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CHEM1612 - Pharmacy Week 11: Kinetics – Arrhenius Equation

CHEM1612 - Pharmacy Week 11: Kinetics – Arrhenius Equation. 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 11: Kinetics – Arrhenius Equation

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  1. CHEM1612 - PharmacyWeek 11: Kinetics – Arrhenius Equation 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. reaction intermediate Reaction Mechanism 2 NO2 (g) + F2 (g)  2 NO2F (g) reaction (1) • A reaction mechanism is a series of elementary reactions (or steps) that add up to give a detailed description of a chemical reaction Step 1 NO2 + F2 NO2F +F slow Step 2 NO2 + F  NO2F fast 2 NO2 + F2 2 NO2F Overall • An elementary reaction is not made up of simpler steps. • For elementary reactions the stoichiometric coefficients are equal to the exponents in the rate law: e.g. for step 2 the rate law is rate = k2 [NO2][F]. k1 k2

  4. Rate-Determining Step Overall reaction is: 2 NO2 (g) + F2 (g)  2 NO2F (g) • A rate-determining (or rate-limiting) step is an elementary reaction that is the slowest step in the mechanism. e.g. Step 1: NO2 + F2 NO2F + F (slow) • The exponents in the overall rate law(overall reaction) are the same as the stoichiometric coefficients of the species involved in the rate-limiting elementary process (if no intermediates are involved; otherwise their concentrations need to be expressed in terms of the reactants used), in this case : For the overall reaction (1) Rate = k [NO2][F2]

  5. Reaction Mechanism - Example 2 The following elementary steps constitute a proposed mechanism for a reaction: Step 1 A + B X fast k1[A][B] =k-1[X] Step 2 X + C Y slow Step 3 Y D fast • Overall reaction? A + B + C D • What are the reaction intermediates? X and Y • Show the mechanism is consistent with the rate law: rate = k [A] [B][C] Rate = k [X][C] = k1/k-1[A][B][C] = k [A] [B][C] k1 k-1 k2 k-2 k3 k-3

  6. Reaction Mechanism - Example 3 For the reaction: NO2 + CO  NO + CO2 • The experimentally determined rate equation is: rate = k[NO2]2 • Show the rate expression is consistent with the mechanism: Step 1 2 NO2N2O4 fast equilibrium Step 2 N2O4  NO + NO3 slow Step 3 NO3 + CO  NO2 + CO2 fast Overall k1 k-1 k2 k3

  7. Reaction Mechanism – Example 4 • A proposed mechanism for the reaction: 2NO(g) + Br2(g)  2NOBr(g) consists of two elementary reactions: NO + Br2 NOBr2 fast equilibrium NOBr2 + NO  2NOBr slow Task: Confirm that this mechanism is consistent with the stoichiometry of the reaction and the observed rate law: Rate = k[NO]2[Br2]

  8. The Oscillating Iodine Reaction • Three solutions are added together quickly: iodate IO3- in acid,malonic acid, and H2O2. • The resultant solution oscillates in colour several times until it finally stops. • The overall reaction is: • IO3- + 2 H2O2 +CH2(CO2H) + H+ → ICH(CO2H)2 + 2 O2 + 3 H2O

  9. Temperature vs. Rate • Low temperature: slow reactions Egg cooking time at 80oC ~ 30 min Bacterial growth at 4oC = slow • High temperature: faster reactions Egg cooking time at 100oC ~ 5 min Bacterial growth at 30oC = fast

  10. Collision Theory A+ B → C • Reactants must collide to react - Orientation and energy factors slow down the reaction • Not all collisions produce a reaction - Need enough energy • Not all collisions are effective, they need a particular orientation - Orientation does not depend on T

  11. Activated complex (TS) Ea Potential Energy A + B C+ D Activated complex (TS) Ea Potential Energy C+ D A + B Reaction Progress Activation Energy • Consider: A + BC + D • For reaction to occur: colliding molecules must have energy greater than Ea • Ea = activation energy • Increasing temperature increases no. of molecules with energy greater than Ea Exo Endo

  12. Orientation and Collision NO2Cl + Cl· → NO2 + Cl2 Only a small fraction of the possible collision geometries can result in a reaction. Orientation is independent of T.

  13. Collision Theory The reaction constant k depends on several factors: k = Z·p·f Z·p = A (frequency factor) f= exp(– Ea / RT ) where Ea = activation energy, Rgas constant, T temperature (K) Z: collision frequency p: orientation probability factor (the fraction of collisions with proper orientation)f: fraction of collisions with sufficient energy Ea Potential Energy Reactants Products

  14. Reaction Rate Depends on Temperature Chemiluminescent reaction The Cyalume stick glows more brightly in the hot water beaker because the reaction is faster. The glow lasts for longer in a cold beaker, because the reaction is slower.

  15. The Arrhenius Equation The Arrhenius equation describes the temperature dependence of the rate constant, k: k = A k ~ 0 k Ea

  16. Rate and Temperature • Consider the decomposition of H2O2: 2 H2O2(aq) 2 H2O(l) + O2(g) k(s–1) T(K) 7.77 x 10–6 298 • Rate increases exponentiallywith temperature. Rate constant, k Temperature

  17. Using Arrhenius Equation To calculate Ea, rearrange the Arrhenius equation (k = A e – Ea / R T) as: ln k = ln A – Ea / R T If k1 and k2 are the rate constants at two temperatures T1 and T2 respectively, then we can also calculate Ea easily: ln k1 = ln A – Ea / R T1 ln k2 = ln A – Ea / R T2 And subtracting the two, the term (ln A) disappears, so we get:

  18. Arrhenius Equation – Example 1 • For the reaction: • 2 NO2(g)  2NO(g) + O2(g) • The rate constant k = 1.00 · 10-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. • What are the values for A and k at 273 K? • What is the value of T when k = 1.00· 10-11 s-1? • Method: Make use of, and rearrange k = A e – Ea / R T

  19. Arrhenius Equation – Example 1 • (a) Find the value of A (independent of T): • A = keEa / R T • = 1.00· 10-10 s-1 · exp [111000 J mol-1 / (8.314 J mol-1 K –1· 300 K)] = 2.13 ·109 s-1 (three significant figures) • Calculate the value of k at 273 K: • k = 2.13 · 109 s-1exp (– 111000 J mol-1) / (8.314 J mol-1 K –1·273 K) = 1.23 · 10-12 s-1 (three significant figures) • Calculate the temperature when k = 1.00· 10-11s-1 • T = Ea / [R· ln (A/k)] = 111000 J mol-1 / (8.314·46.8) J mol-1 K-1 = 285 K (three significant figures)

  20. Distribution of Collision Energy Increasing the temperature increases the number of collisions with sufficient energy to react, i.e. with energy > Ea. Blackman Figure 14.13

  21. Endothermic reaction Ea (rev) Ea (forw) Ea (forw) C + D Ea (rev) A + B 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

  22. Arrhenius Equation – Example 2 If a reaction has a rate constant k of 2.0 · 10-5 s-1 at 20.0C and 7.32·10-5 s-1 at 30.00C, what is the activation energy ? Answer: ln {(2.0 · 10-5)/ (7.32 · 10-5)} = - (Ea/8.314) (1/293 - 1/303.00) Ea = 96 kJ mol-1 (TWO SIGNIFICANT FIGURES)

  23. Determining Ea Ideally you would require many more than two values to determine Ea. lnk = lnA – Ea / R T Plot lnk versus 1/T and get a line with a slope of –Ea/R.

  24. Arrhenius Equation – Example 3 • The rate constant of a particular reaction triples when the temperature is increased from 25 C to 35 C. Calculate the activation energy, Ea, for this reaction. ln (1/3) = - (Ea / 8.314)(1/298 - 1/308) -1.099 = - Ea(1.310 x 10-5) Ea = 8.4·104 J mol-1 or 84 kJ mol-1

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