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Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
Abc Kinetics
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Abc Kinetics
Abc Kinetics
Abc Kinetics
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Abc Kinetics

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  • 1. Kinetics Reaction Rates
  • 2. Kinetics • The study of reaction rates. • Thermodynamics will tell us whether or not the reaction will occur • Spontaneous reactions (ΔG = neg) are reactions that will happen - but we can’t tell how fast. • Diamond will spontaneously turn to graphite – eventually. • Kinetics is about how fast
  • 3. Review- Collision Model of Chemical Reactions Particles must collide with enough energy to break bonds EFFECTIVE COLLISIONS
  • 4. Factors that affect the rate of reaction Increase number of effective collisions  increase rate of reaction • Temperature • Concentration • Surface area • Pressure • Agitation • Catalyst
  • 5. Reaction Rate • Rate = Conc. of A at t2 -Conc. of A at t1 t2- t1 • Rate =∆[A] t • Change in concentration per unit time • Consider the reaction N2(g) + 3H2(g)  2NH3(g)
  • 6. • As the reaction progresses the concentration H2 goes down C o n c e n t r a t i o n Time [H[H22]] N2 + 3H2 → 2NH3 Δ[H2]/time = neg
  • 7. • As the reaction progresses the concentration N2 goes down 1/3 as fast Δ[H2]/time = neg C o n c e n t r a t i o n Time [H[H22]] [N[N22]] N2 + 3H2 → 2NH3 Δ[N2]/time = neg Δ[H2]/time = neg
  • 8. • As the reaction progresses the concentration NH3 goes up 2/3 times C o n c e n t r a t i o n Time [H[H22]] [N[N22]] [NH[NH33]] N2 + 3H2 → 2NH3 Δ[NH3]/time = pos
  • 9. Reaction Rates are always positive • Over time – the concentration of products increases – the concentration of reactants decreases • So Δ[P]/time is positive • But Δ[R]/time is negative, so the rate would be expressed as - Δ[R]/time
  • 10. If rate is measured in terms of concentration of Hydrogen -Δ[H2] - mol H2 time L time and N2 + 3H2  2NH3 Then using the stoichiometry of the reaction -Δmol H2 1 mol N2 L time 3 mol H2 Then -Δmol H2 2 mol NH3 L time 3 mol H2
  • 11. Calculating Rates • Average rates are taken over a period of time interval • Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time
  • 12. Average Rate – Find the slope of the line between the two points Note: slope = C o n c e n t r a t i o n Time [H[H22]] tt Note: slope = Δy / Δx Δconcentration / Δtime THE RATE
  • 13. Instantaneous Rate – Find the slope of the line tangent to that point C o n c e n t r a t i o n Time ∆∆[H[H22]]  tt d[Hd[H22]] dtdt
  • 14. 12_291 0.0003 70s O2 0.0025 0.005 0.0075 0.0100 0.0006 70s 0.0026 110 s NO2 NO 50 100 150 200 250 300 350 400 Concentrations(mol/L) Time (s) ∆[NO2 ] ∆t INSTANTANEOUS RATES 2NO2  2NO + O2
  • 15. C4H9Cl + H2O → C4H9OH + HCl
  • 16. Rate Laws • Reactions are reversible. • As products accumulate they can begin to turn back into reactants. • Early on the rate will depend on only the amount of reactants present. • We want to measure the reactants as soon as they are mixed. • In addition, as the reaction progresses the concentration of reactants decrease • This is called the initial rate method.
  • 17. • Rate laws show the effect of concentration on the rate of the reaction • The concentration of the products do not appear in the rate law because this is an initial rate. • The order (exponent) –must be determined experimentally, –cannot be obtained from the equation Rate LawsRate Laws
  • 18. • You will find that the rate will only depend on the concentration of the reactants. (Initially) • Rate = k[NO2]n • This is called a rate law expression. • k is called the rate constant. • n is the order of the reactant -usually a positive integer. 2 NO2 2 NO + O2
  • 19. NO2  NO + ½ O2 The data for the reaction at 300ºC Plot the data to show concentration as a function of time. Time (s) [NO2] (mol/L) 0.00 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380
  • 20. NO2  NO + ½ O2 To determine the rate law for the reaction, we must find the exponent in the equation. Rate = k [NO2]x Experiment [NO2] M Rate, M s-1 1 0.060 0.0360 2 0.030 0.0090 3 0.020 0.0040
  • 21. NO2  NO + ½ O2 To determine the rate law for the reaction, we must find the exponent in the equation. Rate = k [NO2]x Exp [NO2] M Rate, M s-1 1 0.060 0.0360 2 0.030 0.0090 3 0.020 0.0040 2x [NO2] 4x rate
  • 22. NO2  NO + ½ O2 The exponent for the concentration of NO2 must be 2 to show the effect of changing the concentration of NO2 on the rate. Rate = k [NO2]2 Exp [NO2] M Rate, M s-1 1 0.060 0.0360 2 0.030 0.0090 3 0.020 0.0040 3x [NO2] 9x rate
  • 23. NO2  NO + ½ O2 Rate = k [NO2]x Pick two experiments to plug into the equation. Experiment [NO2] M Rate, M s-1 1 0.060 0.0360 2 0.030 0.0090 3 0.020 0.0040
  • 24. NO2  NO + ½ O2 Use the rate and one of the experiments to calculate the rate constant. Experiment [NO2] M Rate, M s-1 1 0.060 0.0360 2 0.030 0.0090 3 0.020 0.0040
  • 25. 2ClO2 + 2OH- → ClO3 - + ClO2 - + H2O • Determine the rate law for the reaction. • Calculate the rate constant. • What is the overall reaction order? Experiment [ClO2 - ] M [OH- ] M Rate, M s-1 1 0.060 0.030 0.0248 2 0.020 0.030 0.00276 3 0.020 0.090 0.00828
  • 26. Types of Rate Laws • Differential Rate Law - describes how rate depends on concentration. Method to determine- change initial concentration and measure effect on rate • Integrated Rate Law - describes how concentration depends on time. Method to determine- measure the concentration of reactants as function of time • For each type of differential rate law there is an integrated rate law and vice versa.
  • 27. Determining Rate Laws • The first step is to determine the form of the rate law (especially its order). • Must be determined from experimental data. • For this reaction 2 N2O5 (aq) 4NO2 (aq) + O2(g) • The reverse reaction won’t play a role because the gas leaves
  • 28. [N[N22OO55] (mol/L)] (mol/L) Time (s)Time (s) 1.001.00 00 0.880.88 200200 0.780.78 400400 0.690.69 600600 0.610.61 800800 0.540.54 10001000 0.480.48 12001200 0.430.43 14001400 0.380.38 16001600 0.340.34 18001800 0.300.30 20002000 Now graph the data
  • 29. 0 0.2 0.4 0.6 0.8 1 1.2 0 200 400 600 800 1000 1200 1400 1600 1800 2000 • To find rate we have to find the slope at two points • We will use the tangent method. [N[N22OO55]] (mol/L)(mol/L) Time (s)
  • 30. 0 0.2 0.4 0.6 0.8 1 1.2 0 200 400 600 800 1000 1200 1400 1600 1800 2000 At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4 (200 - 600) -400
  • 31. 0 0.2 0.4 0.6 0.8 1 1.2 0 200 400 600 800 1000 1200 1400 1600 1800 2000 At .40 M the rate is (.52 - .32) = 0.20 =- 2.5 x 10 -4 (1000-1800) -800
  • 32. • At 0.8 M rate is 5.0 x 10-4 • At 0.4 M rate is 2.5 x 10-4 • Since the rate is twice as fast when the concentration is twice as big the rate law must be.. First power • Rate = -[N2O5] = k[N2O5]1 = k[N2O5] t • We say this reaction is first order in N2O5 • The only way to determine order is to run the experiment.
