HO_CHEM2056_kinetics_2013_L4-6

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern