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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 ( Δ G = 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 t 2 -Conc. of A at t 1 t 2 - t 1
    • Rate =   [A]  t
    • Change in concentration per unit time
    • Consider the reaction
    • N 2 (g) + 3H 2 (g)  2NH 3 (g)
  • 6.
    • As the reaction progresses the concentration H 2 goes down
    Concentration Time [H 2 ] N 2 + 3H 2 -> 2NH 3 Δ [H 2 ]/time = neg
  • 7.
    • As the reaction progresses the concentration N 2 goes down 1/3 as fast
    Δ [H 2 ]/time = neg Concentration Time [H 2 ] [N 2 ] N 2 + 3H 2 -> 2NH 3 Δ [N 2 ]/time = neg Δ [H 2 ]/time = neg
  • 8.
    • As the reaction progresses the concentration NH 3 goes up 2/3 times
    Concentration Time [H 2 ] [N 2 ] [NH 3 ] N 2 + 3H 2 -> 2NH 3 Δ [NH 3 ]/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 [H 2 ] - mol H 2
    • time L time
    • and N 2 + 3H 2  2NH 3
    • Then using the stoichiometry of the reaction
    • - Δ mol H 2 1 mol N 2
    • L time 3 mol H 2
    • Then
    • - Δ mol H 2 2 mol NH 3
    • L time 3 mol H 2
  • 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 = Concentration Time  [H 2 ]  t Note: slope = Δ y / Δ x Δ concentration / Δ time THE RATE
  • 13.
    • Instantaneous Rate – Find the slope of the line tangent to that point
    Concentration Time  [H 2 ]  t d[H 2 ] dt
  • 14. INSTANTANEOUS RATES 2NO 2  2NO + O 2
  • 15.  
  • 16.  
  • 17. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + HCl
  • 18. 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 .
  • 19.
    • 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 Laws
  • 20.
    • You will find that the rate will only depend on the concentration of the reactants. (Initially)
    • Rate = k [NO 2 ] 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 NO 2 2 NO + O 2
  • 21. NO 2  NO + ½ O 2
    • The data for
    • the reaction at 300 ºC
    • Plot the data to show concentration as a function of time.
    0.00380 300.0 0.00481 200.0 0.00649 100.0 0.00787 50.0 0.01000 0.00 [NO 2 ] (mol/L) Time (s)
  • 22. NO 2  NO + ½ O 2
    • To determine the rate law for the reaction, we must find the exponent in the equation.
    • Rate = k [NO 2 ] x
    0.0040 0.020 3 0.0090 0.030 2 0.0360 0.060 1 Rate, M s -1 [NO 2 ] M Experiment
  • 23. NO 2  NO + ½ O 2
    • To determine the rate law for the reaction, we must find the exponent in the equation.
    • Rate = k [NO 2 ] x
    2x [NO 2 ] 4x rate 0.0040 0.020 3 0.0090 0.030 2 0.0360 0.060 1 Rate, M s -1 [NO 2 ] M Exp
  • 24. NO 2  NO + ½ O 2
    • The exponent for the concentration of NO 2 must be 2 to show the effect of changing the concentration of NO 2 on the rate.
    • Rate = k [NO 2 ] 2
    3x [NO 2 ] 9x rate 0.0040 0.020 3 0.0090 0.030 2 0.0360 0.060 1 Rate, M s -1 [NO 2 ] M Exp
  • 25. NO 2  NO + ½ O 2
    • Rate = k [NO 2 ] x
    • Pick two experiments to plug into the equation.
    0.0040 0.020 3 0.0090 0.030 2 0.0360 0.060 1 Rate, M s -1 [NO 2 ] M Experiment
  • 26. NO 2  NO + ½ O 2
    • Use the rate and one of the experiments to calculate the rate constant.
    0.0040 0.020 3 0.0090 0.030 2 0.0360 0.060 1 Rate, M s -1 [NO 2 ] M Experiment
  • 27. 2ClO 2 + 2OH -  ClO 3 - + ClO 2 - + H 2 O
    • Determine the rate law for the reaction.
    • Calculate the rate constant.
    • What is the overall reaction order?
    0.00828 0.090 0.020 3 0.00276 0.030 0.020 2 0.0248 0.030 0.060 1 Rate, M s -1 [OH - ] M [ClO 2 - ] M Experiment
  • 28. 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.
  • 29. 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 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g)
    • The reverse reaction won’t play a role because the gas leaves
  • 30.