  • 33. The method of Initial Rates • This method requires that a reaction be run several times. • The initial concentrations of the reactants are varied. • The reaction rate is measured just after the reactants are mixed. • Eliminates the effect of the reverse reaction.
  • 34. An example • For the reaction BrO3 - + 5 Br- + 6H+  3Br2 + 3 H2O • The general form of the Rate Law is Rate = k[BrO3 - ]n [Br- ]m [H+ ]p • We use experimental data to determine the values of n,m,and p
  • 35. Initial concentrations (M) Rate (M/s) BrOBrO33 -- BrBr-- HH++ 0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10-4-4 0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10-3-3 0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10-3-3 0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10-3-3 Calculate the rate law and k
  • 36. Integrated Rate Law • Expresses the reaction concentration as a function of time. • Form of the equation depends on the order of the rate law (differential). • Changes Rate = ∆[A]n ∆t • We will only work with n = 0, 1, and 2
  • 37. First Order • For the reaction 2N2O5  4NO2 + O2 • We found the Rate = k[N2O5]1 • If concentration doubles rate doubles. • If we integrate this equation with respect to time we get the Integrated Rate Law [N2O5]0 = initial concentration of N2O5 [N2O5]t = concentration of N2O5 after some time (t) ln = natural log = - ktln ( )[N2O5]0 [N2O5]t
  • 38. First Order rate = k[N2O5] • For the reaction 2N2O5  4NO2 + O2 • Rearrange this equation to y = mx + b form – Since- log of quotient equals difference of logs • SO. ln[N2O5]t = - k t + ln[N2O5]0 - ln[N2O5]t=ln ( )[N2O5]0 [N2O5]t = ktln ( )[N2O5]0 [N2O5]t ln[N2O5]t y mx b+=
  • 39. First Order • Linear form integrated rate law • ln[N2O5]t = - k t + ln[N2O5]0 = ktln ( )[N2O5]0 [N2O5]t y mx b+= ln[N2O5]t t m = slope = -k
  • 40. • Graphing natural log of concentration of reactant as a function of time • If you get a straight line, you may determine it is first order • Integrated First Order First Order = - ktln ( )[R]0 [R]t
  • 41. Half Life • The time required to reach half the original concentration. • [R]t = [R]0/2 when t = t1/2 • If the reaction is first order = - ktln ( )[R]0 [R]t Since [R]t = [R]0/2 [R]t = [R]0 equals 2 ln (2) = kt1/2
  • 42. Half Life • t1/2 = 0.693 / k • The time to reach half the original concentration does not depend on the starting concentration. • An easy way to find k
  • 43. • The highly radioactive plutonium in nuclear waste undergoes first-order decay with a half-life of approximately 24,000 years. How many years must pass before the level of radioactivity due to the plutonium falls to 1/128th (about 1%) of its original potency?
  • 44. Second Order • Rate = -∆[R]/∆t = k[R]2 • integrated rate law • 1/[R]t = kt + 1/[A]0 • y= 1/[A] m = k • x= t b = 1/[A]0 • A straight line if 1/[A] vs t is graphed • Knowing k and [A]0 you can calculate [A] at any time t
  • 45. Second Order Rate = k [R]2 integrated rate law 1/[R]t = k t + 1/[A]0 y = m x + b t 1/[R] Slope = k
  • 46. Second Order Half Life • [A] = [A]0 /2 at t = t1/2 1 2 0 2[ ]A = kt + 1 [A]1 0 2 2 [ [A] - 1 A] = kt 0 0 1 t k[A]1 = 1 0 2 1 [A] = kt 0 1 2
  • 47. Zero Order Rate Law • Rate = k[A]0 = k • Rate does not change with concentration. • Integrated [A] = -kt + [A]0 • When [A] = [A]0 /2 t = t1/2 • t1/2 = [A]0 /2k
  • 48. • Most often when reaction happens on a surface because the surface area stays constant. • Also applies to enzyme chemistry. Zero Order Rate Law
  • 49. Time C o n c e n t r a t i o n
  • 50. Time C o n c e n t r a t i o n ∆[A]/∆t ∆t k = ∆[A]
  • 51. Zero order First order Second order Rate law Graph [A] vs time Integrated rate law
  • 52. Zero order First order Second order Linear graph with time Integrated rate law Rearrange to y=mx+b
  • 53. Summary of Rate Laws
  • 54. Reaction Mechanisms • The series of steps that actually occur in a chemical reaction. • Kinetics can tell us something about the mechanism • A balanced equation does not tell us how the reactants become products.
