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Rates of Reaction Chemical Kinetics


Title: Chapter 14 - Rates of Reaction Chemical Kinetics Author: Jeffrey A. Hurlbut Last modified by: mine Created Date: 2/6/2002 3:14:27 AM Document presentation format – PowerPoint PPT presentation

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Title: Rates of Reaction Chemical Kinetics

Rates of ReactionChemical Kinetics
I. Background on Rates Mechanisms
  • Chemists study reactions. Some of what we study
  • - Major Minor Products
  • - Reactants
  • - Completion
  • - Effects of Catalysts
  • - Separation and Purification of Products
  • - Effects of Variables on Rxn Speed (rate) on
  • - Rates of the Reaction
  • - Mechanism of the Reaction
  • This chapter deals with rates mechanisms of
  • Mechanism Step-by-step progress of the chemical
  • Rate How fast the reaction proceeds (usually
    ? M / Time)

I. Background on Rates Mechanisms
  • Study of rates is useful since the results will
  • 1) Indicate how to manipulate factors to
    control the reaction
  • 2) Lead to the mechanism of the reaction
  • 3) Indicate time needed to get a given amount
    of product
  • 4) Indicate amount of product in a given amount
    of time
  • Study of mechanisms important. With the
    mechanism we can
  • 1) Predict products of similar reactions
  • 2) Better understand the reaction
  • 3) Accurately manipulate the reaction for a
    desired result
  • 4) Organize and simplify the study of organic
  • OH
  • Example CH3-CH-CH3 H I-
    -------) CH3-CH-CH3 H2O

I. Background on Rates Mechanisms
  • Main Factors which influence reaction rate
  • Concentrations of Reactants - Rates usually
    increase as reactant concentrations increase.
  • Reaction Temperature - An increase in temperature
    increases the rate of a reaction.
  • Presence of a Catalyst (not all rxns have
  • A catalyst is a substance which increases the
    rate of a reaction without being consumed in the
    overall reaction.
  • The concentration of the catalyst or its surface
    area (if insoluble) are variables which influence
    the rate.
  • Some catalysts are incredibly complex - like
    enzymes and others are quite simple H
    H2O CH2 CH2 ------) CH3-CH2-OH H
  • Type of Reactants
  • Surface Area of Insoluble Reactant

II. Rates
  • Reaction Rate either the increase in M of
    product per unit time or the decrease in M of
    reactant per unit time ?M / ?T
  • Note X moles X / Liter
  • Example H Catalyst
  • Sucrose H2O --------------) Glucose
  • Rate rate of formation of either product.
  • Rate ? M of glucose / ?sec ?glucose
    / ?sec
  • or
  • Rate rate of disappearance of either
  • Rate - ?sucrose / ?sec
  • In order to obtain rate, we need a way to measure
    ?M of any reactant or product with respect to

II. Rates
  • Example 2 N2O5 -----) 4 NO2 1 O2
  • If we want to equalize the rates then
  • Rate ?O2 1/4 ?NO2 - 1/2
  • ?t ?t ?t
  • - divide by balancing coefficients when we
    equalize rates.
  • Various Rates can be determined 1)
    instantaneous rate at a given time 2) average
    rate over a long period of time or 3) the
    initial rate rate at the beginning of the rxn
    (this is used the most).
  • On next slide the ?O2 versus time is plotted
    for a reaction.
  • Note 1) how the rate changes with time 2) that
    rate is the tangent at a given point on the curve.

II. Rates
II. Rates
II. Rates
  • Example of average rate calculation for
  • I- ClO- -----) Cl- IO-
  • Given the I- and time in seconds, then what is
    the average rate?
  • Time (s) M I-
  • 2.00 0.00169
  • 8.00 0.00101
  • - Rate - ?M / ?T
  • - Rate - (0.00101 - 0.00169) M / (8.00 - 2.00)
    s 6.8x10-4 M / 6.00s
  • - Rate 1.1 x 10-4 M/s

II. Rates
  • Need to obtain the change in M of a given reagent
    per change in time could follow any parameter
    related to concentration.
  • Examples of what one might follow to obtain
  • - a change in pressure (if gas produced or
    consumed in the rxn)
  • - a change in pH (if acidity changes in the rxn)
  • - a change in absorbance of electromagnetic
    radiation (EMR)
  • Usually measure absorbance of Visible or UV EMR
    at a given ? - caused by a change in reactant or
    product concentration.
  • A ?bc at a given ? (wavelength) This is
    Beers Law
  • A absorbance use spectrophotometer to
    measure has no units.
  • ? molar absorptivity a constant _at_ given ?
    has units of M-1cm-1
  • b pathlength of EMR through sample usually
    1.00 cm cuvette used.
  • c concentration in M
  • A plot of A versus M _at_ given ? will yield a
    straight line and the equation
  • A Ebc intercept If follow ?A then
    can convert to ?M get rate.

III. Rate Law rate k x Am x Bn
for A B
  • Rate Law relates the rate to temperature
  • Rate law is given in terms of REACTANTS only
  • k rate constant handles the temperature
  • The exponents are the order handle the
    concentration variables.
  • General form of the rate law for a A b
    B c C d D
  • rate k x Am x Bn
  • - Order for A is m order for B is n
    Overall order is m n
  • - m n are determined experimentally.
  • - k, also determined experimentally units
    depend upon overall order.

III. Rate Law - Rate Constant Units
  • Note Assume time is in seconds (s). Rate k
  • Solve for k plug in units k Rate / Ax
  • (this may be useful for one of the online HW
  • Overall Rxn Order, x Units for k
  • zero Ms-1
  • first s-1
  • second M-1s-1
  • third M-2s-1

III. Rate Law
  • Example 2 NO2 1 F2 ----) 2 NO2F
    Rate k NO2n x F2m
  • In the laboratory, the overall rate was found to
    be second order.
  • n m 2 Possibilities n2 m0
    n0 m2 n1 m1
  • Experiments demonstrated that n1 and m1 How?
    By running the reaction at least three times
    1 getting rate at certain initial
    concentrations of NO2 F2 2 getting rate
    when keeping NO2 the same doubling F2 3
    getting rate when keeping F2 the same
    doubling NO2.
  • They found that doubling NO2 doubled the rate
    doubling F2 doubled the rate so, both
    coefficients had to be 1.
  • Rate k NO21 F21

III. Rate Law - Rate Constant
  • Determination of the rate constant, k. You are
  • 1) aA bB ? cC dD
  • 2) Rate kA0B1 (0 1 were determined
  • 3) M of A B 2.00 moles/L Rate 2.50 x
    10-2 M/s
  • Determine the value for k give complete rate
  • Rate kA0B1 k Rate
  • A0B1
  • k Rate 2.50 x 10-2 M/s 2.50 x
    10-2 M/s
  • A0B1 2.00 M0 x 2.00 M1
    1 x 2.00 M1
  • k 1.25 x 10-2 s-1
  • Rate (1.25 x 10-2 s-1) A0 B1
    (completed rate expression)

III. Rate Law
  • Rate Law for a reaction is found experimentally
    except for a single step in a mechanism
    (elementary reaction). Assume rxn is NOT
    elementary unless it is one step in a mechanism.
  • Given 1NO2 1CO -----) 1NO 1CO2
  • By experiment, the rate law was found to be
  • rate kNO22CO0 or rate kNO22
    ( Note CO0 1 )
  • The order WRT each reactant the overall order
  • 2nd order WRT NO2 0th order WRT CO
    2nd order overall

IV. Order
  • The concentration variables are handled by the
    exponents - the order.
  • The orders are determined experimentally except
    for one case An
  • elementary reaction.
  • Elementary reactions are one step reactions which
    are the individual steps in a mechanism. (For
    an elementary reaction only the balancing
    coefficients determine the order.) - Important
  • Example 1 for a multistep reaction 1 CH2Br2
    2 KI ---) 1 CH2I2 2 KBr
  • Experimentation found that m n were both
    first order then
  • rate k CH2Br2 1 KI1
  • Example 2 for an elementary reaction 2 O3
    ---) 3 O2 (assume elementary)
  • No need for experimentation order comes from
    balancing coefficients
  • rate kO32

IV. Determination of Order
  • Order - from units of k If you are given the
    units of the rate constant for a reaction, then
    you will know the overall order (slide 14). Not
    too common.
  • Order by Method 1 - from altering M Measure
    initial rates keeping one reactant constant and
    change the concentration of another observe the
    rates calculate order as illustrated in the next
    few slides.
  • Order by Method 2 - from integrated rate
    expression Use calculus integrate the rate
    expression between the limits of time 0 time
    t. By plotting out the variables of these
    integrated rate expressions you can determine the
    order. This will be shown in the lecture, and
    you will be doing this in the kinetics lab.

Order by Method 1 - from altering M Measure
initial rates keeping one reactant constant and
change the concentration of another observe the
rates calculate order as illustrated in the next
few slides.
IV. Determination of Order by varying M
  • Example 1 Determine the order for rate
    expression for
  • 2N2O5 ---) 4NO2 1O2 rate k
  • Exp 1 Rate 4.8x10-6 Ms-1 at 1.0x10-2 M
  • Exp 2 Rate 9.6x10-6 Ms-1 at 2.0x10-2 M
  • Order Note that when N2O5 doubles, the rate
    doubles. Since
  • rate a N2O5m rate doubles when N2O5
    doubles, the value of
  • must be 1 the order is 1.
  • - rate a N2O5m rate doubles when
    N2O5 doubles, then
  • go from 1m 1 to 2m 2 m has to be
  • Rate law rate k x N2O51

IV. Determination of Order by varying M
  • Summary
  • Effects of doubling reagent M while keeping
    others constant
  • Rate remains the same 0th order M0
  • Rate doubles 1st order M1
  • Rate quadruples 2nd order M2
  • Rate increases eightfold 3rd order M3

IV. Determination of Order by varying M
  • Example 2 2 NO Cl2 -----) 2 NOCl -
    Calculate order of Rxn
  • Exp Initial NO Initial Cl2 Initial Rate,
  • 1 0.0125 0.0255 2.27x10-5
  • 2 0.0125 0.0510 4.57x10-5
  • 3 0.0250 0.0255 9.08x10-5
  • Rate kNOmCl2n
  • a) calculate n From 1 2 - double Cl2 keep
    NO constant
  • rate increases by factor of 2.01 n 1
  • b) calculate m From 1 3 - double NO keep
    Cl2 constant
  • rate increases by factor of 4.00 m 2
  • Rate kNO2Cl21
  • 2nd order wrt NO 1st order wrt Cl2
    3rd order overall

  • Order by Method 2 - from integrated rate
  • Integrate the rate expression between the limits
    of time 0 and time t. By plotting out the
    variables of these integrated rate expressions
    you can determine the order. You will be doing
    this in the kinetics lab.

IV. Determination of Order by Integrated Rate
  • Summary on use of logarithms
  • Log involves s to the base 10 Log 10x
  • Ln Natural log uses s to the base e Ln ex
  • Ln A/B Ln A - Ln B (or Log)
  • Ln A x B Ln A Ln B (or Log)
  • Ln Ab b Ln A (or Log)
  • To obtain either log or ln use the appropriate
    calculator function.
  • Log 2.1x10-4 - 3.68 (note significant figure
    change - see below)
  • (-4.0000000. 0.32 -3.68 cut off at
    first doubtful digit)
  • To remove Ln Log use the inverse ex 10x
    functions on cal.
  • Inverse log 3.00 or 103.00 1.0 x 103
  • Inverse ln 3.00 or e3.00 20.

IV. Order Integrated Rate Law - First Order Rxns
  • 1) 1st Order Reactions aA -----) Products
    If 1st order, then
  • -?A/?t kA1 (rate expression)
  • This plot for first order data only gives minimal

IV. Order Integrated Rate Law - First Order Rxns
  • 1) 1st Order Reactions aA -----) bB -?A/?t
  • if we integrate from time t to 0, we get the
  • Y mX b
  • lnAt - kt lnAo or
    lnAt/Ao -kt
  • where At M of A at time t Ao M
    of A at t 0
  • - A plot of lnAt versus t gives a straight line
    (Y mX b)
  • b Slope (m) - k
  • Note Only linear for 1st order

Time, t
Example of an integrated rate plot for a 1st
order reaction
Slope -k rise/run
Must be 1st order since plot of lnN2O5 vs t is
linear. Can get k from the slope.
IV. Order Example using 1st order integrated
  • Example 2N2O5 ---) 4NO2 O2 rate k
    N2O51 (1st order)
  • Given k 4.80x10-4s-1 N2O5to 1.65x10-2
    M what is N2O5 at 825 s?
  • ln At - k x
    t ln Ao
  • ln N2O5 - 4.80x10-4s-1 x 825 s
    ln 1.65x10-2
  • ln N2O5 - 0.396
    - 4.104
  • ln N2O5 - 4.500
  • Take inverse ln or anti ln of both sides get
  • inverse ln -4.500 or e -4.500
  • N2O5 0.0111 M

IV. Order Integrated Rate Law - First Order Rxns
  • Half-Life (t1/2) of 1st Order Reaction
  • t1/2 time it takes for Ao to decrease to 1/2
    initial M ½Ao
  • ln At / Ao -kt ln 1/2Ao / Ao
  • ln 1/2 -kt1/2 -0.693 -kt1/2 t1/2
    0.693 / k
  • Note 1) Time for 1/2 to disappear is
    independent of A for 1st order reaction. 2)
    This is an easy way to calculate 1st order rate
    constant, k.
  • Example If t1/2 189 sec for 1st order
    decomposition of 1.0 mole of H2O2, then how much
    H2O2 will be left after 378 sec?
  • Note 378/189 2 Goes through
    two half lives
  • 1.0 mol ---) 0.50 mol ---) 0.25 mol

IV. Order Integrated Rate Law - First Order Rxns
  • Example Given a) k 3.66x10-3s-1 for
    decomposition of H2O2 and b) H2O2o 0.882 M.
  • 1) t1/2
  • 2) How much will be left after one half-life?
  • (Note Reaction must be 1st order examine units
    for k)
  • 1) t1/2 0.693/k
  • t1/2 0.693 / 3.66x10-3s-1 189 s
  • 2) M of H2O2 cut in half in one half-life
    (t1/2) will go from 0.882 to 0.441 M in 189 s.

IV. Order Integrated Rate Law - Second Order Rxns
  • 2) Second Order Reactions
  • - Assume that aA -----) Products is 2nd
  • Rate - ?A / ? t k A2
  • Integrate rate expression from time t to 0 get
  • 1/At k t 1/Ao So, a plot of
    1/At vs t should give a straight line with
    slope k and y intercept 1/Ao
  • t1/2 1 Note Now t1/2 depends on
    initial M
  • k x A0
  • Note can tell if reaction is 2nd order from
    1/A vs t plot.

IV. Order Integrated Rate Law - Second Order
Rxns Example
Plot of lnNO2 vs t is not linear not 1st
IV. Order Integrated Rate Law - 0th Order
  • 3) 0th Order Reactions Assume A ---) B
    is 0th order
  • Rate -kA0
  • Rate -k
  • - Integrated Rate Equation for a 0th order
  • At -k x t A0
  • - a plot of At versus t will give a straight
  • - Again, if you let At 1/2 Ao then t
  • - t1/2 A0 / 2k

IV. Order Integrated Rate Law - Summary
? Rate when double M None
Double Quadruple
IV. Order Integrated Rate Law - Summary
A? B ?A/?t kAn
Note slope -k or k in each case
0th Order n0 At - kt Ao
Time, t
1st Order n1 lnAt - kt lnAo
2th Order n2 1/At kt 1/Ao
Time, t
V. Temperature
  • A collision needs to occur before a reaction can
    take place, the rate constant ( rate) of the
    reaction depends upon the
  • 1) collision frequency (temperature)
  • 2) number of collisions having enough energy for
    rxn (Ea)
  • 3) orientation of particles upon collision
  • Ea energy of activation minimum energy of
    collision in order for the reaction to take
  • Ea ?H can be represented by Potential Energy
    Diagram can draw for one step or for several
    steps in a mechanism.

V. Temperature Reaction Rate A)
Potential Energy Diagram for an Elementary
What is Ea for reverse reaction?
V. Temperature Reaction Rate Arrhenius
  • Arrhenius Equation relates rate constant (k),
    temperature (T), energy of activation (Ea in
    J/mole), orientation factor.
  • k A e-Ea/RT R gas constant use R
    8.31 J/(Kmole)
  • Take ln of both sides ln k -(Ea/R) 1/T
    ln A
  • Y m X b
  • Measure k at several temperatures and make plot
    of ln k
  • versus 1/T. Slope of the curve - Ea/R
    (will give Ea)
  • Note A is a constant includes orientation
  • Note Page 549 contains a form of the Arrhenius
    equation which is useful for some online HW

V. Temperature Reaction Rate Arrhenius
EquationData below from 4 experiments - detn of
rate constant, k, at 4 temperatures for a rxn
ln A
Slope -Ea/R Can now determine Ea
ln k
1/T (in K-1)
ln k -(Ea/R) 1/T ln A
From k Ae-Ea/RT Y m X
b Use R 8.31J/(K.mol)
VI. Mechanisms A) Introduction
  • Mechanism step by step progress of chemical
  • Most likely mechanism is determined
    experimentally from a study of rate data.
  • The mechanism consists of one or more elementary
    reactions which add up to give you the overall
  • Species which is generated then consumed in the
    mechanism is called an intermediate Species
    which is added, consumed then regenerated is a
  • Step with largest Ea (slowest step) is called the
    rate determining step governs overall reaction

VI. Mechanisms B) Example 1 - Information
  • Overall Rxn O3 2NO2 -----) O2 N2O5
  • Suggested two-step mechanism (from
  • Step 1) O3 NO2 -----) NO3 O2
  • Step 2) NO3 NO2 -----) N2O5
  • rate k O31NO21 - all from slow first step
  • Notes a) Two elementary reactions (NOTE
    balancing coefficients orders in an
    elementary rxn)
  • b) Steps add up to give overall rxn
  • c) NO3 is an intermediate (produced used
  • d) There is no catalyst
  • e) Slowest step governs the overall rate
  • mechanism is useful will give us a) practical
    data, b) rate law, c) theoretical data, d)
    understanding of reaction

VI. Mechanisms B) Example 2 - Calculate Rate
  • Determine a) general rate expression b)
    complete rate law from the following mechanism
    Note can directly get the order for an
    elementary rxn from the balancing coefficients.
  • 1) 1I2 2I (fast equilibrium)
  • 2) 2I 1H2 2HI (slow)
  • a) Overall Rxn from addn of steps 1I2 1H2
  • General rate law rate kI2xH2y
  • b) Complete rate expression from mechanism
  • From step 2 (rxn rate slow step rate)
    rate k2 Io2 x H21
  • From step 1 Keq Io2/I21 Io2
    KeqI21 Substitute into above
  • rate k2 Keq I21 H21
  • rate k I21 H21

IV. Mechanisms C) Catalysts
  • Catalyst A chemical which speeds up a reaction
    without being consumed in the reaction.
  • - They operate by lowering the Ea for the rate
    determining step.
  • - One example is Pt which speeds up the
    following rxn
  • CO 1/2 O2 -----) CO2
  • - Pt can be used in catalytic converter for
    your car exhaust.
  • Most famous catalysts are proteins called
  • - Enzymes extremely specific biochemical
    catalysts that allow complex reactions to take
    place in living systems under mild conditions.
  • - Enzymes are very complex, well designed, and
    usually have molecular weights in the tens of
  • - Their mode of operation uncovered only 60
    years ago.

VI. Mechanisms C) Catalysts
A catalyst speeds up the rxn by lowering the Ea
provides a different mechanism with a lower Ea
VI. Mechanisms C) Catalysts
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