The energy of oxidation of 11 g glucose = kj = kg/m 2 s 2

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Amber Ball
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1 Chem 350 thermo problems Key 1. How many meters of stairway could a 70kg man climb if all the energy available in metabolizing an 11 g spoonful of sugar to carbon dioxide and water could be converted to work? The energy of oxidation of 11 g glucose = kj = kg/m 2 s 2 (from 2870 kj/mol *1mol/180.16g *11g) x = w / (mg) = kg/m 2 s 2 / 70 kg / 9.8 m s 2 = 255 m. 2. A mathematical statement of the First Law of Thermodynamics is ΔU = q + w. This holds for all processes. Assume that the only type of work done is PV work. Show that: ΔU =+w for an adiabatic process that ΔU = 0 for an isolated system that ΔU = q for a isochoric process that ΔH = 0 for an adiabatic process For an adiabatic process q = 0, so U = w. In addition to exchanging no heat with its surroundings, an isolated system can do no work; hence U = 0. For a process that occurs at constant volume, w = 0, since only p-v work can be done; thus, U = q. From Eqn. 2.7, H = qp. If, in addition to occurring at constant pressure, the process is also adiabatic, q = 0; giving H = A 60 kg person drinks 0.25kg of water at 62 o C. Assume body tissues have a specific heat capacity of 0.8 kcal/kg. By how many degrees will the hot drink raise the person s body temperature from the normal 37 o C. Explain how arriving at the answer involves the First Law of Thermodyamics. The person and the water are initially at different temperatures. When they come into contact, heat is transferred from the water to the body until thermal equilibrium is attained. The quantity of heat lost by the water must equal the quantity of heat gained by the body, assuming that no heat is lost to the surroundings. The final temperature of the body and water system, Tf, must lie between the initial temperature of the body and the initial temperature of the water. H = qp = Cp T = cimi T, where ci is the specific heat of material i and mi is the mass of this material. q < 0 for the water since it loses heat. For the water, Tf 62 o C = q/(1.0 kcal kg 1 K kg) For the body tissues, Tf 37 o C = +q/(0.8 kcal kg 1 K 1 60 kg)

2 We have two different equations and two unknowns. Because the number of equations equals the number of unknowns, a unique solution can be found. Rearranging the equations in terms of q, summing to eliminate q, and solving for Tf gives 37.1 o C. 4. Explain in structural and thermodynamic terms how the unfolding of a protein is like the melting of an organic crystal. The unfolding of a protein is like the melting of an organic crystal in several ways. In a crystal, individual units, e.g. individual urea molecules, are packed very close to each other, forming a lattice. As urea molecules are uncharged, a urea crystal must be held together by van der Waals interactions. The situation is similar in the core of a protein, where the atoms of amino acid side chains are in close contact. Moreover, just as the relative position of individual units is effectively constant in a crystal, so too the core of a protein molecule is relatively rigid under native conditions. When heat is added, vibrations increase, and when these are large enough the orderly array breaks down. If the crystal is pure, melting occurs over a very narrow range of temperatures. Similarly, the vibrations of amino acid side chains increase in magnitude when a protein solution is heated. There may be larger-scale motions as well, as the atoms of a protein form a complete, covalently-linked unit. When so much heat has been added that the van der Waals forces are no longer great enough to prevent large-scale fluctuations, the protein unfolds, often cooperatively. This results to a greater or lesser degree in the complete solvation of the polypeptide chain. Exceptions to this rule do occur and these will be discussed later. 5. For HEW (hen egg white lysozyme) calorimetric studies have shown that ΔCp,d = 1.5 kcal/molk and at ph 4.75 and ΔHd = 52 kcal/mol (at room temperature). Calculate the enthalpy change on folding (difference between folded and unfolded state) a) at 78 o C (the transition temperature) b) at -10 o C Hd(T2) = Hd(T1) + Cp,d(T2 T1). Using the data given in the statement of the problem, we have a) Hd(78 o C) = Hd(25 o C) + Cp,d(78 o C 25 o C) =52kcalmol calmol 1 K 1 53K = 130,000 cal mol 1 b) Hd( 10 o C) = Hd(25 o C) + Cp,d( 10 o C 25 o C) = 52 kcal mol cal mol 1 K 1 35 K = 500 cal mol 1 We see from part b) that the enthalpy of unfolding can be negative. This means that, under suitable conditions heat is absorbed on protein folding. As will become clear, however, the entropy of folding is so favorable under such conditions that the protein