    • Now graph the data
    [N 2 O 5 ] (mol/L) Time (s) 1.00 0 0.88 200 0.78 400 0.69 600 0.61 800 0.54 1000 0.48 1200 0.43 1400 0.38 1600 0.34 1800 0.30 2000
  • 31.
    • To find rate we have to find the slope at two points
    • We will use the tangent method.
    [N 2 O 5 ] (mol/L) Time (s)
  • 32. At .80 M the rate is (.88 - .68) = 0.20 =- 5.0x 10 -4 (200 - 600) -400
  • 33. At .40 M the rate is (.52 - .32) = 0.20 =- 2.5 x 10 -4 (1000-1800) -800
  • 34.
    • 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 = -  [N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ]  t
    • We say this reaction is first order in N 2 O 5
    • The only way to determine order is to run the experiment.
  • 35. 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.
  • 36. An example
    • For the reaction BrO 3 - + 5 Br - + 6H +  3Br 2 + 3 H 2 O
    • The general form of the Rate Law is Rate = k[BrO 3 - ] n [Br - ] m [H + ] p
    • We use experimental data to determine the values of n,m,and p
  • 37.
    • Calculate the rate law and k
    Initial concentrations (M) Rate (M/s) BrO 3 - Br - H + 0.10 0.10 0.10 8.0 x 10 -4 0.20 0.10 0.10 1.6 x 10 -3 0.20 0.20 0.10 3.2 x 10 -3 0.10 0.10 0.20 3.2 x 10 -3
  • 38. 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
  • 39. First Order
    • For the reaction 2N 2 O 5  4NO 2 + O 2
    • We found the Rate = k[N 2 O 5 ] 1
    • If concentration doubles rate doubles.
    • If we integrate this equation with respect to time we get the Integrated Rate Law
    [N 2 O 5 ] 0 = initial concentration of N 2 O 5 [N 2 O 5 ] t = concentration of N 2 O 5 after some time (t) l n = natural log = - kt l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t
  • 40. First Order rate = k[N 2 O 5 ]
    • For the reaction 2N 2 O 5  4NO 2 + O 2
    • Rearrange this equation to y = mx + b form
      • Since- log of quotient equals difference of logs
    • SO. l n[N 2 O 5 ] t = - k t + l n[N 2 O 5 ] 0
    - l n[N 2 O 5 ] t l n[N 2 O 5 ] t y mx b + = = l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t = kt l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t
  • 41. First Order
    • Linear form integrated rate law
    • l n[N 2 O 5 ] t = - k t + l n[N 2 O 5 ] 0
    y m x b + = l n[N 2 O 5 ] t t m = slope = -k = kt l n ( ) [N 2 O 5 ] 0 [N 2 O 5 ] t
  • 42.
    • 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 = - kt l n ( ) [R] 0 [R] t
  • 43. Half Life
    • The time required to reach half the original concentration.
    • [R] t = [R] 0 /2 when t = t 1/2
    • If the reaction is first order
    Since [R] t = [R] 0 /2 [R] t = [R] 0 equals 2 l n (2) = kt 1/2 = - kt l n ( ) [R] 0 [R] t
  • 44. Half Life
    • t 1/2 = 0.693 / k
    • The time to reach half the original concentration does not depend on the starting concentration.
    • An easy way to find k
  • 45.
    • 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?
  • 46. 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
  • 47. 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
  • 48. Second Order Half Life
    • [A] = [A] 0 /2 at t = t 1/2
  • 49. 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 = t 1/2
    • t 1/2 = [A] 0 /2k
  • 50.
    • Most often when reaction happens on a surface because the surface area stays constant.
    • Also applies to enzyme chemistry.
    Zero Order Rate Law
  • 51. Time Concentration
  • 52. Time Concentration  A]/  t  t k =  A]
  • 53. Rate law Graph [A] vs time Integrated rate law Second order First order Zero order
  • 54. Linear graph with time Integrated rate law Rearrange to y=mx+b Second order First order Zero order
  • 55. Summary of Rate Laws
  • 56. 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.
  • 57.
    • 2NO 2 + F 2 2NO 2 F
      • Rate = k[NO 2 ][F 2 ]
    • 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
  • 58.
    • 2NO 2 + F 2 2NO 2 F
    • Rate = k[NO 2 ][F 2 ]
    • The proposed mechanism is
    • NO 2 + F 2  NO 2 F + F (slow)
    • F + NO 2  NO 2 F (fast)
    • F is called an intermediate It is formed then consumed in the reaction
    Reaction Mechanisms
  • 59.
    • 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
  • 60.
    • 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.
