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1 Chemical Kinetics Chapter 14 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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2 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 B rate = - [A] tt rate = [B] tt [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.
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3 A B rate = - [A] tt rate = [B][B] tt
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4 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector [Br 2 ] Absorption red-brown t 1 < t 2 < t 3
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5 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) average rate = - [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial slope of tangent slope of tangent slope of tangent instantaneous rate = rate for specific instance in time
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6 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) PV = nRT P = RT = [O 2 ]RT n V [O 2 ] = P RT 1 rate = [O 2 ] tt RT 1 PP tt = measure P over time
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8 Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. rate = [B] tt rate = - [A] tt 1 2 aA + bB cC + dD rate = - [A] tt 1 a = - [B] tt 1 b = [C] tt 1 c = [D] tt 1 d
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9 Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = - [CH 4 ] tt = - [O 2 ] tt 1 2 = [H 2 O] tt 1 2 = [CO 2 ] tt
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10 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. aA + bB cC + dD Rate = k [A] x [B] y Reaction is xth order in A Reaction is yth order in B Reaction is (x + y)th order overall
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11 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) 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 ]
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12 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. 1
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13 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) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 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 k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]
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14 First-Order Reactions A product rate = - [A] tt rate = k [A] k = rate [A] = 1/s or s -1 M/sM/s M = [A] tt = k [A] - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 [A] = [A] 0 e −kt ln[A] = ln[A] 0 - kt
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15 2N 2 O 5 4NO 2 (g) + O 2 (g) Graphical Determination of k
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16 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 ln [A] 0 [A] 0 /2 k = t½t½ ln 2 k = 0.693 k = 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 ? t½t½ ln 2 k = 0.693 5.7 x 10 -4 s -1 = = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )
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17 A product First-order reaction # of half-lives [A] = [A] 0 /n 1 2 3 4 2 4 8 16
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18 Second-Order Reactions A product rate = - [A] tt rate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 = [A] tt = k [A] 2 - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] 0 + kt t ½ = t when [A] = [A] 0 /2 t ½ = 1 k[A] 0
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19 Zero-Order Reactions A product rate = - [A] tt rate = k [A] 0 = k k = rate [A] 0 = M/s [A] tt = k - [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 t ½ = [A] 0 2k2k [A] = [A] 0 - kt
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20 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln 2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0
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21 Exothermic ReactionEndothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. A + B AB C + D + +
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22 Temperature Dependence of the Rate Constant E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor ln k = - EaEa R 1 T + lnA (Arrhenius equation) Alternate format:
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23 Alternate Form of the Arrhenius Equation At two temperatures, T 1 and T 2 or
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24 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. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! 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 +
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25 2NO (g) + O 2 (g) 2NO 2 (g) Mechanism:
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26 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 + 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
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27 Unimolecular reactionA productsrate = k [A] Bimolecular reactionA + B productsrate = k [A][B] Bimolecular reactionA + A productsrate = 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.
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28 Sequence of Steps in Studying a Reaction Mechanism
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29 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: Step 1:NO 2 + NO 2 NO + NO 3 Step 2:NO 3 + CO NO 2 + CO 2 What is the equation for the overall reaction? NO 2 + CO NO + CO 2 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
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30 A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. EaEa k rate catalyzed > rate uncatalyzed E a < E a ′ UncatalyzedCatalyzed
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31 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
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32 N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process
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33 Ostwald Process Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g) 4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) + H 2 O (l) HNO 2 (aq) + HNO 3 (aq)
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34 Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter
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35 Enzyme Catalysis
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36 Binding of Glucose to Hexokinase
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37 rate = [P] tt rate = k [ES] Enzyme Kinetics
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