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CHEM1612 - Pharmacy Week 12: Kinetics – Catalysis

CHEM1612 - Pharmacy Week 12: Kinetics – Catalysis. Dr. Siegbert Schmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: siegbert.schmid@sydney.edu.au. Unless otherwise stated, all images in this file have been reproduced from:

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CHEM1612 - Pharmacy Week 12: Kinetics – Catalysis

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  1. CHEM1612 - PharmacyWeek 12: Kinetics – Catalysis Dr. SiegbertSchmid School of Chemistry, Rm 223 Phone: 9351 4196 E-mail: siegbert.schmid@sydney.edu.au

  2. Unless otherwise stated, all images in this file have been reproduced from: Blackman, Bottle, Schmid, Mocerino and Wille,Chemistry, John Wiley & Sons Australia, Ltd. 2008      ISBN: 9 78047081 0866

  3. Endothermic reaction Ea (rev) Ea (forw) Ea (forw) C + D Ea (rev) A + B Larger Ea smaller k lower rate Energy Landscape in Chemical Reactions A + B C + D Exothermic reaction Activated state A + B Figure from Silberberg, “Chemistry”, McGraw Hill, 2006. C + D Forward reaction is faster than reverse Reverse reaction is faster than forward

  4. By the way: why the product? 2×2 =4 elementary reaction A + B→ C rate = k [A][B] Why does rate depend on the product of reactant concentrations? Rateproportional to the number of collisions of A and B No. collisions = product of the number or particles present 3×2 =6 3×3=9

  5. Transition State • If reactants come together with enough energy and the right orientation, they combine to form a transition state (or activated complex). • This species is half-way between the reactants and the products but is not neither. • Transition states are very unstable (cannot be isolated). Blackman Figure 14.10

  6. CH3Br + OH- CH3OH + Br - Transition State Nature of the transition state in the reaction between CH3Br and OH-. Figure from Silberberg, “Chemistry”, McGraw Hill, 2006. transition state or activated complex

  7. Reaction Energy Diagram Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

  8. Transition state in elementary steps k = A e – Ea / R T Ea1 > Ea2, therefore Ea1 is the slow step and Ea2 is the fast step. Two transition states. Blackman Figure 14.11

  9. Transition state in elementary steps Step 1 NO2 + F2 →NO2F + F Step 2 NO2 + F → NO2F Overall 2 NO2 + F2 → 2NO2F Figure from Silberberg, “Chemistry”, McGraw Hill, 2006.

  10. Catalysis • A catalyst increases the rate of a chemical reaction. • A catalyst is not consumed or changed in the overall process. • A catalyst provides an alternative reactionpathway of lower activation energy: more molecules have the minimum energy required for successful reaction and the reaction proceeds at a faster rate. k A + BC + D

  11. Catalysis • A catalyst speeds both the forward and reverse reaction, so does NOT affect the position of the equilibrium. • Does not change the equilibrium constant Keq = k1/k-1; even though k1 and k-1 may be much larger for the catalyzed reaction. • Does not change the G0 for the reaction. • A catalyst can be homogeneous(one phase with reactants and products) or heterogeneous (more than one). • Many catalysed reactions are zero-order. • A small quantity of catalyst affects the reaction rate for a large amount of reactant.

  12. Demo: Catalytic Decomposition of Hydrogen Peroxide Manganese dioxide MnO2 is used to catalyse the decomposition of H2O2. Prepare a slurry of MnO2 and NaOH, and add H2O2. The rapid decomposition of H2O2 occurs with production of a grey foam: 2 H2O2 → 2 H2O + O2

  13. Catalysis • Let’s look at the decomposition of H2O2 again. 2 H2O2(aq) 2 H2O(l) + O2(g) Catalyst Ea (kJ mol-1) Rel. rate of reaction None 75.3 1 I– 56.5 2.0 x 103 Pt 49.0 4.1 x 104 Catalase 8.0 6.3 x 1011 Bombardier beetle (Brachinus fumans)

  14. Homogeneous Catalysis • Let’s closely examine the reaction of H2O2 with I–: • Rate law: rate = k[H2O2][I–]. • Reaction occurs in two steps: Step 1: H2O2 + I– H2O+ IO– Step 2: H2O2 + IO– H2O+ O2(g) +I– • Note that I– regenerated during reaction and it does not appear in overall reaction. • I– acts as a homogeneous catalyst for H2O2 decomposition.

