## 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