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a A + b B c C + d D. 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.
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aA + bB cC + dD 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 xth order in A reaction is yth order in B reaction is (x +y)th order overall 13.2
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
F2(g) + 2ClO2(g) 2FClO2(g) 1 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. rate = k [F2][ClO2] 13.2
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) rate k = 2.2 x 10-4 M/s = [S2O82-][I-] (0.08 M)(0.034 M) 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) = 0.08/M•s 13.2
A product rate = [A] M/s D[A] - M = k [A] Dt [A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt D[A] rate = - Dt First-Order Reactions rate = k [A] = 1/s or s-1 k = [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 13.3
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? 0.88 M ln 0.14 M = 2.8 x 10-2 s-1 ln ln[A]0 – ln[A] = k k [A]0 [A] [A]0 = 0.88 M ln[A] = ln[A]0 - kt [A] = 0.14 M kt = ln[A]0 – ln[A] = 66 s t = 13.3
[A]0 ln t½ [A]0/2 0.693 = = = = k 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 ln2 0.693 = k k 5.7 x 10-4 s-1 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 = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s-1) 13.3
A product # of half-lives [A] = [A]0/n First-order reaction 1 2 2 4 3 8 4 16 13.3
A product rate = [A]2 M/s D[A] 1 1 - M2 = k [A]2 = + kt Dt [A] [A]0 t½ = D[A] rate = - Dt 1 k[A]0 Second-Order Reactions rate = k [A]2 = 1/M•s k = [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 13.3
A product rate [A]0 D[A] - = k Dt [A]0 t½ = D[A] 2k rate = - Dt Zero-Order Reactions rate = k [A]0 = k = M/s k = [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 13.3
Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]0 = [A]0 t½ = t½ t½ = ln2 2k k 1 k[A]0 Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions [A] = [A]0 - kt rate = k 0 ln[A] = ln[A]0 - kt 1 rate = k [A] 2 rate = k [A]2 13.3