Copyright 2011 Pearson Education, Inc. Chapter 13 Chemical Kinetics Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

Copyright 2011 Pearson Education, Inc.

Chapter 13Chemical Kinetics

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA


Chemical Kinetics

• The speed of a chemical reaction is called its reaction rate

• The rate of a reaction is a measure of how fast the reaction makes productsor uses reactants

• The ability to control the speed of a chemical reaction is important

2Tro: Chemistry: A Molecular Approach, 2/e


Defining Rate

• Rate is how much a quantity changes in a given period of time

• The speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour)so the rate of your car has units of mi/hr



Defining Reaction Rate

• The rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of timeor product concentration increases

• For reactants, a negative sign is placed in front of the definition



Reaction Rate Changes Over Time

• As time goes on, the rate of a reaction generally slows down because the concentration of the reactants

decreases

• At some time the reaction stops, either because the reactants run out or because the system has reached equilibrium



Reaction Rate and Stoichiometry• In most reactions, the coefficients of the balanced

equation are not all the same

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

• For these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another for the above reaction, for every 1 mole of H2 used, 1

mole of I2 will also be used and 2 moles of HI made therefore the rate of change will be different

• To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient



Average Rate

• The average rate is the change in measured concentrations in any particular time periodlinear approximation of a curve

• The larger the time interval, the more the average rate deviates from the instantaneous rate



Instantaneous Rate

• The instantaneous rate is the change in concentration at any one particular timeslope at one point of a curve

• Determined by taking the slope of a line tangent to the curve at that particular pointfirst derivative of the function

for you calculus fans



H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is

Using [HI], the instantaneous rate at 50 s is



Example 13.1: For the reaction given, the [I] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the

average rate in the first 10 s and the D[H+].H2O2 (aq) + 3 I(aq) + 2 H+

(aq) I3(aq) + 2 H2O(l)

10

solve the rate equation for the rate (in terms of the change in concentration of the given quantity)

solve the rate equation (in terms of the change in the concentration for the quantity to find) for the unknown value

Tro: Chemistry: A Molecular Approach, 2/e


Practice – If 2.4 x 102 g of NOBr (MM 109.91 g) decomposes in a 2.0 x 102 mL flask in 5.0 minutes,

find the average rate of Br2 production in M/s2 NOBr(g) 2 NO(g) + Br2(l)



2 mol NOBr:1 mol Br2, 1 mol = 109.91g, 1 mL=0.001 L

Solve:

Conceptual Plan:

Relationships:

240.0 g NOBr, 200.0 mL, 5.0 min.

D[Br2]/Dt, M/s

Given:

Find:

Practice – If 2.4 x 102 g of NOBr decomposes in a 2.0 x 102 mL flask in 5.0 minutes, find the average rate of Br2 production

2 NOBr(g) 2 NO(g) + Br2(l)

12

D MD mole Br2

RateD min D s }

5.45 M Br2, 3.0 x 102 s

D[Br2]/Dt, M/s

240.0 g NOBr, 0.2000 L, 5.0 min.

D[Br2]/Dt, M/s



Measuring Reaction Rate• To measure the reaction rate you need to be

able to measure the concentration of at least one component in the mixture at many points in time

• There are two ways of approaching this problem1. for reactions that are complete in less than 1

hour, it is best to use continuous monitoring of the concentration

2. for reactions that happen over a very long time, sampling of the mixture at various times can be used when sampling is used, often the reaction in the

sample is stopped by a quenching technique



Factors Affecting Reaction Rate: Nature of the Reactants

• Nature of the reactants means what kind of reactant molecules and what physical condition they are in small molecules tend to react faster than large molecules gases tend to react faster than liquids, which react faster

than solids powdered solids are more reactive than “blocks”

more surface area for contact with other reactantscertain types of chemicals are more reactive than others

e.g. potassium metal is more reactive than sodiumions react faster than molecules

no bonds need to be broken14Tro: Chemistry: A Molecular Approach, 2/e


Factors Affecting Reaction Rate:Temperature

• Increasing temperature increases reaction ratechemist’s rule of thumb—for each 10 °C rise in

temperature, the speed of the reaction doublesfor many reactions

• There is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius, which will be examined later



Factors Affecting Reaction Rate:Catalysts

• Catalysts are substances that affect the speed of a reaction without being consumed

• Most catalysts are used to speed up a reaction; these are called positive catalysts catalysts used to slow a reaction are called negative

catalysts.

