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Chemical Kinetics

Chemical Kinetics. Rates of chemical reactions and how they can be measured experimentally and described mathematically. So far, we have worked with reactions that occur almost instantaneously. Precipitations Ba 2+ (aq) + SO 4 2- (aq)  BaSO 4 (s) Acid-base reactions

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Chemical Kinetics

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  1. Chemical Kinetics Rates of chemical reactions and how they can be measured experimentally and described mathematically

  2. So far, we have worked with reactions that occur almost instantaneously • Precipitations • Ba2+ (aq) + SO42- (aq)  BaSO4 (s) • Acid-base reactions • HCl(aq) + NaOH (aq) NaCl (aq)+ H2O (l)

  3. Lots of reactions are much slower… • Rusting • 4Fe (s) + 3O2 (g)  2Fe2O3 (s) • Formation of ammonia • N2 (g) + 3H2 (g)  2NH3 (g) • Formation of diamond • C (graphite)  C (diamond)

  4. In this chapter we will… • Explore the factors that affect rates of reactions • Quantify the influence of the above factors on the rates of reactions • Determine what happens at the molecular level in reactions

  5. REACTION RATES • The change in molar concentration of a reactant or a product per unit time • Units are Moles/L s

  6. In an experiment, the decomposition of N2O5 in 100 mL of a 2.00 M CCl4 solution produced 0.500 L of oxygen at STP after 200 minutes. Calculate the concentration of unreacted N2O5 at this time. 2N2O5 (in CCl4) 4NO2 (in CCl4) + O2 (g) • ? Moles oxygen produced • ? Moles N2O5 reacted • ? Moles N2O5 initially • ? Moles N2O5 left over • ? Moles/L

  7. 0.500 L x 1mole = 0.02232 mol O2 22.4 L • 0.02232 mol O2 x 2N2O5 = 0.04464 mol N2O5 1 O2 • Initially… (0.100L) (2.00M) = 0.200 mol N2O5 • 0.200 mol – 0.04464 mol = 0.155 mol left over • M = 0.155 mol/ 0.100 L = 1.55 M

  8. Following this procedure we can calculate the concentration of unreacted N2O5 at any stage of the reaction and plot the data as shown This gives the average rate over time, D[N2O5]/Dt To find the rate at any instant, draw a tangent to the curve and determine its slope Mol/L Time

  9. If you know the rate of one substance in a reaction, you can determine the rate of any other substance in the reaction by using a mole ratio 2N2O5 (in CCl4)  4NO2 (in CCl4) + O2 (g) N2O5 decomposes at a rate of 0.23M/s, what is the rate of formation of NO2 ? 0.23M N2O5 x 4 NO2 = 0.46 M/s NO2 s 2 N2O5

  10. Rate Laws • Have the general form: rate = k[A]n • [A] = concentration of reactants and catalyst • k = rate constant • n = order of reactant (an integer or fraction)

  11. Initial Rate Method • To determine the form of the rate law, one MUST use experimental data • Do a series of experiments in which the initial concentrations of the reactants are varied one at a time and record the initial rates of reaction

  12. (OH-) I- (aq) + OCl- (aq)  OI- (aq) + Cl- (aq) initial M Rate (mol/Ls) I- OCl- OH- • 0.01 0.01 0.01 6.1 x 10-4 • 0.02 0.01 0.01 12.2 x 10-4 • 0.01 0.02 0.01 12.3 x 10-4 • 0.01 0.01 0.02 3.0 x 10-4

  13. Rate = k[I-]x [OCl-]y [OH-]z 1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z 2: 12.2 x 10-4 = k[0.02]x[0.01]y[0.01]z Solve for x… 0.5 = 0.5x X = 1

  14. 1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z 3: 12.3 x 10-4 = k[0.01]x[0.02]y[0.01]z Solve for y 0.5 = 0.5y Y = 1

  15. 1: 6.1 x 10-4 = k[0.01]x[0.01]y[0.01]z 4: 3.0 x 10-4 = k[0.01]x[0.01]y[0.02]z Solve for z 2 = 0.5z (ohhh…this is tricky!) Log 2 = log 0.5z Z log 0.5 = log 2 Z = log 2/log 0.5 Z = -1

  16. Rate = k[I-][OCl-]/[OH-] The reaction is first order with respect to I- and OCl-, and inverse first order with respect to [OH-]. What is the overall reaction order? (sum of exponents) What are the units of k for this reaction? (plug units into the rate law and solve)

  17. YOU TRY! Do your “You Try!” section now.

  18. Integrated Rate Law Method First Order Rate Laws rate = k [A]1 = - D [ A ] / D t or rate = -D[A] = k D t [A]

  19. Integrating both sides gives… -ln ( [A]t / [A]0 ) = kt ln ( [A]0 / [A]t ) = kt ln [A]t = -kt + ln [A]0 slope = - k ln[A]t time