3 does fold. 6. When glucose is combusted 6.73 kcal are given off per mole of glucose oxidized at 25 o C. a) What is ΔU at this temperature? b) What is ΔH at this temperature? c) Suppose glucose is fed to a culture of bacteria and 400 kcal/mol of glucose is catabolized. Is there a discrepancy between your answer for b? If so explain. H= U+ (nrt) U= 673kcalmol 1 n=0, so H= 673kcal mol 1. The bacteria give off less heat in consuming glucose than is measured in a bomb calorimeter because they do not completely oxidize the carbohydrate to carbon dioxide and water; some of the combustion energy is used for growth and therefore not given off to the environment as heat. 7. A protein called lactalbumin is a close homolog of hen egg white lysozyme. Unlike lysozyme, lactalbumin binds Ca 2+ with high affinity. The measured enthalpy of binding, however, is much smaller in magnitude than the enthalpy of hydration. Explain. Analysis of the crystallographic structure of baboon α-lactalbumin has revealed that several water molecules remain bound to the calcium ion in the binding site. That is, the binding process removes only some of the water molecules in the solvation shell. The enthalpy of solvation of a particular water molecule depends on how many molecules are already bound to the ion. The enthalpy change is greatest for the first water molecule and progressively smaller for subsequent ones. The enthalpy difference between the bound and unbound states of calcium in α-lactalbumin is therefore only a few kilocalories per mole. Second Law 8. Suppose 45 J is transferred from a heat source at 375 K to a heat sink at 25 o C. Calculate the maximum work that can be done. Efficiency of a heat engine = w max /q transferred = (1 T cold /T hot ) = 1 298K / 375K = w max = q transferred efficiency = 45 J = 9.2 J. 9. One calorie is produced for every J of work done. If 1 cal of heat is available can J of wok be accomplished with it? Explain.

4 No. The transfer of x J of heat cannot produce x J of work. Heat transfer is always attended by an increase in entropy, which corresponds to the fraction of heat transferred that is unavailable to do work. 10. (3.8) For a protein, suppose that DH d =10 kcal/mol at room temperature. The melting temperature is 68 C and the constant pressure heat capacity ΔC P = 1650cal/molK. Calculate the following quantities and explain their significance: a) ΔS d at T m b) ΔS d at 37 C c) ΔS d at 15 C d) At what temperature will ΔS d =0? H d (68 o C) = 10,000 cal mol 1 + 1,650 cal mol 1 o C 1 (68 o C 25 o C) = 80,950 cal mol 1. S d (Tm) = H d (T m )/T m = 80,950 cal mol 1 / (68 K K) = 237 cal mol 1 K 1. This is the increase in entropy on protein unfolding. S d (37 o C) = S d (T m ) + 1,650 cal mol 1 K 1 ln(310 K / 341 K) = cal mol 1 K cal mol 1 K 1 = 80.1 cal mol 1 K 1 The change in entropy is strongly dependent on temperature. S d (15 o C) = S d (T m ) + 1,650 cal mol 1 K 1 ln(288 K / 341 K) = cal mol 1 K cal mol 1 K 1 = 41.3 cal mol 1 K 1 At 15 o C, the entropy change for this protein is less than zero. In other words, there is an decrease in entropy on protein unfolding! This would probably come about from an increase in the order of water molecules solvating hydrophobic amino acid side chains. 0 = S d (Tm) + 1,650 cal mol 1 K 1 ln(t / 341 K) T = 341 K exp( cal mol 1 K 1 / 1,650 cal mol 1 K 1 ) = 295 K. At around 295 K, the various contributions to the entropy of unfolding cancel each other out; there is no difference in entropy between the folded state and the unfolded state. 11. Is it possible for heat to be taken into a system and converted into work with no other change in the system or surroundings? Explain If the heat transfer is reversible, S = q/t ( S > q/t when the transfer is irreversible). The quantity of heat transferred, q, is positive for the system but negative for the surroundings. So the total change in entropy is Ssystem + Ssurroundings = q/t + ( q)/t = 0; there is no entropy change for a reversible transfer heat at constant temperature.

