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1 CHEM-UA 652: Thermodynamics and Kinetics Problem set #8: due 4/17/2014 Practice problems: 11.13, 11.14, 11.15, 11.22, 12.18, 12.16 Additional Practice Problem: i. Derive the expression given in class for the boiling point elevation starting from equalization of chemical potentials between the solution and vapor phases, and show that the elevation of the boiling point is predicted as part of your derivation. Graded problems 1. A solution is prepared by mixing liquid benzene and liquid carbon tetrachloride in equal amounts at 25◦ C. When equilibrium is established, the vapor above the solution is collected and analyzed for its content. Determine to at least three significant figures the mole fraction of each gas in the vapor using only the following data: The standard enthalpies of formation for benzene in the gas and liquid phases are 82.93 kJ/mol, 49.03 kJ/mol, respectively. The standard entropies for benzene in the gas and liquid phases are 269.2 J/mol·K, 172.8 J/mol·K, respectively. The standard enthalpies of formation for carbon tetrachloride in the gas and liquid phases are -102.9 kJ/mol and -135.44 kJ/mol, respectively. The standard entropies for carbon tetrachloride in the gas and liquid phases are 309.74 J/mol·K and 216.4 J/mol·K, respectively. 2. When BaCl2 and K2 SO4 solutions are mixed together, a solid precipitate, BaSO4 , is formed according to the equation: BaCl2 (aq) + K2 SO4 (aq) −→ BaSO4 (s) + 2KCl(aq) In a certain experiment, 0.2000 L of a 1.6000 M BaCl2 solution are combined with 0.3000 L of a K2 SO4 solution of unknown molarity. The final solution is analyzed and found to contain unreacted Barium ions. In addition, the boiling point of the solution is measured and found to be 101.4245 o C, and the mass of the final solution is determined to be 599.2300 g. What is the molarity of the K2 SO4 solution? 3. The activity coefficient γ of a nonideal gas is defined to be a correction factor, which, when multiplied by the observed pressure of the gas gives the pressure it would have if it were an ideal gas. Consider the gas phase reaction: H2 (g) + F2 (g) ⇀ ↽ 2HF(g) whose equilibrium constant at a high temperature is K = 115.00. Suppose that H2 , F2 and HF gases are to be treated as nonideal gases with activity coefficients 1.111, 1.180 and 1.250, respectively. A 50 L vessel contains HF gas at an observed partial pressure of 2.4 atm and H2 gas at an observed partial pressure of 4.5 atm. 2 a. It is desired to add enough fluorine gas to the vessel to obtain a total ideal gas pressure of 10 atm. To what observed partial pressure must fluorine gas be added? b. What will be the observed partial pressures of each gas in the vessel after equilibrium is established? c. What will be the total ideal gas pressure in the container at equilibrium? d. Suppose the volume of the vessel is doubled. What effect does this have on the equilibrium? (Think carefully about this one!) Determine the sign of the Gibbs free energy change for this volume doubling process. 4. In 1918, the German chemist Fritz Haber won the Nobel prize for the nitrogen fixation process, through which ammonia for nitrogen-based fertilizers is synthesized from its elemental components, N2 and H2 gases, via the equilibrium N2 (g) + 3H2 (g) ⇀ ↽ 2NH3 (g) In this problem, assume all gases can be treated as ideal gases. a. Suppose a reaction vessel is filled with N2 and H2 gases to partial pressures of 3.0 atm and 2.0 atm, respectively, but suppose the the temperature at which the reaction occurs is unknown. When equilibrium is established in the vessel, it is observed that the total pressure in the vessel is 4.50 atm. What is the temperature in the vessel? The following data might be helpful: ∆Hf◦ (NH3 (g)) = −46.11 kJ/mol ∆S ◦ (N2 (g)) = 191.50 J/mol ∆S ◦ (NH3 (g)) = 192.34 J/mol ∆S ◦ (H2 (g)) = 130.57 J/mol b. What would the total pressure in the vessel be if the same reaction were carried out at 25◦ C? You may assume that the standard enthalpy and entropy are roughly temperature independent. c. Despite the high yield at 25◦ C, the reaction proceeds very slowly at this temperature. For this reason, high temperatures are used as a compromise between yield and speed. However, in order that yields not be unreasonably low, very high pressures must also be used. Suppose the reaction is carried out at 800 K, and we wish to use the same ratio, 3/2, of initial partial pressures of N2 and H2 gases. What would the initial pressures of these gases need to be in order that the yield of NH3 at equilibrium be 15% of what it would be if the reaction went to completion? You may assume that the standard enthalpy and entropy are roughly temperature independent. 5. Recall that the inhibition constant Ki for an enzyme E complexed with an inhibitor I is the equilibrium constant for the dissociation of the enzyme-inhibitor complex EI into free enzyme and free inhibitor in solution at body temperature: EI(aq) ⇀ ↽ E(aq) + I(aq) 3 a. Suppose a medical patient is diagnosed with a viral infection, the treatment for which is the administration of a drug I that inhibits and enzyme E produced by the virus that is essential in the viral replication cycle. The viral load is such that the concentration of the enzyme is 1.0×10−5 M in the patient’s body. It is known that the drug has an inhibition constant of 2.0×10−8 . You need to calculate a drug dose such that the at least 99.9% of the enzyme is inhibited at equilibrium for the dose to be considered effective. If the drug has a molar mass of 698.14 g/mol, how many milligrams of the drug should be given to the patient (in pure crystalline form) such that the dose is effective? Assume the patient’s blood volume is 3.79 L. Warning: There is no credit for killing your patient with an overdose of the drug. b. Over a period of time, the drug is metabolized, and the concentration of the enzyme returns to the original 1.0×10−5 M, so that a new dose of the drug is needed. However, it is found that the virus has created a mutant enzyme E′ that is more resistant to the drug, such that 60% of the total enzyme concentration (1.0×10−5M) is the mutant, while just 40% is the original enzyme E. The inhibition constant of the drug I for the mutant enzyme is 50 times larger than that for the origianl enzyme E. Remembering that 99.9% of the total enzyme content must be in the form of enzyme-inhibitor complexes for the dose to be effective, how many milligrams of the drug must be administered to the patient? Assume the patient’s blood volume is 3.79 L. Warning: There is no credit for killing your patient with an overdose of the drug.