University:
Amherst College
Course:
CHEM 165 | Foundations of Thermodynamics and Kinetics
Academic year:

2009
Views:

350

Pages:

3
Author:

Trnovkamil6o

Chemistry 12 Assignment 13-Solutions Fall 2009 Chapter 15: Chemical Kinetics Problems: #56, 60, 61, 62, 63, 64, 65 Temperature Dependence of Rate Constants and the Collision Model 56. H,O'(aq) + OH(aq) + 2 H₂O(1) should have the faster rate. H,O" and OH will be electro- statically attracted to each other; Cet and Hg, will repel each other (so E, is much larger). 60. In E RT T Since the rate doubles, then k₂=2k. In 2.00 - 1 1 ). 8.3145 JK mol-1 298 K 308 K E-5 5.3×10 J/mol = 53 kJ/mol 61. From the Arrhenius equation in logarithmic form (Ink-E/RT + In A), a graph of In k vs. 1/T should yield a straight line with a slope equal to -E/R and a y-intercept equal to In A. a. slope-E/R, E,1.10×10 Kx 8.3145 J K mol 9.15 10 J/mol = 91.5 kJ/mol b. The units for A are the same as the units for k (s'). y-intercept In A, A53.54 × 104 s c. Ink-E/RT+ In A or k = A exp(-E/RT) k=3.54x10x exp -9.15 x 10 J/mol 8.3145 J K mol x 298 K 3.24 x 101 62. A graph of in k vs. 1/T should be linear with slope = -E/R. T(K) 1/T (K) k (s') In k 338 2.96 x 10-3 4.9 x 103 -5.32 318 3.14 x 10 5.0 x 10 -7.60 298 3.36 x 10 3.5 x 10 -10.26 63. The Arrhenius equation is: kA exp (-E/RT) or in logarithmic form, Ink-E/RT + In A. Hence, a graph of In k vs. 1/T should yield a straight line with a slope equal to -E/R since the logarithmic form of the Arrhenius equation is in the form of a straight line equation, y=mx+b. Note: We carried one extra significant figure in the following In k values in order to reduce round off erroог. T (K) 1/T (K) 195 5.13 x 10 230. 4.35 x 10 260. 3.85 x 10 298 3.36 x 10 369 2.71 x 10 k (L mol¹ s¹) 1.08 x 10 2.95 x 10° 5.42 x 10 12.0 × 10 35.5 x 10° In k 20.80 21.81 22.41 23.21 24.29 24.00 In k 23.00 22.00 21.00 2.00 3.00 4.00 5.00 6.00 From the "eyeball" line on the graph: 20.95-23.65 5.00×10-3.00×10 1000 T(K) 2.00×10-3 slope = -2.70 -1.35x10 K= -E R E1.35 x 10° K × 8.3145 J K¹ mol¹1.12 10 J/mol = 11.2 kJ/mol From the best straight line (by calculator): slope-1.43 x 10 K and E, 11.9 kJ/mol

Chemical Kinetics Assignment 13 - Solutions

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