  • 55. • 2NO2 + F2 2NO2F Rate = k[NO2][F2] • The overall equation is a summary. • In order for this reaction to occur as written, three molecules must collide simultaneously • An unlikely event. • So a reaction mechanism is proposed showing a series of likely steps Reaction Mechanisms
  • 56. • 2NO2 + F2 2NO2F • Rate = k[NO2][F2] • The proposed mechanism is • NO2 + F2  NO2F + F (slow) • F + NO2  NO2F (fast) • F is called an intermediate It is formed then consumed in the reaction Reaction Mechanisms
  • 57. • Each of the two reactions is called an elementary step . • The rate for a reaction can be written from its molecularity. The reaction rate reflects the stoichimetry. • Molecularity is the number of pieces that must come together. • Elementary steps add up to the balanced equation Reaction Mechanisms
  • 58. • Unimolecular step involves one molecule - Rate is first order. • Bimolecular step - requires two molecules - Rate is second order • Termolecular step- requires three molecules - Rate is third order • Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.
  • 59. • A  products Rate = k[A] • A+A  products Rate= k[A]2 • 2A  products Rate= k[A]2 • A+B  products Rate= k[A][B] • A+A+B  products Rate= k[A]2 [B] • 2A+B  products Rate= k[A]2 [B] • A+B+C  products Rate= k[A][B][C] Molecularity and Rate Laws For an elementary step – the rate law reflects the stoichiometry
  • 60. Reaction Mechanisms • Proposed series of elementary steps. – Describing each collision • The rate of the reaction is determined by the rate limiting step. • The overall experimentally determined rate law must be consistent with the rate law that reflects the rate limiting step.
  • 61. • The proposed mechanism is NO2 + F2  NO2F + F (slow) rate = k [NO2][F2] F + NO2  NO2F (fast) rate = k [F][NO2] Since the first step is the slow step, it is the rate limiting step and thus the rate of the overall reaction depends upon the rate of this step. Note: The rate of this step is also the rate of the overall reaction. 2NO2 + F2  2NO2F Rate = k[NO2][F2]
  • 62. The steps must add up to the overall reaction. • 2 NO2 Cl → 2 NO2 + Cl2 • Proposed steps • NO2 Cl → NO2 + Cl (slow) NO2 Cl + Cl → NO2 + Cl2 (fast) • Write the overall rate law that would be consistent with this proposed reaction mechanism.
  • 63. How to get rid of intermediates • They can’t appear in the rate law. • Slow step determines the rate and the rate law • Use the reactions that form them • If the reactions are fast and irreversible the concentration of the intermediate is based on stoichiometry. • If it is formed by a reversible reaction set the rates equal to each other.
  • 64. Formed in reversible reactions • 2 NO + O2 2 NO2 • Mechanism • 2 NO N2O2 (fast) • N2O2 + O2 2 NO2 (slow) • rate = k2[N2O2][O2] • k1[NO]2 = k1[N2O2] (equilibrium) • Rate = k2 (k1/ k-1)[NO]2 [O2] • Rate =k [NO]2 [O2]
  • 65. Formed in fast reactions • 2 IBr I2+ Br2 • Mechanism • IBr I + Br (fast) • IBr + Br I + Br2 (slow) • I + I I2 (fast) • Rate = k[IBr][Br] but [Br]= [IBr] because the first step is fast • Rate = k[IBr][IBr] = k[IBr]2
  • 66. Collision theory • Molecules must collide to react. • Concentration affects rates because collisions are more likely. • Must collide hard enough. • Temperature and rate are related. • Only a small number of collisions produce reactions.
