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