Chap 16 Web

Factors that Influence Reaction Rate

Expressing the Reaction Rate

Average, Instantaneous, Initial Reaction rates

Rate & Concentration

The Rate Law and its Components

Determining the Initial Rate

Reaction Order Terminology

Determining Reaction Orders

Determining the Rate Constant

Chap 16 - Chemical Kinetics

04/08/23 1

Integrated Rate Laws: Concentration changes over time

First, Second, and Zero-Order Reactions

Reaction Order

Reaction Half-Life

The Effect of Temperature on Reaction Rate

Explaining the Effects of Concentration and Temperature

Collision Theory

Transition State Theory

Chap 16 - Chemical Kinetics

04/08/23 2

Reaction Mechanisms: Steps in the Overall reaction

Elementary Reactions

The Rate-Determining Step

The Mechanism and the Rate Law

Catalysis: Speeding up a Chemical Reaction

Homogeneous Catalysis

Heterogeneous Catalysis

Chap 16 - Chemical Kinetics

04/08/23 3

Chemical Kinetics is the study of: Chemical Reaction Rates The changes in Chemical Concentration of

reactants as a function of time Chemical reactions range from very fast to very

slow Under a given set of conditions each reaction has

its own rate Factors that influence reaction rate:

Concentration Physical state (surface area) Temperature (frequency & energy of particle

collisions

Chemical Kinetics

04/08/23 4

Factors That Influence Reaction Rate Concentration

Molecule must Collide to React Reaction rate is proportional to the

concentration of the reactants

Rate collision frequency concentration Physical State

Molecules must Mix to Collide The more finely divided a solid or liquid

reactant:The greater its surface area per unit volumeThe more contact it makes with the other

reactantThe faster the reaction occurs

Chemical Kinetics

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Temperature

Molecules must collide with enough energy to react

At a higher temperature, more collisions occur in a given time

Raising the temperature increases the reaction rate by increasing the number and energy of the collisions

Rate Collision Energy Temperature

Chemical Kinetics

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A fundamental question addressed in chemical reactions is “how fast does the reaction occur?”

Kinetics is the study of the rate of chemical reactions; rate is a time dependent process

Rate units are concentration over time Consider the reaction A B Reactant concentrations [A] decrease while product

concentrations [B] increase

Note: Reaction rate is positive, but the concentration of A at t2 (A2) is always less than the concentration of A at t1 (A1), thus, the change in concentration (final – initial) of reactant A is always negative

04/08/23 7

Chemical Kinetics

2 1

2 1

conc A - conc Achange in concentration of A Δ(conc A)Rate of Reaction = - = - = -

change in time t - t Δt

Consider the reaction:

A + B C Concentrations of both reactants ([A] & [B])

decrease at the same rate

Chemical Kinetics

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change in concentrationRate =

change in time

Δ A Δ BRate = - = -

Δt Δt

Indicates “Change in”

Brackets [ ] indicate concentration

Reaction - Butyl Chloride (C4H9Cl) and water (H2O)

CH3(CH2)2CH2-Cl(l) + H2O(l) CH3(CH2)2CH2-OH(l) + HCl

04/08/23 9

Chemical Kinetics

-4ΔA 0.0905 mol / L - 0.1000 mol / LEx. rate = - = - = 1.9×10 mol / sec

Δt 50.0 s - 0.0 s

Chemical Kinetics

04/08/23 10

Butyl Chloride (C4H9Cl) When plotting When plotting Concentration Concentration

versus Timeversus Time for a chemical for a chemical reaction, the tangent at any reaction, the tangent at any point on the curve (drawn point on the curve (drawn through the concentration through the concentration points) defines the points) defines the instantaneous rateinstantaneous rate of the of the reactionreaction

The The average rateaverage rate of a of a reaction over some time reaction over some time interval is determined through interval is determined through triangulation of concentration triangulation of concentration plot (slope of hypotenuse of plot (slope of hypotenuse of right triangle)right triangle)

The rate of the reaction The rate of the reaction decreasesdecreases over time as the over time as the reactants are consumedreactants are consumed

Rate of reaction of the Products The rate of reaction for the formation of the

products is the same as for the reactants, but opposite, that is the concentrations are increasing

Chemical Kinetics

04/08/23 11

2 4 3 2 4 2

32 4 2 4 2

C H + O C H O + O

Δ[O ]Δ[C H ] Δ[C H O] Δ[O ]Rate = - = - =

+ = + Δt Δt Δ

t t

Δ

The rate of change of ethane (C2H4) and Ozone (O3) is the same, but exactly opposite for acetaldehyde (C2H4O) and oxygen (O2)

Product concentration increases at the same rate that the reactant concentrations decrease

The curves have the same shape, but are inverted

The Rate expression must be consistent with stoichiometry

When the stoichiometric molar ratios are not 1:1, the reactants disappear and the products appear, but at different rates

Chemical Kinetics

04/08/23 12

2 2H (g) + I (g) = 2HI(g)

22 Δ I Δ HIΔ[H ] 1Rate = - = - =

Δt Δt 2 Δt

For every molecule of H2 that disappears, one molecule of I2 disappears and 2 molecules of HI appear

The rate of H2 decrease is the same as the rate of I2 decrease, but both are only half the rate of HI increase

Chemical Kinetics

04/08/23 13

aA + bB = cC + dD

Δ A Δ B Δ C Δ D1 1 1 1Rate = - = - = + = +

a Δt b Δt c Δt d Δt

Summary equation for any reaction

The rate of a reaction is dependent on the concentration of reactants

The average reaction rate is the change in reactant (or) product concentration over a change in time, t

The instantaneous rate at a time, t, is obtained from the slope of the tangent to a concentration vs. time curve at a given time, t

As reactant concentrations decrease, the reaction rates decrease with time

Product concentrations increase at the same rate as the reactants relative to the stoichiometric ratios

The rate of a reaction depends on the following variables: reactant concentration temperature presence and concentration of a catalyst surface area of solids, liquids or catalysts

Chemical Kinetics

04/08/23 14

Write an expression defining equivalent rates for the loss of NO2 and the formation of NO in the following reaction with respect to the rate of formation of O2.

2 NO2(g) 2 NO(g) + O2(g)

Sample Problems

04/08/23 15

2 2Δ O Δ NOΔ NO1 1Rate = = = -

Δt 2 Δt 2 Δt

The dependence of reaction rate on concentrations is expressed mathematically by the rate law

The rate law expresses the rate as a function of reactant concentrations, product concentrations, and temperature

In the following development, only the reactants appear in the rate law

For a general reaction at a fixed temperature:

aA + bB + … cC + dD + …

the rate law has the form:

Rate = k[A]m[B]n ...

Note: The Stoichiometric Coefficients – a, b, c – are not used in the rate equation and are not related to the reaction order terms – m, n, p, etc.

Chemical Kinetics –The Rate Law

04/08/23 16

Components of the rate law

aA + bB + … products

Rate = -[A]/t = k[A]m[B]

n[C]

p

[A] & [B] = concentrations of reactants (M)

[C] = concentration of catalyst (M), if used

k = rate constant

m, n, & p = Reaction Orders

Note: Reaction Orders are not related to the Stoichiometric coefficients in chemical equation)

Rate Law

04/08/23 17

The Rate Law The rate constant “k” is a proportionality constant “k” changes with temperature; thus it determines

how temperature affects the rate of the reaction The exponents (m, n, p, etc.) are called reaction

orders, which must be determined experimentally Reaction orders define how the rate is affected by

the reactant concentration If the rate doubles when [A] doubles, the rate

depends on [A] raised to the first power, i.e., m =1 (a 1st order reaction)

If the rate Quadruples when [B] doubles, the rate depends on [B] raised to the second power, i.e., n = 2 (a 2nd order reaction)

Chemical Kinetics

04/08/23 18

Units of the Rate Constant k change depending on the overall Reaction Order

Rate Constant - Units

04/08/23 19

Overall Reaction Order Units of k (t in seconds)

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

1 1/s (or s-1)

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

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

The Rate Law

If the rate does not change even though [A] doubles, the rate does not depend on the concentration of A and m = 0.

The Stoichiometric coefficients, a, b, c, etc. in the general balanced equation are not necessarily related in any way to the reaction orders m, n, etc.

