chapter 14: chemical kinetics · 2018. 11. 4. · most catalysts work by lowering the activation...

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Chapter 14: Chemical Kinetics NOTE THIS CHAPTER IS #2 TOP TOPICS ON AP EXAM!!! NOT ONLY DO YOU NEED TO FOCUS ON THEORY (and lots of MATH) BUT YOU MUST READ THE FIGURES TOO!!! Ch 14.1 ~ Factors that Affect Reaction Rates SG Questions Read p. 574-577. Answer the following questions 1. Define chemical kinetics? 2. What is the definition of a reaction rate? 3. What is a reaction mechanism? 4. What are the four factors that affect a reaction (rxn) rate? 5. Why do each of the factors affect the rate? Explain. 6. What do rxn rates depend on? 7. Describe the relationship between the frequency of collusions and the rxn rate. Chemical kinetics is the study of the speed or rate of a reaction under various conditions. Reaction Rates depend on collusions between molecules. Increase collusions = increase rates of the rxn THE COLLISION THEORY OF REACTION RATES - a reaction will only take place if ALL of the following conditions are met: 1. Particles must collide. 2. Only two particles may collide at one time. 3. Proper orientation of colliding molecules so that atoms can come in contact with each other to become products. 4. The collision must occur with enough energy so that the electrons can rearrange and form new bonds. 5. This new collision product is at the peak of the activation energy hump and is called the activated complex or the transition state. At this point, the activated complex can still either fall to reactants or to products. If these conditions are not met, then no reaction will occur…..its amazing that we have reactions occurring at all!

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  • Chapter 14: Chemical Kinetics

    NOTE – THIS CHAPTER IS #2 TOP TOPICS ON AP EXAM!!! NOT ONLY DO YOU NEED TO FOCUS ON THEORY (and lots of MATH) BUT YOU MUST READ THE FIGURES TOO!!!

    Ch 14.1 ~ Factors that Affect Reaction Rates SG Questions – Read p. 574-577. Answer the following questions

    1. Define chemical kinetics? 2. What is the definition of a reaction rate? 3. What is a reaction mechanism? 4. What are the four factors that affect a reaction (rxn) rate? 5. Why do each of the factors affect the rate? Explain. 6. What do rxn rates depend on? 7. Describe the relationship between the frequency of collusions and the rxn rate.

    Chemical kinetics is the study of the speed or rate of a reaction under various

    conditions.

    Reaction Rates – depend on collusions between molecules.

    Increase collusions = increase rates of the rxn

    THE COLLISION THEORY OF REACTION RATES - a reaction will only take place if ALL of the following conditions are met:

    1. Particles must collide. 2. Only two particles may collide at one time. 3. Proper orientation of colliding molecules so that atoms can come in contact with each other to

    become products.

    4. The collision must occur with enough energy so that the electrons can rearrange and form new bonds.

    5. This new collision product is at the peak of the activation energy hump and is called the activated complex or the transition state. At this point, the activated complex can still either

    fall to reactants or to products.

    If these conditions are not met, then no reaction will occur…..its amazing that we have reactions occurring at all!

  • Factors that Affect Reaction Rates – VERY IMPORTANT!!!

    There are several important factors which affect rates of reactions:

    1. Nature of the reactants--Some reactant molecules react in a hurry, others react very

    slowly.

    Physical state- solids vs. liquids vs. gases.

    example: When a solid reacts, only the particles on the surface of the solid are available for reaction. If the solid is broken up into smaller pieces its surface area gets larger and more particles are available for collision, therefore the reaction rate increases.

    Chemical identity - What is reacting? Usually ions of

    opposite charge react very rapidly. Usually, the more

    bonds between reacting atoms in a molecule, the slower the

    reaction rate.

    2. Concentration of reactants--more molecules, more

    collisions.

    Increasing the reactants concentration by putting more reactants into the same space increases the collision frequency, leading to a faster rate of reaction.

    A similar effect is observed when increasing the pressure in a gaseous reaction. Increasing the pressure of a gas can be achieved by reducing its volume whilst leaving the number of particles the same. (Boyle’s Law)

    3. Temperature—heat them up & speed them up; the faster

    they move, the more likely they are to collide.

    An increase in temperature produces more successful

    collisions that are able to overcome the needed activation

    energy, therefore, a general increase in reaction rate with

    increasing temperature.

    4. Catalysts—a chemical added to speed up the reaction. Catalysts are substances that increase the rate of a reaction whilst remaining chemically unchanged.

  • Catalysts usually work by providing an alternative reaction route that requires a lower activation energy (Ea).

    Most catalysts work by lowering the activation energy needed for the reaction to proceed, therefore,

    more collisions are successful and the reaction rate is increased.

    Remember! The catalyst is not part of the chemical reaction and is not used up during the reaction.*

    (May be homogeneous or heterogeneous catalysts.)

    Ch 14.2 ~ Reaction Rates SG Questions – Read p. 577-581. Answer the following questions

    1. How do we define the speed of a chemical rxn? 2. What units are used to describe a rxn rate?

    3. Consider the rxn A B, what is the average rate of disappearance of reactant A? (Don’t forget +/- signs)

    4. Consider the rxn A B, what is the average rate of appearance of product B? 5. Before rxn starts, how much of reactant do you have? How much product? 6. Can a rate be negative? If not, why?

    7. Write a rate expression for 2 HCl (aq) H2 (g) + Cl2 (g). 8. What happens to a rxn as it proceeds? 9. What is the Instantaneous Rate of of a reaction? 10. How is the instantaneous rate determined? 11. What is the instantaneous rate called at time=0?

    Reaction Rates

    • The speed of a reaction is defined as the change that occurs per unit time. • The speed of the chemical reaction is its reaction rate.

    For the reaction A → B, there are two ways of measuring rate:

    • the rate of disappearance of reactant A .

    [A] = concentration of A ∆ = change ∆ concentration of A = (concentration of A at final time) - (concentration of A at initial time) Note the negative sign! This reminds us that rate is being expressed in terms of the disappearance of a reactant.

    t

    A

    ][A Rate Average

  • t

    [B]B rate Average

    s

    M023.0

    s 0s 20

    M 0.000.46MRate Avg

    t

    D

    dt

    C

    ct

    B

    bt

    A

    a

    1111Rate

    • the rate of appearance of product B • The most useful unit for concentration is molarity (M). • The units for rate are M/s.

    Time (seconds) Concentration A Concentration B 0 1.00 M 0M 20 0.54 M 0.46M 40 0.30 M 0.30 M

    RATES ARE ALWAYS POSITIVE!!!

    Rate is not constant, it decreases with time!!!

    Reaction Rates and Stoichiometry

    For the reaction: aA + bB → cC + dD

    Example: 2HI(g) →H2(g) + I2(g)

    s 002

    )s 0at B of (Conc)20at B of (ConcB) (Conc Rate Avg

    s

    tst

    t

    ttt

    22 IHHI

    2

    1Rate

  • Ch. 14.3 ~ Concentration and Rate Laws SG Questions – Read p. 581-587. Answer the following questions

    1.What is a rate law? 2. What is the rate law for a reaction? 3. What does the constant “k” represent and what does it tell us? 4. When does the value of “k” change? 5. What do the exponents in the rate law represent? 6. How do we determine the overall reaction order? 7. Given the rate law, Rate = k[NH4

    +][NO2

    -], what would happen to the rate if the

    concentration of NH4 doubles? 8. How do we determine a rate law?

    9. What are the units for a rate constant? Does it vary? If so, provide some examples.

    Concentration and Rate

    In general, rates:

    • increase when reactant concentration is increased. • decrease as the concentration of reactants is reduced.

    • We often examine the effect of concentration on reaction rate by measuring the way in which reaction rate at the beginning of a reaction depends on reactants.

    • The overall concentration dependence of reaction rate is given in a rate law

    Rate law = k[A]

    x[B]

    y

    k = rate constant (depends on TEMPERATURE!)

    [A] = concentration of reactant A

    [B] = concentration of reactant B

    x, y = reaction orders (zero order, 1st order, 2

    nd order, etc….)

    Overall order of the reaction = sum (add) of all the individual orders.

    The following should be noted: VERY IMPORTANT!!!

    1. It is not possible to determine anything about the order of a reaction from the balanced

    chemical equation. Orders must be found experimentally (need data). Orders can be

    whole numbers or fractions.

    2. Units of the rate constant (k) varies and are often the subject of many AP questions.

    3. Rate constant (k) is dependent on TEMPERATURE!!

    4. A reactant that has no effect on the rate is said to have an zero order. Zero orders have no

    effect on the rate and since any number raised to the power of 0 is equal to 1, then don’t

    write it in the rate law equation.

  • HOW TO SOLVE FOR ORDERS USING DATA: MY RULES!!

    1. Look for two trials where the concentration of a reactant was held constant.

    2. Next, focus on the other reactant. Ask yourself how it’s concentration changed for the same two

    trials. Was it doubled? Was it tripled? Was it halved?

    Once you have determined the factor by which the concentration of the other reactant was changed,

    determine how that affected the rate for those same two trials. Expect easy math!

    Did changing the concentration have zero effect on the rate? If so, then it is zero order.

    Did the rate double when the concentration of the other reactant was doubled? If so, it is first order.

    Did the rate quadruple as a result of the reactant’s concentration doubling? If so, it is second order.

    Did the rate increase by a factor of eight? If so, it is third order.

    THINK of the concentration doubling as the number “two”.

    Rate = k[reactant] becomes

    Rate = k[2]m

    If the Rate doubled think…… 2 = k[2]m

    and more simply

    2 = [2]m

    so m = 1 (1st order for reactant)

    If the Rate quadrupled….. 4 = [2]m

    so m = 2 (2nd

    order for reactant)

    Experiment Number Initial Rate (M/s)

    Initial concentration

    [A]

    Initial concentration

    [B]

    1 0.50 × 10−2

    0.50 0.20

    2 0.50 × 10−2

    0.75 0.20

    3 0.50 × 10−2

    1.00 0.20

    4 1.00 × 10−2

    0.50 0.40

    5 1.50 × 10−2

    0.50 0.60

  • (Remember k depends on TEMPERATURE…so no matter what

    experiment you choose to calculate k, it will be the same for all

    experiments until Temperature changes!!! Keep this always in

    mind because you know chemistry – things don’t always stay the

    same.)

    Examples

    1. For the reaction between X and Y data has been collected in the table below. Determine the

    orders of reaction with respect to each reactant, the overall order, the rate equation, a

    value for the rate constant (k) and the units of the rate constant.

    Experiment Starting [X] Starting [Y] Rate (M/s)

    1 0.00500 0.0250 1.00

    2 0.0100 0.0250 4.00

    3 0.00500 0.0125 1.00

    Rate = k[X]a[Y]

    b

    2. For the reaction between X and Y data has been collected in the table below. Deduce the

    orders of reaction with respect to each reactant, the overall order, the rate equation, a

    value for the rate constant (k) and the units of the rate constant.

    Experiment Starting [X] Starting [Y] Rate (M/s)

    1 0.00500 0.0250 1.00

    2 0.0100 0.0250 2.00

    3 0.00500 0.0125 0.500

  • Units of Rate Constants (k)– They change depending on the orders!!! • For example, for a reaction that is second order overall: • Units of rate are:

    • Thus the units of the rate constant are:

    Just Remember……

    • Rate depends on concentration!!!!

    • Rate constant (k) IS affected by temperature and by the presence of a catalyst!!!!

    Ch. 14.4 ~ The Change of Concentration with Time SG Questions – Read p. 587-593. Answer the following questions

    1. What does a first-order reaction depend on? 2. What is the first order integrated rate law? 3. What are some examples of what you can solve for using the first order integrated law? 4. What does the integrated rate law’s equation take form of? 5. What are the units for an integrated rate law? 6. Look at Figure 14.8. We infer about the line of the graph of a first order reaction? 7. What is the second-order integrated rate law? 8. What is the zero-order integrated rate law? 9. What is the half-life of a reaction? 10. What is the equation used to find the half life of a first order rxn? Second order? 11. In a first order rxn, what occurs to the concentration of the reactant? 12. What is the relationship between the initial concentration of a reaction towards its half-

    life? 13. What do the slopes of a first order rxn and a zero order rxn have in common?

    2Mk(M/s)

    11

    22ion)concentratof(Units

    rate)of(Unitsk of Units sM

    M

    sM

  • The Change of Concentration with Time – Involves Calculus!!

    Integrated rate law—data table contains concentration and time data. Use graphical methods to

    determine the order of a given reactant. The value of the rate constant k is equal to the slope.

    Not nearly as hard as it sounds!

    INTEGRATED RATE LAW: CONCENTRATION/TIME RELATIONSHIPS.

    Set up your axes so that time is always on the x-axis. Plot the concentration of the reactant on the y-axis of

    the first graph. Plot the natural log of the concentration (ln [A], NOT log[A]) on the y-axis of the second

    graph and the reciprocal of the concentration on the y-axis of the third graph. You are in search of linear

    data!

    Zero order First order Second order k = negative slope k = negative slope k = positive slope

    “A” is reactant A and Ao is the initial concentration of reactant A at time zero [the y-intercept].

    Remember from basic Algebra….. y = mx + b

    zero order [A] = −kt + [Ao]

    first order ln[A] = −kt + ln [Ao]

    second order 1/[A] = kt + 1/[Ao]

    Also recognize that slope = k, since the rate constant is NEVER negative. If you are asked to write

    the rate law, it is simply Rate = k[A]order you determined from analyzing the graphs

    **** Can use pressure (gases only) instead of concentration!!!!

  • Zero-Order reaction

    First-Order reaction

    Integrated Rate law (Concentrations)

    ln[A]t = - kt + ln[A]0

    Integrated Rate law (Pressures)

    ln[A]t = - kt + ln[A]0

    Half-life - t½ , is the time required for the concentration of a reactant to decrease to half its original

    value.

    t1/2 = ln ½ = 0.693

    k k

    ln [A]t

    ln [A]0

    k = - slope

    t

    Does NOT depend on initial

    [Concentration ]

    lnP

    t

    [A]t

    [A]0

    k = - slope

    t

    Using

    pressure

    (gases only)

    l

    n

    P

  • Second-Order Reaction

    Integrated Rate Law

    1 = kt + 1

    [A]t [A]0

    Half-life

    t1/2 = 1

    k[A]0

    *****SEE GRAPHING INSTRUCTIONS FOR YOUR CALCULATOR*****

    Lets do some housekeeping first!

    Press ON CLEAR to clear your “home” screen. Next, let us clear your lists.

    Press 2nd

    + and your screen will look like this:

    Press 4, then ENTER

    It will say “Done”. Press Y= to double check that all list are clear.

    Then press 2nd

    MODE to exit once you confirm all list are clear.

    1

    [A]t k

    1

    [A]0

    t

    Dependent of initial [concentration ]

    k

  • Check your MODE

    Press MODE…check that all the following are highlighted: Then press 2nd

    MODE to exit once you confirm all seetings.

    TO ENTER DATA INTO LISTS

    Press STAT , then 5 “setup editor”, ENTER ….should say “Done”

    STAT 1 “edit”

    Let’s enter this example set of data:

    [N2O5] Time (s)

    0.1000 0

    0.0707 50

    0.0500 100

    0.0250 200

    0.0125 300

    0.00625 400

    • Time is the independent variable and goes into L1 (to be plotted on the x-axis). Hit ENTER

    after each number and the cursor moves down the lists.

    • If you mess up, just use your arrow keys to navigate to the goof and type over it! When all the data is entered, screen looks like this:

    DATA TABLE

    L1 = x-axis (time)

    L2 = Molarity [M] or pressure (gases)

    L3 = ln concentration (move cursor on L3 you need “ln L2” so plug in ln 2nd

    2 )

    L4 = 1/ [A] (move cursor on L4 you need “1/L2”… so plug in 1 / 2nd

    2 )

  • Hopefully your screen will look like this.

    SETTING UP GRAPH

    2nd

    y= : stat plot

    1: ENTER

    Turn ON

    Type:

    Mark: or +

    DETERMINING ORDERS – Time to graph! (I always graph 1st order first)

    X list: L1 (time) LINEAR means that the rate is 1st order

    Y list: L3 (ln M) If curve – NOT 1st order

    OR

    X list: L1 (time) LINEAR means that the rate is 2nd

    order

    Y list: L4 (1/ M) If curve – NOT 2nd

    order.

    Press GRAPH

    If you don’t see a graph, it means that you axis window

    is not set properly.

    Press ZOOM 9: Zooms in on graph

    if straight linear line than its “1st order”

  • if curved than not 1st order….. so then if you graphed 2nd order it will be linear straight line

    DETERMINING SLOPE – Rate Constant (k)

    If 1st order slope = -k

    If 2nd

    order slope = +k

    Press STAT

    move cursor to CALC

    4 “linreg (ax + b),

    Then type “L1 , L3 , ” for 1st order. (If second order, type “L1 ,

    L4 , “)

    Press VARS

    move cursor to Y-VARS, ENTER to “1:Function”

    Press ENTER to “Y1”

    Press ENTER

    Remember Alg class y = mx+b where m is slope

    In chemistry, y = ax+b where a is the slope

    a = slope = k (units are 1/s if 1st order…. Units for second order are 1/M s)

    Since graph ln[conc.] vs. time showed a first order reaction,

    Rate = k[N2O5] AND…

    The k is to the slope

    Rate = 6.93 x 10 –3

    1/s [N2O5]

    Let’s graph 2nd

    order so you can see the difference in linear vs. curve!

  • Ch. 14.5 ~ Temperature and Rate SG Questions – Read p. 593-599. Answer the following questions

    1.What happens to the rate as temperature increases? 2.What is the collision model? 3.What happens as concentration increases, according to the collusion model? 4.What can the kinetic energy of molecules be used for?

    5. Define Activation Energy (Ea). 6.Draw Figure 14.15.

    7. Draw Figure 14.17. Label where the Ea, and ΔE within the energy profile. Also label where the activation complex is.

    8. What does the activation complex tell us? Explain. 9. What is the relationship between the rate of a rxn and the amount of Ea is has? 10. Draw the reverse energy profile for Figure 14.17. What should change about it? 11. How can we tell, from the energy profile, when the reaction is endothermic or exothermic? 12. Copy Figure 14.18!!!! Was on the AP test last year!

    Muy importante! Understand it.

    13. Write the Arrhenius Equation. What does it tell us? 14. What is the “manipulated” version of the Arrhenius

    equation?

    Temperature and Rate

    • Most reactions speed up as temperature increases. • Example: Chemiluminescent light sticks (PRODUCES LIGHT) • Two light sticks are placed in water, one at room temperature and one in ice. • The one at room temperature is brighter than the one in ice. • Its luminescence also fades more quickly. The higher the temperature, the faster the reaction and the brighter the light. • As temperature increases, the rate increases.

    How is the relationship between temperature and rate reflected in the rate expression?

    • The rate law has no temperature term in it, so the rate constant must depend on temperature.

  • The Collision Model – Essay Free Response!!!

    Rates of reactions are affected by concentration and temperature.

    • In order for molecules to react, they must collide.

    • The greater the number of collisions = the faster the rate. • The more molecules present, the greater the probability of collision and the faster the rate. • The higher the temperature, the more energy available to the molecules and the more frequently the molecules collide. • Thus, reaction rate should increase with an increase in temperature.

    Activation Energy – IMPORTANT!!!

    Molecules must posses a minimum amount of energy to react. Why? • In order to form products, bonds must be broken in the reactants.

    • Bond breakage requires energy.(endothermic)

    Activation energy, Ea, is the minimum energy required to initiate a chemical reaction. • The activation energy is the difference in energy between reactants and the activated complex (The species at the top of the barrier – highest point on curve). • The rate depends on the magnitude of the Ea. • In general, the lower the Ea, the faster the rate!!!! (And faster the rate constant (k)) • Notice that if a forward reaction is exothermic , then the reverse reaction is endothermic • ∆Erxn has no effect on reaction rate.

  • Activation Graphs

    Ea

    Reactants

    Potential Energy

    E

    Products

    Reaction Pathway

    Activated Complex

    Products

    Potential Energy Ea ----------------

    Reactants E

    ----------------

    Reaction Pathway

    Activated Complex

    E = - (Exothermic)

    E = + (Endothermic)

  • The Arrhenius Equation

    • Arrhenius discovered that most reaction-rate data obeyed an equation based on three factors: • the number of collisions per unit time, • the fraction of collisions that occur with the correct orientation, and • the fraction of the colliding molecules that have an energy equal to or greater than ∆Ea. • From these observations Arrhenius developed the Arrhenius equation.

    (use this equation if Temp doesn’t change) y = mx + b k is the rate constant Ea is the activation energy R is the gas constant (8.314 J/K∙mol) ( Remember 1 J = 1kg m

    2 /s

    2)

    T is the temperature in K. A = frequency factor

    A graph of ln k (y-axis) vs 1/T (x-axis)

    Slope = –Ea/R

    y-intercept of ln A.

    We can graph this in calculator…..

    DATA TABLE

    L1 = T

    L2 = k (rate constant)

    L3 = 1/T (move cursor on L3 so plug in 1 / 2nd

    1 )

    L4 = ln k (move cursor on L4 so plug in ln 2nd

    2 )

    Time to graph! – Will always produce a linear line!!!

    X list: L3 (1/T)

    Y list: L4 (ln k)

    Slope: Then type “L3 , L4 , ” for 1

    st order.

    **If temperature changes, use a different Arrhenius Aequation**

    ART

    Ek a lnln

    ln k

    ln A

    Slope k = - Ea/R

    1/T

    122

    1 11lnTTR

    Ea

    k

    k

  • REMEMBER: k is controlled by Temperature!!!

    So, if T changes, k changes (As you can see we now have 2 k’s and 2 T’s)

    Ch. 14.6 ~ Reaction Mechanisms SG Questions – Read p. 599-606. Answer the following questions 1. Define Reaction Mechanism. What does it describe? 2. Define and give an example of an Elementary Reaction. 3. How many elementary reactions can be found in a complete reaction? 4. What is the relationship between the molecularity of a reaction and its elementary reactions? 5. What is the molecularity of a reaction when there is only one single molecule involved? Two?

    Three? Which one is the most common? Least common?

    6. Copy and Read Figure 14.20. 7. How do we determine when a reactant is an intermediate? What can we observe from the

    elementary steps to help figure it out?

    8. How do we determine if a reaction is elementary? 9. What is the relationship between a reaction’s elementary steps and rate law? 10. Copy Table 14.3. Important to know! 11. What does the slowest elementary step reveal about a reaction’s rate? 12. If Rate=k[NaCl] for the reaction 2 NaCl → 2 Na(s) + Cl2(g), then what is the rate of the reverse

    reaction?

    Reaction Mechanisms

    • The balanced chemical equation provides information about substances present at the beginning and end of the reaction.

    • The reaction mechanism is the process by which the reaction occurs. • Mechanisms provide a picture of which bonds are broken and formed during the course of a reaction.

    Elementary Reactions

    • Elementary reactions are any processes that occur in a single step. • The number of molecules present in an elementary step is the molecularity of that elementary step. • Unimolecular reactions involve one molecule. • Bimolecular elementary reactions involve the collision of two molecules. • Termolecular elementary reactions involve the simultaneous collision of three molecules. [very rare!]

  • Multistep Mechanisms

    • A multistep mechanism consists of a sequence of elementary steps. • The elementary steps must add to give the balanced chemical equation. • Some multistep mechanisms will include intermediates.

    INTERMEDIATES: - appear in an elementary step but are neither a reactant nor product. - are formed in one elementary step and consumed in another. - They are not found in the overall balanced equation.

    REMEMBER: RATE LAWS ARE DETERMINED EXPERIMENTALY!! HOWEVER… The rate laws of the elementary steps determine the overall rate law of the reaction. • The rate law of an elementary step is determined by its molecularity (BALANCED EQUATION). ***Can use coefficients to determine order if it states “rxn occurs as single elementary rxn”*** • Unimolecular processes are first order. • Bimolecular processes are second order.

    • Termolecular processes are third order.

    The Rate Determining Step for a Multistep Mechanism

    • Most reactions occur by mechanisms with more than one elementary step. • Often one step is much slower than the others. • The slow step limits the overall reaction rate. • This is called the rate-determining step (rate-limiting step) of the reaction. • This step governs the overall rate law for the overall reaction.

    ELEMENTARY STEP MOLECULARITY RATE EXPRESSION

    A→ products unimolecular rate = k[A]

    A + B → products bimolecular rate = k[A][B]

    A + A → products bimolecular rate = k[A]2

    2 A + B → products* termolecular* rate = k[A]2[B]

  • 2k

    2k

    1

    1

    k

    k

    1k

    Mechanisms with a Slow Initial Step

    • Consider the reaction: NO2(g) + CO(g) →NO(g) + CO2(g)

    The experimentally derived rate law is: Rate = k[NO2]2

    • We propose a mechanism for the reaction: • Step 1: NO2(g) + NO2(g) NO3(g) + NO(g) slow step

    • Step 2: NO3(g) + CO(g) NO2(g) + CO2(g) fast step Note that NO3 is an intermediate.

    SO…… Rate = k[NO2]

    2

    Mechanisms with a Fast Initial Step

    • Consider the reaction: 2NO(g) + Br2(g) → 2NOBr(g)

    The experimentally determined rate law is:Rate = k[NO]2[Br2]

    • Consider the following proposed mechanism: • Step 1: NO (g) + Br2(g) NOBr2(g) fast step

    •Step 2: NOBr2(g) + NO(g) 2NOBr(g) slow step The theoretical rate law for this mechanism is based on the rate-determining (slow) step, So…..Rate = k2[NOBr2][NO]

  • Ch 14.7 ~ Catalysis SG Questions – Read p. 606-609. Answer the following questions 1. Define Catalysis? 2. What is the difference between a homogeneous and heterogeneous catalyst? Give different

    examples than those provided in the text.

    3. How do can we distinguish which reactant and products are catalysts or intermediates in a reaction?

    4. Copy Figure 14.23. Notice what the catalyst does to the activation energy In a reaction. What does it do? Explain why.

    Catalysis

    A catalyst is a substance that changes the rate of a chemical reaction without itself undergoing a permanent chemical change in the process.

    There are two types of catalysts:

    • homogeneous and • heterogeneous.

    Catalysts are common in the body, in the environment, and in the chemistry lab!

    Homogeneous Catalysis

    • A homogeneous catalyst is one that is present in the same phase as the reacting molecules. • For example, hydrogen peroxide decomposes very slowly in the absence of a catalyst: 2H2O2(aq) →2H2O(l) + O2(g)

    • In the presence of bromide ion, the decomposition occurs rapidly in acidic solution:

    2Br

    –(aq) + H2O2(aq) + 2H

    +(aq) → Br2(aq) + 2H2O(l)

    Br2(aq) + H2O2(aq) → 2Br–(aq) + 2H

    +(aq) + O2(g)

    • Br2 is the intermediate (gets produced and quickly comsumed) Br

    – is a catalyst because it appreas as a reactant in the 1

    st rxan, disappears, and then is

    regenerated (reproduced) at the end of the 2nd

    reaction. • The net reaction is still: 2H2O2(aq) →2H2O(l) + O2(g)

  • How do catalysts increase reaction rates? 1. Catalysts speed up the reaction 2. Catalysts operate by lowering the overall activation energy for a reaction. 3. Catalysts can operate by increasing the number of effective collisions. 4. Catalyst provide a completely different mechanism for the reaction.

    Heterogeneous Catalysis

    • A heterogeneous catalyst exists in a different phase than the reactants.

    • Often we encounter a situation involving a solid catalyst in contact with gaseous reactants and gaseous products (example: catalytic converters in cars) or with reactants in a liquid.