  • 61.
    • 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
  • 62. 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.
  • 63.
    • The proposed mechanism is
    • NO 2 + F 2  NO 2 F + F (slow) rate = k [NO 2 ][F 2 ]
    • F + NO 2  NO 2 F (fast) rate = k [F][NO 2 ]
    • 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.
    2NO 2 + F 2  2NO 2 F Rate = k[NO 2 ][F 2 ]
  • 64. The steps must add up to the overall reaction.
    • 2 NO 2 Cl -> 2 NO 2 + Cl 2
    • Proposed steps
    • NO 2 Cl -> NO 2 + Cl (slow) NO 2 Cl + Cl -> NO 2 + Cl 2 (fast)
    • Write the overall rate law that would be consistent with this proposed reaction mechanism.
  • 65. 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.
  • 66. Formed in reversible reactions
    • 2 NO + O 2 2 NO 2
    • Mechanism
    • 2 NO N 2 O 2 (fast)
    • N 2 O 2 + O 2 2 NO 2 (slow)
    • rate = k 2 [N 2 O 2 ][O 2 ]
    • k 1 [NO] 2 = k 1 [N 2 O 2 ] (equilibrium)
    • Rate = k 2 (k 1 / k -1 )[NO] 2 [O 2 ]
    • Rate =k [NO] 2 [O 2 ]
  • 67. Formed in fast reactions
    • 2 IBr I 2 + Br 2
    • Mechanism
    • IBr I + Br (fast)
    • IBr + Br I + Br 2 (slow)
    • I + I I 2 (fast)
    • Rate = k[IBr][Br] but [Br]= [IBr] because the first step is fast
    • Rate = k[IBr][IBr] = k[IBr] 2
  • 68. 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.
  • 69. Potential Energy Reaction Coordinate Reactants Products
  • 70. Potential Energy Reaction Coordinate Reactants Products Activation Energy E a
  • 71. Potential Energy Reaction Coordinate Reactants Products Activated complex
  • 72. Potential Energy Reaction Coordinate Reactants Products  E }
  • 73. Potential Energy Reaction Coordinate 2BrNO 2NO + Br Br---NO Br---NO 2 Transition State
  • 74. 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.
  • 75. 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.
  • 76. Arrhenius
    • Number of collisions with the required energy = ze -E a /RT
    • z = total collisions
    • e is Euler’s number (inverse of ln)
    • E a = activation energy
    • R = ideal gas constant (in J/K mol)
    • T is temperature in Kelvin
  • 77. 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.
  • 78. No Reaction O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br O N Br
  • 79. Arrhenius Equation
    • k = zpe -E a /RT = Ae -E a /RT
    • A is called the frequency factor = zp
    • k is the rate constant
    • ln k = -(E a /R)(1/T) + ln A
    • Another line !!!!
    • ln k vs 1/T is a straight line
    • With slope E a /R so we can find E a
    • And intercept ln A
  • 80. Arrhenius Equation- Another line !!!!
    • l n k = -(E a /R) (1/T) + ln A
    • y = m x + b
    l n k 1/T slope = -(E a /R)
  • 81. Activation Energy and Rates The final saga
  • 82. 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.
  • 83.
    • This reaction takes place in three steps
  • 84.
    • First step is fast
    • Low activation energy
     E a
  • 85. Second step is slow High activation energy E a 
  • 86. E a Third step is fast Low activation energy 
  • 87. Second step is rate determining
  • 88. Intermediates are present
  • 89. Activated Complexes or Transition States
  • 90. 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.
  • 91. 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
  • 92.
    • Hydrogen bonds to surface of metal.
    • Break H-H bonds
    Heterogenous Catalysts Pt surface H H H H H H H H
  • 93. Heterogenous Catalysts Pt surface H H H H C H H C H H
  • 94. Heterogenous Catalysts
    • The double bond breaks and bonds to the catalyst.
    Pt surface H H H H C H H C H H
  • 95. Heterogenous Catalysts
    • The hydrogen atoms bond with the carbon
    Pt surface H H H H C H H C H H
  • 96. Heterogenous Catalysts Pt surface H C H H C H H H H H
  • 97. Homogenous Catalysts
    • Chlorofluorocarbons (CFCs) catalyze the decomposition of ozone.
    • Enzymes regulating the body processes. (Protein catalysts)
  • 98. 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
  • 99. Catalysts and rate.
    • Rate increases until the active sites of catalyst are filled.
    • Then rate is independent of concentration
    Concentration of reactants Rate
  • 100.  
  • 101.