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Chapter 12 Chemical Kinetics The area of chemistry that concerns reaction rates. Goals: To understand the steps (reaction mechanism) by which a reaction.

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Presentation on theme: "Chapter 12 Chemical Kinetics The area of chemistry that concerns reaction rates. Goals: To understand the steps (reaction mechanism) by which a reaction."— Presentation transcript:

1 Chapter 12 Chemical Kinetics The area of chemistry that concerns reaction rates. Goals: To understand the steps (reaction mechanism) by which a reaction takes place. Allows us to find ways to facilitate the reaction

2 Reaction Rate Kinetics deals with the speed (rate) at which changes occur. The quantity that changes is the amount or concentration of a reactant or a product. Change in concentration (conc.) of a reactant or product per unit time.

3 Continued…. 2NO 2(g)  2NO (g) + O 2(g) The concentration of the reactant (NO 2 ) decreases with time and the concentration of the products (NO and O 2 ) increase with time. A change can be positive (increase) or negative (decrease) leading to a positive or negative reaction rate. However, we will always define the rate as a positive quantity.

4 Figure 12.1 Definition of Rate (Conc. vs.Time)

5 Continued…. The concentration of NO 2 decreases with time,  [NO 2 ] is a negative quantity. The reaction rate is a positive quantity The concentration of reactants always decrease with time. The rate expression involving reactant will include a negative sign.

6 Continued…. 2NO 2(g)  2NO (g) + O 2(g) The rate can also be defined in terms of the products. In doing so we must take into account the coefficients in the balance equation for the reaction. Stoichiometry determines the relative rates. Rate of consumption = rate of production = 2(rate of production of NO 2 of NO of O 2 )

7 Rate Laws 2NO 2(g)  2NO (g) + O 2(g) The reaction rate will depend on the concentrations of the reactants Rate = k[NO 2 ] n The above expression shows how the rate depends on the concentration of reactants is called a rate law. k = rate constant (proportionality constant) n = rate order (integer or fraction including zero)

8 Types of Rate Laws Differential Rate Law: expresses how rate depends on concentration. Integrated Rate Law: expresses how concentration depends on time.

9 Continued…. We consider reactions where the reverse reaction is unimportant, rate laws involve only concentrations of reactants. 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. Experimental convenience dictates which types of rate law is determined experimentally.

10 Method of Initial Rates Initial Rate: the “instantaneous rate” just after the reaction begins. The initial rate is determined in several experiments using different initial concentrations.

11 Figure 12.3 A Plot of the Concentration of N 2 O 5 as a Function of Time for the Reaction

12 Overall Reaction Order Sum of the order of each component in the rate law. rate = k[H 2 SeO 3 ][H + ] 2 [I  ] 3 The overall reaction order is = 6.

13 First-Order Rate Law For aA  Products in a 1st-order reaction, Integrated first-order rate law is ln[A] =  kt + ln[A] o

14 Continued…. ln[A] = -kt + ln[A] o The equation shows how the concentration of A depends on time. If the initial concentration of A and the rate constant k are known, the concentration of A at any time can be calculated. The above equation is the equation of straight line of the form y = mx + b, where a plot of y versus x is a straight line with slope m and intercept b.

15 Continued…. The reaction is first order in A if a plot of ln[A] versus t is a straight line. The integrated rate law for a first order reaction can also be expressed in terms of a ratio of [A] and [A] o as follow:

16 Figure 12.4 A Plot of In(N 2 O 5 ) Versus Time

17 Half-Life of a First-Order Reaction The time required for a reactant to reach half its original concentration is called the half-life of a reactant and is designated by the symbol t 1/2. t 1/2 = half-life of the reaction, k = rate constant For a first-order reaction, the half-life does not depend on concentration.

18 Figure 12.5 A Plot of (N 2 O 5 ) Versus Time for the Decomposition Reaction of N 2 O 5

19 Second-Order Rate Law For aA  products in a second-order reaction, Integrated rate law is A plot of 1/[A] versus t will produce a straight line with a slope equal to k The above equation shows how [A] depends on time and can be used to calculate [A] at any time t, provided k and [A] o are known

20 Half-Life of a Second-Order Reaction When one half-life of the second order reaction has elapsed (t = t 1/2 ), by definition, [A] = [A] o /2 then the integrated rate law becomes t 1/2 = half-life of the reaction, k = rate constant, A o = initial concentration of A The half-life is dependent upon the initial concentration.

21 Figure 12.6 (a) A Plot of In(C 4 H 6 ) Versus t (b) A Plot of 1/(C 4 H 6 ) Versus t

22 Zero-Order Rate Laws The rate law for a zero-order reaction is Rate = k[A] o = k(1) = k For a zero-order reaction, the rate is constant. It does not change with concentration as it does for first-order or second-order reactions. The integrated rate law for a zero-order reaction is [A] = -kt + [A] o

23 continued… [A] = -kt + [A] o In this case a plot of [A] versus t gives a straight line of slope –k. [A] = [A] o /2, when t = t 1/2 [A] o /2 =-kt 1/2 + [A]o Solving for t 1/2 gives, t 1/2 = [A] o /2k

24 Figure 12.7 A Plot of (A) Versus t for a Zero-Order Reaction

25 Rate Laws for Reactions with More Than One Reactant A + B + C  Product Rate = k[A] n [B] m [C] p For such reaction, concentration of one reactant remain small compared with the concentrations of the others. So the rate law reduce to Rate = k`[A] n Where, k` = k[B] m [C] o p and [B] o >>[A] o and [C] o >>[A] o The value of n can be obtained by determining whether a plot of [A] versus t is linear (n = 0), a plot of ln[A] versus t is linear (n = 1), or a plot of 1/[A] versus t is linear (n = 2). The value of k` is determined from the slope.

26 A Summary 1.Simplification: Conditions are set such that only forward reaction is important. 2.Two types of rate law: differential rate law and integrated rate law 3.Which type? Depends on the type of data collected - differential and integrated forms can be interconverted.

27 A Summary (continued) 4.Most common: method of initial rates. 5.Concentration v. time: used to determine integrated rate law, often graphically. 6.For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions).

28 Reaction Mechanism 4 The series of steps by which a chemical reaction occurs. 4 A chemical equation does not tell us how reactants become products - it is a summary of the overall process. 4 The purpose for studying kinetics is to learn as much as possible about the steps involved in a reaction.

29 Reaction Mechanism (continued) The reaction has many steps in the reaction mechanism. The rate law for this reaction is known from experiment to be Rate = k[NO 2 ] 2 The balanced equation tells us the reactants, the products, the stoichiometry. NO 2(g) + NO 2(g) NO 3(g) + NO(g) NO 3(g) + CO (g) NO 2(g) + CO 2(g)

30 Often Used Terms Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product. (neither a reactant nor a product) Molecularity: the number of species that must collide to produce the reaction indicated by that step. Elementary Step: A reaction whose rate law can be written from its molecularity. uni, bi and termolecular The sum of the elementary steps must give the overall balanced equation The mechanism must agree with the experimentally determined rate law.

31 Rate-Determining Step Multistep reaction often have one step that is much slower than all the others. Reactants can become products only as fast as they can get through this slowest step. The overall reaction can be no faster than the slowest or rate determining step. In a multistep reaction, it is the slowest step. It therefore determines the rate of reaction. Overall rate = k 1 [NO 2 ] 2

32 Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why? Arrhenius: An activation energy (threshold energy) must be overcome to produce a chemical reaction. 2BrNO (g)  2NO (g) + Br 2(g)

33 Figure Change in Potential Energy

34 Arrhenius Equation 4 Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy). 4 Orientation of reactants must allow formation of new bonds necessary to produce products.

35 Figure Several Possible Orientations for a Collision Between Two BrNO Molecules

36 Arrhenius Equation (continued) k = rate constant, A = frequency factor E a = activation energy, T = temperature R = gas constant ln(k) = -Ea/R(1/T) + ln(A) A plot of ln(k) versus 1/T gives a straight line, slope = -Ea/R and intercept = ln(A)

37 Figure Plot of In(k) Versus 1/T for the Reaction 2N 2 O 5 (g)    g) + O 2 (g)

38 Arrhenius Equation (continued) Activation energy (E a ) can also be calculated from the values of k at only two temperatures At temperature T 1, where the rate constant is k 1, At temperature T 2, where the rate constant is k 2, E a can be calculated from k 1 and k 2 at temperature T 1 and T 2

39 Catalysis Catalyst: A substance that speeds up a reaction without being consumed it self. Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

40 Figure Energy Plots for a Catalyzed and an Uncatalyzed Pathway for a Given Reaction

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