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Ch. 13: Chemical Kinetics Dr. Namphol Sinkaset Chem 201: General Chemistry II.

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Presentation on theme: "Ch. 13: Chemical Kinetics Dr. Namphol Sinkaset Chem 201: General Chemistry II."— Presentation transcript:

1 Ch. 13: Chemical Kinetics Dr. Namphol Sinkaset Chem 201: General Chemistry II

2 I. Chapter Outline I.Introduction II.The Rate of a Chemical Reaction III.Reaction Rate Laws IV.Integrated Rate Laws V.Temperature and Rate VI.Reaction Mechanisms VII.Catalysis

3 I. Introduction Some reactions are quick (explosions) while others are slow (rusting of iron). Knowing the rate of a reaction and what factors influence it allow chemists to plan accordingly. If we understand what contributes to the rate, we can control the reaction.

4 I. Introduction Balanced equations only give net change. Equations tell us nothing about how the reaction happens. One possibility for the above: simultaneous collision of 6 molecules. Unlikely, so reaction must occur in a series of small steps that leads to final products. We study mechanisms later in the chapter. C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g)

5 II. Reaction Rates Rates are generally change of something divided by change in time. Reaction rates are no different. The rate of a reaction can be written with respect to any compound in that reaction. However, there can only be one numerical value for a rate of reaction.

6 II. Average Rates of Reaction H 2(g) + I 2(g)  2HI (g)

7 II. General Reaction Rates aA + bB  cC + dD

8 II. Some Rate Data If we plot average rate data as a function of time, we see that the reaction rate constantly changes. Thus, rate depends on concentration of reactants!

9 III. Rate Laws If the rate depends on concentration of reactants, then we should be able to write an equation. A rate law describes the mathematical relationship between the concentration of reactants and how fast the reaction occurs.

10 III. A Simple Rate Law Consider a decomposition reaction where A  products If the reverse reaction is negligible, then the rate law is: Rate = k[A] n.  k is called the rate constant  n is called the reaction order

11 III. Reaction Orders The reaction order, n, determines how the rate depends on the concentration of the reactant. For the previous reaction, if…  n = 0, zero order, rate is independent of [A]  n = 1, first order, rate is directly proportional to [A]  n = 2, second order, rate is proportional to the square of the [A]

12 III. Reaction Orders and Rate The rate law for the decomposition can then be either:  Rate = k[A] 0 = k  Rate = k[A] 1  Rate = k[A] 2 Each will have a different type of curve when graphed.

13 III. Determining Orders Reaction orders can only be determined by experiment!! Reaction orders are not related to the stoichiometry of a reaction! If reaction orders match a reaction’s stoichiometry, it is just a coincidence. Therefore, orders cannot be determined without experimental data!

14 III. Sure-fire Method [A] (M)Initial Rate (M/s) 0.100.015 0.200.060 0.400.240 For the reaction, A  Products, we have the following data:

15 III. More Complex Reactions What if we have a more complicated reaction like: aA + bB  cC + dD? Writing the general rate law is easy. Simply include all reactants, each with its own order.  Rate = k[A] m [B] n If there are more reactants, there are more terms in the rate law.

16 III. Example Reaction 2H 2(g) + 2NO (g)  N 2(g) + 2H 2 O (g) After looking at experimental data, the rate law was found to be Rate = k[H 2 ][NO] 2. We say the reaction is 1 st order in H 2, 2 nd order in NO, and 3 rd order overall. Note that Rate always has units of M/s, so the units on k will depend on the rate law. What are the units of k for the rate law above?

17 III. Steps for Finding Rate Law 1)Pick two solutions where one reactant stays same, but another changes. 2)Write rate law for both w/ as much information as you have. 3)Ratio the two and solve for an order. 4)Repeat for another pair of solutions. 5)Use any reaction to get value of k.

18 III. Sample Problem [CHCl 3 ] (M)[Cl 2 ] (M)Initial Rate (M/s) 0.010 0.0035 0.0200.0100.0069 0.020 0.0098 0.040 0.027 Determine the complete rate law for the reaction CHCl 3(g) + Cl 2(g)  CCl 4(g) + HCl (g) using the data below.

19 III. Sample Problem [NO] (M)[H 2 ] (M)Initial Rate (M/s) 0.10 0.00123 Sometimes, rate laws can be found by inspection. Determine the rate law for the reaction 2NO (g) + 2H 2(g)  N 2(g) + H 2 O (g) using the data below.

20 IV. Concentration and Time Study and elucidation of rate laws allow the prediction of when a reaction will end. An integrated rate law for a chemical reaction is a relationship between the concentrations of reactants and time. Integrated rate laws depend on the order of the reaction; thus, we examine each separately. We will only consider reactions with one reactant.

21 IV. 1 st Order Integrated Rate Law

22 Notice this equation is in y = mx + b form. A plot of ln[A] vs. t for a 1 st order reaction yields a straight line with m = -k and b = ln[A] 0.

23 IV. 2 nd Order Integrated Rate Law

24 Again, this equation is in y = mx + b form. A plot of 1/[A] vs. t yields a straight line with slope equal to k and y-intercept equal to 1/[A] 0.

25 IV. Zero Order Integrated Rate Law

26 Yet again in y = mx + b form! Plot of [A] vs. t results in a straight line with slope equal to -k and b = [A] 0.

27 IV. Reaction Half Lives The half-life, t 1/2, of a reaction is the time required for the concentration of a reactant to decrease to half its initial value. Half life equations depend on the order of the reaction.

28 IV. 1 st Order Reaction Half Life

29 Notice that the half life doesn’t depend on reactant concentration! Unique for 1 st order. The half life for a 1 st order reaction is constant.

30 IV. 1 st Order Half Lives

31 IV. 2 nd Order Reaction Half Life

32 For 2 nd order, the half life depends on initial concentration. As concentration decreases, half life gets longer and longer.

33 IV. Zero Order Reaction Half Life

34 We see that for zero order reactions, the half life depends on concentration as well.

35 V. Temperature and Rate In general, rates of reaction are highly sensitive to temperature. If Rate = k[A] n, where does the temperature factor in? It’s in the constant k! Generally, increasing temperature increases k.

36 V. The Arrhenius Equation Note that R is the gas constant, and T is temperature in kelvin.

37 V. Parameters of Arrhenius Eqn. We can describe the physical meanings of the aspects of the Arrhenius equation by considering a specific reaction.

38 V. Activation Energy To get to product state, reactant must go through high-energy activated complex, or transition state. Even though reaction is exo overall, it must go through an endo step. Higher E a means slower reaction.

39 V. Frequency Factor The frequency factor represents the number of approaches to the activation barrier per unit time. For this reaction, it represents how often the NC part of the molecule vibrates. Note that not all approaches result in reaction due to not having enough energy. A frequency factor of 10 9 /s means that there are 10 9 vibrations per second of the NC group.

40 V. Exponential Factor The exponential factor is a number between 0 and 1 that represents the fraction of molecules that successfully react upon approach. An exponential factor of 10 -7 means that 1 out of every 10 7 molecules has enough energy to cross the energy barrier.

41 V. Exponential Factor & Temp Since exponential factor = e -Ea/RT, temperature has a huge influence. As T  0, the factor goes to 0, and as T  ∞, the factor goes to 1. Thus, higher temperatures mean more successful approaches because the molecules have more energy to overcome the activation barrier.

42 V. Finding A and E a

43 V. Arrhenius Plots If we have kinetic data at various temperatures, we can plot ln k vs. 1/T. We should get a straight line with m = -E a /R and b = ln A.

44 V. Two-Point Form

45 V. Sample Problem The decomposition of HI has rate constants of k = 0.079 1/M·s at 508 °C and k = 0.24 1/M·s at 540 °C. What is the activation energy of this reaction in kJ/mole?

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