Throwing Light on Reaction Dynamics: H + HBr The thermal reaction - PowerPoint PPT Presentation

Throwing Light on Reaction Dynamics: H + HBr

The thermal reaction of hydrogen gas ( H 2 ) and bromine gas ( Br 2 ) to form hydrogen bromide vapor ( HBr ) is a classic reaction: 22 +2rHBrHB ææƨ Energetics ( thermodynamics ) tells us that Energetics ( thermodynamics ) does NOT the equilibrium constant is large, that is, tell us how this reaction occurs or how favors the formation of products because rapidly equilibrium is established. the Br-Br bond is so weak in comparison to For answers to those questions, we need the others. kinetics and dynamics .

Kinetic Studies of H 2 + Br 2 M. Bodenstein and S. Lind, Z. physikal. Chem . 57 , 168 (1906). Rate of reaction: 1222 1d[]=k[][BrdtrHB]H2 Great surprise and mystery

Reaction Mechanism ææÆ 1 2k +M2+rBrBM(1) Initiation: ææÆææÆ 23 2k2k HHH++(Br Propagation: ææÆ -1 k2 B2+MBrr+M(4) Termination:

Kinetic Studies of H 2 + Br 2 M. Bodenstein and S. Lind, Z. physikal. Chem . 57 , 168 (1906). Rate of reaction: 22212 k[][]1d[]=2dt1+HBrHBk[

Reaction Mechanism Christiansen, Dansk. V. d. Math. Phys. Medd. 1 , 14 (1919) Herzfeld, Ann. Phys . 59 , 635 (1919) M. Polanyi, Z. El. Ch . 26 , 10 (1920)

Reaction Mechanism ææÆ 1 2k +M2+rBrBM(1) Initiation: ææÆææÆ 23 2k2k HHH++(Br Propagation: ææÆææÆ -2-3 kk22 BrBHBr Inhibition: ææÆ -1 k2 B2+MBrr+M(6) Termination:

H + HBr Æ H 2 + Br + + How does such a simple reaction occur? Try to make a movie in your mind of how the reaction takes place.

H + HBr Æ H 2 + Br + + How much does the H 2 product vibrate and rotate? Can a measurement of the rotation and vibration tell us about the mechanism of a chemical reaction?

The Transition State 113 10m10fs10m/s DistanceTimeRate = Reagents Products

The Born-Oppenheimer Approximation 0.511.521.522.53-100-80-60-40-100-80-60-40 H-H separation (Å) •Separate wave equation; calculate the electronic energy as a function of nuclear geometry (PES). V (kcal/mol) •Calculate motion along that PES (using classical or quantum H-Br separation (Å) mechanics).

Asymptotic Approach The product vibrational energy can be related to the location of the transition state. A + BC _ AB + C 0123401234 0123401234 r B-C r B-C r A-B r A-B

H + HBr Æ H 2 + Br •H 2 is described by two quantum numbers: v' , j' . v’ : • Those quantum numbers describe the asymptotic state. • We measure partial cross sections for forming individual quantum states: _ ( v' , j' ). Classical Barrier Height: 1.7 kcal/mol j’ : Reaction Exoergicity: 19.1 kcal/mol Total Reactive Cross Section: 1.2 Å 2

Experimental Protocol

Experimental Results H + HBr Æ H 2 ( v' =2, j' ) + Br

Possible State Distributions Purely Statistical ()() ,, vibrottrans PopvjDegeneracyvjDegenDegenDeg 121 vibrottranstrans DegenD

Different Approach Geometries Perpendicular Collinear

Kinematically Constrained State Distribution Valentini and co-workers: translational energy is needed to surmount the reaction barrier Perpendicular Collinear

Video of PES Video - Low >> Video - High >>

Kinematically Constrained State Distribution Max. Min. mass-weighted E int collinear PES E trans 221122 TQQ ʈÁ˜Á˜Ë¯ = 1ACABBC mmcos(mm)(mm)forthereactionA Q 1 E total The internal energy is ß(masses) kinematically constrained Q 2

C. A. Picconatto, A. Srivastava, and J. J. Valentini, J. Chem. Phys. 114, 1663-1671 (2001). The minimum translational energy of the products is 2transtrans EcosE ¢=b

Reaction ( E trans ) j ' max,meas j ' max,model j ' max ( E avail) H+HCl (37 kcal/mol) H 2 ( v '=0) 11 11 15 7 H 2 ( v '=1) 7 13 P. M. Aker, G. J. Germann, and J. J. Valentini, J. Chem. Phys. 90 , 4795 (1989).

Reaction ( E trans ) j ' max,meas j ' max,model j ' max ( E avail) H+HBr (37 kcal/mol) H 2 ( v '=0) 15 19 13 H 2 ( v '=1) 12 17 11 H 2 ( v '=2) 5 8 15 P. M. Aker, G. J. Germann, and J. J. Valentini, J. Chem. Phys. 90 , 4795 (1989).

Reaction ( E trans ) j ' max,meas j ' max,model j ' max ( E avail) H+HI (37 kcal/mol) H 2 ( v '=0) 17 19 23 H 2 ( v '=1) 17 17 21 H 2 ( v '=2) 15 15 19 H 2 ( v '=3) 11 11 17 H 2 ( v ’=4) 5 7 14 P. M. Aker, G. J. Germann, and J. J. Valentini, J. Chem. Phys. 96 , 2756 (1992).

H + HBr Æ H 2 ( v' =2, j' ) + Br E coll = 53.0 kcal/mol Statistical Kinematic Limit

Quasiclassical Trajectory Method (QCT) 0.511.521.522.53-100-80-60-40-100-80-60-40 H-H separation (Å) • Find potential quantum mechanically • Take derivative to yield forces • Solve Newton’s V (kcal/mol) equations of motion • Find position of each atom at all times H-Br separation (Å) • Bin into quantum states

Experimental and Theory H + HBr Æ H 2 ( v' =2, j' ) + Br E coll = 53.0 kcal/mol

H + HBr Æ H 2 ( v' =2, j' ) + Br E coll = 53.0 kcal/mol

Kinematically Forbidden Quantum States E coll = 53 kcal/mol

Kinematically Forbidden Quantum States Collinear Perpendicular H – Br separation H – Br separation H – H separation H – H separation

Transition State Geometry The transition state is the configuration for which potential energy reaches a _ maximum in a reactive trajectory. _ is the angle between the H–Br bond axis and line connecting the attacking H-atom with the HBr center of mass.

Quasiclassical Trajectory Calculations Kinematically Kinematically Allowed Excluded Cos ( _ ts ) Number H 2 Internal Energy (kcal/mol)

Spectroscopy of the Transition State Collinear: Kinematically Excluded Cos ( _ ts ) Kinematically Allowed Perpendicular H 2 Internal Energy (kcal/mol)

Mechanism for Forming Internally Cold H 2

Mechanism for Forming Internally Cold H 2 v’ = 1, j’ = 3 _ = 15° E int = 16.7 Transition State

Mechanism for Forming Internally Hot H 2

Mechanism for Forming Internally Hot H 2 v’ = 4, j’ = 13 _ = 70° E int = 70.3 Transition State

Direct Trajectory

Direct Trajectory v’ = 0, j’ = 2 _ = 19° E int = 11.6 Transition State

Direct Trajectory

Direct Trajectory v’ = 0, j’ = 14 _ = 25° E int = 38.2 Transition State

Direct Trajectory

Direct Trajectory v’ = 3, j’ = 5 _ = 37° E int = 47.0 Transition State

Medium-Lifetime Transition State

Medium-Lifetime Transition State v’ = 0, j’ = 1 _ = 86° E int = 8.2 Transition State

Medium-Lifetime Transition State

Medium-Lifetime Transition State v’ = 3, j’ = 9 _ = 80° E int = 56.1 Transition State

Long-Lifetime Transition State

Long-Lifetime Transition State v’ = 7, j’ = 3 _ = 22° E int = 73.8 Transition State

Long-Lifetime Transition State

Long-Lifetime Transition State v’ = 0, j’ = 19 _ = 113° E int = 56.1 Transition State

What Did We Learn? Using this principle, we Kinematically have identified two simple Excluded Cos ( _ ts ) pathways for the reaction H + HBr Æ H 2 ( v ’, j ’) + Br: Kinematically Allowed H 2 Internal Energy (kcal/mol) • Internally cold H 2 results • Internally hot H 2 results from collinear transition from bent transition states. states

Overarching Conclusion At sufficiently high energies some significant fraction of reactions do not proceed along or close to the minimum energy path. Prediction This behavior is general and more common than realized before.

Acknowledgments Drew Pomerantz, Jon Camden, Albert Chiou, Florian Ausfelder Navdeep Chawla William Hase National Science Foundation

H 2 CO + h ν Æ H 2 + CO: Production of Cold H 2 D. Townsend, S. A. Lahankar, S. K. Lee, S. D. Chambreau, A. G. Suits, X. Zhang, J. Rheinecker, L. B. Harding, and J. M. Bowman, Science 306 , 1158-1161 (2004).

H 2 CO + h ν Æ H 2 + CO: Production of Hot H 2 D. Townsend, S. A. Lahankar, S. K. Lee, S. D. Chambreau, A. G. Suits, X. Zhang, J. Rheinecker, L. B. Harding, and J. M. Bowman, Science 306 , 1158-1161 (2004).

O + CH 3 Six possible sets of exothermic products: _ H =-70 kcal/mol CH 3 + O Æ H + H 2 CO H 2 + HCO _ H =-84 H + HCOH _ H =-13 H 2 + COH _ H =-44 H + H 2 + CO _ H =-70 CH + H 2 O _ H =-10 Seakins and Leone (1992) reported the detection of CO (v) from this reaction using FTIR emission spectroscopy. They estimated the CO branching fraction to be 0.40 ± 0.10.

O + CH 3 Both experimental and theoretical studies confirm the existence of a CO producing channel in the reaction of CH 3 with O atoms [T. P. Marcy, R. R. Diaz, D. Heard, S. R. Leone, L. B. Harding, and S. J. Klippenstein, J. Phys. Chem. A 2001, 105, 8361-8369]. The mechanism involves the elimination of H 2 from an energy-rich CH 3 O radical forming HCO, followed by the decomposition of HCO to form the observed CO (v) product. The most unusual feature of this mechanism is that there appears to be no saddle point for the direct elimination of H 2 from CH 3 O.

O + CH 3 The methoxy radical is formed with 90 kcal/mol of excess energy, or 60 kcal/mol above its lowest barrier for decomposition (CH bond cleavage).