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ChE 553 Lecture 21

ChE 553 Lecture 21. Estimating Activation Barriers On Metal Surfaces. Objective. Learn how to estimate heats of reaction for reactions on surfaces Learn how to estimate activation barriers for reactions on surfaces. Example CH 3 OH CO+2H 2.

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ChE 553 Lecture 21

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  1. ChE 553 Lecture 21 Estimating Activation Barriers On Metal Surfaces

  2. Objective • Learn how to estimate heats of reaction for reactions on surfaces • Learn how to estimate activation barriers for reactions on surfaces

  3. Example CH3OHCO+2H2 Figure 5.14 The Mechanism of Methanol Decomposition on Pt(111).

  4. How can I estimate The activation Energy For CH3OHadCH3Oad + Had? • Use polayni or blowers-masel to estimate activation barrier • Need a way to estimate the heat of reaction

  5. First consider CH3OHad • From Nist webbook Hf =– 49.3 in the gas phase • From Lect 4 • For polar physisorbed molecules (Not CO) works out to be ~5-8 Kcal/mole • Hf= (-49.3 kcal/mole)- 7 kcal=-56 kcal/mol

  6. Next Consider CH3Oad • From Nist Webbook Hf=+ 4 kcal/mol • Next need to estimate energy of Pt-methoxy bond

  7. Benziger’s Method Heat of formation of oxygen double bond to surface Hf,ad= Hf,gas- D(M-O)*(nb/2) nb = number of bonds to the surface

  8. Summary • For Methoxy • Hf=+4kcal/mole-84 kcal/mole*(1/2)=-38 kcal/mole • For hydrogen atoms • Hf= -9 kcal/mole • Therefore ΔHr= -38-9-(-56)=+9 kcal/mole

  9. Next Calculate Activation Energy Via The Polayni relationship • Ea= 12+0.7*(9)=18 kcal/mole • Reaction seen when T>18/0.07=257 K • Actually starts to appear at 230 K • Rate of catalytic cycle significant when T>18/0.05=360 K • actually need 380 K.

  10. Next Consider Formation Of A CH2OH intermediate instead • From Nist Webbook Hf,gas= -2 kcal/mole • Hf,ad= -2 -130/4= -34.5 kcal/mol • ΔHr=-34.5-9-(-56)=12.5 kcal/mol

  11. Pictures of molecule • Notice that carbon can never get close to the surface • We will show later: • Ea= Ea0 + (energy of bonds that break when getting to transition state)- (energy of new bonds that form in getting to the transition state). = Ea0 + ΔHr

  12. Implications • For OH bond scission Ea=Ea0 +(energy loss of broken OH bond)- (energy gain of new hydrogen-surface bond)-(energy gain of new oxygen-surface bond) • For CH bond scission Ea=Ea0 +(energy loss of broken CH bond)- (energy gain of new hydrogen-surface bond)-(energy gain of new carbon-surface bond) • OH Bond scission has lower effective barrier (~20 kcal/mole lower)

  13. For CH bond scission • Ea=12 + (0.7)*12.5+20=41 kcal/mole

  14. What Is The Next Step? Hydrogen loss Two possibilities: • Carbon binds to surface • Carbon binds to oxygen forming formaldehyde (H2C=O) (β elimination)

  15. If Formaldehyde Forms via a β elimination • Hf,gas of formaldehyde= -27.7 mole • Hf,ad of formaldehyde= -27.7-6=-34 mole • Hf,Ad of hydrogen= -9 kcal/mole • Hf, Ad of methoxy= -38 kcal/mol • ΔHr= -34 -9- (-38)= -5 kcal/mole • Ea=15+0.3*(-5)= 14 kcal/mole

  16. Alternative is a formyl di-radical • Not stable in gas phase • How to estimate gas phase energy? • Methanol-CH bond-OH bond +2 H atoms = -49.3 +93+103-2*(52)=+43 kcal/mol • Hf,ad=+43-(oxygen)-carbon • Hf,ad=+43-(84/2)-(130/4)= -31.5

  17. Calc Ea Ea= (intrinsic barrier)+0.3*Hr+(proximity) Ea= 12+0.3*(-31.5)+20=22.5 Higher! So formaldehyde forms next

  18. The result is the catalytic cycle we noted previously Figure 5.14 The Mechanism of Methanol Decomposition on Pt(111).

  19. Summary • Can use the same procedures outlined in earlier to estimate reactions on metal surfaces • Numbers are approximately correct! • Key weakness: cannot estimate heat of reaction accurately – leads to ~10-20% errors in Ea

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