HW2SolnME525SP2011

HW2SolnME525SP2011 - 1 ME 525 Homework #2 Due Tuesday,...

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1 ME 525 Homework #2 Due Tuesday, February 8, 2011 Prof. Lucht (E-mail address: [email protected]) 1. Turns, Problem 4.10. Solution: At equilibrium the forward and backward rates of the reaction are equal:         43 33 44 fr f r kC H M kC H HM k CH H M CH H H M C H  But the concentrations and mole fractions are related by 3 4 0 3 0 4 0 00 0.003691 i i u CH H CH H f rC H u C H u CH H P CH f Pf r P ru u xP X RT xx k CH H PP P k CH x x P P K k Kk k K kR T R T      Calculating the rate coefficient for the reverse reaction we obtain     55 6 5 2 2.82 10 exp 9835 / 2.82 10 (1500)exp 9835 /1500 6.01 10 r kTT m kmol s 
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2 Substituting in the expression for f k we obtain    53 6 5 2 3 1.01325 10 / 6.01 10 0.003691 8314 1500 18.0 f f Jm m k kmol s J K kmol K m k kmol s     2. The decomposition of ozone (O 3 ) can be modeled as the following sequence of elementary reactions: O 3 M k 1    O 2 O M log 10 A 14.63 , b 0, E A 22.18 kcal gmol O 3 O k 2   O 2 O 2 log 10 A 13.06 , b E A 4.57 kcal gmol [ A ] cm 3 gmol sec (a) Calculate the rate constants for the two forward reactions as written above at a temperature of 800 K. (b) Write a differential equation for the change in [O] with respect to time and solve the equation for the case of constant ozone and diatomic oxygen concentrations. Neglect reverse reactions. (c) Calculate the steady-state concentration of O atoms in a gas mixture with a temperature of 800 K, a pressure of 2 atm, and mole fractions (initially) of 0.2 for ozone and 0.8 diatomic oxygen. Neglect reverse reactions. (d) Based on your answer to part (c), at what rate in gmol/(cm 3 -sec) does the ozone concentration decrease when the oxygen atom concentration reaches its steady state value? Assume that the concentrations of ozone and diatomic oxygen have not yet changed significantly for this calculation, and that the reactions occur in a constant volume reactor. Neglect reverse reactions. Solution: (a) The forward rate constants are evaluated as follows:
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3  1 2 3 14.63 14 1 3 13.06 13 1 exp 0 4.1868 22.18 92.86 92,860 4.1868 4.57 19.1 19,100 8.314 10 4.266 10 10 1.148 10 b A u A A u E k T AT b for both reactions RT kcal kJ Ek J J gmol kcal kcal kJ J J gmol kcal J R gmol K cm A gmol s cm A gmo         3 3 14 8 1 3 3 13 11 2 92,860 4.266 10 exp 3.69 10 8.314 800 19,100 1.148 10 exp 6.50 10 8.314 800 ls cm J cm k gmol s gmol s J K gmol K cm J cm k gmol s gmol s J K gmol K    (b) The differential equation for the concentration of atomic oxygen is given by      13 23 dO kO M kO O O dt where k O M and k O 
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This note was uploaded on 04/17/2011 for the course ME 525 taught by Professor Lucth during the Spring '11 term at Purdue.

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HW2SolnME525SP2011 - 1 ME 525 Homework #2 Due Tuesday,...

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