#Chem 162-2008 homework 4th week

#Chem 162-2008 homework 4th week - Chem 162-2007 Homework...

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Chem 162-2007 Homework 4 th Week CHAPTER 13 51. Chemical reactions occur as a result of molecular collisions, and the frequency of molecular collisions can be calculated with the kinetic-molecular theory. Calculations of the rates of chemical reactions, however, are generally not very successful. Explain why. The rates of chemical reactions depend on the percentage of chemical reactions that have enough energy to overcome the energy of activation barrier. It is not possible to determine that with any reasonable degree of accuracy. The rates of chemical reactions also depend on the percentage of chemical reactions that are sterically oriented in such a manner that can possibly form the desired molecule. This, also, is not possible to determine with any reasonable degree of accuracy. 55. Rate constants for the first-order decomposition of acetonedicarboxylic acid are k = 4.75 x 10 -4 s -1 at 293K and k = 1.63 x 10 -3 s -1 at 303K. CO(CH 2 COOH) 2 (aq) → CO(CH 3 ) 2 (aq) + CO 2 (g) Acetonedicarboxylic Acetone acid What is the activation energy of this reaction? ln (k 2 /k 1 ) = -(E a /R)[(1/T 2 ) - (1/T 1 )] 1 o (1) k = 4.75 x 10 -4 s -1 t = 293K (2) k = 1.63 x 10 -3 s -1 t = 303K ?E a ln(1.63 x 10 -3 s -1 /4.75 x 10 -4 s -1 ) = -(E a /8.314)[(1/303) - (1/293)] E a = 9.10 x 10 4 J = 91.0 kJ 57. The decomposition of ethylene oxide at 652K is a first-order reaction with k = 0.0120 min -1 and an activation energy of 218 kJ/mol. (CH 3 ) 2 O(g) → CH 4 (g) + CO(g) Calculate (a) the rate constant of the reaction at 525K and (b) the temperature at which the rate constant k = 0.0100 min -1 . 1 o T 1 = 652K k 1 = 0.0120 min -1 E a = 218000 J/mol T 2 = 525K k 2 = ? 1
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(a) ln (k 2 /k 1 ) = -(E a /R)[(1/T 2 ) - (1/T 1 )] ln (k 2 /0.0120 min -1 ) = -(218000Jmol -1 /8.314Jdeg -1 mol -1 )[(1/525K) - (1/652K)] ln (k 2 /0.0120) = -(218000/8.314)[(1/525) - (1/652)] (((ln (k 2 /0.0120)) = ((-(218000/8.314)) x [(1/525) - (1/652)])),X) k = 7.15 x 10 -7 min -1 (b) 1 o T 1 = 652K k 1 = 0.0120 min -1 E a = 218000 J/mol T 2 = ? k 2 = 0.0100 min -1 ln (k 2 /k 1 ) = -(E a /R)[(1/T 2 ) - (1/T 1 )] ln (0.0100min -1 /0.0120 min -1 ) = -(218000Jmol -1 /8.314Jdeg -1 mol -1 )[(1/T 2 ) - (1/652K)] T2 = 649K 59. It is easy to see how a bimolecular elementary reaction in a reaction mechanism can occur as a result of a collision between two molecules. How do you suppose a unimolecular process is able to occur? A unimolecular process occurs by one molecule decomposing. 61. The following is proposed as a plausible reaction mechanism: A + B → I (slow) I + B → C + D (fast) What is (a) the net reaction described by this mechanism, and (b) a plausible rate law for the reaction? The net reaction is the sum of the two equations: A + 2B → C + D Plausible rate law depends on the slow step: Rate = k[A][B] 65. The following reaction exhibits the rate law: Rate = k[NO] 2 [Cl 2 ]. 2 NO(g) + Cl
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This note was uploaded on 09/23/2009 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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#Chem 162-2008 homework 4th week - Chem 162-2007 Homework...

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