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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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