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Chemical Kinetics © 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the.

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Presentation on theme: "Chemical Kinetics © 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the."— Presentation transcript:

1 Chemical Kinetics © 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

2 Chemical Kinetics © 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI (g)  H 2 (g) + I 2 (g) In such a case, Rate = − 1212  [HI]  t =  [I 2 ]  t

3 Chemical Kinetics © 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry To generalize, then, for the reaction aA + bBcC + dD Rate = − 1a1a  [A]  t = − 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t =

4 Chemical Kinetics * consider the reaction : 4NO 2(g) + O 2(g) → 2 N 2 O 5 (g), if rate of disappearance of O 2 at Particular moment = 0.024 M/s, whate is the rate of N 2 O 5 being formed. Rate = -1/4∆ [NO 2 ]/∆t = - ∆ [O 2 ]/∆t = 1/2 ∆ [N 2 O 2 ]/∆t = - ∆ [O 2 ]/∆t = 0.024 M/s = 1/2 ∆ [N 2 O 2 ]/∆t [N 2 O 2 ]/∆t = 2x0.024 0.048 M/s * cosider the following reaction : 4 PH 3 → P 4(g) + 6 H 2(g). If molecular H 2 is being fomed at rate of 0.078 M/s. a) At whate rate is P 4 being formed : ∆ [P 4 ]/∆t = 1/6 x 0.078 = 0.013 M/s b) At whate rate is PH 3 reacting : - ∆ [PH 3 ]/∆t = 4/6 x 0.078 = 0.052 M/s © 2009, Prentice-Hall, Inc.

5 Chemical Kinetics © 2009, Prentice-Hall, Inc. Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

6 Chemical Kinetics © 2009, Prentice-Hall, Inc. Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH 4 + ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

7 Chemical Kinetics © 2009, Prentice-Hall, Inc. Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO 2 − ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

8 Chemical Kinetics © 2009, Prentice-Hall, Inc. Concentration and Rate This means Rate  [NH 4 + ] Rate  [NO 2 − ] Rate  [NH 4 + ] [NO 2 − ] which, when written as an equation, becomes Rate = k [NH 4 + ] [NO 2 − ] This equation is called the rate law, and k is the rate constant. Therefore,

9 Chemical Kinetics © 2009, Prentice-Hall, Inc. Rate Laws A rate law shows the relationship between the reaction rate and the concentrations of reactants. The exponents tell the order of the reaction with respect to each reactant. Since the rate law is Rate = k [NH 4 + ] [NO 2 − ] the reaction is First-order in [NH 4 + ] and First-order in [NO 2 − ].

10 Chemical Kinetics © 2009, Prentice-Hall, Inc. Rate Laws Rate = k [NH 4 + ] [NO 2 − ] The overall reaction order can be found by adding the exponents on the reactants in the rate law. This reaction is second-order overall.

11 Chemical Kinetics © 2009, Prentice-Hall, Inc. Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us ln [A] t [A] 0 = −kt Where [A] 0 is the initial concentration of A, and [A] t is the concentration of A at some time, t, during the course of the reaction.

12 Chemical Kinetics © 2009, Prentice-Hall, Inc. Integrated Rate Laws Manipulating this equation produces… ln [A] t [A] 0 = −kt ln [A] t − ln [A] 0 = − kt ln [A] t = − kt + ln [A] 0 …which is in the form y = mx + b


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