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Ch. 14: Chemical Kinetics Dr. Namphol Sinkaset Chem 201: General Chemistry II
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  • Ch. 14: Chemical Kinetics

    Dr. Namphol Sinkaset Chem 201: General Chemistry II

  • 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

  • I. Introduction

    • Molecules/atoms are constantly moving; in some collisions, electrons in one are attracted to nuclei in another.

    • Some reactions are quick (explosions) while others are slow (rusting of iron).

    • If we understand what contributes to the rate, we can control the reaction.

  • I. Introduction

    • Temperature influences how fast reactions occur. • Biochemical reactions are subject to the same

    rules as other chemical reactions.

  • I. Introduction

    • We will begin by examining the collision model of chemical reactions.

    • Three factors that influence how fast reactions occur: Reactant concentrations Temperature Structure and orientation of colliding

    particles

  • 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.

  • II. Average Rates of Reaction

    H2(g) + I2(g) 2HI(g)

  • II. General Reaction Rates

    aA + bB cC + dD

  • 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!

  • II. Measuring Rates

    • Need a chemical “handle” to see how fast a reaction takes place.

    • One way is to measure how quickly the color of a solution changes.*

  • 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.

  • 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

  • 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]

  • 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.

  • 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!

  • III. Sure-fire Method

    [A] (M) Initial Rate (M/s)

    0.10 0.015 0.20 0.060 0.40 0.240

    For the reaction, A Products, we have the following data:

  • 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.

  • III. Example Reaction • 2H2(g) + 2NO(g) N2(g) + 2H2O(g) • After looking at experimental data, the rate

    law was found to be Rate = k[H2][NO]2. • We say the reaction is 1st order in H2, 2nd

    order in NO, and 3rd 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?

  • 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.

  • III. Sample Problem 14.1

    [CHCl3] (M) [Cl2] (M) Initial Rate (M/s)

    0.010 0.010 0.0035 0.020 0.010 0.0069 0.020 0.020 0.0098 0.040 0.040 0.027

    Determine the complete rate law for the reaction CHCl3(g) + Cl2(g) CCl4(g) + HCl(g) using the data below.

  • III. Sample Problem 14.2

    [NO] (M) [H2] (M) Initial Rate (M/s)

    0.10 0.10 0.00123 0.10 0.20 0.00246 0.20 0.10 0.00492

    Sometimes, rate laws can be found by inspection. Determine the rate law for the reaction 2NO(g) + 2H2(g) N2(g) + H2O(g) using the data below.

  • 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.

  • IV. 1st Order Integrated Rate Law

  • IV. 1st Order Integrated Rate Law

    • Notice this equation is in y = mx + b form. • A plot of ln[A] vs. t for a 1st order reaction

    yields a straight line with m = -k and b = ln[A]0.

  • IV. 2nd Order Integrated Rate Law

  • IV. 2nd Order Integrated Rate Law

    • 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.

  • IV. Zero Order Integrated Rate Law

  • IV. Zero Order Integrated Rate Law

    • 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.

  • IV. Reaction Half Lives

    • The half-life, t1/2, 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.

  • IV. 1st Order Reaction Half Life

  • IV. 1st Order Reaction Half Life

    • Notice that the half life doesn’t depend on reactant concentration!

    • Unique for 1st order. • The half life for a 1st order reaction is…*

  • IV. 1st Order Half Lives

  • IV. 2nd Order Reaction Half Life

  • IV. 2nd Order Reaction Half Life

    • For 2nd order, the half life depends on initial concentration.

    • As concentration decreases, half life…*

  • IV. Zero Order Reaction Half Life

  • IV. Zero Order Reaction Half Life

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

    • As concentration decreases, half life…*

  • IV. Rate Law Summary

  • V. Temperature and Rate

    • In general, rates of reaction are highly sensitive to temperature – a 10 °C increase in T increases rate 2x to 3x.

    • If Rate = k[A]n, where does the temperature factor in?

    • It’s in the constant k! • Generally, increasing temperature

    increases k.

  • V. The Arrhenius Equation

    • Note that R is the gas constant, and T is temperature in kelvin.

  • V. Parameters of Arrhenius Eqn.

    • We can describe the physical meanings of the aspects of the Arrhenius equation by considering a specific reaction.

  • 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 Ea means…*

  • V. Frequency Factor • The frequency factor (A) 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 109/s means that there are 109 vibrations per second of the NC group.

  • 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 107 molecules has enough energy to cross the energy barrier.

  • 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.

  • V. Finding A and Ea

  • 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 = -Ea/R and b = ln A.

  • V. Two-Point Form

  • V. Sample Problem 14.3

    • 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?

  • V. The Collision Model

    • In the collision model of a reaction (2 reactants), two factors influence whether a reaction will occur. Energy of the collision Orientation of the

    collision

  • V. Collision Theory & Arrhenius Equation

    • p is the orientation factor, a number between 0 and 1 (1 means any orientation will work for the reaction).

    • z is the collision frequency, i.e. how many collisions per second.

  • V. Effective/Ineffective Collisions

  • VI. How Reactions Occur

    • Most chemical reactions occur through several small steps, not one big step.

    • A chemical equation typically shows the overall reaction, not the intermediate steps.

    • e.g. H2(g) + 2ICl(g) 2HCl(g) + I2(g) only shows what’s at the beginning and what you end up with.

  • VI. Reaction Mechanisms

    • A reaction mechanism is a series of individual chemical steps through which an overall chemical reaction occurs.

    • A proposed mechanism for the reaction H2(g) + 2ICl(g) 2HCl(g) + I2(g) is:

    Step 1 H2(g) + ICl(g) HI(g) + HCl(g) Step 2 HI(g) + ICl(g) HCl(g) + I2(g)

  • VI. Elementary Steps • The reactions in a mechanism are called

    elementary steps; what’s implied in these steps is exactly what happens.

    • Proposed reaction mechanisms must add up to the overall reaction!

    • Does the previous mechanism add up? • Species that are formed in one step and then

    consumed in another are known as intermediates. What is/are the intermediate(s) in the previous mechanism?

  • VI. Elementary Step Rate Laws

    • Elementary steps are characterized by their molecularity, i.e. the # of reactant particles involved in the step.

    • Rate laws for elementary steps can be written directly from their stoichiometry!

    • e.g. If A + B C + D is an elementary step, then the rate law for this step is: Rate = k[A][B].

  • VI. Energy Diagram, 2-Step Mechanism

  • VI. The Rate-Determining Step

    • The slow step in the mechanism will determine the overall rate of reaction.

    • This step is known as the rate-determining step.

    • It’s the bottleneck of the reaction.

  • VI. Valid Mechanisms

    • Valid mechanisms satisfy 2 criteria: Elementary steps add up to overall

    reaction. Rate law predicted by mechanism must be

    consistent with experimental rate law. • Note that a valid mechanism is not a

    proven mechanism.

  • VI. Example • Consider the reaction: NO2(g) + CO(g)

    NO(g) + CO2(g). • Experimentally, Rate = k[NO2]2. This

    implies it’s not a single-step reaction. Why?

    • Is the mechanism below valid?

    NO2(g) + NO2(g) NO3(g) + NO(g) Slow NO3(g) + CO(g) NO2(g) + CO2(g) Fast

  • VI. Rate Laws w/ Intermediates

    • Rate laws must always be written from the rate-determining step.

    • However, rate laws cannot contain intermediates.

    • Rate laws from other steps can be used to substitute for intermediates.

    • We look at fast first steps.

  • VI. Fast 1st Steps

    • When the 1st step is fast, its products will build up and reverse reaction starts.

    • Eventually, an equilibrium is set up. • Thus, for A + B C + D (Fast), we can

    write A + B C + D. Rate = k[A][B] and Rate = k-1[C][D]. At equilibrium, k[A][B] = k-1[C][D]. This can be used to rewrite rate laws.

  • VI. Sample Problem 14.4

    • What is the overall reaction and rate law for the mechanism below? Identify the intermediates as well.

    Cl2(g) 2Cl(g) Fast Cl(g) + CHCl3(g) HCl(g) + CCl3(g) Slow CCl3(g) + Cl(g) CCl4(g) Fast

  • VII. Catalysts

    • We know we can change reaction rates by changing the temperature or changing reactant concentrations.

    • However, there are limits to these tactics. What are these limits?*

    • If available, can use catalysts, substances that increase reaction rate, but aren’t used up in the reaction.

  • VII. Catalytic Destruction of O3

    Uncatalyzed: O3(g) + O(g) 2O2(g)

    Catalyzed: Cl(g) + O3(g) ClO(g) + O2(g) ClO(g) + O(g) Cl(g) + O2(g)

    Atomic chlorine from photodissociated CFC’s is the catalyst.

    O3(g) + O(g) 2O2(g)

  • VII. How Do Catalysts Work?

    • Catalysts provide a lower-energy mechanism for the reaction.

  • VII. Types of Catalysts

    • There are homogeneous and heterogeneous catalysts.

  • VII. Hydrogenation of Ethene

  • VII. Biological Catalysts • A biological catalyst is called and enzyme. • An enzymes has an active site into which a

    specific substrate fits – like a lock and key.

  • VII. Sucrose Glucose + Fructose

    Ch. 14: Chemical KineticsI. Chapter OutlineI. IntroductionI. IntroductionI. IntroductionII. Reaction RatesII. Average Rates of ReactionII. General Reaction RatesII. Some Rate DataII. Measuring RatesIII. Rate LawsIII. A Simple Rate LawIII. Reaction OrdersIII. Reaction Orders and RateIII. Determining OrdersIII. Sure-fire MethodIII. More Complex ReactionsIII. Example ReactionIII. Steps for Finding Rate LawIII. Sample Problem 14.1III. Sample Problem 14.2IV. Concentration and TimeIV. 1st Order Integrated Rate LawIV. 1st Order Integrated Rate LawIV. 2nd Order Integrated Rate LawIV. 2nd Order Integrated Rate LawIV. Zero Order Integrated Rate LawIV. Zero Order Integrated Rate LawIV. Reaction Half LivesIV. 1st Order Reaction Half LifeIV. 1st Order Reaction Half LifeIV. 1st Order Half LivesIV. 2nd Order Reaction Half LifeIV. 2nd Order Reaction Half LifeIV. Zero Order Reaction Half LifeIV. Zero Order Reaction Half LifeIV. Rate Law SummaryV. Temperature and RateV. The Arrhenius EquationV. Parameters of Arrhenius Eqn.V. Activation EnergyV. Frequency FactorV. Exponential FactorV. Exponential Factor & TempV. Finding A and EaV. Arrhenius PlotsV. Two-Point FormV. Sample Problem 14.3V. The Collision ModelV. Collision Theory & Arrhenius EquationV. Effective/Ineffective CollisionsVI. How Reactions OccurVI. Reaction MechanismsVI. Elementary StepsVI. Elementary Step Rate LawsVI. Energy Diagram, 2-Step MechanismVI. The Rate-Determining StepVI. Valid MechanismsVI. ExampleVI. Rate Laws w/ IntermediatesVI. Fast 1st StepsVI. Sample Problem 14.4VII. CatalystsVII. Catalytic Destruction of O3VII. How Do Catalysts Work?VII. Types of CatalystsVII. Hydrogenation of EtheneVII. Biological CatalystsVII. Sucrose Glucose + Fructose


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