Chem 162 2012 final exam solutions 7 6 Chem 162 2012 Final Exam Answers Chapter

# Chem 162 2012 final exam solutions 7 6 chem 162 2012

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Chem 162-2012 final exam + solutions 7 6 Chem 162-2012 Final Exam + Answers Chapter 14 – Chemical Kinetics Reaction Mechanisms, Activation Energy & Catalysts Calculations The rate of a certain reaction triples when the temperature is raised from 25 °C to 37 °C. Calculate the activation energy of this reaction. A. 70.3 molB. 52.5 kJmolC. 31.2 kJmolD. 86.5 kJmol kJ
Chem 162-2012 final exam + solutions 8 7 Chem 162-2012 Final Exam + Answers Chapter 14 – Chemical Kinetics Rates, Rate Constants, Reaction Orders, Half-Lives Calculations 2HgCl 2 (aq) + C 2 O 4 2- 2Cl - + 2CO 2 (g) + Hg 2 Cl 2 (s) Experiment Initial [HgCl 2 ] M Initial [C 2 O 4 2- ] M Rate M/min 1 0.0836 M 0.202 0.52x10 -4 2 0.0836 M 0.404 2.08x10 -4 3 0.0418 0.404 1.04x10 -4 4 0.0627 0.606 ? Find the initial rate for experiment 4. A. 8.75x10 -3 M/min B. 1.20x10 -4 M/min C. 6.00x10 -4 M/min D . 3.51x10 -4 M/min E. 1.25x10 -4 M/min Use the “Initial Rate Method”. Rate = k[HgCl 2 ] m [C 2 O 4 2- ] n Exp. 2: 2.08x10 -4 = k[0.0836] m [0.404] n Exp. 1: 0.52x10 -4 = k[0.0836] m [0.202] n 4 = 2 n n = 2 Exp. 2: 2.08x10 -4 = k[0.0836] m [0.404] n Exp. 3: 1.04x10 -4 = k[0.0418] m [0.404] n 2 = 2 m m = 1 Rate = k[HgCl 2 ] 1 [C 2 O 4 2- ] 2 Find k using data from experiment 1, 2 or 3. Experiment 1: 0.52x10 -4 = k[0.0836][0.202] 2 (0.52x10 -4 ) = (k x [0.0836] x ([0.202] 2 )) k = 0.01524 Rate = 0.01524 x [HgCl 2 ]1[C 2 O 4 2- ] 2 Now find the rate of experiment 4: Experiment 4 Rate = 0.01524 x [0.0627] 1 [0.606] 2 Rate = 3.51 x 10 -4
Chem 162-2012 final exam + solutions 9 8 Chem 162-2012 Final Exam + Answers Chapter 14 – Chemical Kinetics Reaction Mechanisms, Activation Energy & Catalysts Concepts Which mechanism for the overall reaction A C is consistent with the reaction profile below? A B is a rapid, reversible step (which can be called a “fast” step), and B C is a relatively slow step. A B B C