  • 67. P o t e n t i a l E n e r g y Reaction Coordinate Reactants Products
  • 68. P o t e n t i a l E n e r g y Reaction Coordinate Reactants Products Activation Energy Ea
  • 69. P o t e n t i a l E n e r g y Reaction Coordinate Reactants Products Activated complex
  • 70. P o t e n t i a l E n e r g y Reaction Coordinate Reactants Products ∆E }
  • 71. P o t e n t i a l E n e r g y Reaction Coordinate 2BrNO 2NO + Br Br---NO Br---NO 2 Transition State
  • 72. Terms • Activation energy - the minimum energy needed to make a reaction happen. • Activated Complex or Transition State - The arrangement of atoms at the top of the energy barrier.
  • 73. Arrhenius • Said that reaction rate should increase with temperature. • At high temperature more molecules have the energy required to get over the barrier. • The number of collisions with the necessary energy increases exponentially.
  • 74. Arrhenius • Number of collisions with the required energy = ze-Ea/RT • z = total collisions • e is Euler’s number (inverse of ln) • Ea = activation energy • R = ideal gas constant (in J/K mol) • T is temperature in Kelvin
  • 75. Problem with this • Observed rate is too small • Due to molecular orientation- they have to be facing the right way • written into equation as p the steric factor.
  • 76. O N Br O N Br O N Br ONBr ONBr O NBr O N BrONBr No Reaction O NBr O NBr
  • 77. Arrhenius Equation •k = zpe-Ea/RT = Ae-Ea/RT • A is called the frequency factor = zp • k is the rate constant • ln k = -(Ea/R)(1/T) + ln A • Another line !!!! • ln k vs 1/T is a straight line • With slope Ea/R so we can find Ea • And intercept ln A
  • 78. Arrhenius Equation- Another line !!!! ln k = -(Ea/R)(1/T) + ln A y = m x + b ln k 1/T slope = -(Ea/R)
  • 79. Activation Energy and Rates The final saga
  • 80. Mechanisms and rates • There is an activation energy for each elementary step. • Activation energy determines k. • k = Ae- (E a/RT) • k determines rate • Slowest step (rate determining) must have the highest activation energy.
  • 81. • This reaction takes place in three steps
  • 82.  Ea First step is fast Low activation energy
  • 83. Second step is slow High activation energy Ea 
  • 84. Ea Third step is fast Low activation energy 
  • 85. Second step is rate determining
  • 86. Intermediates are present
  • 87. Activated Complexes or Transition States
  • 88. Catalysts • Speed up a reaction without being used up in the reaction. • Enzymes are biological catalysts. • Homogenous Catalysts are in the same phase as the reactants. • Heterogeneous Catalysts are in a different phase as the reactants.
  • 89. How Catalysts Work • Catalysts allow reactions to proceed by a different mechanism - a new pathway. • New pathway has a lower activation energy. • More molecules will have this activation energy. • Does not change E • Show up as a reactant in one step and a product in a later step
  • 90. Pt surface HH HH HH HH • Hydrogen bonds to surface of metal. • Break H-H bonds Heterogenous Catalysts
  • 91. Pt surface HH HH Heterogenous Catalysts C HH C HH
  • 92. Pt surface HH HH Heterogenous Catalysts C HH C HH • The double bond breaks and bonds to the catalyst.
  • 93. Pt surface HH HH Heterogenous Catalysts C HH C HH • The hydrogen atoms bond with the carbon
  • 94. Pt surface H Heterogenous Catalysts C HH C HH H HH
  • 95. Homogenous Catalysts • Chlorofluorocarbons (CFCs) catalyze the decomposition of ozone. • Enzymes regulating the body processes. (Protein catalysts)
  • 96. Catalysts and rate • Catalysts will speed up a reaction but only to a certain point. • Past a certain point adding more reactants won’t change the rate. • Zero Order
  • 97. Catalysts and rate. Concentration of reactants R a t e • Rate increases until the active sites of catalyst are filled. • Then rate is independent of concentration

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