  15. A B Heterogeneous Catalysis • Heterogeneous catalysis: most important industrially. • Catalytic converters: First converter (A): Rh Catalyses: 2 NO(g) N2(g) + O2(g) Second converter (B): Pt/Pd Catalyses: 2 CO(g) + O2(g) 2 CO2(g) Catalytic converter

  16. The metal-catalyzed formation of ammonia Fe N2(g) + 3H2(g) → 2NH3(g) • Both substrates must bind to a free active site on the Fe surface before the reaction can proceed. • Increasing the concentration of either gas cannot increase the rate of reaction (i.e. rate independent of concentration).

  17. Enzymes • Catalysts of biological reactions • Complex 3D structure • Huge molar mass • Active site attracts substrates through intermolecular forces • Haber process (500 atmand 450 °C; Nitrogenase(1 atm and 25°C) Enzyme-substrate complex of elastase and small peptide

  18. Active site Structure of the enzyme Hexokinase from X-ray data. Enzymatic Catalysis • All enzymes are proteins, but not all proteins are enzymes. • Enzymes mustpossess catalytic activity. • The part of the enzyme tertiary structure that is responsible for the catalytic activity is known as the “active site”. • Each enzyme catalyses a singlechemical reaction on a specific • substrate molecule with high selectivity.

  19. Michaelis-Menten mechanism: enzyme-substrate complex ES. Enzymatic Catalysis • Efficient: rate enhancements of 108 to 1020 possible • Specific (one enzyme per reaction) • Low tolerance to temperature and pH changes • Lock-and-key model (E. Fischer, 1894) • Induced fit model (D. Koshland, 1958) Product + enzyme

  20. Enzymatic Catalysis • Hexokinasealters its conformation to fit around the substrate molecule (D-glucose). • Enzyme and substrate adapt to accommodate one another. • “Enzymes are molecules that are complementary in structure to the transition states of the reactions they catalyze”. • Linus Pauling (1948)

  21. Enzymatic Catalysis • Enzymes can distinguish between enantiomers: • Only one of the enantiomers can be used as a substrate for this enzyme.

  22. Free energy E – E = E D a(uncat) a(cat) a Enzymatic Catalysis ….. uncatalyzed reaction catalyzed reaction Reaction co-ordinate If DEa= 10 kJ mol-155-fold rate acceleration (at 25°C). If DEa= 20 kJ mol-13000-fold rate acceleration (at 25°C). If DEa= 40 kJ mol-1107-fold rate acceleration (at 25°C).

  23. Enzymatic Catalysis k = A e – Ea / R T • The Arrhenius equation indicates that in order to increase the rate of a reaction: • The temperature must be increased, • Ea must be decreased, and/or • The reactants must be positioned so as to maximise the reaction efficiency. • Increasing the temperature is not an option for most biological reactions, so the remaining options are exploited by Nature.

  24. Summary CONCEPTS • Elementary reactions, reaction mechanisms • Dependence of reaction rate on temperature and orientation • Arrhenius equation and its implications • Activation energy and transition states • Catalysis CALCULATIONS • Express reaction rate in terms of reactant/product concentrations • Derive rate law of a reaction from experimental data on reactant consumption/product formation • Derive rate law of a reaction, knowing its elementary steps • Calculate k, A, Tor Ea for a reaction using Arrhenius equation

  25. The Ozone System • In the atmosphere: • O2 + hn (<200 nm)  O + O k1 • O + O2  O3 assume very fast • O3 + hn(210-300 nm)  O2 + O k2 • O3 + O  2 O2 k3 • Kinetics very complicated since UV intensity will vary so much in time and place. • At equilibrium, rate of ozone creation and destruction will be the same: steady state approximation.

  26. The Ozone Hole • Chlorofluorocarbons provide an additional pathway for ozone decomposition. • CF2Cl2 + hn CF2Cl• + Cl• • Cl• + O3  ClO• + O2 removing ozone • ClO• + O•  Cl• + O2 • Cl is regenerated during the reaction • Cl will stick around until eventually reacts to HCl and is precipitated out.

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