5 This tells us that because the inside of a living organism is effectively uniform in temperature, exchanges of heat within the organism cancel out and result in no change in entropy. The situation is different at the system-surroundings border, which in the case of humans is the skin. Here heat is readily lost to the surroundings. To maintain body heat, we must eat 12.What are the units of K eq? Explain The units depend on the chemical reaction. When the number of reactants is the same as the number of products, K eq is unitless. 13. Calculate DG o for K eq values of 0.001,0.01, 0.1, 1, 10, 100 and Which reactions are spontaneous at STP? G o = RTlnK eq. For instance, G o = J K 1 mol K ln(0.001) = 17.1 kj mol 1. K eq G o (4.10) When a photon in the visible range is absorbed in the retina by rhodopsin, the photoreceptor in rod cells, 11-cis-retinal is converted to the all trans isomer. Light energy is transformed into molecular motion. The efficiency of photos to initiate the reaction is about 20% at 500 nm (57kcal/mol). About 50% of the absorbed energy is available for the next signaling step. This process takes about 10 ms. In the absence of light, spontaneous isomerization of 11-cis-retinal is very slow, on the order of /yr. Experimental studies have shown that the equilibrium energetics of retinal isomerization are DS o =4.4 cal/molk and DH o = 150 cal/mol. Calculate the standard state free energy and the equilibrium constant. K eq = exp( G o / RT) = exp( ( H o T S o ) / RT) = exp( (150 cal mol K 4.4 cal mol 1 K 1 ) / ( cal mol 1 K K)) = Calculate DG when the concentrations of glucose-1-phosphate and glucose-6- phosphate are maintained at 0.01mM and 1mM, respectively. Comment on the meaning of the sign of DG and what that means to the reaction you have written.

6 G for the conversion for the conversion at 298 K is 7.3 kj mol 1. This value can be obtained from the net G ' = G + RTln([G6P]/[G1P]) = 7.3 kj mol 1 + (8.314 J mol 1 K K ln(1 mm/0.01 mm) = 4.1 kj. The direction of the reaction has been reversed by adjusting the concentrations of reactants and products. Control of the concentration of a metabolite is an important means of regulating the flux of matter through biochemical reaction pathways. 16. Show that for a reaction at room temperature that yields 1 mol of water ΔG o = ΔG o kj/mol (remember [H 2 O] in an aqueous solution is 55.5 M) We require that the reaction occur in dilute aqueous solution and yield n H 2 O molecules: A + B C + D + nh 2 O Because the solution is dilute, [H 2 O] = 55.5 M G ' = G + nrtln[h 2 O] = G + 1 mol J K 1 mol K ln (55.5 M) = 9.96 kj mol The equilibrium constant for Glu - + NH 4 + <=> Gln + H 2 O is /M at ph 7 and 310 K; the reaction lies far to the left. The synthesis of gln from glu is made energetically favorable by coupling it to hydrolysis of the terminal phophodiester bond of ATP. The equilibrium constant for the coupled reaction, which is known from experiments with glutamine synthase is Calculate the phosphate bond energy in ATP at ph7 and 310 K. K G = [Gln]/([Glu ][NH4 + ]) = M 1. K coupled = [Gln][ADP][Pi]/([ATP][Glu ][NH4 + ]) Combining the two reactions produces an equilibrium that is the product of the two individual reactions = K G K ATP = K ATP = [ADP][Pi]/[ATP] = K coupled /K G = 1200/( M 1 ) = M. G ATP = RTlnKATP = cal mol 1 K K ln( ) = 7.9 kcal mol 1.

7 18. The concentration of creatine in urine is 40 fold greater than in serum. Calculate the free energy change per molecule required for transfer of creatine from blood to urine at 37 o C. [creatine u ] 40[creatines] G = G + RTln[creatine] The transfer is from blood to urine, so G = G u G s = G u + RTln[creatine] u (G s + RTln[creatine] s ) = (G u G s ) + RT(ln[creatine] u ln[creatine] s ) = (G u G s ) + RTln([creatine] u /[creatine] s ). We assume that the standard state free energy of creatine does not depend on whether it is in urine or serum. This is not a bad approximation because creatine is soluble and the concentration of water is high in urine and serum. So we are left with G = RTln([creatine] u /[creatine] s ) RTln(40) cal mol 1 K K 2.3 kcal mol Calculate Keq for the hydrolysis of the following compounds at neutral ph and 25C: phophoenolpyruvate ΔG o = kj/mol pyrophosphate ΔG o = kj/mol glucose-1-phosphate ΔG o = kj/mol (you can find the needed reaction by looking up glycolysis) G ' is larger than G by about 10 kj mol 1 when one mole of water molecules is produced by the reaction. This must be taken into account because our known quantities are primes and the unknowns are not. G ' = G + nrtln[h 2 O] G = RTlnKeq = G ' nrtln[h 2 O]. In hydrolysis water molecules are consumed, so n, the number of water molecules produced, is negative. The hydrolysis of phosphoenolpyruvate yields pyruvate and phosphate, and one molecule of water is consumed. Using the result of the previous problem and plugging in known values gives: Keq = exp( ( 61.9 kj mol 1 ( 1) 9.96 kj mol 1) / ( kj mol K) = The calculation is similar for pyrophosphate: Keq = exp( ( 33.5 kj mol kj mol 1) / ( kj mol K) = And for glucose-1-phosphate: Keq = exp( ( 20.9 kj mol kj mol 1) / ( kj mol K) = 82.7.

8 Note that the driving force for hydrolysis of phosphoenolpyruvate is much greater than that of glucose 1-phosphate. Phosphoenolpyruvate is a product of Reaction 9 of glycolysis, the metabolism of glucose. Glucose 1-phosphate is a direct product of glycogen breakdown. Conversion to glucose 6-phosphate enables the sugar to enter the glycolytic pathway. Pyrophosphate is a product of DNA synthesis (and other biochemical reactions), as only one phosphate group of a nucleotide is incorporated into the sugar phosphate backbone of the nucleic acid. 20. The citric acid cycle is the common mode of oxidative degradation in eukaryotes and prokaryotes. Two components of the citric acid cycle are a-ketoglutarate and isocitrate. If [NAD ox ]/[NAD red ] = 8 [a-ketoglutarate] = 0.1mM [isocitrate] = 0.02mM Assume 25 C and ph 7.0. a) Calculate ΔG. b) Is this reaction a likely site for metabolic control? Explain G ' for the conversion of isocitrate into α-ketoglutarate is 21 kj mol 1. This reaction involves NAD+ and NADH. G ' = G because water is not involved in the reaction; there is no change in ph if it is assumed that the carbon dioxide produced by the reaction does form carbonic acid. G = G + R T ln([α-kg][nadh]/([nad+ ][IC])) = 21 kj mol J mol K ln((0.1 mm) / 8 / 0.02 mm) = 22 kj mol 1 < 0. This reaction is a site for metabolic control because G is negative. Moreover isocitrate dehydrogenase, the enzyme that catalyses this reaction, is strongly inhibited in vitro by NADH, a product of the reaction. In contrast, most of the other reactions of the citric acid cycle have G 0.

reduction kj/mol

reduction kj/mol 1. Glucose is oxidized to water and CO 2 as a result of glycolysis and the TCA cycle. The net heat of reaction for the oxidation is -2870 kj/mol. a) How much energy is required to produce glucose from