Ch 1 Chemical Kinetics

Copyright 2011 Pearson Education, Inc.

Chapter 1

Chemical Kinetics


Kinetics: Rates and Mechanisms of Chemical Reactions

1.1 Expressing Reaction Rate: Differential rate law

Graphical representation of reaction rate

1.2 The Rate Law : Order of reaction, Initial rate method

Integrated rate law, rate constant, half -life

1.3 Collision Theory: Activation energy, effective collisions

1.6 Arrhenius Equation: Determination of rate constant and

activation energy

1.7 Reaction Mechanism: Intermediates, molecularity,

rate determining step

1.4 Factors affecting reaction rate: Concentration, temperature, catalyst,

nature of reactants

1.5 Transition State Theory: Activated complex


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 products, or how fast the

reactants are consumed.

• Chemical kinetics is the study of the factors

that affect the rates of chemical reactions

Such as temperature, reactant concentrations

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


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 time

or product concentration increases

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

the definition



06250061

87Rate

tt

XX

t

XRate

12

12

.

250061

84Rate

tt

AA

t

ARate

12

12

.

at t = 0

[A] = 8

[B] = 8

[C] = 0

at t = 0

[X] = 8

[Y] = 8

[Z] = 0

at t = 16

[A] = 4

[B] = 4

[C] = 4

at t = 16

[X] = 7

[Y] = 7

[Z] = 1

250061

04Rate

tt

CC

t

CRate

12

12

.

06250061

01Rate

tt

ZZ

t

ZRate

12

12

.



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



Average Rate

• The average rate is the change in measured

concentrations in any particular time period

linear approximation of a curve



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



H2 I2 HI

Stoichiometry tells us that for

every 1 mole/L of H2 used,

2 moles/L of HI are made.

Assuming a 1 L container, at

10 s, we used 0.181 moles of

H2. Therefore the amount of

HI made is 2(0.181 moles) =

0.362 moles.

At 60 s, we used 0.699 moles

of H2. Therefore the amount

of HI made is 2(0.699 moles)

= 1.398 moles.

The average rate is

the change in the

concentration in a

given time period

In the first 10 s, the

[H2] is −0.181 M,

so the rate is



average rate in a given

time period = slope of

the line connecting the

[H2] points; and ½ +slope

of the line for [HI]

the average rate for the

first 10 s is 0.0181 M/s







Instantaneous Rate

• The instantaneous rate is the change in

concentration at any one particular time

Determined by taking the slope of a line tangent to the

“concentartion vs time” curve at any particular point.

• The initial rate is the reaction rate at the

beginning of a reaction

slope of a line tangent to the curve at time t = 0

No product formed yet, reversed rate = 0



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



Problem 1.1: For the reaction given, the [I] changes from1.000 M to 0.868 M

in the first 10 s. Calculate the average rate in the first 10 s, and [H+]/t.

H2O2 (aq) + 3 I(aq) + 2 H+(aq) I3

(aq) + 2 H2O(l)

13

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/s.

2 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.

[Br2]/t, 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)

15

M mole Br2

Rate min s

}

5.45 M Br2, 3.0 x 102 s

[Br2]/t, M/s

240.0 g NOBr, 0.2000 L, 5.0 min.

[Br2]/t, M/s



Practice: Consider the reaction:

8A(g) + 5B(g) 8C(g) + 6D(g)

If [C] is increasing at the rate of 4.0 mol L¯1s¯1, at what rate is

[B] changing?

A. -0.40 mol L¯1s¯1

B. -2.5 mol L¯1s¯1

C. -4.0 mol L¯1s¯1

D. -6.4 mol L¯1s¯1

E. None of these choices is correct, since its rate of

change must be positive.

Tro: Chemistry: A Molecular Approach 16


In general, the differential rate law for the reaction

aA + bB cC + dD

rate = 1

a - = -

[A]

t

1

b

[B]

t

1

c

[C]

t = +

1

d

[D]

t = +

The numerical value of the rate depends upon the substance that

serves as the reference. The rest is relative to the balanced

chemical equation.


SOLUTION:

Expressing Rate in Terms of Changes in Concentration with Time

PROBLEM 1.2: Because it has a nonpolluting product (water vapor), hydrogen

gas is used for fuel aboard the space shuttle and may be used

by earthbound engines in the near future.

2H2(g) + O2(g) 2H2O(g)

(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.

(b) When [O2] is decreasing at 0.23 mol/L*s, at what rate is [H2O] increasing?