The components of the Rate Law – rate, reaction orders, rate constant – must be determined experimentally; they cannot be deduced or inferred from the balanced stoichiometric equation

The Rate Law

04/08/23 20

The Rate Law - ExamplesNO(g) + O3(g) NO2(g) + O(g)

Rate = k[NO]1[O3]1

Reaction is 1st order with respect to NO, m=1Rate depends on [NO] raised to 1st powerReaction is 1st order with respect to O3, n=1

The overall reaction is 2nd order, m + n = 1 + 1 = 2

2NO(g) + 2H2(g) N2(g) + 2H2O(g)Rate = k[NO]2[H2]1

Reaction is 2nd order in NO and 1st order in H2

Overall reaction is 2 + 1 = 3rd orderNote:[NO] coefficient (2) is not related to the

[NO] reaction order (2)

Rate Law

04/08/23 21

The rate law – Examples

(CH3)3C–Br(l) + H2O(l) (CH3)C–OH(l) + H+(aq) + Br-

(aq)

Rate = k[(CH3)3CBr]1[H2O]0 or

Rate = k[(CH3)3CBr]1

Reaction is first order in 2-bromo-2-methyl propaneReaction is zero order (n=0) in water [H2O]0

Note: zero order reaction order terms, ex. [H2O]0 can be eliminated from the overall rate equation, i.e. any term raised to the “0” power is equal to 1

[H2O]0 = 1

Rate Law

04/08/23 22

The Rate Law

CHCl3(g) + Cl2(g) CCl4(g) + HCl(g)

Rate = k[CHCL3][Cl2]1/2

The reaction order means that the rate depends on the square root of the Chlorine (CL2) concentration

If the initial Cl2 concentration is increased by a factor of 4, while the initial concentration of CHCl3 is kept the same, the rate increases by a factor of 2, the square root of the change in Cl2

Rate Law

04/08/23 23

The Rate Law

Negative reaction orders are used when the law includes the product(s)

If the O2 concentration doubles, the reaction proceeds at one half (1/2) the rate

Rate Law

04/08/23 24

3 2

22 -1 3

3 2 12

2O (g) 3O (g)

[O ]Rate = [O ] [O ] =

[O ]k k

Overall reaction order = sum of exponents in rate equation

Order of Rxn Possible Expression of Rate Law

1 k[A]

2 k[A]2

2 k[A][B]

3 k[A]2[B]

3 k[A][B][C]

Rate Law – Reaction Order

04/08/23 25

What is the reaction order of acetaldehyde and the overall order in the following reaction

CH3CHO(g) CH4(g) + CO(g)

Rate = k[CH3CHO]3/2

Ans: 3/2 order in CH3CHO

Overall order: 3/2

Practice Problem

04/08/23 26

Experiments are performed for the reaction

A B + C

and the rate law has the been determined to be of the form

Rate = k[A]x

Determine the value of the exponent “x” for each of the following:

a. [A] is tripled and you observe no rate change

Ans: x = 0 k[3A]0

b. [A] is doubled and the rate doubles

Ans: x = 1 k[2A]1

c. [A] is tripled and the rate increases by a factor of 27

Ans: x = 3 k[3A]3

Practice Problem

04/08/23 27

Concentration Exponents (reaction orders) must be determined experimentally because the stoichiometric balanced equation with its reaction coefficients, does not indicate the mechanism of the reaction

Experimentally, the reaction is run with varying concentrations of the reactants, while observing the change in rate over time

The initial rate of reaction is observed, where the rate is linear with time (instantaneous rate = average rate); usually just when the reaction begins

Experimental Rate Law

04/08/23 28

Initial rates of reaction from experiments on the reaction:O2(g) + 2NO)g) 2 NO2)g)

Rate = k[O2]m[NO]n

Determine “m” & “n” from experimental data

Experimental Rate Law

04/08/23 29

Initial Reactant Concentration (mol/L)

