. From the following thermodynamic data, calculate G ' for the reduction of two CO 2 by H 2 to form acetate. Is the G ' for this reaction sufficient to drive ATP synthesis? CO 2 (g) + 8H + + 8e - CH 4 (g) + 2H 2 O E ' = V CH 3 CO 2 H (aq) CH 4 + CO 2 (g) G ' = kj mol -1 CH 3 CO 2 H (aq) CH 3 CO H+ G ' = kj mol-1 2H + + 2e - H 2 E ' = V 3

4 CO 2 (g) + 8H + + 8e - CH 4 (g) + 2H 2 O G ' = -8F(-0.244) = CH 4 + CO 2 (g) CH 3 CO 2 H (aq) G ' = kj mol -1 CH 3 CO 2 H (aq) CH 3 CO H+ G ' = kj mol -1 4H 2 8H + + 8e - G ' = -8F (0.414) = total 2CO 2 + 4H 2 CH 3 CO H + + 2H 2 O G ' = kj mol -1 since G for ATP synthesis is about 60 kj mol -1, this reaction is sufficient to drive ATP synthesis 4

5 1c (5 pts) A Daniell-type cell is constructed from Cu 2+ /Cu 0 and Zn 2+ /Zn 0 half cells at 298 K. The Cu 2+ concentration is 1.00 M, while due to an equipment malfunction with a balance, the concentration of Zn 2+ is not known with certainty, but may be 0.02 M. The potential of this cell is measured to be V with the Cu electrode positive. Using the Nernst equation, is the measured potential consistent (within 2%) with the concentration of Zn 2+ = 0.02 M in the Zn 2+ /Zn 0 half cell? Daniell cell from class notes and demonstration oxidation Zn Zn2+ + 2e- E = V reduction Cu2+ + 2e- Cu E = V total Cu 2+ (1 M) + Zn Zn 2+ (X M) + Cu E = V Nernst equation E = E RT nf 1.15 = = ln X 2+ ( ( X M )) Cu 0 Cu 2+ 1 M ( ) ( ( ))( Zn 0 ) ln Zn ( ) ln X 1 X = e 3.89 = 0.020M yes, the measured potential is consistent with (Zn 2+ ) = M 5

6 2 I'm more positive it's more negative The energetics of oxidation-reduction reactions can be significantly influenced by protonation and molecular structure. As illustrations of these effects, we will explore the reduction potentials of several dicarboxylic acids (as an aside, some of the early thermodynamic characterizations of these compounds were conducted in Kerckhoff Laboratories (see Huffman and Fox, J. Amer. Chem. Soc. 60, 1400 (1938)). Contemporary values for the free energies of formation ( G f at ph 0) of succinic acid, fumaric acid and maleic acid in aqueous solution are tabulated below, in both acid (unionized) and carboxylate (ionized) forms. compound G f (aq. soln) kj mol -1 succinic acid succinate fumaric acid fumarate maleic acid not determined maleate Note: the free energies of formation of H + and e - at ph 0 are each defined as 0 kj mol -1. Succinic acid is the reduced form of these dicarboxylic acids, while fumaric and maleic represent the two conformationally distinct oxidized forms (with trans- and cis- versions of the double bond, respectively). 2a (5 pts) written: The fumarate/succinate and maleate/succinate half cell reactions may be Calculate the E reduction potentials for these two half cell reactions (in V) from the free energies of formation tabulated above. Is fumarate or maleate the stronger oxidant? In 1-2 sentences provide an explanation for this conclusion based on the electrostatic interactions between carboxylate groups in the reactants and products. 6

7 G = G f G f products reac tan ts = nfe fumarate/succinate half cell G = = -2F E ; E = V maleate/succinate G = = -2F E ; E = V maleate is the stronger oxidant since the driving force for reduction of maleate is greater than for fumarate, as reflected in the higher reduction potential ( V vs V) the cis configuration of the carboxylate groups in maleate leads to greater electrostatic repulsion that destabilizes the oxidized form relative to fumarate. 7

8 2b (8 pts) The fumaric acid / succinic acid half cell reaction may be written: Calculate the E reduction potential for this half-cell reaction (in V) from the free energies of formation tabulated at the beginning of problem 2. Is it more thermodynamically favorable to reduce the double bond when the carboxyl groups are both protonated, or when they are both ionized? Based on this calculation, are the pka's for succinic acid higher or lower than the pka's for fumaric acid? Explain your reasoning in 1-2 sentences. fumaric acid/succinic acid G = = -2F E ; E = V The increased E for the protonated form means that protons are bound more tightly to the reduced form (succinic acid) than the oxidized form (fumaric acid) and hence stabilize the reduce form, increasine E. Consequently, the pkas for succinic acid are greater than for fumaric acid (ie - it is harder to ionize succinic acid than fumaric acid, which requires higher ph). succinic acid pkas = 4.21 and 5.63 maleate pkas = 3.03 and

9 3 Multiple equilibria and cooperativity One of the unheralded accomplishments of the genomics/proteomics/whateveromics revolution is the discovery that MIT students are characterized by a dimeric hemoglobin (Hb MIT ) that is less effective at transporting oxygen, presumably as an adaptation to smaller brains that need less oxygen. MIT students transferring to CIT have a variant hemoglobin (Hb CIT ), also a dimer, that has superior oxygen transport properties. Evaluation of the macroscopic dissociation constants for the binding of oxygen to Hb MIT and Hb CIT gave the following results: hemoglobin (Pa) (Pa) Hb MIT Hb CIT a (5 pts) Is the binding of oxygen to Hb MIT characterized by positive, negative, or nocooperativity? Is the binding of oxygen to Hb CIT characterized by positive, negative, or no-cooperativity? Explain briefly. What does this say (very briefly) about the culture of collaboration and cooperation at MIT and CIT? Hb MIT Hb CIT / = 10 5 > 4 negative cooperativity / = 10 5 < 4 positive cooperativity cooperativity at CIT is times more positive than at MIT 9

10 3b (8 pts) Assume that the partial pressures of oxygen in the lungs and brain are 10 4 and 10 3 Pa, respectively. Calculate the average number of oxygen molecules bound to Hb MIT and Hb CIT in the lungs and brain (ie, calculate ν at these two oxygen concentrations for both proteins), and from these numbers determine how many moles O 2 per mole Hb would be released in the brain. Which hemoglobin would be most efficient at delivering oxygen from the lungs to this vital organ (at least at CIT), the brain? ( ) + 2( O 2 ) 2 ( ) + ( O 2 ) 2 K ν = 2 O 2 + O 2 Hb 4 Pa, ν = 1.01 Hb 3 Pa, ν = 0.99 = = 0.02 mole O 2 per mole Hb Hb 4 Pa, ν = 1.82 Hb 3 Pa, ν = 0.18 = = 1.64 mole O 2 per mole Hb Hb CIT is ~80x more effective as Hb MIT at transferring O 2 from the lungs to the brain (1.64 vs 0.02 moles O 2 per mole of the relevant Hb) 10

11 3c (7 pts) Calculate the O 2 concentration (in Pa) where Hb MIT and Hb CIT are each half saturated (ie ν = 1). For both Hb MIT and Hb CIT, calculate the fraction of each protein with 0, 1 and 2 bound O 2 molecules at this oxygen concentration. (hint: the O 2 concentration at half saturation point can be calculated from the values of and by solving for (O 2 ) when ν = 1, using the general expression for ν for a two-site binding problem discussed in lecture 15 (5/19/15)) hemoglobin fraction with 0 fraction with 1 fraction with 2 bound O 2 bound O 2 bound O 2 Hb MIT Hb CIT when ν = 1, ( O 2 ) = = Pa (for both Hbs) Hb MIT fraction with 0 O 2 bound = fraction with 1 O 2 bound = fraction with 2 O 2 bound = ( P) ( P) + ( PL) + ( PL 2 ) = ( P) = P T ( PL) = = P T P T ( ) PL 2 = K 2 P T P T + + = = ( = fraction with 0 O 2 bound) Hb CIT fraction with 0 O 2 bound = fraction with 1 O 2 bound = fraction with 2 O 2 bound =

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13 4 Cooperativity and disproportionation reactions We have considered cooperativity in the context of ligand (L) binding to macromolecules (P). For a two site system, the binding equilibria may be described by the following scheme: Eq. 4.1 PL 2 PL+L P + 2L where and are the macroscopic dissociation constants, defined as: Eq. 4.2 = ( P) ( L) and K ( PL) 2 = ( PL) ( L) ( PL 2 ) One way to evaluate the extent of cooperativity (positive or negative) in ligand binding is in terms of a disproportionation reaction characterized by an equilibrium constant K dp : Eq PL P + PL ( 2 K dp = P) ( PL 2 ) PL ( ) 2 4a (5 pts) derive an expression for K dp (eq. 4.3) in terms of and (eq. 4.2) K dp = P ( ) ( PL 2 ) = ( P) ( L) ( PL 2 ) ( PL) 2 ( PL) ( PL) ( L) = 13

14 4b (8 pts) i. What is the value of K dp for a system where the binding sites are identical and noninteracting? ii. Under conditions where an average of one L ligand is bound per P, what are the relative concentrations of the P species with 0, 1 and 2 bound ligands when the binding sites are identical and non-interacting? i for identical and non-interacting sites K dp = 1 = = 1 4 ii for identical and non-interacting sites with an average of one L per P (f = 0.5 per binding site), the probability of having P species with 0, 1 and 2 bound ligands = 25%, 50%, 25% this can also be solved from K dp : let a 0 = the total amount of P present a 0 = (P) + (PL) + (PL 2 ) let x = (P) = (PL 2 ) at equilibrium, so (PL) = a 0-2x K dp = P ( ) ( PL 2 ) x 2 = ( PL) 2 ( a 0 2x) = x 2 = a 0 2 4a 0 + 4x 2 a 0 = 4x ( ) ( ) x = 1 a 0 4 = P = PL 2 P T P T and ( PL) P T = a 0 2x ( ) a 0 =

15 4c (4 pts) For succinic acid, = 2.3 x 10-6 M and = 6.4 x 10-5 M. i. What is the value of K dp for succinic acid? ii. Is this an example of positive, negative, or no cooperativity? i. K dp = = ii, = ~ 28 negative cooperativity

16 5. The Argonaut's dream A kilogram of gold is worth ~$40,000, coincidentally about a year's tuition at Caltech The ocean contains an estimated gm of gold, which would easily cover four years here, including living expenses and season tickets to all Caltech sporting events. The problem is that the ocean has a lot of water (~10 21 liter), so that the concentration of gold is quite dilute (10-8 gm per liter or ~5 x M). Capitalizing on your Ch 25 education, you contemplate a transport-based approach for extracting a kg of gold from the ocean, based on a hypothetical polymeric ligand with such a high affinity for Au (and completely specific for Au, without affinity for any other element) that it quantitatively absorbs any gold species colliding with the surface. (10 pts) This approach is based on the diffusion-to-capture model where a suitably sized spherical sample of this ligand is suspended in the ocean, absorbing every Au that reaches the surface. If the sphere radius = 0.1 m and the diffusion constant of Au is 10-9 m 2 s -1, how long would it take to accumulate 1 kg of Au by this approach? Assume that the only way gold reaches the ligand surface is through diffusion (ie - there is no mixing, stirring, etc.). All you have to do is sit back, relax and wait for the riches to come to you. Express your answer in both seconds and years. (6.02 x 0 x pts) Should you quit your day job and enjoy a lavish lifestyle financed by Ch25 insights? use the diffusion current expression derived for the diffusion to capture model: I = 4πaDC o C o = 10 8 gm L -1 =10-8 kg m -3 I = 4π kg I ( )( 10 8 ) = kg s -1 ( ) 10 9 = s = yr 16

17 6 Having a meltdown A radioisotope of iodine, I-131, is generated during the fission of uranium and plutonium, as occurs in nuclear reactors and during explosion of nuclear bombs. As a component of fallout, it has significant health consequences due to absorption by the thyroid. The worldwide release of I-131 exceeds Curies, with the largest single release from the Chernobyl accident in 1986 (5x10 7 Curies); over 10 8 Curies were released from the Nevada Test Site between 1952 and (see The health consequences of this fallout motivated Linus Pauling to advocate for an end to the testing of nuclear weapons, leading to his Nobel Peace Prize in 1963 (and general estrangement from the Caltech community as this was not a popular position during the Cold War.) 6a (5 pts) The half-life of I-131 is 8.03 days. How many grams of I-131 are present in Curies? Remember, 1 Curie = 3.7 x Bq. k = ln2 t 1/2 = dn dt ln = s 1 = kn = atoms/s N = atoms/s = atoms s atoms 131 gm/mol = gm N A atoms/mol 17

18 6b (8 pts) The FDA has an intervention level for I-131 in milk of ~10-9 Curie per liter. If a "typical" milk sample has a total concentration of 100 µg iodine per liter, what fraction of the total iodine in milk in such a sample corresponds to I-131? Assume the iodine in milk has an average atomic weight of moles of I-131 in 10-9 Curie: k = s 1 dn = kn = s 1 /s dt 37 N = s = atoms atoms N A atoms/mol = mol moles of iodine in 100 ug per liter milk 10 4 gm 126.9gm / mol = mol fraction of total I as I-131 at the intervention level = mol mol 18

19 6c (5 pts) In humans, iodine is essential for the production of thyroid hormones that regulate metabolism. The enzyme thyroperoxidase (E) catalyzes the iodination of the protein thyroglobulin (T) that serves as the precursor to thyroid hormones (P). A plausible reaction mechanism involves the ordered binding of iodide (I - ) to E, followed by the binding of T to yield P: 1/ E + I 1/K 1 EI + T EI T k E + P Assuming that the catalytic step (rate constant k) is rate-determining so that all the binding steps can be assumed to be at equilibrium, derive the pre-equilibrium, initial velocity expression for the rate of formation of P, and express it in the following form: v V max = a + b I ( I ) T ( ) + c T ( ) ( ) + I ( ) ( ) T where a, b and c are functions of the dissociation constants and (and with V max = k[e T ]). or a =, b =, c = 0 = (E)(I) (EI) and = (EI)(T ) (EIT ) (EI) = (I) (E) and (EIT ) = (T ) (EI) = (I) (T ) (E) v = (EIT ) = V max E T = (EIT ) (E) + (EI) + (EIT ) = (I)(T ) + (I) + (I)(T ) (I) (T ) 1 + (I) + (I) (T ) 19

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21 7 Relax - the end of Ch25 is in sight!! 7a (2 pts) In one or two sentences, describe Manfred Eigen's contributions to chemical kinetics. development of relaxation methods (T-jump, P-jump) to measure kinetics of fast reactions, specifically the rate of proton transfer reactions 7b (5 pts) For the following reaction, the equilibrium constant, K, equals 2 and the relaxation time, τ, equals (1/3) s. From this data, determine the numerical values of the rate constants k 1 and k -1 in units of s -1. k A 1 B k 1 K = k 1 k 1 = 2 k 1 = 2k 1 1 τ = 3s 1 = k 1 + k 1 = 3k 1 k 1 = 1s 1 and k 1 = 2s 1 21

Chemistry 25 (Spring term 2015) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!!

Name Chemistry 25 (Spring term 2015) Final Examination for NON-SENIORS GOOD LUCK AND HAVE A GREAT SUMMER!!! Distributed Thursday, June 4 Due by Thursday, June 11 at 1 pm to 362 Broad Conditions Open this