CHM 132 - Exam 2 - 2005 Key

CHM 132 - Exam 2 - 2005 Key - Chemistry 132 Hour...

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Chemistry 132 Hour Examination 2 March 17, 2005 Name ( PRINT)________ SOLUTIONS ________ Student number____________________________ Workshop Leader__________________________ 1. (35 points ) ____________________ 2. (30 points ) ____________________ 3. (35 points ) ____________________ 4. (30 points ) ____________________ Extra Credit ____________________ TOTAL (130 points ) Show your work neatly and circle your answers . Give correct units for answers requiring them. The last page of the exam gives constants, formulas, and a table of reduction potentials
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Name________________________ 2 1. (35 points) Joe Blow was working in the Chemistry stockroom. He inadvertently placed a gallon jug of hydrochloric acid (HCl) in the same cabinet with aqueous ammonia (NH 4 OH). The next morning, he looked in the cabinet and found a white solid covering every surface. He realized that this solid was NH 4 Cl, and decided to learn about the thermochemistry of the mess he had created. He speculated that the solid was formed by the following reaction: HCl (g) + NH 3 (g) NH 4 Cl (s) Substance H f ° ° (kJ mole -1 ) S ° (J K -1 mole -1 ) HCl ( g) -92.3 186.9 NH 3 (g) -45.9 192.8 N H 4 Cl (s) -314.6 94.9 a) (6 points) Calculate H ° and S ° for the reaction What can you say about the relative bond strengths of reactants and products? H ° = H f ° ° (NH 4 Cl) - H f ° ° (HCl) - H f ° ° (NH 3 ) = -314.6 –(-92.3) –(-45.9) kJ / mole = -176.4 kJ / mole The product bonds are stronger than the reactant bonds. S ° = S ° (NH 4 Cl) - S ° (HCl) - S ° (NH 3 ) = 94.9 –186.9 –192.8 J /(K-mole) = -284.8 J/ (K-mole) b) (4 points) Calculate G ° at 298 K G ° = H ° - T S ° = -176.4 kJ / mole – (298 K) ( -284.8 J/K- mole) × 10 -3 kJ/J = -176.4 + 84.9 = -91.5 kJ/mole c) (6 points) Calculate K eq at T = 298 K. ln K eq = - G ° / RT = - (-91,500 J/mole) / (8.31 J / K-mole × 298 K) = 36.95 K eq = e 36.95 = 1.1 × 10 16
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Name________________________ 3 1. (continued) d) (7 points) Calculate the enthalpic contribution to K eq at T = 298 K. Rationalize (a few words) the magnitude of this term. Enthalpic contribution = e - H ° / RT Exponent = - H ° /RT = -(-176,400)/(8.31 × 298) = +71.23 Enthalpic factor = e +71.23 = 8.6 × 10 30 Stronger product bonds drive this term e) (7 points) Calculate the entropic contribution to K eq at T = 298 K. Rationalize (a few words) the magnitude of this term. Entropic contribution = e
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This test prep was uploaded on 04/17/2008 for the course CHM 132 taught by Professor Farrar during the Spring '08 term at Rochester.

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CHM 132 - Exam 2 - 2005 Key - Chemistry 132 Hour...

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