  20. Half-life • time it takes for the reactant concentration to reach one half of its initial value • symbol t ½ • for first order: t ½ = (ln 2) / k

  21. Integrated Rate Law Method Second Order Rate Laws rate = k [A]2 = - D [ A ] /D t integrating gives 1 / [A]t = kt + 1/[A]0 t1/2 = 1 / k[A]0

  22. 2nd order graph

  23. Integrated Rate Law Method Zero Order Rate Laws rate = k [A]0 = k integrating gives [A]t = -kt + [A]0 t1/2 = [A]0 / 2k

  24. Rate data was plotted, what is the order of NO2?

  25. What is the order with respect to A?

  26. YOU TRY! • Do your “You Try!” section now.

  27. Collision Theory The rate of a reaction depends on the 1. concentration of reactants 2. temperature 3. presence/absence of a catalyst

  28. The value of the rate constant is dependent on the temperature How can we explain the effect of temperature on the rate of a reaction?

  29. NO(g) + Cl2(g) --> NOCl(g) + Cl- (g) At 25oC: k= 4.9 x 10-6 L/mol s At 35oC: k= 1.5 x 10-5 L/mol s k has increased by a factor of THREE! WHY?

  30. The Collision Theory states that in order to react, molecules have to collide…. • with the proper orientation • with an energy at least equal to Ea activation energy (Ea ) = required minimum energy for a reaction to occur

  31. k for a reaction depends on 3 things: • Z = collision frequency (# collisions/second) higher temperature means more collisions • f = fraction of collisions that occur with E > Ea *this factor changes rapidly with T f = e –(Ea/RT) • p = fraction of collisions that occur with the proper orientation (independent of T)

  32. Overall: k = pfz = A e -(Ea/RT) • where A = pz • Arrhenius Equation: ln k = -Ea1 + ln A R T • ln k2 = Ea1 1 k1 R T1 T2

  33. YOU TRY! Calculate the activation energy for the reaction: 2HI (g) --> H2 (g) + I2 (g) GIVEN: k at 650. K = 2.15 x 10-8 L/mol s and k at 700. K = 2.39 x 10-7 L/mol s R = 8.3145 J / mol K 1.82 x 105 J

  34. Transition State Theory Two reactants come together to form an Activated Complex, or TRANSITION STATE which then separates to form the products.

  35. Potential Energy Diagram2NO + Cl2 2NOCl PE Reaction 

  36. In the activated complex, the N—Cl bond has partially formed, while the Cl—Cl bond has partially broken.

  37. Breaking bonds requires an input of energy while forming bonds RELEASES energy. • If E reactants > E products then the reaction is EXOTHERMIC • If E reactants < E products then the reaction is ENDOTHERMIC

  38. Catalysis Uncatalyzed Catalyzed

  39. A catalyst increases the rate of reaction by… LOWERING THE ACTIVATION ENERGY • Homogeneous Catalysis- catalyst is in the same phase as the reactants • Heterogeneous Catalysis- catalyst is in a different phase than the reactants

  40. The Haber Process: N2 + 3 H2 2NH3

  41. Reaction Mechanisms The overall balanced equation usually represents the SUM of a series of simple reactions called ELEMENTARY STEPS because they represent the progress of the reaction at the molecular level. The sequence of elementary steps is called the REACTION MECHANISM

  42. NO2 (g) + CO (g)  NO (g) + CO2 (g)Rate = k [NO2]2 The above reaction actually takes place in two steps: 1. NO2 + NO2 NO3 + NO SLOW 2. NO3 + CO  NO2 + CO2 FAST

  43. Intermediate- • Unimolecular step: • Bimolecular step: • Termolecular step:

  44. The reaction mechanism must satisfy two requirements: 1) Sum of elementary steps must be the overall reaction 2) The rate law indicated by the mechanism must match the experimentally determined rate law

  45. 2 H2O2(aq)  O2(g) + 2 H2O(l)by experiment Rate = k[H2O2] [I-] H2O2(aq) + I-(aq)  IO-(aq) + H2O(l) SLOW H2O2(aq) + IO-(aq)  I-(aq) + H2O(l) + O2(g)

  46. The overall rate of the reaction is controlled by the slow step also known as the….. RATE DETERMINING STEP

  47. H2O2(aq) + I-(aq)  IO-(aq) + H2O(l) SLOWH2O2(aq) + IO-(aq)  I-(aq) + H2O(l) + O2(g) When the slow step is used to determine the rate law, we get: What is the intermediate in this reaction? What is the catalyst?

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