reduction kj/mol

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1 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 kj/mol. a) How much energy is required to produce glucose from water and CO 2? Is this value thermodynamically favorable? b) Predict qualitatively, whether the oxidation or the reduction is spontaneous. Explain your answer. c) Plants efficiently fix CO 2 to form glucose. In this biosynthetic pathway, if photons of 700nm wavelength are absorbed by the chlorophyll photosystem synthetic complex, how many photons are required to fix one mole of CO 2? d) Given your answer to c and the constraints of the second law of thermodynamics comment on the number of photons actually used in CO 2 fixation. Explain your answer. a) The first thing is to write write and balance reaction: C 6 H 12 O 6 + 6H 2 O à 6CO H 2 O oxidation kj/mol Reverse reaction 6CO H 2 O à C 6 H 12 O 6 + 6H 2 O reduction kj/mol Formation of glucose is simply the reverse reaction so the energy is equivalent in magnitude but opposite in sign. The formation of glucose is not thermodynamically favorable b) The heat of reaction for the oxidation is favorable (-2870 kj/mol ). Since 7 moles are being converted to 12 moles and the products are 12 moles of gas entropy should be favorable in the direction of oxidation. Therefore the oxidation reaction should be spontaneous. c) E=hv = hc/λ convert 700nm wavelength to 700E-9m E= 6.626E-34 m 2 kg/s * 2.997E8m/s *1/700E-9 E=2.837E-19 J for a single photon 2870 kj/mol = 2870E3 J for one mole of glucose 2870E3 J/ 2.837E-19J = E25 photons for a mole of glucose a single mole of glucose corresponds to 6 moles of CO E25 /6 = 1.686E24 ( a bit more than a mole of photons for a mole of gas) d) Since the second law of thermodynamics prohibits perfect energy transfer much of the light energy would be lost (not used to fix CO 2 ). Therefore it would take more than the calculated E24 photons to fix a mole.

2 2. Giant sequoias, an indigenous species to California, live up to 3500 years and can reach heights in excess of 300 m. a) Calculate the work that must be done in transporting a single water molecule to the top of a 300 m sequoia. b) Show (explain) how your answer involves the first law of thermodynamics (use equations and words). a) W=mgh mass of one water molecule 18.0g/mol x 1mol NA b) first law U=q+w constant as work is done to the surroundings heat must be absorbed into the system a) w=mgh mass single water molecule 18g/mol * 1mol/6.023E23 = 2.99E-23 g = 2.99E-26 kg w = 2.99E-26 kg * 300 m * 9.8 m/s 2 = E-23 kgm 2 /s 2 = E-23 J actually = E-23 J as the work is done against gravity b)the conservation of energy indicates that energy is conserved forms can change but total energy is invariant. In this case the work involved in moving the molecule up the tree is balanced by the heat gained by the system. ΔU = q + w since ΔU is 0 (the water molecule does not change internal energy) 0 = q + w and -w = q

3 3. The entropy change for an nucleic acid was investigated upon thermal unfolding under constant pressure (d refers to denaturation). The following values were measured: ΔH 0 d = 9.2 kcal/mol, T m = 72 0 C, ΔC p = 1200 cal/mol K. a) Calculate the enthalpy of denaturation at T m b) Calculate ΔS d at T m - Is the unfolding at T m entropically favorable? Rationalize/explain c) Calculate ΔS d at 15 0 C- Is the unfolding entropically favorable? Rationalize/explain a) constant pressure q=h therefore ddh = dcp dt H (72) H (25) = 1200 cal/mol K * (72-25)K Solve for H (72) H (72) = 9.2kcal/mol cal/mol K * (72-25)K = kcal/mol b) all at 72C S = H/T S (72) = 65780cal / ( ) K = 177 cal/k this number is >0 indicating increase in entropy and therefore entropically favorable c) DS (15) = ds (72) + dcpln (288K/335) = 177 cal/k cal/mol * ln(288/345) = cal/molk this number is <0 indicating a decrease in entropy and therefore not entropically favorable. Since entropy is highly dependent upon temperature at low temperature the decrease in entropy is significant moving from a favorable process at higher temperatures to an unfavorable process when cold.

4 4. The following reaction is catalyzed by the enzyme creatine kinase: Creatine phosphate + ADP ----> Creatine + ATP Under the listed conditions (1 atm, 37 o C, ph 7.0, pmg 3.0 and an ionic strength of 0.25 M) with the concentrations of all reactants at 1mM and products equal to 10 mm: ΔG 335 for this reaction is kj/mole. ΔG = kj/mol a) What is the equilibrium constant for this reaction at 335K? Is this reaction favorable? Explain b) What is the standard free energy change, ΔG 0, for this reaction at 335 K? Is this reaction favorable under standard conditions? Explain c) What is the standard enthalpy change, ΔH 0, for this reaction? d) What is the standard entropy change, ΔS 0, for this reaction? Is this change favorable? Explain e) Given your answers is this reaction driven by entropy or enthalpy or both? Explain your rationale a) Keq= [creatine][atp]/[creatine-p][adp] 10mm(10mm)/1mM(1mM) = 100 (can keep in mm since units cancel) yup more products than reactants by a long shot so the Keq>>1 b) ΔG = ΔG 0 + RTlnKeq J/mol = ΔG J/molK * 335K *ln J/mol = ΔG J/mol ΔG 0 = -26,126 J/mol = kj/mol yes very favorable at 335 K (more so than at 298 K) c) Have G 0 values at more than one temperature and want the enthalpy - use Gibbs Helmholtz G2/T2 - G1/T1 = ΔH (1/T2 1/T1) /335 - (-12600/298) = ΔH (1/335 1/298) J/molK J/molK = ΔH 1/K J/mol = ΔΗ ΔH= 9633 J/mol or 9.63 kj/mol d) ΔS ΔG=ΔH-TΔS J/mol = 9633 J/mol K* ΔS ΔS = J/molK very favorable the contribution of entropy out weighs the contribution of enthalpy in this reaction.

5 5. A diatomic ideal gas (0.5 moles) is expanded from a volume of 1.5 L to 3 L. a) Under a constant external pressure of 1 atm, at a temperature of 25 0 C, calculate the work. Is the work done on the system or by the system? Explain practically and mathematically. b) If the expansion in part a) is carried out incrementally allowing the system to come to equilibrium at each step rather than in a single step expansion, calculate the work. Again is work done on the system or by the system? c) Compare the work in part a to part b. Explain any difference or if no difference explain why. d) If the expansion in part b is allowed to occur with a change in temperature (final temperature 40 0 C) under a pressure of 1 atm, calculate the work, heat, internal energy change and enthalpy change for the process. e) Can the values for ΔE and ΔH calculate in part c be applied to the process in part b? Why or why not? Explain. a) w = -P(V 2 -V 1 ) this is a single step transformation under a constant external pressure w= 1 atm(3-1.5 )L = -1.5 Latm *101.3 J/Latm = J The negative sign indicates work done by the system on the surroundings b) w= nrt ln V 2 /V 1 this is a reversible path maximum work = 0.5mol*8.314J/molK (298K) ln (3/1.5) = J The negative sign indicates work done by the system on the surrounding c) part b reversible path maximum work entire area under the curve value is larger sign agrees d) ΔU =q p pdv ideal gas 0.5 mol diatomic C v = df*nr/2 and datiomic df = 5 C v = 5*0.5mol*8.314J/molK/2 = J/K Cp = C v +nr = 10.39J/K +0.5mol (8.314J/molK) = J/K ΔU = C v (ΔT) = J/K (40-25)K = J in 0.5 mol or J/mol ΔH = C p (ΔT) = J/K (40-25)K =218 J in 0.5 mol or 436 J/mol Δq = ΔH = 436 J/mol (constant pressure) or 218 J ΔU = q + w = J/mol = 436 J/mol + w w = J/mol (expansion!!!) e) yes internal energy and enthalpy are state functions path independent cannot be applied for q and w and as you see in parts a and b w is different than in d

6 6. Given the standard enthalpy of combustions of: methane gas ( kj/mol), hydrogen gas ( kj/mol), C graphite ( kj/mol), the bond dissociation enthalpy of hydrogen gas (435.8 kj/mol), and the latent heat of sublimation of graphite (718.4 kj/mol) evaluate the bond enthalpy of the C-H bond in methane. Clearly explain using a brief diagram (description) of the path chosen. Hint: CH4 (g) à C(g) + 4H(g) would represent bond formation Combustion of methane all kj/mol CH4 + 2O2 -à CO2 + 2H2O H2 + O2à 2 H2O C (s)+ O2 à CO H2à2H C(s) à C(g) What is C-H? energy Want C(g) + 4 H(g) à CH4 (g) Path devised CH4 + 2O2 -à CO2 + 2H2O CO2 à C (s)+ O (reversed) C(s) à C(g) H2O à 2H2 + O (reversed) 2H2à4H x Net Reaction CH4 à C + 4H all gas This describes making methane from 1 C and 4 H Total for 4 bonds = 344 kj/mol for one C-H bond