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    Kinetics ppt Kinetics ppt Presentation Transcript

    • Chemical Kinetics Chapter 13 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.
    • Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time ( M /s).  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative . A B rate = -  [A]  t rate =  [B]  t
    • A B rate = -  [A]  t rate =  [ B ]  t
    •  [Br 2 ]   Absorption red-brown t 1 < t 2 < t 3 Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) time 393 nm light Detector
    • instantaneous rate = rate for specific instance in time Br 2 ( aq ) + HCOOH ( aq ) 2Br - ( aq ) + 2H + ( aq ) + CO 2 ( g ) average rate = -  [Br 2 ]  t = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent
    • rate  [Br 2 ] rate = k [Br 2 ] = rate constant = 3.50 x 10 -3 s -1 k = rate [Br 2 ]
    • PV = nRT 2H 2 O 2 ( aq ) 2H 2 O ( l ) + O 2 ( g ) P = RT = [O 2 ] RT n V [O 2 ] = P RT 1 rate =  [O 2 ]  t RT 1  P  t = measure  P over time
    •  
    • Reaction Rates and Stoichiometry Two moles of A disappear for each mole of B that is formed. 2A B rate =  [B]  t rate = -  [A]  t 1 2 a A + b B c C + d D rate = -  [A]  t 1 a = -  [B]  t 1 b =  [C]  t 1 c =  [D]  t 1 d
    • Write the rate expression for the following reaction: CH 4 ( g ) + 2O 2 ( g ) CO 2 ( g ) + 2H 2 O ( g ) rate = -  [CH 4 ]  t = -  [O 2 ]  t 1 2 =  [H 2 O]  t 1 2 =  [CO 2 ]  t
    • The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. Rate = k [A] x [B] y Reaction is x th order in A Reaction is y th order in B Reaction is (x +y)th order overall a A + b B c C + d D
    • rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ] F 2 ( g ) + 2ClO 2 ( g ) 2FClO 2 ( g )
    • rate = k [F 2 ][ClO 2 ] Rate Laws
      • Rate laws are always determined experimentally.
      • Reaction order is always defined in terms of reactant (not product) concentrations.
      • The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
      F 2 ( g ) + 2ClO 2 ( g ) 2FClO 2 ( g ) 1
    • rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 = 0.08/ M • s rate = k [S 2 O 8 2- ][I - ] Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- ( aq ) + 3I - ( aq ) 2SO 4 2- ( aq ) + I 3 - ( aq ) 2.2 x 10 -4 0.017 0.16 3 1.1 x 10 -4 0.017 0.08 2 2.2 x 10 -4 0.034 0.08 1 Initial Rate ( M /s) [I - ] [S 2 O 8 2- ] Experiment k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M /s (0.08 M )(0.034 M )
    • First-Order Reactions rate = k [A] k = = 1/s or s -1 [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t =0 A product rate = -  [A]  t rate [A] M / s M =  [A]  t = k [A] - [A] = [A] 0 e − kt ln[A] = ln[A] 0 - kt
    • Graphical Determination of k 2N 2 O 5 4NO 2 ( g ) + O 2 ( g )
    • ln[A] = ln[A] 0 - kt kt = ln[A] 0 – ln[A] t = = 66 s [A] 0 = 0.88 M [A] = 0.14 M The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? ln[A] 0 – ln[A] k ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x 10 -2 s -1 =
    • First-Order Reactions The half-life , t ½ , is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 ) ln [A] 0 [A] 0 /2 k = t ½ ln 2 k = 0.693 k = t ½ ln 2 k = 0.693 5.7 x 10 -4 s -1 =
    • First-order reaction 1 2 3 4 2 4 8 16 A product # of half-lives [A] = [A] 0 / n
    • Second-Order Reactions rate = k [A] 2 k = = 1/ M • s [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t =0 t ½ = t when [A] = [A] 0 /2 A product rate = -  [A]  t rate [A] 2 M / s M 2 =  [A]  t = k [A] 2 - 1 [A] = 1 [A] 0 + kt t ½ = 1 k [A] 0
    • Zero-Order Reactions rate = k [A] 0 = k k = = M /s [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t = 0 t ½ = t when [A] = [A] 0 /2 [A] = [A] 0 - kt A product rate = -  [A]  t rate [A] 0  [A]  t = k - t ½ = [A] 0 2 k
    • Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt [A] = [A] 0 - kt Order Rate Law Concentration-Time Equation Half-Life 1 [A] = 1 [A] 0 + kt t ½ ln 2 k = t ½ = [A] 0 2 k t ½ = 1 k [A] 0
    • Exothermic Reaction Endothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + +
    • Temperature Dependence of the Rate Constant E a is the activation energy (J/mol) R is the gas constant (8.314 J/K •mol) T is the absolute temperature A is the frequency factor (Arrhenius equation) Alternate format: ln k = - E a R 1 T + ln A
    • Alternate Form of the Arrhenius Equation At two temperatures, T 1 and T 2 or
    • Importance of Molecular Orientation effective collision ineffective collision
    • Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism . N 2 O 2 is detected during the reaction! 2NO ( g ) + O 2 ( g ) 2NO 2 ( g ) Elementary step: NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 +
    • Mechanism: 2NO ( g ) + O 2 ( g ) 2NO 2 ( g )
    • Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step.
      • The molecularity of a reaction is the number of molecules reacting in an elementary step.
      • Unimolecular reaction – elementary step with 1 molecule
      • Bimolecular reaction – elementary step with 2 molecules
      • Termolecular reaction – elementary step with 3 molecules
      Elementary step: NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 +
    • rate = k [A] rate = k [A][B] rate = k [A] 2 Rate Laws and Elementary Steps
      • Writing plausible reaction mechanisms:
      • The sum of the elementary steps must give the overall balanced equation for the reaction.
      • The rate-determining step should predict the same rate law that is determined experimentally.
      The rate-determining step is the slowest step in the sequence of steps leading to product formation. Unimolecular reaction A products Bimolecular reaction A + B products Bimolecular reaction A + A products
    • Sequence of Steps in Studying a Reaction Mechanism
    • The experimental rate law for the reaction between NO 2 and CO to produce NO and CO 2 is rate = k [NO 2 ] 2 . The reaction is believed to occur via two steps: What is the equation for the overall reaction? What is the intermediate? NO 3 What can you say about the relative rates of steps 1 and 2? rate = k [NO 2 ] 2 is the rate law for step 1 so step 1 must be slower than step 2 Step 1: NO 2 + NO 2 NO + NO 3 Step 2: NO 3 + CO NO 2 + CO 2 NO 2 + CO NO + CO 2
    • CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 • • Chemistry In Action: Femtochemistry CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 2 CH 2 CH 2 2
    • A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. rate catalyzed > rate uncatalyzed Uncatalyzed Catalyzed E a k E a < E a ′
    • In heterogeneous catalysis , the reactants and the catalysts are in different phases. In homogeneous catalysis , the reactants and the catalysts are dispersed in a single phase, usually liquid.
      • Haber synthesis of ammonia
      • Ostwald process for the production of nitric acid
      • Catalytic converters
      • Acid catalysis
      • Base catalysis
    • Haber Process N 2 ( g ) + 3H 2 ( g ) 2NH 3 ( g) Fe/Al 2 O 3 /K 2 O catalyst
    • Ostwald Process Pt catalyst Pt-Rh catalysts used in Ostwald process 4NH 3 ( g ) + 5O 2 ( g ) 4NO ( g ) + 6H 2 O ( g ) 2NO ( g ) + O 2 ( g ) 2NO 2 ( g ) 2NO 2 ( g ) + H 2 O ( l ) HNO 2 ( aq ) + HNO 3 ( aq )
    • Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter
    • Enzyme Catalysis
    • Binding of Glucose to Hexokinase
    • rate = k [ES] Enzyme Kinetics rate =  [P]  t