Answer

The Arrhenius equation can be written as

so a plot of ln k vs. 1/T (with T in units of K) gives the activation energy.

The shaded columns in the tables are the data plotted.

Then, E_a = –slope × R (R = 8.314 J/mol·K)

All answers are to 3 significant figures

Data Set 0

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 1

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 2

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 3

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 4

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 5

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 6

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 7

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 8

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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

 

Data Set 9

T (°C) T (K) 1/T (K^–1) k (s^–1) ln(k) (ln(s^–1))

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

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