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22.5 The temperature dependence of reaction rates

22.5 The temperature dependence of reaction rates

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22.5 The temperature dependence of reaction rates

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  1. 22.5 The temperature dependence of reaction rates • Arrhenius equation: A is the pre-exponential factor; Ea is the activation energy. The two quantities, A and Ea, are called Arrhenius parameters. • In an alternative expression lnk = lnA - one can see that the plot of lnk against 1/T gives a straight line.

  2. 25 20 15 Series1 10 5 0 0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 Example: Determining the Arrhenius parameters from the following data: T/K 300 350 400 450 500 k(L mol-1s-1) 7.9x106 3.0x107 7.9x107 1.7x108 3.2x108 Solution: 1/T (K-1) 0.00333 0.00286 0.0025 0.00222 0.002 lnk (L mol-1s-1) 15.88 17.22 18.19 18.95 19.58 The slope of the above plotted straight line is –Ea/R, so Ea = 23 kJ mol-1. The intersection of the straight line with y-axis is lnA, so A = 8x1010 L mol-1s-1

  3. The interpretation of the Arrhenius parameters • Reaction coordinate: the collection of motions such as changes in interatomic distance, bond angles, etc. • Activated complex • Transition state • For bimolecular reactions, the activation energy is the minimum kinetic energy that reactants must have in order to form products.

  4. Applications of the Arrhenius principle Temperature jump-relaxation method: consider a simple first order reaction: A ↔ B at equilibrium: After the temperature jump the system has a new equilibrium state. Assuming the distance between the current state and the new equilibrium state is x, one gets [A] = [A]eq + x; [B] = [B]eq - x;

  5. 22.6 Elementary reactions • Elementary reactions: reactions which involves only a small number of molecules or ions. A typical example: H + Br2 → HBr + Br • Molecularity: the number of molecules coming together to react in an elementary reaction. • Molecularity and the reaction order are different !!! Reaction order is an empirical quantity, and obtained from the experimental rate law; Molecularity refers to an elementary reaction proposed as an individual step in a mechanism. It must be an integral.

  6. An elementary bimolecular reaction has a second-order rate law: A + B → P • If a reaction is an elementary bimolecular process, then it has second-order reaction kinetics; However, if the kinetics are second-order, then the reaction might be complex.

  7. 22.7 Consecutive elementary reactions • An example: 239U → 239Np →239Pu • Consecutive unimolecular reaction A → B → C The rate of decomposition of A is: • The intermediate B is formed from A, but also decays to C. The net formation rate of B is therefore: • The reagent C is produced from the unimolecular decay of B:

  8. Integrated solution for the first order reaction (A) is: • Then one gets a new expression for the reactant B: the integrated solution for the above equation is: when assuming [B]0 = 0. • Based on the conservation law [A] + [B] + [C] = [A]0

  9. Example. In an industrial batch process a substance A produces the desired compound B that goes on to decay to a worthless product C, each step of the reaction being first-order. At what time will B be present in the greatest concentration? Solution: At the maximum value of B Using the equation 25.7.6 and taking derivatives with respect to t: In order to satisfy = 0 tmax = The maximum concentration of B can be calculated by plugging the tmax into the equation.

  10. Steady State Approximation • Assuming that after an initial induction period, the rates of change of concentrations of all reaction intermediates are negligibly small. • Substitute the above expression back to the rate law of B 0 ≈ [B] = (k1/ k2)[A] • Then • The integrated solution of the above equation is [C] ≈ (1 - )[A]0

  11. Self-test 22.8 Derive the rate law for the decomposition of ozone in the reaction 2O3(g) → 3O2(g) on the basis of the following mechanism O3 → O2 + O k1 O2 + O → O3 k1’ O + O3 → O2 + O2 k2 Solution: First write the rate law for the reactant O3 and the intermediate product O Applying the steady state approximation to [O] Plug the above relationship back to the rate law of [O3]

  12. Rate determining step

  13. Simplifications with the rate determining step • Suppose that k2 >> k1, then k2 – k1 ≈ k2 therefore concentration C can be reorganized as [C] ≈ (1 - )[A]0 • The above result is the same as obtained with steady state approximation

  14. Kinetic and thermodynamic control of reactions • Consider the following two reactions A + B → P1 rate of formation of P1 = k1[A][B] A + B → P2 rate of formation of P2 = k2[A][B] • [P1]/[P2] = k1/k2 represents the kinetic control over the proportions of products.

  15. Problems 22.6 The gas phase decomposition of acetic acid at 1189 K proceeds by way of two parallel reactions: • CH3COOH → CH4 + CO2, k1 = 3.74 s-1 • CH3COOH → H2C=C=O + H2O, k2 = 4.65 s-1 What is the maximum percentage yield of the ketene CH2CO obtainable at this temperature.

  16. Pre-equilibrium • Consider the reaction: A + B ↔ I → P when k1’ >> k2, the intermediate product, I, could reach an equilibrium with the reactants A and B. • Knowing that A, B, and I are in equilibrium, one gets: and • When expressing the rate of formation of the product P in terms of the reactants, we get

  17. Self-test 22.9: Show that the pre-equilibrium mechanism in which 2A ↔ I followed by I + B → P results in an overall third-order reaction. Solution: write the rate law for the product P Because I, and A are in pre-equilibrium so [I] = K [A]2 Therefore, the overall reaction order is 3.

  18. Kinetic isotope effect • Kinetic isotope effect: the decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope. • Primary kinetic isotope effect: the kinetic isotope effect observed when the rate-determining step requires the scission of a bond involving the isotope: with • Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product: with

  19. Kinetic isotope effect

  20. 22.8 Unimolecular reactions • The Lindemann-Hinshelwood mechanism A reactant molecule A becomes energetically excited by collision with another A molecule: A + A → A* + A The energized molecule may lose its excess energy by collision with another molecule: A + A* → A + A The excited molecule might shake itself apart to form products P A* → P The net rate of the formation of A* is

  21. If the reaction step, A + A → A* + A, is slow enough to be the rate determining step, one can apply the steady-state approximation to A*, so [A*] can be calculated as Then The rate law for the formation of P could be reformulated as Further simplification could be obtained if the deactivation of A* is much faster than A*  P, i.e., then in case

  22. The equation can be reorganized into • Using the effective rate constant k to represent • Then one has

  23. The Rice-Ramsperger-Kassel (RRK) model • Reactions will occur only when enough of required energy has migrated into a particular location in the molecule. • s is the number of modes of motion over which the energy may be dissipated, kb corresponds to k2

  24. The activation energy of combined reactions • Consider that each of the rate constants of the following reactions A + A → A* + A A + A* → A + A A* → P has an Arrhenius-like temperature dependence, one gets Thus the composite rate constant also has an Arrhenius-like form with activation energy, E = E1 + E2 – E1’ Whether or not the composite rate constant will increase with temperature depends on the value of E, if E > 0, k will increase with the increase of temperature

  25. Combined activation energy

  26. Theoretical problem 22.20 The reaction mechanism A2↔ A + A (fast) A + B → P (slow) Involves an intermediate A. Deduce the rate law for the reaction. • Solution: