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Chemistry 125: Lecture 69 April 14, 2011 Measuring Bond Energies This For copyright notice see final page of this file.

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Presentation on theme: "Chemistry 125: Lecture 69 April 14, 2011 Measuring Bond Energies This For copyright notice see final page of this file."— Presentation transcript:

1 Chemistry 125: Lecture 69 April 14, 2011 Measuring Bond Energies This For copyright notice see final page of this file

2 Are Bond Energies “ Real ” ? Bond Dissociation Energies

3 BondDissn Energies 99 90 113 89 105 111 89 115 111 123 136.2 127 84 85 91 97 74 122857254 5946 51 67 56 58 57 72 74 73 84 63 92 94 best values as of 2003

4 Presentation by Prof. G. Barney Ellison University of Colorado, Boulder

5 D 0 (RH) = Bond Dissociation Energy Definition of D 0   100 kcal mol -1 or 4 eV

6 How could specific bond energies be measured? Consider methyl alcohol CH 3 O-H  CH 3 O + H or H-CH 2 OH  H +  CH 2 OH or CH 3 -OH  CH 3 + OH

7 One way to measure BDE: Acidity/Negative Ion Cycle CH 3 OH + F – CH 3 O – + HF K eq k1k1 k -1 He Monitor time Experiment of Veronica Bierbaum HF F–F– CH 3 OH Add CH 3 OH to a flowing stream of He containing F – and see how much CH 3 O - has formed at various times later. This gives k 1. growth of CH 3 O – m/z 31 CH 3 O – CH 3 ON=O HF growth of F – m/z 19 Add HF to a flowing stream of He containing CH 3 O – and see how much F - has appeared downstream at various times later. This gives k -1. – CH 2 OH never !

8 c) Acidity/Negative Ion Cycle CH 3 OH + F – CH 3 O – + HF  acid H 298 (CH 3 O-H) = DH 298 (CH 3 O-H) + IE(H) – EA(CH 3 O) K eq k1k1 k -1 “ known ” ? Bierbaum The equilibrium constant K eq = k 1 /k -1 gives the difference in acidity between CH 3 OH and HF. Since the acidity of HF is known [370.424 ± 0.003 kcal mol -1 ], this experiment determines the energy required for acid dissociation of CH 3 OH. CH 3 OH CH 3 O – + H + Acid which can be thought of as: CH 3 OH CH 3 O + H -Electron Affinity Ionization Energy Dissn to find

9 Anion Photoelectron Spectrum Measures Electron Affinity as Electron Binding Energy: CH 3 O — +   0  CH 3 O + e — (KE) CH 3 O (no extra vibration) CH 3 O – (no extra vibration) laser light energy Measured Kinetic Energy of ejected “photoelectron” Electron Binding Energy If the product radical is vibrating, the photo-electron kinetic energy will be smaller and the measured electron binding energy will be larger.

10 EA(CH 3 O) = (2.540 - 0.968) = 1.572 ± 0.004 eV no extra vibration  0.968 eV Engelking, Ellison, Lineberger, J. Chem. Phys. 69, 1826 (1978) CH 3 O — +   0  CH 3 O + e — Electron Kinetic Energy/eV Photoelectron counts

11 c) Acidity/Negative Ion Cycle  acid H 298 (CH 3 OH) = DH 298 (CH 3 O-H) + IE(H) – EA(CH 3 O)  acid H 298 ( CH 3 OH ) = 381.9 ± 0.5 kcal mol -1 (Bierbaum) IE(H) = 13.59844 eV or (at 298K) 313.6 kcal mol -1 EA( CH 3 O ) = 1.572 ± 0.004 eV or 36.3 ± 0.5 kcal mol -1 (Ellison et al.) DH 298 (CH 3 O-H) = 104.6 ± 0.6 kcal mol -1

12 c) Acidity/Negative Ion Cycle Problems ? Can’t apply to H-CH 2 OH any base you can think of always gets most acidic proton CH 3 OH  CH 3 O – + H + electron on O atom (good)  – CH 2 OH + H + electron on C atom (bad)  acid H 298 (H-CH 2 OH) = DH 298 (H- CH 2 OH ) + IE(H) – EA( CH 2 OH ) no gas-phase [CH 2 OH] –  CH 2 OH + e – can’t measure  acid H 298 (H-CH 2 OH) & can’t measure EA(CH 2 OH)

13 c) Acidity/Negative Ion Cycle Problems ? Can’t apply to H-CH 2 OH any base you can think of always gets most acidic proton CH 3 OH  [CH 3 O] – + H +  [CH 2 OH] – + H +  acid H 298 (H-CH 2 OH) = DH 298 (H- CH 2 OH ) + IE(H) – EA( CH 2 OH ) no gas-phase [CH 2 OH] –  CH 2 OH + e – can’t measure  acid H 298 (H-CH 2 OH) & can’t measure EA(CH 2 OH) However: CH 3 OH + C CH 2 OH + HC measure K equi via k 1 and k -1 to extract  rxn H 298  DH 298 (H-CH 2 OH) - DH 298 (HC ) DH 298 (HC ) = 103.15 ± 0.03 kcal mol -1 DH 298 (H-CH 2 OH) = 96.1 ± 0.2 kcal mol -1

14 Uses of heats of formation,  f H 298 (R) ? What is the C-O bond in methanol? CH 3 -OH  CH 3 OH What is ?  or 

15 Uses of heats of formation,  f H 298 (R) ? What is the C-O bond in methanol? CH 3 -OH  CH 3 OH DH 298 (CH 3 -H) =  f H 298 (CH 3 ) +  f H 298 (H) -  f H 298 (CH 4 ) Radical kinetics/PIMS studies  DH 298 (CH 3 -H) = 104.99 ± 0.03 kcal -1 &  f H 298 (H) is known from D 0 (H 2 ) Classical thermochemistry finds  f H 298 (CH 4 ) J. B. Pedley, R. D. Naylor, and S. P. Kirby, Thermochemistry of Organic Compounds; 2 nd ed.; Chapman and Hall: New York, 1986.  f H 298 (CH 3 ) = 35.05 ± 0.07 kcal mol -1 BDE(H 2 O)  f H 298 (OH) = 8.91 ± 0.07 kcal mol -1 Pedley et al provides  f H 298 (CH 3 OH) DH 298 (CH 3 -OH) =  f H 298 (CH 3 ) +  f H 298 (OH) -  f H 298 (CH 3 OH) DH 298 (CH 3 -OH) = 92.1 ± 0.1 kcal mol -1

16 Ellison I

17 Ellison II

18 c) Acidity/Negative Ion Cycle  acid H 298 (CH 3 OH) = DH 298 (CH 3 O-H) + IE(H) – EA(CH 3 O)  acid H 298 ( CH 3 OH ) = 381.9 ± 0.5 kcal mol -1 (Bierbaum) IE(H) = 13.59844 eV or (at 298K) 313.6 kcal mol -1 EA( CH 3 O ) = 1.572 ± 0.004 eV or 36.3 ± 0.5 kcal mol -1 (Ellison et al.) DH 298 (CH 3 O-H) = 104.6 ± 0.6 kcal mol -1

19 Bond Energies of Alcohols/kcal mol -1 DH 298 (CH 3 O-H) = 104.6 ± 0.7  CH 3 O + H DH 298 (CH 3 CH 2 O-H) = 104.7 ± 0.8 DH 298 ((CH 3 ) 2 CHO-H) = 104.6 ± 0.7 DH 298 ((CH 3 ) 3 CO-H) = 106.3 ± 0.7

20 Bond Energies of Alcohols/kcal mol -1 DH 298 (CH 3 O-H) = 104.6 ± 0.7  CH 3 O + H DH 298 (CH 3 CH 2 O-H) = 104.7 ± 0.8 DH 298 ((CH 3 ) 2 CHO-H) = 104.6 ± 0.7 DH 298 ((CH 3 ) 3 CO-H) = 106.3 ± 0.7 DH 298 (C 6 H 5 O-H) = 85.8 ± 0.1

