Unformatted text preview: sion Based on the observed value for the preexponential factor for the
following reaction, use the collision theory to estimate the steric factor at
628 K:
H +C H C H
2 A = 1.26 x 106 L mol‐1 s‐1 2 4 2 6 NA = 6.022x1023 mol‐1 (H2) = 0.27 nm2, (C2H4) = 0.64 nm2 T = 628 K kb = 1.380 × 10‐23 J K‐1 J = 1 kg m2 s‐2 Molar mass of H2 = 2.016 g mol‐1
Molar mass of C2H4 = 28.05 g mol‐1
To find mass of compound: multiply by 1.669x10‐27 mol = (1/NA)×10‐3 Collision crosssection * is the reactive crosssection
* = P (P ≤ 1)
L Grondahl CHEM2056 1/ 2 8k T A P N A b L Grondahl CHEM2056 Example: Estimating a steric factor (P)
Based on the observed value for the preexponential factor for the
following reaction, use the collision theory to estimate the steric factor at
628 K:
H +C H C H
2 2 4 2 6 Try problem 4 on ‘Practice problem sheet week 6’.
L Grondahl CHEM2056 L Grondahl CHEM2056 Harpoon mechanism; P > 1 Estimation of P
for the harpoon mechanism Gas‐phase reaction, P = 4.8
‐
K + Br2 K+ + Br2 KBr + Br e The harpoon extends the cross‐section
for the encounter of K and Br2 L Grondahl CHEM2056 Ionization energy, I, of K Electron affinity, Eea, of Br2 ‘harpoon line’:
Coulombic attraction between ions “harpoon” Calculation of distance at which it is energetically favourable for the electron to leap from K to Br2
Contributions to the energy of interaction (E) Coulombic interaction energy between the ions K+ and Br2 is: e2/4πε0r ‐ e = 1.6×10‐19 C (elementary charge); ε0 = 8.85×10‐12 J‐1C2m‐1 (vacuum permittivity); r is the separation between K+ and Br2‐ L Grondahl CHEM2056 Estimation of P Example calculation for: K + Br2 for the harpoon mechanism It can be shown that theseparation at which electron transfer can occur, r*, can be found from the following relationship I – Eea = e2/4πε0r* I = 7.0×1019 J, Eea = 4.2×1019 J, d = 400 pm = 4×1010 m e = 1.6×1019 C; ε0 = 8.85×1012 J1C2m1 Calculate P The reactive cross‐section for this process: The steric factor: 2 e2
P 4 d(I E ) 4.2 0
ea C2
Units: 1 2 1 J C m m( J ) * = πr*2 * ( r *)2 e2
P 2 4 d ( I E ) d
0
ea 2 d = r(K) + r(Br2) L Grondahl CHEM2056 Interpretation of the energy requirement
for 1st order reactions L Grondahl CHEM2056 The Lindemann Mechanism
Activation
k1
A + A A* + A If molecules in a unimolecular reaction gains sufficient enery through collision; then why is the kinetics not second order? k1 k1 Deactivation
A + A* isomerisation k1 A+A Unimolecular decay
k2 k2
A* P d [ A*] k1[ A]2 k 1[ A][ A*] k 2 [ A*]
dt
L Grondahl CHEM2056 A*: energized molecule L Grondahl CHEM2056 The Lindemann Mechanism The Lindemann Mechanism Using the steady‐state approximation on A*:
Activation
k1
A + A A* + A
Deactivation
A + A* k1 A+A At High pressure the molecule concentration is high and therefore the deactivation step will dominate, k1[ A]2
k1[ A] k2 d [ P]
k k [ A]2 k2 [ A*] 1 2
dt
k 1[ A] k2 Unimolecular decay
k2
A* HIGH vs LOW pressure (recall PA/RT = nA/V): d [ A*] k1[ A]2 k 1[ A][ A*] k 2 [ A*] 0
dt
[ A*] d [ P]
k k [ A]2 12
dt
k 1[ A] k2 k 1[ A] k 2 ; so
HIGH
LOW d [ P ] k1k 2 [ A] dt
k 1 FIRST ORDER At Low pressure the decay reaction will dominate P k 1[ A] k 2 ; so d [ P] k1[ A...
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This document was uploaded on 04/10/2014.
 Three '14
 Mole, Kinetics

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