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Ex-3-3530-S10-soln - CHEM 3530 Exam 3 April 2 2010 Name...

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Unformatted text preview: CHEM 3530 - Exam 3 - April 2, 2010 Name fiM/Q’n. C Multiple Choice Answers Please put your answers in clear letters. The number of correct answers will be determined solely from the letters that appear on this page. 1. 8 11, fl - 2._Q_12_é__ 2.11 12.5; 1.__C_ 11._1_3__ 2.__/_4__1._C___ a. D 12. L 1.__8__11.fi_ 8. i. 18- __C_____ 2._g_ 12.11.. CHEM 3530 - Exam 3 — April 2, 2010 (57) PART I. MULTIPLE CHOICE (Circle the ONE correct answer) For #1 - #2: When 45 grams of glucose (M=180) is added to 120 grams of water (M=18), the density of the solution is 1.25 g/mL. 1. The Molarity of the above solution is (A) 1.5 Molar (B) 1.9 Mol (C) 2.6 Molar (D) 2.1 Molar 2. The molality of the above solution is (A) 1.5 molal (B) 1.9 molal (C) 2.6 molal 3. How many grams of sucrose (M=342) are required to prepare 2.5 L of a 0.40 Molar sucrose solution? 342 grams (B) 171 grams (C) 137 grams (D) 55 grams 4. The vapor pressure of water (M=18) at 50 °C is 90 torr. When 90 grams of glucose (M=180) is dissolved in 90 grams of water (M=18), the vapor pressure of the solution at 50 °C is approximately: (A) 8torr (B) 76 torr (D) 99 torr 5. The weak electrolyte, A83 (M=80) dissociates according to the equilibrium: AB3 = A3“ + 38'. When 40 grams of A83 are dissolved in 150 grams of H20 (Kf = 1.9 °C), A33 is 70% dissociated. The freezing point of the solution is: (B) -15.2°c (C) -10.8°C (D) 6.3 °c 6. When 41 grams of the strong electrolyte, Ca(NO3)2 (M=164) is dissolved in 250 grams of water (Kf=1 .9 °C/m), the freezing point of the solution is (A) -17.1°c (B) +5.7 °c (C) 4.9%: 7. The osmotic pressure of an aqueous solution of sucrose is 6.40 bar at 30 °C. How many moles of sucrose are contained in 250 mL of this solution? (A) 3.8x10‘3 mol B) 0.064 mo (C) 0.25 mol (D) 6.4x10“ mol 8. The normal boiling point of pure CC|4(I) is 77 °C and the boiling point elevation constant is 5.0 °C/m. When 5 grams of an unknown compound is placed in 100 grams of CCl4, the boiling point of the solution is 84.5 °C. The Molar Mass of the unknown compound is approximately: (A) 3.3 g/mol (B) 64 g/mol (C) 128 g/mol m For #9 - #12: Consider the gas phase equilibrium, 2 N02(g) + 1/2 02(9) <—> N205(g). The enthalpy change for this reaction is —170 kJ and the equilibrium constant is 2.5 at 150 °C. 9. For the above reaction, if N2(g) is added at a constant total pressure of 5 bar, (A) the equilibrium will move to the left and K will decrease (B) the equilibrium will move to the right and K will i (B) the equilibrium will move to the right and K will increase (C) the equilibrium will move to the left and K will remain constant (D) the equilibrium will move to the right and K will remain constant 11. The standard state Gibbs energy change, ArGO, for the above reaction at 150 °C is approximately : A) 52 kJ\\ (B) -1.4 kJ (o) -3200 kJ (D) +1.4 kJ 12. What is K for the reaction, 2 N205(g) e) 4 N02(g) + 02(9)? (A) 6.25 (B) 0.20 (D) 0.80 13. For the gas phase reaction, 2 A(g) <—> 8(9) + C(g), AGo = -8.0 kJ at 100 °C. What is AG at 100 °C when P(A) = 0.3 bar, P(B) = P(C) = 5.0 bar? (A) +5.7 kJ +9.4 kJ 3 (o) +5.2 kJ (D) 254 kJ 14. For the gas phase equilibrium, 2A(g) + B(g) <—> 2C(g), the equilibrium constant is K = 1x10'3. If one puts A and B into a vessel with initial pressures, Pin;t(A) = Pinn(B) = 3. bar, what is the pressure of C at equililbrium? NOTE: You may assume that very little A and B react. (A) 0.045 bar (B) 