• Homogeneous = present in same phase• Heterogeneous = present in different phase• How catalysts work will be examined later



Factors Affecting Reaction Rate:Reactant Concentration

• Generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule

contactconcentration of gases depends on the partial

pressure of the gas higher pressure = higher concentration

• Concentrations of solutions depend on the solute-to-solution ratio (molarity)



The Rate Law• The rate law of a reaction is the mathematical

relationship between the rate of the reaction and the concentrations of the reactantsand homogeneous catalysts as well

• The rate law must be determined experimentally!!• The rate of a reaction is directly proportional to the

concentration of each reactant raised to a power• For the reaction aA + bB products the rate law

would have the form given belown and m are called the orders for each reactantk is called the rate constant



Reaction Order• The exponent on each reactant in the rate law

is called the order with respect to that reactant• The sum of the exponents on the reactants is

called the order of the reaction • The rate law for the reaction

2 NO(g) + O2(g) ® 2 NO2(g)

is Rate = k[NO]2[O2]

The reaction is

second order with respect to [NO],

first order with respect to [O2],

and third order overall19Tro: Chemistry: A Molecular Approach, 2/e


Rate = k[NO][O3]

Solve:

Conceptual Plan:

Relationships:

[NO] = 1.00 x 10−6 M, [O3] = 3.00 x 10−6 M, Rate = 6.60 x 10−6 M/s k, M−1s−1

Given:

Find:

Example: The rate equation for the reaction of NO with ozone is Rate = k[NO][O3]. If the rate is 6.60 x 10−5 M/sec when [NO] =

1.00 x 10−6 M and [O3] = 3.00 x 10−6 M, calculate the rate constant

Rate, [NO], [O3] k



Practice – The rate law for the decomposition of acetaldehyde is Rate = k[acetaldehyde]2. What is the rate of the reaction when the

[acetaldehyde] = 1.75 x 10−3 M and the rate constant is 6.73 x 10−6 M−1•s−1?



Rate = k[acetaldehyde]2

Solve:

Conceptual Plan:

Relationships:

[acetaldehyde] = 1.75 x 10−3 M, k= 6.73 x 10−6 M−1•s−1

Rate, M/s

Given:

Find:

Practice – What is the rate of the reaction when the [acetaldehyde] = 1.75 x 10−3 M and the rate constant is

6.73 x 10−6 M−1•s−1?

k, [acetaldehyde] Rate



Finding the Rate Law:the Initial Rate Method

• The rate law must be determined experimentally• The rate law shows how the rate of a reaction

depends on the concentration of the reactants• Changing the initial concentration of a reactant

will therefore affect the initial rate of the reaction

23

if for the reaction A → Products

then doubling the initial concentration of A doubles the initial reaction rate

then doubling the initial concentration of A does not change the initial reaction rate

then doubling the initial concentration of A quadruples the initial reaction rate



Rate = k[A]n

• If a reaction is Zero Order, the rate of the reaction is always the samedoubling [A] will have no effect on the reaction rate

• If a reaction is First Order, the rate is directly proportional to the reactant concentrationdoubling [A] will double the rate of the reaction

• If a reaction is Second Order, the rate is directly proportional to the square of the reactant concentrationdoubling [A] will quadruple the rate of the reaction



Determining the Rate Law When there Are Multiple Reactants

• Changing each reactant will effect the overall rate of the reaction

• By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined

• In examining results, we compare differences in rate for reactions that only differ in the concentration of one reactant



Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)

given the data below

Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO]

does not26Tro: Chemistry: A Molecular Approach, 2/e
















n = 2, m = 0




given the data belowSubstitute the concentrations and rate for any experiment into the rate law and solve for k

Expt.Number

Initial [NO2], (M)

Initial [CO], (M)

Initial Rate(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Rate k[NO2]2

for expt 1

0.0021 Ms k 0.10 M 2

k 0.0021 M

s

0.01 M20.21 M 1 s 1



Practice – Determine the rate law and rate constant for the reaction NH4

+ + NO2− ® N2 + 2 H2O




Practice – Determine the rate law and rate constant for the reaction NH4

+ + NO2− ® N2 + 2 H2O

given the data below Rate = k[NH4

+]n[NO2]m



Finding the Rate Law:Graphical Methods

• The rate law must be determined experimentally• A graph of concentration of reactant vs. time can

be used to determine the effect of concentration on the rate of a reaction

• This involves using calculus to determine the area under a curve – which is beyond the scope of this course

• Later we will examine the results of this analysis so we can use graphs to determine rate laws for several simple cases



Reactant Concentration vs. TimeA Products

35

insert figure 13.6



Integrated Rate Laws

• For the reaction A Products, the rate law depends on the concentration of A

• Applying calculus to integrate the rate law gives another equation showing the relationship between the concentration of A and the time of the reaction – this is called the Integrated Rate Law



Half-Life

• The half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactant to fall to ½ its initial value

• The half-life of the reaction depends on the order of the reaction



Zero Order Reactions• Rate = k[A]0 = k

constant rate reactions• [A] = −kt + [A]initial

• Graph of [A] vs. time is straight line with slope = −k and y–intercept = [A]initial

• t ½ = [Ainitial]/2k• When Rate = M/sec, k = M/sec

[A]initial

[A]

time

slope = −k



First Order Reactions

• Rate = k[A]1 = k[A]• ln[A] = −kt + ln[A]initial

• Graph ln[A] vs. time gives straight line with slope = −k and y–intercept = ln[A]initial

used to determine the rate constant

• t½ = 0.693/k

• The half-life of a first order reaction is constant

• When Rate = M/sec, k = s–1



ln[A]initial

ln[A]

time

slope = −k



The Half-Life of a First-Order Reaction Is Constant



Rate Data for C4H9Cl + H2O ® C4H9OH + HCl



C4H9Cl + H2O C4H9OH + 2 HCl







slope = −2.01 x 10−3

k =2.01 x 10−3 s-1



Second Order Reactions

• Rate = k[A]2

• 1/[A] = kt + 1/[A]initial

• Graph 1/[A] vs. time gives straight line with slope = k and y–intercept = 1/[A]initial

used to determine the rate constant

• t½ = 1/(k[A0])

• When Rate = M/sec, k = M−1s−1



1/[A]initial

1/[A]

time

slope = k



Rate Data For2 NO2 ® 2 NO + O2



Rate Data For2 NO2 ® 2 NO + O2



Rate Data Graphs for2 NO2 ® 2 NO + O2




2.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4

4.2

4.4

4.6

4.8

0 50 100 150 200 250

ln(p

ress

ure

)

Time (hr)

ln(PNO2) vs. Time





Copyright 2011 Pearson Education, Inc.53Tro: Chemistry: A Molecular Approach, 2/e


Example 13.4: The reaction SO2Cl2(g) SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10−4 s−1 at a given set of conditions.

Find the [SO2Cl2] at 865 s when [SO2Cl2]initial = 0.0225 M

54

the new concentration is less than the original, as expected

[SO2Cl2]init = 0.0225 M, t = 865, k = 2.90 x 10-4 s−1

[SO2Cl2]

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

[SO2Cl2][SO2Cl2]init, t, k



Practice – The reaction Q 2 R is second order in Q. If the initial [Q] = 0.010 M and after 5.0 x 102 seconds

the [Q] = 0.0010 M, find the rate constant



Practice – The reaction Q 2 R is second order in Q. If the initial [Q] = 0.010 M and after 5.0 x 102 seconds

the [Q] = 0.0010 M, find the rate constant

56

[Q]init = 0.010 M, t = 5.0 x 102 s, [Q]t = 0.0010 M

k

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

k[Q]init, t, [Q]t



Graphical Determination of the Rate Law for A Product

• Plots of [A] vs. time, ln[A] vs. time, and 1/[A] vs. time allow determination of whether a reaction is zero, first, or second order

• Whichever plot gives a straight line determines the order with respect to [A]if linear is [A] vs. time, Rate = k[A]0

if linear is ln[A] vs. time, Rate = k[A]1

if linear is 1/[A] vs. time, Rate = k[A]2



Practice – Complete the Table and Determine the Rate Equation for the

Reaction A ® 2 Product

58

What will the rate be when the [A] = 0.010 M?



Practice – Complete the Table and Determine the Rate Equation for the

Reaction A ® 2 Product






Practice – Complete the table and determine the rate equation for the

reaction A ® 2 Product

Because the graph 1/[A] vs. time is linear, the reaction is second order,

63

Rate k[A]2

k slope of the line 0.10 M–1 s –1



Relationship Between Order and Half-Life

• For a zero order reaction, the lower the initial concentration of the reactants, the shorter the half-life t1/2

= [A]init/2k • For a first order reaction, the half-life is

independent of the concentration t1/2 = ln(2)/k

• For a second order reaction, the half-life is inversely proportional to the initial concentration – increasing the initial concentration shortens the half-life t1/2 = 1/(k[A]init)



Example 13.6: Molecular iodine dissociates at 625 K with a first-order rate constant of 0.271 s−1. What is the half-life of this

reaction?

65

the new concentration is less than the original, as expected

k = 0.271 s−1

t1/2

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

t1/2k



Practice – The reaction Q 2 R is second order in Q. If the initial [Q] = 0.010 M and the rate constant is 1.8 M−1•s−1 find the length of time for [Q] = ½[Q]init



Practice – The reaction Q 2 R is second order in Q. If the initial [Q] = 0.010 M and the rate constant is 1.8 M−1•s−1 find the length of time for [Q] = ½[Q]init

67

[Q]init = 0.010 M, k = 1.8 M−1s−1

t1/2, s

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

t1/2[Q]init, k



• Changing the temperature changes the rate constant of the rate law

• Svante Arrhenius investigated this relationship and showed that:

R is the gas constant in energy units, 8.314 J/(mol•K)where T is the temperature in kelvins

A is called the frequency factor, the rate the reactant energy approaches the activation energy

Ea is the activation energy, the extra energy needed to start the molecules reacting

The Effect of Temperature on Rate




Activation Energy and theActivated Complex

• There is an energy barrier to almost all reactions

• The activation energy is the amount of energy needed to convert reactants into the activated complexaka transition state

• The activated complex is a chemical species with partially broken and partially formed bondsalways very high in energy because of its partial

bonds



The Arrhenius Equation:The Exponential Factor

• The exponential factor in the Arrhenius equation is a number between 0 and 1

• Tt represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier the higher the energy barrier (larger activation energy), the fewer

molecules that have sufficient energy to overcome it• That extra energy comes from converting the kinetic

energy of motion to potential energy in the molecule when the molecules collide increasing the temperature increases the average kinetic energy

of the molecules therefore, increasing the temperature will increase the number of

molecules with sufficient energy to overcome the energy barrier therefore increasing the temperature will increase the reaction rate




Arrhenius Plots• The Arrhenius Equation can be algebraically

solved to give the following form:

this equation is in the form y = mx + bwhere y = ln(k) and x = (1/T)

a graph of ln(k) vs. (1/T) is a straight line

(−8.314 J/mol∙K)(slope of the line) = Ea, (in Joules)

ey–intercept = A (unit is the same as k)



Example 13.7: Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g)

given the following data:









76

slope, m = 1.12 x 104 Ky–intercept, b = 26.8



Arrhenius Equation:Two-Point Form

• If you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:



Example 13.8: The reaction NO2(g) + CO(g) CO2(g) + NO(g) has a rate constant of 2.57 M−1∙s−1 at 701 K and 567 M−1∙s−1

at 895 K. Find the activation energy in kJ/mol

78

most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable

T1 = 701 K, k1 = 2.57 M−1∙s−1, T2 = 895 K, k2 = 567 M−1∙s−1

Ea, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

EaT1, k1, T2, k2



Practice – It is often said that the rate of a reaction doubles for every 10 °C rise in

temperature. Calculate the activation energy for such a reaction.

79

Hint: make T1 = 300 K, T2 = 310 K and k2 = 2k1



Practice – Find the activation energy in kJ/mol for increasing the temperature 10 ºC doubling the rate

80

most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable

T1 = 300 K, T1 = 310 K, k2 = 2 k1

Ea, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

EaT1, k1, T2, k2



Collision Theory of Kinetics• For most reactions, for a reaction to take place,

the reacting molecules must collide with each other on average about 109 collisions per second

• Once molecules collide they may react together or they may not, depending on two factors –

1. whether the collision has enough energy to "break the bonds holding reactant molecules together"; and

2. whether the reacting molecules collide in the proper orientation for new bonds to form



Effective CollisionsKinetic Energy Factor

For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide the activated complex can form



Effective CollisionsOrientation Effect



Effective Collisions

• Collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions

• The higher the frequency of effective collisions, the faster the reaction rate

• When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed – the activated complex



Collision Theory and the Frequency Factor of the Arrhenius Equation

• The Arrhenius Equation includes a term, A, called the Frequency Factor

• The Frequency Factor can be broken into two terms that relate to the two factors that determine whether a collision will be effective



Collision Frequency

• The collision frequency is the number of collisions that happen per second

• The more collisions per second there are, the more collisions can be effective and lead to product formation



Molecular Interpretation of Factors Affecting Rate – Temperature

• Increasing the temperature raises the average kinetic energy of the reactant molecules

• There is a minimum amount of kinetic energy needed for the collision to be converted into enough potential energy to form the activated complex

• Increasing the temperature increases the number of molecules with sufficient kinetic energy to overcome the activation energy



Molecular Interpretation of Factors Affecting Rate – Concentration

• Reaction rate generally increases as the concentration or partial pressure of reactant molecules increasesexcept for zero order reactions

• More molecules leads to more molecules with sufficient kinetic energy for effective collisiondistribution the same, just bigger curve



Molecularity

• The number of reactant particles in an elementary step is called its molecularity

• A unimolecular step involves one particle• A bimolecular step involves two particles

though they may be the same kind of particle

• A termolecular step involves three particlesthough these are exceedingly rare in elementary

steps



Catalysts• Catalysts are substances that affect the rate of a

reaction without being consumed• Catalysts work by providing an alternative

mechanism for the reactionwith a lower activation energy

• Catalysts are consumed in an early mechanism step, then made in a later step

mechanism without catalyst

O3(g) + O(g) 2 O2(g) V. Slow

mechanism with catalyst

Cl(g) + O3(g) O2(g) + ClO(g) Fast

ClO(g) + O(g) O2(g) + Cl(g) Slow



Molecular Interpretation of Factors Affecting Rate – Catalysts

• Catalysts generally speed a reaction• They give the reactant molecules a different

path to follow with a lower activation energyheterogeneous catalysts hold one reactant

molecule in proper orientation for reaction to occur when the collision takes placeand sometimes also help to start breaking bonds

homogeneous catalysts react with one of the reactant molecules to form a more stable activated complex with a lower activation energy



Catalysts

• Homogeneous catalysts are in the same phase as the reactant particlesCl(g) in the destruction of O3(g)

• Heterogeneous catalysts are in a different phase than the reactant particlessolid catalytic converter in a car’s exhaust system



Types of Catalysts



Enzymes• Because many of the molecules are large and

complex, most biological reactions require a catalyst to proceed at a reasonable rate

• Protein molecules that catalyze biological reactions are called enzymes

• Enzymes work by adsorbing the substrate reactant onto an active site that orients the substrate for reaction

94

1) Enzyme + Substrate Enzyme─Substrate Fast

2) Enzyme─Substrate → Enzyme + Product Slow


Documents

Copyright 2011 Pearson Education, Inc. Chapter 13 Chemical Kinetics Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA