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CHM 112
Section 1

Homework Assignment #5
Due February 3, 2006 by 8:05 a.m.

Practice (not to be turned in): Review Question 13.13, Problem 13.55.

Consider the reaction

A(g) products

Choose the data set corresponding the to last digit in your Student ID Number.

Find the activation energy in units of kJ/mol by plotting the data.

The Arrhenius equation can be written as

so a plot of ln k vs. 1/T gives the activation energy.
The shaded columns in the tables are the data plotted.

Then,

Ea = –slope × R
R = 8.314 J/molK

All answers are to 3 significant figures

data set 0
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
1.22×106
14.015
5
278
0.00360
1.32×106
14.094
10
283
0.00353
1.43×106
14.171
15
288
0.00347
1.54×106
14.244
20
293
0.00341
1.65×106
14.316
25
298
0.00336
1.77×106
14.384
30
303
0.00330
1.89×106
14.451
35
308
0.00325
2.01×106
14.516
40
313
0.00319
2.14×106
14.578
45
318
0.00314
2.28×106
14.638
50
323
0.00310
2.41×106
14.697


slope = –1210 so

Ea = –(–1210) × (8.314) = 10000 J/mol = 10.0 kJ/mol



data set 1
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
2.98×104
10.302
5
278
0.00360
3.49×104
10.461
10
283
0.00353
4.07×104
10.614
15
288
0.00347
4.72×104
10.761
20
293
0.00341
5.44×104
10.904
25
298
0.00336
6.24×104
11.041
30
303
0.00330
7.13×104
11.175
35
308
0.00325
8.11×104
11.304
40
313
0.00319
9.19×104
11.428
45
318
0.00314
1.04×105
11.549
50
323
0.00310
1.17×105
11.666


slope = –2410 so

Ea = –(–2410) × (8.314) = 20000 J/mol = 20.0 kJ/mol



data set 2
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
5.46×102
6.302
5
278
0.00360
6.92×102
6.540
10
283
0.00353
8.70×102
6.769
15
288
0.00347
1.09×103
6.990
20
293
0.00341
1.34×103
7.204
25
298
0.00336
1.65×103
7.411
30
303
0.00330
2.02×103
7.610
35
308
0.00325
2.45×103
7.804
40
313
0.00319
2.95×103
7.991
45
318
0.00314
3.54×103
8.172
50
323
0.00310
4.22×103
8.348


slope = –3620 so

Ea = –(–3620) × (8.314) = 30100 J/mol = 30.1 kJ/mol



data set 3
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
8.88
2.184
5
278
0.00360
12.2
2.501
10
283
0.00353
16.6
2.806
15
288
0.00347
22.2
3.102
20
293
0.00341
29.6
3.387
25
298
0.00336
38.9
3.662
30
303
0.00330
50.8
3.929
35
308
0.00325
65.8
4.186
40
313
0.00319
84.4
4.436
45
318
0.00314
107.5
4.678
50
323
0.00310
135.9
4.912


slope = –4820 so

Ea = –(–4820) × (8.314) = 40100 J/mol = 40.1 kJ/mol



data set 4
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
0.14
–1.999
5
278
0.00360
0.20
–1.603
10
283
0.00353
0.30
–1.221
15
288
0.00347
0.43
–0.852
20
293
0.00341
0.61
–0.495
25
298
0.00336
0.86
–0.151
30
303
0.00330
1.2
0.182
35
308
0.00325
1.7
0.504
40
313
0.00319
2.3
0.816
45
318
0.00314
3.1
1.118
50
323
0.00310
4.1
1.411


slope = –6030 so

Ea = –(–6030) × (8.314) = 50100 J/mol = 50.1 kJ/mol



data set 5
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
1.98×10–3
–6.223
5
278
0.00360
3.19×10–3
–5.747
10
283
0.00353
5.05×10–3
–5.288
15
288
0.00347
7.86×10–3
–4.846
20
293
0.00341
1.21×10–2
–4.418
25
298
0.00336
1.82×10–2
–4.005
30
303
0.00330
2.72×10–2
–3.605
35
308
0.00325
4.00×10–2
–3.219
40
313
0.00319
5.82×10–2
–2.844
45
318
0.00314
8.36×10–2
–2.482
50
323
0.00310
1.19×10–1
–2.130


slope = –7230 so

Ea = –(–7230) × (8.314) = 60100 J/mol = 60.1 kJ/mol



data set 6
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
2.83×10–5
–10.474
5
278
0.00360
4.92×10–5
–9.920
10
283
0.00353
8.40×10–5
–9.384
15
288
0.00347
1.41×10–4
–8.868
20
293
0.00341
2.32×10–4
–8.369
25
298
0.00336
3.76×10–4
–7.887
30
303
0.00330
5.99×10–4
–7.421
35
308
0.00325
9.40×10–4
–6.970
40
313
0.00319
1.45×10–3
–6.533
45
318
0.00314
2.22×10–3
–6.110
50
323
0.00310
3.35×10–3
–5.700


slope = –8440 so

Ea = –(–8440) × (8.314) = 70200 J/mol = 70.2 kJ/mol



data set 7
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
3.94×10–7
–14.746
5
278
0.00360
7.43×10–7
–14.113
10
283
0.00353
1.37×10–6
–13.501
15
288
0.00347
2.47×10–6
–12.911
20
293
0.00341
4.37×10–6
–12.341
25
298
0.00336
7.58×10–6
–11.790
30
303
0.00330
1.29×10–5
–11.257
35
308
0.00325
2.16×10–5
–10.741
40
313
0.00319
3.56×10–5
–10.242
45
318
0.00314
5.78×10–5
–9.759
50
323
0.00310
9.23×10–5
–9.290


slope = –9640 so

Ea = –(–9640) × (8.314) = 80100 J/mol = 80.1 kJ/mol



data set 8
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
5.41×10–9
–19.035
5
278
0.00360
1.10×10–8
–18.321
10
283
0.00353
2.20×10–8
–17.633
15
288
0.00347
4.27×10–8
–16.969
20
293
0.00341
8.11×10–8
–16.328
25
298
0.00336
1.51×10–7
–15.708
30
303
0.00330
2.74×10–7
–15.109
35
308
0.00325
4.90×10–7
–14.529
40
313
0.00319
8.59×10–7
–13.967
45
318
0.00314
1.48×10–6
–13.423
50
323
0.00310
2.51×10–6
–12.896


slope = –10800 so

Ea = –(–10800) × (8.314) = 89800 J/mol = 89.8 kJ/mol



data set 9
T (oC)
T (K)
1/T (K–1)
k
ln(k)
0
273
0.00366
7.34×10–11
–23.335
5
278
0.00360
1.62×10–10
–22.543
10
283
0.00353
3.48×10–10
–21.778
15
288
0.00347
7.28×10–10
–21.040
20
293
0.00341
1.49×10–9
–20.328
25
298
0.00336
2.96×10–9
–19.639
30
303
0.00330
5.76×10–9
–18.973
35
308
0.00325
1.10×10–8
–18.328
40
313
0.00319
2.05×10–8
–17.705
45
318
0.00314
3.74×10–8
–17.100
50
323
0.00310
6.73×10–8
–16.515


slope = –12100 so

Ea = –(–12100) × (8.314) = 100000 J/mol = 100. kJ/mol