0.063 bar (C) 0.082 bar 0.164 bar‘ 15. A hypothetical biochemical reaction is written as: A(aq) ) + 2H“(aq) -> B(aq). The Gibbs energy difference using the Physical Chemists’ Standard State is AG°= -20 kJ. What is the Gibbs energy difference, AG°’, for this reaction using the Biochemists’ (or Biological) Standard State? (A) -60 kJ (B) -100 kJ (D) +20 kJ For #16 - #17: We learned in class that the average number of ligands bound to a protein, R, is related to the ligand concentration, L, by: R = "[L] where n is the maximum number of bound ligands and K + [L] K is the dissociaton constant For a given protein, it was determined from equilibrium dialysis experiments that the Dissociation Constant is 0.15 and the maximum number of bound ligands is 80. 16. The slope of the double reciprocal plot, 1IR vs. 1/[L], is: (A) 1.25x10'2 (B) 5.33x1o3 (C) 8.33x10'2 ’0) 1.88x10‘3 17. The intercept of the double reciprocal plot, 1/R vs. 1/[L], is: (B) 5.33x1o3 (o) 8.33x10‘2 (D) 1.88x10'3 18. Which of the following statements is/are true concerning the binding of 02 to Myoglobin (Mb) and Hemoglobin (Hb)? M0) in resting muscle tissue (in which the 02 pressure is 40 torr), both Mb and Hb retain most of their bound 02 V6?) In muscle tissue after exercise (in which the 02 has dropped to 20 torr), Mb retains most of its bound 02, but Hb has depleted its 02 reserve Xfiii) The curve of 02 saturation vs. P02 in Hb is sigmoidal because the first bound 02 makes it more difficult for additional 0 molecules to bind to Mb (A) ii only (B) i only (D) i and iii 19. The 5th. step in the glycolysis cycle is GAP + Pi ‘——‘ DPG with AG°’ = +6.3 kJ/mol. The 6th. step is DPG + ADP ‘-—‘ 3PG + ATP with AG°’ = -18.8 kJ/mol. The 6th. step drives the 5th. step by removing DPG as it is formed. This is an example of a step driven by A Tandem Reacti - - (B) Coupled Reactions (C) PhysiologicalConcentrations (D) Endergonic Reaction Imbalance PART II. PROBLEMS (Show work for partial credit) (6) 1. Calculate the mole fraction of valine, 05H11N02 (M = 117.) in a 0.45 molal aqueous solution of valine. flow a?” ”1" ”5’3”"? W %/ féfim ! fl/ [fli— 2’7‘4/ ’7 " ”f " . (10) . 2. Two solutions are placed in contact through a semipermeable membrane. The initial concentrations of the two solutions are: Solution A: 0.80 M. K+P' [dissociated potassium salt of a protein] Solution B: 0.50 M KCl Calculate the equilibrium concentrations of K+ and Cl‘ in both solutions. [3 ,4— fl'?’ soy/x, {($1.55sz I; at/Lfl I 5 ]) [K77 ; 6’, QM T V i i L _ 5/0 f-Axix/é 79>?) 7/ @5529? [6/17 aflfl/Z‘WX’ a ’ V / [5925: 5 ' 5-52: 2?? 5:42:65“: 5-1%“? X ’3’ (12) 3. For the equilibrium reaction, A = B, the van’t Hoff plot [ln(K) vs. 1/T] below was obtained. Calculate ArH° and ArS° for this reaction. (15) 4. For the gas phase equilibrium, PCl5(g)— - PCl3(g) + Cl2(g), the equilibrium constant IS K= 4..5at150°C 3&5” x X (9) a) If one puts PCl5(g) into a container with an initial pressure of P0 = 3.0 bar, what are the pressures of PCI5(g) and PCl3(g) at equilibrium. NOTE: You cannot assume that only a small amount of PCl5 dissociates. /<: 605(3 @1— P %SV /L)£’—‘> X: {/151ng >le%f\)(fl C20 70;: 3 )< ”9'5”“sz €26 (6) b) What is AG, in kJ, for this reaction at 150 °C when the partial pressures of the three gases are: P(PCl5) = 0.8 bar and P(PCl3) = 2. 2 bar and P(Cl2) = 2.2 bar? 15/; . /5.:% =5 «5:2?7 J" 45. ”1-2; AK “SEE/WK (.9523 K3 % Constants and Conversion Factors R = 8.31 J/mol-K = 8.31 kPa-L/mol-K bar = 100 kPa 1 kPa = 7.50 torr J —b+\ll)2 —4ac 2a Note: If ax2+bx+c=0, then x: ...
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