Experiment

O2 NOInitial Rate

Mol/Ls

1 1.10x10-2 1.30x10-2 3.21x10-3

2 2.20x10-2 1.30x10-2 6.40x10-3

3 1.10x10-2 2.60x10-2 12.8x10-3

4 3.30x10-2 1.30x10-2 9.60x10-3

5 1.10x10-2 3.90x10-2 28.8x10-3

Rate equations from two applicable experiments are combined,

depending on the reactant order to be determined

Select the 1st two experiments where the effect of doubling the concentration of O2 is observed at constant temperature

Experimental Rate Law

04/08/23 30

n2 2 2

2 1 1

k[O ] [NO]Rate2=

Rate1 k[O ] [NO]

m

m n

K is constant and [NO] does not change

2

1

m

2 2 2

22 1

[O ] [O ]Rate2=

Rate1 [O ][O ]

m

m

Substitute rate values and concentration values

-3 -2

-3 -26.40 x 10 mol/ l•s 2.20 x 10 mol/ L

=3.21 x 10 mol/ l•s 1.10 x 10 mol/ L

m

=1(rounded)m

Reaction is 1st order in O2:

When [O2] doubles, the rate doubles

1.99=2.00m

log(1.99) = log(2.00)m

Determining the Rate Constant (k) The rate data from any one of the experiments in the

previous table can be used to compute the rate constant

Using the first experiment:

Experimental Rate Law

04/08/23 31

3 2 2 =1.73 x 10 L / mol •s (Note units for overall reaction order = 3)k

-2 -22[O ]=1.10 x 10 mol / L [NO]=1.30 x 10 mol / L

-3Rate 1=3.21 x 10 mol / L•s

2Rate 1 = [O ] [NO]m nk

2

Rate 1 =

[O ] [NO]m nk

m=1; n=2 (From previous calculations) (Overall reaction order = 1 + 2 = 3)

1k

-3

2-2 -2

3.21 x 10 mol / L•s =

1.10 x 10 mol / L 1.30 x 10 mol / L

Concentration changes over time Previous notes assume that time is not a variable

and the rate or concentration for a reaction is at a given instant in time

By using time as a factor in the reaction, the rate law can be integrated

“How long will it take

for x moles per liter of reactant ‘A’ to be used up?”

“What are the concentrations of ‘A’ after ‘y’ minutes of the reaction”

Integrated Rate Laws

04/08/23 32

Integration of Rate Equation

04/08/23 33

1

o

t 0

0 t

0

t

Δ A- = k A

ΔtRearrange Equation

Δ A 1 1- = kΔt dx = kdt dx = ln x

A x x

-ln A = kt + C (at t = 0, C = ln[A ])

-ln A - A = kt

ln A - A = kt

A ln = kt or

A

0

t

A ktlog =

A 2.303

First Order Reaction

Second Order Reaction

Integration of Rate Equation

04/08/23 34

2

n+1 -2+1

2 2 n

0

t 0

Δ A- = k A

Δt

Δ A 1 1 x x 1- = kΔt dx = kdt dx = = = -

n + 1 -2 + 1 xx xA

1 1- - = kt + C a t = 0, C =

A A

1 1- - = kt +

A A

t 0

t 0

1 1 - = kt

A A

1 = kt

A - A

Zero Order Reaction

Integration of Rate Equation

04/08/23 35

0

t 0

t 0

Δ A- = k A = k × 1

ΔtΔ A

- = kAt

-Δ A = kAt

-( A - A ) = kt

A - A = - kt

Integrated Rate Law – Straight Line Plot

Integrated Rate Law

04/08/23 36

st

0 t

t 0

For a 1 order reaction

ln[A] - ln[A] = kt

Rearrange into equation for a straight line

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

y = mx + b (mnd

t 0

t 0

= slope; b = y - axis intercept)

For a simple 2 order reaction :

1 1 - = kt

[A] [A]

1 1 = kt +

[A] [A]

y = mx + b (

t 0

t 0

m = slope; b = y - axis intercept)

For a zero - order reaction :

[A] - [A] = - kt

[A] = - kt + [A]

y = mx + b

Integrated Rate Law

04/08/23 37

04/08/23 38

First-Order Concentration vs.Time Graphs

1st Order Reaction Half-Life The half-life of a reaction is the time required

for the reactant concentration to reach ½ its initial value

At fixed conditions, the half-life of a 1st order reaction is a constant, independent of reactant concentration

