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CHM 132 - Final - 2005 Key

CHM 132 - Final - 2005 Key - Chemistry 132 Final...

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Chemistry 132 Final Examination May 6, 2005 Name ( PRINT) _______ SOLUTIONS _______ Workshop Leader________________________ There are 9 questions on this exam. The last three pages of this exam contain a Periodic Table, a list of constants and formulas, and a sheet of scratch paper. Show your work neatly and circle your answers. Give units for all answers requiring them. Score Part 1: Mostly First Law 1. (35 points)_____________ 2. (35 points)_____________ subtotal __________________ Part 2: Second Law, Electrochemistry 3. (20 points)_____________ 4. (25 points)_____________ 5. (25 points)_____________ subtotal __________________ Part 3: Kinetics, Nuclear Decay, Atoms 6. (30 points)_____________ 7. (30 points)_____________ subtotal __________________ Part 4: Chemical Bonding 8. (25 points)_____________ 9. (25 points)_____________ subtotal __________________ Grand total (250 max) __________________
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Name___________________________ 2 1. (35 points) One mole of an ideal gas initially at a pressure of 1.00 atm and a temperature of 300K (point A ) undergoes a process in which the gas is expanded reversibly and isothermally until the pressure of the gas has decreased to 0.250 atm (point B ). a) Sketch the process AB on a P-V diagram b) Calculate the initial and final volumes of the gas for the process AB . V A = RT/P A = (.0821)(300)/1 L = 24.6 L V B = RT/P B = (.0821)(300)/0.25 = 98.5 L c) Calculate q, w, E, and S for the process AB . Process is isothermal, therefore E = 0 w = -RT ln (V B /V A ) = -(8.31)(300) ln (98.5/24.6) = - 3460 J q = -w = + 3460 J . This is q for a reversible process. S = q rev /T = 3460 J/300 K = + 11.5 J/K V 1 atm V A V B 0.25 atm P
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Name___________________________ 3 1. (continued) d) Now consider an irreversible expansion of the gas between points A and B in which the pressure is decreased instantaneously from 1.00 atm to 0.25 atm at constant volume, then the volume is increased to the value corresponding to the same final volume achieved in the reversible expansion. Sketch this process on a PV-diagram below. Now calculate w and S for this irreversible process. How does the irreversible work of expansion compare with the reversible work? The work for this irreversible expansion takes place at constant P ext w = -P ext V = - (0.25 atm)(98.5 – 24.6 L) × 101.3 J /(L-atm) = -1870 J Because the irreversible process occurs between the same initial and final states as calculated in part c), S = + 11.5 J/K , the same as in part c) 1 atm V A V B 0.25 atm P
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Name___________________________ 4 2. (35 points) Dr. Wing Tsang (he’s a real scientist, not just an actor on television), head of the Thermochemistry Division of the National Institute of Standards and Technology (NIST), once again brings you a sample of a substance, which he says has a formula of C 10 H 22 . He says that he needs a measurement of the enthalpy of formation of this substance. In a constant volume calorimeter, you carry out the combustion of 1.42 g of this material to CO 2 ( g ) and H 2 O( l ). Atomic masses: H = 1, C = 12, O = 16. a) Write the balanced chemical reaction corresponding to this combustion process. C 10 H 22 (s) + 15.5 O 2 (g) 10 CO 2 (g) + 11 H 2 O(l) b) The heat capacity of the calorimeter and its contents is 17,990 J deg -1 . (The units on this quantity are correct .) When the combustion reaction is carried out, the temperature of the calorimeter is observed to rise 3.80 ° C. Use this information to calculate E ° for the combustion of the compound.
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