21 Bond Energies of Alcohols/kcal mol -1 DH 298 (CH 3 O-H) = 104.6 ± 0.7  CH 3 O + H DH 298 (CH 3 CH 2 O-H) = 104.7 ± 0.8 DH 298 ((CH 3 ) 2 CHO-H) = 104.6 ± 0.7 DH 298 ((CH 3 ) 3 CO-H) = 106.3 ± 0.7 DH 298 (C 6 H 5 O-H) = 85.8 ± 0.1 DH 298 (HOO-H) = 87.8 ± 0.5  HOO + H DH 298 (CH 3 OO-H) = 88 ± 1 DH 298 (CH 3 CH 2 OO-H) = 85 ± 2 DH 298 (CH 3 ) 3 COO-H) = 84 ± 2

22 Bond Energies of Alcohols/kcal mol -1 DH 298 (CH 3 O-H) = 104.6 ± 0.7 DH 298 (CH 3 CH 2 O-H) = 104.7 ± 0.8 DH 298 ((CH 3 ) 2 CHO-H) = 104.6 ± 0.7 DH 298 ((CH 3 ) 3 CO-H) = 106.3 ± 0.7 DH 298 (C 6 H 5 O-H) = 85.8 ± 0.1 DH 298 (HO-H) = 118.82 ± 0.07  HO + H

23 What is a bond strength? Consider methane: CH 4  C + 4 H  atomization H 298 (CH 4 ) = 397.5 kcal mol -1  D avg H 298 (CH 4 ) = 99.4 kcal mol -1 SpeciesDH 298 /kcal mol -1 Heat of Formation CH 3 -H 104.99 ±0.03  f H 298 [CH 3 ] CH 2 -H110.4 ± 0.2  f H 298 [CH 2 ] CH-H101.3 ± 0.3  f H 298 [CH] C-H80.9 ± 0.2  f H 298 [C] a) No bond equals the “average” C-H bond … careful. b) The sum of the BED’s is 397.5 ± 0.6 kcal mol -1. 1 st Law really works!

24 Boulder Ion Gang, 1980 Leone Bierbaum Herman Ellison DePuy Ferguson Lineberger

25 EA(CH 3 O) = (2.540 - 0.968) = 1.572 ± 0.004 eV no extra vibration  0.968 eV Engelking, Ellison, Lineberger, J. Chem. Phys. 69, 1826 (1978) CH 3 O — +   0  CH 3 O + e — Electron Kinetic Energy/eV Photoelectron counts ?

26 Anion Photoelectron Spectrum Measures Electron Affinity as Electron Binding Energy: CH 3 O — +   0  CH 3 O + e — (KE) CH 3 O (no extra vibration) CH 3 O – (no extra vibration) laser light energy Measured Kinetic Energy of ejected “photoelectron” Electron Binding Energy If the product radical is vibrating, the photo-electron kinetic energy will be smaller and the measured electron binding energy will be larger. Vibrationally excited CH 3 O - gives “hot band” leaving more of   0 energy for ejected electron.

27 EA(CH 3 O) = (2.540 - 0.968) = 1.572 ± 0.004 eV no extra vibration  0.968 eV Engelking, Ellison, Lineberger, J. Chem. Phys. 69, 1826 (1978) CH 3 O — +   0  CH 3 O + e — Electron Kinetic Energy/eV Photoelectron counts transitions to vibrationally excited states of CH 3 O radical transition from vibrationally excited state of CH 3 O - anion (weaker vibration)

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30 End of Lecture 69 April 15, 2011 Copyright © G. B. Ellison 2011. Some rights reserved. Except for cited third-party materials, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0). Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: G. B. Ellison, Chem 125. License: Creative Commons BY-NC-SA 3.0


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