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.

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Presentation transcript:

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 13.2

Double [F2] with [ClO2] constant F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant Rate doubles x = 1 rate = k [F2][ClO2] Quadruple [ClO2] with [F2] constant Rate quadruples y = 1 13.2

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. F2 (g) + 2ClO2 (g) 2FClO2 (g) 1 rate = k [F2][ClO2] 13.2

S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Determine the rate law and calculate the rate constant for the following reaction from the following data: S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq) Experiment [S2O82-] [I-] Initial Rate (M/s) 1 0.08 0.034 2.2 x 10-4 2 0.017 1.1 x 10-4 3 0.16 rate = k [S2O82-]x[I-]y y = 1 x = 1 rate = k [S2O82-][I-] Double [I-], rate doubles (experiment 1 & 2) Double [S2O82-], rate doubles (experiment 2 & 3) k = rate [S2O82-][I-] = 2.2 x 10-4 M/s (0.08 M)(0.034 M) = 0.08/M•s 13.2

First-Order Reactions rate = - D[A] Dt A product rate = k [A] D[A] Dt = k [A] - rate [A] M/s M = 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] = [A]0exp(-kt) ln[A] = ln[A]0 - kt 13.3

2N2O5 4NO2 (g) + O2 (g) 13.3

First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 ln [A]0 [A]0/2 k = t½ ln2 k = 0.693 k = What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½ ln2 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) 13.3

Second-Order Reactions rate = - D[A] Dt A product rate = k [A]2 rate [A]2 M/s M2 = D[A] Dt = k [A]2 - k = = 1/M•s 1 [A] = [A]0 + kt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = 1 k[A]0 13.3

Zero-Order Reactions rate = - D[A] Dt A product rate = k [A]0 = k rate = M/s [A] is the concentration of A at any time t [A] = [A]0 - kt [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = [A]0 2k 13.3

Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions Order Rate Law Concentration-Time Equation Half-Life t½ = [A]0 2k rate = k [A] = [A]0 - kt t½ ln2 k = 1 rate = k [A] ln[A] = ln[A]0 - kt 1 [A] = [A]0 + kt t½ = 1 k[A]0 2 rate = k [A]2 13.3

A + B AB C + D + Exothermic Reaction Endothermic Reaction The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction. 13.4

Temperature Dependence of the Rate Constant k = A • exp( -Ea / RT ) (Arrhenius equation) Ea 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 lnk = - Ea R 1 T + lnA 13.4

lnk = - Ea R 1 T + lnA 13.4

N2O2 is detected during the reaction! 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) + O2 (g) 2NO2 (g) N2O2 is detected during the reaction! Elementary step: NO + NO N2O2 Overall reaction: 2NO + O2 2NO2 + Elementary step: N2O2 + O2 2NO2 13.5

Unimolecular reaction – elementary step with 1 molecule 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. Elementary step: NO + NO N2O2 N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 + 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 13.5

Rate Laws and Elementary Steps Unimolecular reaction A products rate = k [A] Bimolecular reaction A + B products rate = k [A][B] Bimolecular reaction A + A products rate = k [A]2 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. 13.5

Sequence of Steps in Studying a Reaction Mechanism 13.5

What is the equation for the overall reaction? The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps: Step 1: NO2 + NO2 NO + NO3 Step 2: NO3 + CO NO2 + CO2 What is the equation for the overall reaction? NO2+ CO NO + CO2 What is the intermediate? NO3 What can you say about the relative rates of steps 1 and 2? rate = k[NO2]2 is the rate law for step 1 so step 1 must be slower than step 2 13.5

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A • exp( -Ea / RT ) Ea k Uncatalyzed Catalyzed ratecatalyzed > rateuncatalyzed Ea < Ea ‘ 13.6

Haber synthesis of ammonia In heterogeneous catalysis, the reactants and the catalysts are in different phases. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Acid catalysis Base catalysis 13.6