CHM 132 - Exam 2 - 2003 Key

CHM 132 - Exam 2 - 2003 Key - Chemistry 132 Hour...

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Chemistry 132 Hour Examination 2 March 20, 2003 Name ( PRINT)_____ SOLUTIONS ___________ Student number____________________________ Workshop Leader__________________________ 1. (30 points ) ____________________ 2. (25 points ) ____________________ 3. (25 points ) ____________________ 4. (20 points ) ____________________ 5. (25 points ) ____________________ TOTAL (125 points ) Show your work neatly and circle your answers. Give correct units for answers requiring them. Tables of formulas, constants, and reduction potentials are found on the last page of the exam.
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2 Name_________________________ 1. (30 points) Water gas, a mixture of H 2 and CO, is a fuel made by reacting steam with red-hot coke, a by-product of coal distillation. The water gas reaction is an important source of hydrogen gas for fuel cells. H 2 O( g) + C (s) CO (g) + H 2 (g) H f ° (kJ mol -1 )S ° (J K -1 mol -1 ) H 2 O( g ) -242 189 C( s ) 0 6 CO( g ) -110.5 198 H 2 ( g ) 0 131 (a) From the data given, calculate H ° and S ° for production of water gas. H ° = -110.5 –(-242) = + 131.5 kJ mol -1 S ° = 198 + 131 –189 – 6 = + 134 J K -1 mol -1 (b) Based on the value of H ° calculated in (a), the chemical bonds of the products of the reaction are (circle one) s tronger weaker than those of the reactants. Explain. The reaction is endothermic, so potential energy of system is higher. Stronger chemical bonds lower the overall potential energy of the system. (c) Estimate the temperature at which the reaction becomes favorable, i.e., K eq = 1. K eq = 1 G ° = H ° - T S ° =0; therefore T = H ° / S ° T = 131,500 J mol -1 / 134 J K -1 mol -1 = 980 K (d) In what direction must the temperature be changed to improve the yield of hydrogen gas? Explain. The reaction is endothermic, so LeChatelier’s Principle says that an increase in temperature can be compensated for by forming more products i.e., absorbing heat.
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3 Name_________________________ 2. (25 points) An electrolytic cell consists of a pair of inert metallic electrodes in an aqueous solution buffered to pH = 5.0 and containing CuSO 4 at a concentration of 1.00 M. A current of 1.0 amps is passed through the cell for 10.0 hours. (a) Calculate ε for the reaction 2H + + 2e - H 2 at pH = 5.0 ε = ε° - (0.0592/2)log {1/[H + ] 2 }= ε° + (0.0592)log [H + ] = ε° - 0.0592 pH ε = 0.00 - .0592 × (5) v = -0.296 v (b) For the reaction Cu 2+ + 2e - Cu (s) , ε° = +0.34 V With this information and the result from part (a), what product is formed at the cathode of the electrolytic cell?
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CHM 132 - Exam 2 - 2003 Key - Chemistry 132 Hour...

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