HO_CHEM2056_kinetics_2013_L4-6

26x106 lmol1 s1 2 4 2 6 na 6022x1023 mol1

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Unformatted text preview: sion Based on the observed value for the pre-exponential 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 cross-section * is the reactive cross-section * = 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 pre-exponential 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 = 7.0×10-19 J, Eea = 4.2×10-19 J, d = 400 pm = 4×10-10 m e = 1.6×10-19 C; ε0 = 8.85×10-12 J-1C2m-1 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? k-1 k1 Deactivation A + A* isomerisation k-1 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* k-1 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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