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Chapter 12 Chemical Kinetics.

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Presentation on theme: "Chapter 12 Chemical Kinetics."— Presentation transcript:

1 Chapter 12 Chemical Kinetics

2 Really Big Concepts in Chemistry
Thermodynamics describes the changes in the form of energy when a reaction occurs, for example, converting chemical energy to heat. Kinetics describes how quickly or slowly a reaction occurs OR how the reaction occurs over the course of time Equilibrium describes reactions in which the reactants and products coexist Copyright © Cengage Learning. All rights reserved

3 Why study Kinetics? We can uncover the actual steps that occur in a reaction (the mechanism) which will allow us to understand the how and why of a reaction (not just overall yield or energy). If we know the mechanism, then we can find ways to improve the speed of the reaction or the yield of the reaction (i.e. increase efficiency) Copyright © Cengage Learning. All rights reserved

4 12.2 Rate Laws: An Introduction
Reaction Rates Rate Laws: An Introduction Determining the Form of the Rate Law The Integrated Rate Law Reaction Mechanisms 12.6 A Model for Chemical Kinetics 12.7 Catalysis Copyright © Cengage Learning. All rights reserved

5 We will focus on the following reaction: 2 NO2 (g) → 2NO (g) + O2 (g)
The following data shows the concentration of both the reactant and the products as time goes by. Start with pure NO2 at 0.01M and 300°C What do you notice about the data (Table 12.1 pg 554 or the next slide) ? Reactant is decreasing, products are increasing NO is increasing twice as much as O2 for each interval What else??? Copyright © Cengage Learning. All rights reserved

6 The Decomposition of Nitrogen Dioxide
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7 The Decomposition of Nitrogen Dioxide
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8 Copyright © Cengage Learning. All rights reserved

9 Copyright © Cengage Learning. All rights reserved

10 Copyright © Cengage Learning. All rights reserved

11 The Decomposition of Nitrogen Dioxide
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12 Instantaneous Rate = Value of the rate at a particular time.
Can be obtained by computing the slope of a line tangent to the curve at that point. Use Figure 12.1 (pg 554) to compute the instantaneous rate of NO2 at 100s and the rate of NO and O2 at 250s using the tangent lines. Copyright © Cengage Learning. All rights reserved

13 The rate varies: - with time - with each chemical
Therefore we have to be very specific when discussing the rate (Example: The rate of consumption of nitrogen dioxide at 50s) Copyright © Cengage Learning. All rights reserved

14 Shows how the rate depends on the concentrations of reactants.
Rate Law Shows how the rate depends on the concentrations of reactants. For the decomposition of nitrogen dioxide: 2NO2(g) → 2NO(g) + O2(g) Rate = k[NO2]n: k = rate constant n = order of the reactant Copyright © Cengage Learning. All rights reserved

15 Rate Law Rate = k[NO2]n The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate. Copyright © Cengage Learning. All rights reserved

16 Rate Law Rate = k[NO2]n The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation. Copyright © Cengage Learning. All rights reserved

17 Types of Rate Laws Differential Rate Law (rate law) – shows how the rate of a reaction depends on concentrations. Integrated Rate Law – shows how the concentrations of species in the reaction depend on time. Copyright © Cengage Learning. All rights reserved

18 Rate Laws: A Summary Because we typically consider reactions only under conditions where the reverse reaction is unimportant, our rate laws will involve only concentrations of reactants. Because the differential and integrated rate laws for a given reaction are related in a well–defined way, the experimental determination of either of the rate laws is sufficient. Copyright © Cengage Learning. All rights reserved

19 Rate Laws: A Summary Experimental convenience usually dictates which type of rate law is determined experimentally. Knowing the rate law for a reaction is important mainly because we can usually infer the individual steps involved in the reaction from the specific form of the rate law. Copyright © Cengage Learning. All rights reserved

20 Determine experimentally the power to which each reactant concentration must be raised in the rate law. Copyright © Cengage Learning. All rights reserved

21 Method of Initial Rates
The value of the initial rate is determined for each experiment at the same value of t as close to t = 0 as possible. Several experiments are carried out using different initial concentrations of each of the reactants, and the initial rate is determined for each run. The results are then compared to see how the initial rate depends on the initial concentrations of each of the reactants. Copyright © Cengage Learning. All rights reserved

22 Experiments Initial Rate 1 0.1M 0.005M 1.35X10-7 2 0.01M 2.70X10-7 3 0.2M 5.40X10-7

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24 Overall Reaction Order
The sum of the exponents in the reaction rate equation. Rate = k[A]n[B]m Overall reaction order = n + m k = rate constant [A] = concentration of reactant A [B] = concentration of reactant B Copyright © Cengage Learning. All rights reserved

25 Concept Check How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? The exponents in the rate law are not equal to the stoichiometric coefficients unless the reaction actually occurs via a single step mechanism (an elementary step); however, the exponents are equal to the stoichiometric coefficients of the rate-determining step. Copyright © Cengage Learning. All rights reserved

26 Half-Life of Reactions
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29 Rate = k[A] Integrated: ln[A] = –kt + ln[A]o First-Order
[A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

30 First Order Graph - Plot of ln[N2O5] vs Time
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31 Half–life does not depend on the concentration of reactants.
First-Order Half–Life: k = rate constant Half–life does not depend on the concentration of reactants. Copyright © Cengage Learning. All rights reserved

32 Exercise A first order reaction is 35% complete at the end of 55 minutes. What is the value of k? k = 7.8 x 10–3 min–1 Copyright © Cengage Learning. All rights reserved

33 Rate = k[A]2 Integrated: Second-Order
[A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

34 Plot of ln[C4H6] vs Time and Plot of 1/[C4H6] vs Time
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35 Each successive half–life is double the preceding one.
Second-Order Half–Life: k = rate constant [A]o = initial concentration of A Half–life gets longer as the reaction progresses and the concentration of reactants decrease. Each successive half–life is double the preceding one. Copyright © Cengage Learning. All rights reserved

36 Write the rate law for this reaction. rate = k[A]2 b) Calculate k.
Exercise For a reaction aA  Products, [A]0 = 5.0 M, and the first two half-lives are 25 and 50 minutes, respectively. Write the rate law for this reaction. rate = k[A]2 b) Calculate k. k = 8.0 x 10-3 M–1min–1 Calculate [A] at t = 525 minutes. [A] = 0.23 M Copyright © Cengage Learning. All rights reserved

37 Rate = k[A]0 = k Integrated: [A] = –kt + [A]o Zero-Order
[A] = concentration of A at time t k = rate constant t = time [A]o = initial concentration of A Copyright © Cengage Learning. All rights reserved

38 Plot of [A] vs Time Copyright © Cengage Learning. All rights reserved

39 Zero-Order Half–Life: k = rate constant [A]o = initial concentration of A Half–life gets shorter as the reaction progresses and the concentration of reactants decrease. Copyright © Cengage Learning. All rights reserved

40 Concept Check How can you tell the difference among 0th, 1st, and 2nd order rate laws from their graphs? Copyright © Cengage Learning. All rights reserved

41 Rate Laws Copyright © Cengage Learning. All rights reserved

42 Summary of the Rate Laws
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