- 1

2

[H2]

t = -

[O2]

t

= + [H2O]

t

1

2

0.23 mol/L*s = + [H2O]

t

1

2

= 0.46 mol/L*s [H2O]

t

rate = (a)

[O2]

t - = - (b)


The Rate Law

• The rate law of a reaction is the mathematical

relationship between the rate of the reaction and the

concentrations of the reactants

and 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 below n and m are called the orders for each reactant

k 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 overall order of the reaction

• The rate law for the reaction given: 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 overall. 20 Tro: Chemistry: A Molecular Approach, 2/e


SOLUTION:

Determining Reaction Order from Rate Laws

PROBLEM 1.3: For each of the following reactions, determine the

reaction order with respect to each reactant and the

overall order from the given rate law.

(a) 2NO(g) + O2(g) 2NO2(g) rate = k[NO]2[O2]

(b) H2O2(aq) + 3I-(aq) + 2H+(aq) I3-(aq) + 2H2O(l) rate = k[H2O2][I

-]

PLAN: Look at the rate law and not the coefficients of the chemical reaction.

(a) The reaction is 2nd order in NO, 1st order in O2, and 3rd order overall.

(b) The reaction is 1st order in H2O2, 1st order in I- and zero order in H+,

while being 2nd order overall.


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 k

Rate, [NO], [O3] k



Practice:

The rate law for the decomposition of acetaldehyde,

CH3CHO, is Rate = k[CH3CHO]2.

What is the rate of the reaction when the [CH3CHO]

= 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:

[CH3CHO] = 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

[CH3CHO] = 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

25

if for the reaction

A → Products

Rate = k[A]n

then doubling the

initial concentration

of A doubles the

initial reaction rate

then doubling the


of A does not

change the initial

reaction rate. Thus,

n = 0 (Zero Order)

Rate = k

then doubling the


of A quadruples the

initial reaction rate



Proposed rate law: Rate = k[A]n

• If a reaction is Zero Order, the rate of the reaction

is always the same

doubling [A] will have no effect on the reaction rate

• If a reaction is First Order, the rate is directly

proportional to the reactant concentration

doubling [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

concentration

doubling [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 not 28 Tro: Chemistry: A Molecular Approach, 2/e


Example 13.2: Determine the rate law and rate constant

for the reaction NO2(g) + CO(g) NO(g) + CO2(g)

















n = 2, m = 0






Substitute 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 M2 0.21 M1 s1



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


Rate = k[NH4+]n[NO2

]m



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



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



Half-Life

• The half-life, t1/2, of a reaction is the

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



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



Zero Order Reactions

• Rate = k[A]0 = k

• Integrated rate law: [A] = −kt + [A]initial

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

• t ½ = [Ainitial]/2k = [A0]/2k

• When Rate = M/sec, k = M/sec

[A]initial

[A]

time 45 Tro: Chemistry: A Molecular Approach, 2/e


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



PLAN:

SOLUTION:

Determining the Half-Life of a First-Order Reaction

PROBLEM 1.3: Cyclopropane is the smallest cyclic hydrocarbon. It is

thermally unstable and rearranges to propene at 1000oC

via the following first-order reaction:

CH2

H2C CH2(g)

H3C CH CH2 (g)

The rate constant is 9.2 s-1, (a) What is the half-life of the reaction?

(b) How long does it take for the concentration of cyclopropane to

reach one-quarter of the initial value?

Use the half-life equation, t1/2 = 0.693

k

One-quarter of the initial value means two half-lives have passed.

t1/2 = 0.693/9.2 s-1 = 0.075 s (a) 2 t1/2 = 2(0.075 s) = 0.150 s (b)

Copyright 2011 Pearson Education, Inc. 49 Tro: Chemistry: A Molecular Approach, 2/e


Table 1 Units of the Rate Constant k

Overall Reaction Order Units of k (t in seconds)

0 mol/L*s (or mol L-1 s-1)

1 1/s (or s-1)

2 L/mol*s (or L mol -1 s-1)

3 L2 / mol2 *s (or L2 mol-2 s-1)


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

51

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

53

[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

55

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






Conclusion for the determination of the rate

equation for the reaction A 2 Product

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

the reaction is second order,

60

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 half-life is directly proportional to the initial concentration. 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 = constant) 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) 61 Tro: Chemistry: A Molecular Approach, 2/e


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?

62

the new concentration is less than the original,

as expected

k = 0.271 s−1

t1/2

Check:

Solution:

Conceptual

Plan:

Relationships:

Given:

Find:

t1/2 k



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

64

[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



Collision Theory

• 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



Activation Energy

• Molecules must possess a certain minimum kinetic

energy called the activation energy for a reaction

to occur

energy needed to break the bonds of the reacting

molecules

Reaction with low activation energy is easier to take

place, and is faster



Effective Collisions

Kinetic Energy Factor

For a collision to be

effective, the reacting

molecules must have

sufficient kinetic

energy to overcome

the energy barrier.




Orientation Factor

• The proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form

• The more complex the reactant molecules, the less frequently they will collide with the proper orientation




Orientation Effect



Collision Frequency

• The collision frequency is the number of collisions

that happen per unit time

• The more collisions per second there are, the

more collisions can be effective and lead to

product formation




• 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

• Any factor that can increase the frequency of

effective collisions would increase the rate of a

reaction



Factors Affecting Reaction Rate

Under specific conditions, every reaction has its own

characteristic rate, which may be controlled by four

factors:

1. Concentrations of reactants

2. Chemical nature of reactants

3. Temperature

4. The use of a catalyst



Factors Affecting Reaction Rate:

Reactant Concentration

• Generally, the larger the concentration of

reactant molecules, the faster the reaction

increases the frequency of effective collisions

between reactant molecules

concentration 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)



Factors Affecting Rate – Temperature

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

• Increasing the temperature increases the number of molecules with sufficient kinetic energy to overcome the activation energy, leads to increase in frequancy of effective collisions. Hence, increasing the rate of reaction

• The distribution of kinetic energy is shown by the Boltzmann distribution curves




Factors Affecting Rate – Catalyst

• Catalysts are substances that affect the rate of a

reaction without being consumed

• Catalysts generally speed up a reaction

• They give the reactant molecules a different path

to follow with a lower activation energy



Catalysts

• Homogeneous catalysts are in the same phase

as the reactant particles

Cl(g) in the destruction of O3(g)

• Heterogeneous catalysts are in a different

phase than the reactant particles

solid catalytic converter in a car’s exhaust system



Types of Catalysts



Factors Affecting Reaction Rate:

Nature of the Reactants

• Nature of the reactants means what kind of reactant

molecules and what physical conditions 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 reactants

certain types of chemicals are more reactive than others

e.g. potassium metal is more reactive than sodium

ions react faster than molecules

no bonds need to be broken 80 Tro: Chemistry: A Molecular Approach, 2/e


The Transition State Theory

• There is an energy barrier to almost all reactions

• There exists an intermediate stage when reactant

molecules change into products.

• The activated complex or transition state is a

chemical species with partially broken and partially

formed bonds

an intermediate formed between reactant and

product molecules

has highest energy and extremely unstable



Figure 2

Reaction energy diagrams and possible transition states.


Energy Profile of a Catalyzed Reaction

polar stratospheric

clouds contain ice

crystals that catalyze

reactions that release

Cl from atmospheric

chemicals



Reaction progress

Po

ten

tia

l E

ne

rgySOLUTION:

Drawing Reaction Energy Diagrams and Transition States

PROBLEM 1.5: A key reaction in the upper atmosphere is

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

The Ea(fwd) is 19 kJ, and the Hrxn for the reaction is -392 kJ. Draw a

reaction energy diagram for this reaction, postulate a transition state, and

calculate Ea(rev).

PLAN: Exothermic reaction. The reactants are at a higher energy level than the

products and the transition state is slightly higher than the reactants.

O3+O

2O2

Ea= 19 kJ

Hrxn = -392 kJ

Ea(rev)= (392 + 19) kJ =

411kJ

O O

O

O

breaking

bond

forming

bond

transition state


• Changing the temperature changes the rate constant,

k, of the rate law

• The Arrhenius Equation shows the relationship

between absolute temperature, T, and rate constant,

k :

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

where T is the temperature in kelvins

A is called the Arrhenius constant (which includes the

frequency factor and the orientation factor)

Ea is the activation energy, the minimum energy

needed for the molecules to react

The Arrhenius Equation



The Arrhenius Equation:

The Exponential Factor

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

• The larger the activation energy, Ea, the fewer molecules that have sufficient energy to overcome the energy barrier, the smaller the rate constant k.

• Increase in temperature, T, will gives a larger k value therefore increasing the temperature will increase the reaction rate



Arrhenius Plots

• The Arrhenius Equation: The integrated form

this equation is in the linear form y = mx + b

where y = ln(k) and x = (1/T)

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

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

• 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:



Example 13.7: Determine the activation energy and

frequency factor for the reaction O3(g) O2(g) + O(g)




Example 13.7: Determine the Ea and the Arrhenius

constant, A, for the reaction O3(g) O2(g) + O(g)


90

slope, m = 1.12 x 104 K

y–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

92

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:

Ea T1, 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.

93

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

94

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:

Ea T1, k1, T2, k2



Reaction Mechanisms

• We generally describe chemical reactions with an equation listing all the reactant molecules and product molecules

• But the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible

• Most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules

• Describing the series of reactions that occurs to produce the overall observed reaction is called a reaction mechanism

• Knowing the rate law of the reaction helps us understand the sequence of reactions in the mechanism



An Example of a Reaction Mechanism

• Overall reaction:

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)

• Mechanism:

Step 1: H2(g) + ICl(g) HCl(g) + HI(g)

Step 2: HI(g) + ICl(g) HCl(g) + I2(g)

• The reactions (1) and (2) in this mechanism are called the elementary steps, meaning that they cannot be broken down into simpler steps, and that the molecules actually interact directly in this manner without any other steps



Overall: H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)

Step (1): H2(g) + ICl(g) HCl(g) + HI(g)

Step (2): HI(g) + ICl(g) HCl(g) + I2(g)

Intermediates

• Notice that the HI is a product in Step 1, but then a

reactant in Step 2

• Because HI is made but then consumed, HI does

not show up in the overall reaction

• Materials that are products in an early mechanism

step, but then a reactant in a later step, are called

intermediates

Adding equations (1) and (2), removing HI, will give

the overall equation.



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 particles

though these are exceedingly rare in elementary

steps



Rate Laws for Elementary Steps

• Each step in the mechanism is like its own little

reaction – with its own activation energy and own

rate law

• The rate law for an overall reaction must be

determined experimentally

• But the rate law of an elementary step can be

deduced directly from the equation of the step

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)

1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl]

2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]



Rate Laws of Elementary Steps



Rate Determining Step

• In most mechanisms, one step occurs slower than the other steps

• The result is that product formation cannot occur any faster than the slowest step – the step determines the rate of the overall reaction

• We call the slowest step in the mechanism the rate determining step

the slowest step has the largest activation energy

• The rate law of the rate determining step determines the rate law of the overall reaction



Another Reaction Mechanism

The first step in this

mechanism is the rate

determining step.

The first step is slower than the

second step because its

activation energy is larger.

The rate law of the first step is

the same as the rate law of the

overall reaction.

102

NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2

1. NO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 Slow

2. NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] Fast



Validating a Mechanism

• To validate (not prove) a mechanism, two

conditions must be met:

1. The elementary steps must sum to the overall

reaction

2. The rate law predicted by the mechanism must

be consistent with the experimentally observed

rate law



Mechanisms with a Fast

Initial Step

• When a mechanism contains a fast initial step, the rate

limiting step may contain intermediates

• When a previous step is rapid and reaches equilibrium,

the forward and reverse reaction rates are equal – so

the concentrations of reactants and products of the

step are related

and the product is an intermediate

• Substituting into the rate law of the RDS will produce a

rate law in terms of just reactants



An Example

1. 2 NO(g) N2O2(g) Fast

2. H2(g) + N2O2(g) H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2]

3. H2(g) + N2O(g) H2O(g) + N2(g) Fast

k1

k−1

2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2



Example 13.9: Show that the proposed mechanism for the

reaction 2 O3(g) 3 O2(g) matches the observed rate law

Rate = k[O3]2[O2]

−1

1. O3(g) O2(g) + O(g) Fast

2. O3(g) + O(g) 2 O2(g) Slow Rate = k2[O3][O]

k1

k−1



Practice – Mechanism

Determine the overall reaction, the rate determining

step, the rate law, and identify all

intermediates of the following mechanism

1. A + B2 AB + B Slow

2. A + B AB Fast



Practice – Mechanism

Determine the overall reaction, the rate determining

step, the rate law, and identify all intermediates of

the following mechanism

1. A + B2 AB + B Slow

2. A + B AB Fast

reaction 2 A + B2 2 AB

B is an intermediate

The first step is the rate determining step

Rate = k[A][B2], same as RDS


Documents

Ch 1 Chemical Kinetics