Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

Chemical Kinetics: The Rates of Chemical Reactions
Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Chemical Kinetics will now provide information about the arrow! This gives us information on HOW a reaction occurs! Reactants Products

Chemical Kinetics Kinetics is the study of how fast (Rates) chemical reactions occur. Important factors that affect the rates of chemical reactions: reactant concentration or surface area in solids temperature action of catalysts Our goal: Use kinetics to understand chemical reactions at the particle or molecular level.

Rate of Reactions Reactants go away with time.
Products appear with time. The rate of a reaction can be measured by either. In this example by the loss of color with time.

Determining a Reaction Rate
Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc. with time — can be determined from the plot. See Chapter 15 Video Presentation Slide 5 Dye Conc A B Time

Reaction Rate & Stoichiometry
In general for the reaction: aA + bB  cC + dD reactants go away with time therefore the negative sign… Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

The rate of appearance or disappearance is measured in units of concentration vs. time. Rate = = Ms1 or Mmin1 etc... time There are three “types” of rates initial rate average rate instantaneous rate

Reactant concentration (M)
Reaction Rates The concentration of a reactant decreases with time. Reactant concentration (M) Time

Reaction Rates Initial rate Reactant concentration (M) Time

Reaction Rates Reactant concentration (M) Time

Instantaneous rate (tangent line)
Reaction Rates Instantaneous rate (tangent line) Reactant concentration (M) Time

Instantaneous rate (tangent line)
Reaction Rates Initial rate Instantaneous rate (tangent line) During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction. Reactant concentration (M) Time

Problem: Consider the reaction:
Over a period of 50.0 to s, the concentration of NO(g) drops from M to M. a) What is the average rate of disappearance of NO(g) during this time?

Problem: a) RATE= - D[NO] Dt 1 2 b) D[NO]/ Dt Consider the reaction:
Over a period of 50.0 to s, the concentration of NO(g) drops from M to M. What is the rate of rxn? What is the average rate of disappearance of NO(g) during this time? a) RATE= - D[NO] Dt 1 2 (0.0100M  M) 100.0 s  50.0 s = 1.50104 Ms1 b) D[NO]/ Dt = /50.0 M/s = X 10-4M/s

Practice example Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) rate = - D[CH4] Dt = - D[O2] Dt 1 2 = D[CO2] Dt = D[H2O] Dt 1 2

Reaction Conditions & Rate
There are several important factors that will directly affect the rate of a reaction: Temperature The physical state of the reactants Addition of a catalyst All of the above can have a dramatic impact on the rate of a chemical process.

Reaction Conditions & Rate: Temperature
Bleach at 54 ˚C Bleach at 22 ˚C See Chapter 15 Video Presentation Slide 8

Reaction Conditions & Rate: Physical State of Reactants
See Chapter 15 Video Presentation Slide 7

Reaction Conditions & Rate: Catalysts
Catalyzed decomposition of H2O2 2 H2O2  2 H2O + O2 See Chapter 15 Video Presentation Slide 4

Effect of Concentration on Reaction Rate: Concentration
Mg(s) HCl(aq)  MgCl2(aq) + H2(g) See Chapter 15 Video Presentation Slide 6 0.3 M HCl 6 M HCl

The Rate Law Expression
The rate of reaction must be a function of concentration: As concentration increases, so do the number of collisions… As the number of collisions increase, so does the probability of a reaction. This in turn increases the rate of conversion of reactants to products. The relationship between reaction rate and concentration is given by the reaction Rate Law Expression. The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants.

For the general reaction:
Reaction Order For the general reaction: aA + bB  cC + dD Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms1). x and y are the reactant orders determined from experiment. x and y are NOT the stoichiometric coefficients.

Reaction Orders A reaction order can be zero, or positive integer and fractional number. Order Name Rate Law 0 zeroth rate = k[A]0 = k 1 first rate = k[A] 2 second rate = k[A]2 0.5 one-half rate = k[A]1/2 1.5 three-half rate = k[A]3/2 two-thirds rate = k[A]2/3

Reaction Order Overall order: 1 + ½ + 2 = 3.5 = 7/2
or sevenhalves order note: when the order of a reaction is 1 (first order) no exponent is written.

Reaction Constant To find the units of the rate constant, divide the rate units by the Molarity raised to the power of the overall reaction order. k = if (x+y) = 1 k has units of s-1 if (x+y) = 2 k has units of M-1s-1

Reaction Constant The rate constant, k, is a proportionality constant that relates rate and concentration. It is found through experiment that the rate constant is a function temperature. Rate constants must therefore be reported at the temperature with which they are measured. The rate constant also contains information about the energetics and collision efficiency of the reaction.

EXAMPLE: The reaction, 2 NO (g) + 2 H2 (g)  N2 (g) H2O (g) is experimentally found to be first order in H2 and third order in NO a) Write the rate law.

Rate(Ms-1) = k [H2] [NO] EXAMPLE: The reaction,
2 NO (g) + 2 H2 (g)  N2 (g) H2O (g) is experimentally found to be first order in H2 and third order in NO a) Write the rate law. 3 Rate(Ms-1) = k [H2] [NO] b) What is the overall order of the reaction?

Rate(Ms-1) = k [H2] [NO] “4th order” EXAMPLE: The reaction,
2 NO (g) + 2 H2 (g)  N2 (g) H2O (g) is experimentally found to be first order in H2 and third order in NO a) Write the rate law. 3 Rate(Ms-1) = k [H2] [NO] b) What is the overall order of the reaction? Overall order = 1 + 3 = 4 “4th order” c) What are the units of the rate constant?

Rate Law Expression If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is?

Rate Law Expression If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is? one… 1

Determining a Rate Equation
Initial rate Instantaneous rate (tangent line) During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction. Reactant concentration (M)

Determining Reaction Order: The Method of Initial Rates
The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g) From the following experimental data, determine the rate law and rate constant. Trial [NO]o (M) [H2]o (M) Initial Rate (Mmin-1) 1 0.0100 2 0.0200 0.0300 0.144 3 0.0120

The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g) Notice that in Trial 1 and 3, the initial concentration of NO is held constant while H2 is changed. Trial [NO]o (M) [H2]o (M) Initial Rate (Mmin-1) 1 0.0100 2 0.0200 0.0300 0.144 3 0.0120

The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g) This means that any changes to the rate must be due to the changes in H2 which is related to the concentration of H2 & its order! Trial [NO]o (M) [H2]o (M) Initial Rate (Mmin-1) 1 0.0100 2 0.0200 0.0300 0.144 3 0.0120

2NO(g) + 2H2(g)  N2(g) + 2H2O(g) The rate law for the reaction is given by: Rate(M/min) = k [NO]x [H2]y Taking the ratio of the rates of Trials 3 and 1 one finds: Rate (Trial 3) = Rate (Trial 1) Plugging in the values from the data:

2.00 Take the log of both sides of the equation: log 2.00 log(2.00) log(2.00) y = 1 Rate(M/min) = k [NO]x[H2]

Similarly for x: Rate(M/min) = k [NO]x[H2]y k k x = 3

The Rate Law expression is: 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) The order for NO is 3 The order for H2 is 1 The over all order is =4

The Rate constant Rate(M/min) = k [NO]3[H2] To find the rate constant, choose one set of data and solve:

Concentration–Time Relationships: Integrated Rate Laws
It is important know how long a reaction must proceed to reach a predetermined concentration of some reactant or product. We need a mathematical equation that relates time and concentration: This equation would yield concentration of reactants or products at a given time. It will all yield the time required for a given amount of reactant to react.

Integrated Rate Laws For a zero order process where “A” goes onto products, the rate law can be written: A  products = k k has units of Ms1 For a zero order process, the rate is the rate constant!

Integrated Rate Laws A  products = k This is the “average rate”
Zero order kinetics A  products = k This is the “average rate” If one considers the infinitesimal changes in concentration and time the rate law equation becomes: This is the “instantaneous rate”

Integrated Rate Laws Zero order kinetics
where [A] = [A]o at time t = 0 and [A] = [A] at time t = t [A]t  [A]o = k(t  0) = kt [A]t  [A]o = kt

Integrated Rate Laws [A]o  [A]t = kt [A]t = kt + [A]o
Zero order kinetics what’s this look like? [A]o  [A]t = kt [A]t = kt + [A]o rearranging… y = mx + b a plot of [A]t vs t looks like…  the y-intercept is [A]o k has units of M×(time)1 slope = k [A]t (mols/L) Conclusion: If a plot of reactant concentration vs. time yields a straight line, then the reactant order is ZERO! t (time)

Integrated Rate Laws This is the “instantaneous rate”
For a first order process, the rate law can be written: A  products This is the “average rate” If one considers the infinitesimal changes in concentration and time the rate law equation becomes: This is the “instantaneous rate”

Integrated Rate Laws First order kinetics
Taking the exponent to each side of the equation: or Conclusion: The concentration of a reactant governed by first order kinetics falls off from an initial concentration exponentially with time.

Integrated Rate Laws First order kinetics
Taking the natural log of both sides… = ln[A]o  kt rearranging… ln[A]t =  kt + ln[A]o

Integrated Rate Laws First order kinetics ln[A]t =  kt + ln[A]o
y = mx + b so a plot of ln[A]t vs t looks like… k has units of (time)1  the y-intercept is ln[A]o slope = k ln[A]t Conclusion: If a plot of natural log of reactant concentration vs. time yields a straight line, then the reactant order is FIRST! t (time)

Problem: The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s1?

Problem: The decomposition of N2O5(g) following 1st order kinetics.
If 2.56 mg of N2O5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s1? Begin with the integrated rate law for a 1st order process: Wait… what is the volume of the container??? Do we need to convert to moles?

Problem: The decomposition of N2O5(g) following 1st order kinetics. If 2.56 mg of N2O5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s1? You don’t need the volume of the container! Check it out!

Problem: taking the natural log and substituting time in seconds:
The decomposition of N2O5(g) following 1st order kinetics. If 2.56 mg of N2O5 is initially present in a container and 2.50 mg remains after 4.26 min, what is the rate constant in s1? taking the natural log and substituting time in seconds: k = 9.3 105 s1

Integrated Rate Laws Second order kinetics A  Products Rate = k[A]2
k has units of M1s1 Integrating as before we find:

Integrated Rate Laws Second order kinetics y = mx + b
so a plot of 1/[A]t vs t looks like… k has units of M1s1 slope = k 1/[A]t Conclusion: If a plot of one over reactant concentration vs. time yields a straight line, then the reactant order is second!  the y-intercept is 1/[A]o t (time)

Summary of Integrated Rate Laws

Half-life & First-Order Reactions
Half-life of a reaction is the time taken for the concentration of a reactant to drop to one-half of the original value.

Reaction Half-Life: 1st Order Kinetics
at time = t½ For a first order process the half life (t½ ) is found mathematically from: Start with the integrated rate law expression for a 1st order process Bring the concentration terms to one side. Express the concentration terms as a fraction using the rules of ln.

exchange [A] with [A]0 to reverse the sign of the ln term and cancel the negative sign in front of k Substitute the value of [A] at the half-life

The half-life is independent of the initial concentration! So, knowing the rate constant for a first order process, one can find the half-life!

Problem: A certain reaction proceeds through first order kinetics.
The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s?

The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find:

The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find: k = s-1

The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find: k = s-1 =

The half-life of the reaction is 180. s. What percent of the initial concentration remains after 900.s? Using the integrated rate law, substituting in the value of k and 900.s we find: k = s-1 = Since the ratio of [A]t to [A]0 represents the fraction of [A] that remains, the % is given by: 100  = 3.12%

Elements that decay via radioactive processes do so according to 1st order kinetics: Element: Half-life: 238U  234Th +  14C  14N +  131I  131Xe +  4.5  109 years 5730 years 8.05 days

Tritium decays to helium by beta () decay: The half-life of this process is 12.3 years Starting with 1.50 mg of 3H, what quantity remains after 49.2 years. Solution: Begin with the integrated rate law expression for 1st order kinetics.

Recall that that the rate constant for a 1st order process is given by: [3H]t = 1.50 mg  = mg

Notice that 49.2 years is 4 half-lives… After 1 half life: = 0.75 mg remains After 2 half life's: = 0.38 mg remains After 3 half life's: = 0.19 mg remains After 4 half life's: = mg remains

A Microscopic View of Reaction Rates
Arrhenius: Molecules must posses a minimum amount of energy to react. Why? (1) In order to form products, bonds must be broken in the reactants. (2) Bond breakage requires energy. (3) Molecules moving too slowly, with too little kinetic energy, don’t react when they collide. The Activation energy, Ea, is the minimum energy required to initiate a chemical reaction. Ea is specific to a particular reaction.

A Microscopic View of Reaction Rates
When one writes a reaction all that is seen are the reactants and products. This details the overall reaction stoichiometry. How a reaction proceeds is given by the reaction mechanism. See Chapter 15 Video Presentation Slide 9

Activation Energy The reaction of NO2 and CO (to give NO and CO2) has an activation energy barrier of 132 kJ/mol-rxn. The reverse reaction (NO + CO2  NO2 + CO) requires 358 kJ/mol-rxn. The net energy change for the reaction of NO2 and CO is 226 kJ/mol-rxn.

Activation Energy The progress of a chemical reaction as the reactants transform to products can be described graphically by a Reaction Coordinate. In order for the reaction to proceed, the reactants must posses enough energy to surmount a reaction barrier. Transition State Eact reactants Potential Energy HRXN products Reaction Progress

Activation Energy E > Ea E < Ea
The temperature for a system of particles is described by a distribution of energies. At higher temps, more particles have enough energy to go over the barrier. E > Ea Since the probability of a molecule reacting increases, the rate increases. E < Ea

Reaction coordinate diagram
Activation Energy Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, Ea. See Chapter 15 Video Presentation Slide 10 Reaction coordinate diagram

Activation Energy Orientation factors into the equation
The orientation of a molecule during collision can have a profound effect on whether or not a reaction occurs. The reaction occurs only when the orientation of the molecules is just right… When the green atom collides with the green atom on the molecule, a reactive or effective collision occurs.

Activation Energy Orientation factors into the equation
In some cases, the reactants must have proper orientation for the collision to yield products. This reduces the number of collisions that are reactive! See Chapter 15 Video Presentation Slides 10, 13, & 14 When the green atom collides with the red atom on the molecule, this leads to a non-reactive or ineffective collision occurs.

The Arrhenius Equation
Arhenius discovered that most reaction-rate data obeyed an equation based on three factors: (1) The number of collisions per unit time. (2) The fraction of collisions that occur with the correct orientation. (3) The fraction of the colliding molecules that have an energy greater than or equal to Ea. From these observations Arrhenius developed the aptly named Arrhenius equation.

k is the rate constant T is the temperature in K R is the ideal-gas constant (8.314 J/Kmol) Ea is the activation energy A is known the frequency or pre–exponential factor In addition to carrying the units of the rate constant, “A” relates to the frequency of collisions and the orientation of a favorable collision probability Both A and Ea are specific to a given reaction.

Temperature Dependence of the Rate Constant: Increasing the temperature of a reaction generally speeds up the process (increases the rate) because the rate constant increases according to the Arrhenius Equation. Rate (Ms-1) = k[A]x[B]y As T increases, the value of the exponential part of the equation becomes less negative thus increasing the value of k.

Effect of Temperature Reactions generally occur slower at lower T.
In ice at 0 oC Reactions generally occur slower at lower T. Room temperature See Chapter 15 Video Presentation Slides 11 & 12 Iodine clock reaction. H2O2 + 2 I- + 2 H+  2 H2O + I2

Determining the Activation Energy Ea may be determined experimentally. First take natural log of both sides of the Arrhenius equation: ln y = mx + b ln k

Determining the Activation Energy One can determine the activation energy of a reaction by measuring the rate constant at two temperatures: Writing the Arrhenius equation for each temperature: Subtracting k1 from k2 we find that:

Determining the Activation Energy Knowing the rate constants at two temps yields the activation energy. or Knowing the Ea and the rate constant at one temp allows one to find k(T2)

Problem: The activation energy of a first order reaction is 50.2 kJ/mol at 25 °C. At what temperature will the rate constant double? (1) (2) (3)

algebra! Problem: T2 = 308 K (4) (5)
The activation energy of a first order reaction is 50.2 kJ/mol at 25 °C. At what temperature will the rate constant double? (4) algebra! (5) A 10 °C change of temperature doubles the rate!! T2 = 308 K

Catalysis Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier. Dr. James Cusumano, Catalytica Inc. See Chapter 15 Video Presentation Slides 19, 20, & 21 What is a catalyst? Catalysts and the environment Catalysts and society

Catalysis In auto exhaust systems — Pt, NiO 2 CO + O2  2 CO2
2 NO  N2 + O2 See Chapter 15 Video Presentation Slide 22

Catalysis 2. Polymers: H2C=CH2  polyethylene
3. Acetic acid: CH3OH + CO  CH3CO2H 4. Enzymes — biological catalysts

Catalysis MnO2 catalyzes decomposition of H2O2 2 H2O2  2 H2O + O2
Catalysis and activation energy Uncatalyzed reaction Catalyzed reaction See Chapter 15 Video Presentation Slide 4

Iodine-Catalyzed Isomerization of cis-2-Butene

Reaction Mechanisms The overall stoichiometry of a chemical reaction is most often the sum of several steps: (1) 2AB  A2B2 (2) A2B2 + C2  A2B + C2B (3) A2B + C2  A2 + C2B 2AB + A2B2 + A2B + 2C2  A2B2 + A2B + A2 + 2C2B See Chapter 15 Video Presentation Slides 15, 16, 17, & 18 Net: AB + 2C2  A2 + 2C2B The sequence of steps (1-3) describes a possible “reaction mechanism”.

Reaction Mechanisms The overall stoichiometry of a chemical reaction is most often the sum of several steps: (1) 2AB  A2B2 (2) A2B2 + C2  A2B + C2B (3) A2B + C2  A2 + C2B 2AB + A2B2 + A2B + 2C2  A2B2 + A2B + A2 + 2C2B Net: AB + 2C2  A2 + 2C2B The species that cancel out (not part of the overall reaction) are called “reaction intermediates”.

Reaction Mechanisms (1) 2AB  A2B2 (2) A2B2 + C2  A2B + C2B
(3) A2B + C2  A2 + C2B 2AB + 2C2  A2 + 2C2B Each step in the mechanism is called an “elementary step”. The number of reactants in an elementary step is called the “molecularity”. In this example each step (1-3) is a bimolecular process. (2 reactants) A2  2A is a unimolecular process (1 reactant) 2A + B  is a termolecular process (3 reactants)

Reaction Mechanisms slow fast (1) 2AB  A2B2 (2) A2B2 + C2  A2B + C2B (3) A2B + C2  A2 + C2B 2AB + 2C2  A2 + 2C2B The rate of the overall reaction can never be faster than the “slowest step” in the mechanism. If reaction (1) is the slowest of the three steps in the mechanism… Then it is known as the “rate determining step”

Reaction Mechanisms slow fast (1) 2AB  A2B2 (2) A2B2 + C2  A2B + C2B (3) A2B + C2  A2 + C2B 2AB + 2C2  A2 + 2C2B The slowest step controls the rate of the reaction. It determines the rate law! Rate (Ms-1) = k[AB]2 The rate law is not based on the overall reaction: Rate (Ms-1) = k[AB]2[C2]2

Reaction Mechanisms Consider the following reaction:
2NO2(g) + F2(g)  2FNO2(g) If the reaction proceeded by the overall reaction, the rate law for the reaction would be 3rd order overall. The actual rate law is found to be: Rate = k[NO2][F2] Indicating that the slowest step in the mechanism is a bimolecular reaction between NO2 and F2.

Reaction Mechanisms 2NO2(g) + F2(g)  2FNO2(g) Rate = k[NO2][F2]
To explain the observed kinetics, a possible mechanism is proposed:

Reaction Mechanisms 2NO2(g) + F2(g)  2FNO2(g) Rate = k[NO2][F2]
Since the 1st step in the reaction is the slow step, it determines the kinetics and rate law for the reaction:

Reaction Mechanisms Validating a Reaction Mechanism:
A mechanism is a proposal of how the reaction proceeds at the molecular level. The individual elementary steps must sum to yield the overall reaction with correct stoichiometry. The predicted reaction rate law must be in agreement with the experimentally determined rate law. Note that there may be more than one mechanism that is in agreement with the reaction stoichiometry and kinetics.

Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

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Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

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