Integrated Rate Law

04/08/23 39

0

t

11/ 2 t 02

01/ 21

02

1/ 2

1/ 2

[A]ln = kt

[A]

After one half - life, t = t , and [A] = [A]

Substituting

[A]ln = kt

[A]

ln 2 = kt

ln2 0.693t = =

k k

[Recall Radioactivity half-life (Chap 24)]

2nd Order Reaction Half-Life

Integrated Rate Law – Half-LIfe

04/08/23 40

0 0

11/ 2 t 02

0 0 0 0 0 01/ 2

1/ 20

1 1 - = kt

A A

After one half - life, t = t , and [A] = [A]

1 1 1 1 2 1 1- - -

A t A 1 / 2 A A A A At = = = =

k k k k

1t =

k A

Zero Order Reaction Half-Life

Integrated Rate Law – Half-LIfe

04/08/23 41

t 0

11/ 2 t 02

t 0 0 0 01/ 2

01/ 2

[A] - [A] = - kt

After one half - life, t = t , and [A] = [A]

[A] - [A] 1 / 2[A] - [A] -0.5[A]t = = =

-k -k -k

[A]t =

2k

A reaction is first order with respect to A. The first-order rate constant is 2.61 /min. How long will it take the concentration of A to decrease from 0.100 M to 0.00812 M? What is the half-life of the reaction? How long will it take for the concentration of A to decrease by 85%?

Practice Problem

04/08/23 42

t = 0.962min

k = 2.61 / min

0 tln[A] - ln[A] = kt

0

t0 t

[A] 0.100ln ln[A] ln 12.3152709ln[A] - ln[A] 2.510840.00812

t = = = = =k k 2.61 / min 2.61 2.61

1/2ln2 0.693 0.693

t = = = = 0.266 mink k 2.61 / min

0

t0 t

[A] 0.1ln ln[A] ln 6.66667ln[A] - ln[A] 1.897120.1×0.15

t = = = = = = 0.727 mink k 2.61 2.61 2.61

A reaction is second order with respect to B.The second-order rate constant is 1.5 L/molmin

How long will it take the concentration of B to decrease from 0.100 M to 0.025 M?

Practice Problem

04/08/23 43

40 -10t = = 20min

1.5

k = 1.5 L / mol • min

t 0

1 1- = kt

[A] [A]

t 0

1 1-

[A] [A]t =

k1 1

-0.025 mol / L 0.100 mol / Lt =

1.5 L / mol • min

An increase in Temperature (T) generally increases the reaction rate

A 10 oC increase in T usually doubles rate Temperature affects the rate constant (K) of the

rate equation Temperature effect process is described by

Collision Theory Can calculate the effect of T on rate of a reaction

using the Arrhenius Equation

Temperature Dependence of Reaction Rate

04/08/23 44

Two major models explain the observed effects of Concentration & Temperature on reaction rate Collision Theory

Views the reaction rate as a result of particles colliding with a certain frequency and minimum energy

Transition State Theory Close-up view of how the energy of a

collision converts reactant to product

Effects of Concentration & Temperature

04/08/23 45

Why concentrations are “Multiplied” in the Rate Law Consider 2 particles of “A” & 2 particles of “B” Total A-B collisions = 4 (2 x 2)

A1B1 A1B2 A2B1 A2B2 Add additional Particle of “A” Total A-B collisions = 6 (3 x 2)

A1B1 A1B2 A2B1 A2B2 A3B1 A3B2 It is the product of the number of different

particles, not the sum (6 vs 5), that determines the number of collisions (reactions) possible

The number of particles of reactant A (concentration) must be multiplied by the number of particles of Reactant B to account for the total number of collisions (reactions) that occur.

Collision Theory

04/08/23 46

Increasing the temperature of a reaction increases the average speed of particles; thus, the frequency of collision

Most collisions do not result in a “reaction” Collision Theory assumes that, for a reaction to

occur, reactant molecules must collide with an energy greater than some minimum value and with proper orientation

Activation Energy (Ea) The rate constant, k, for a reaction is a function of

3 collision related factors: Z, collision frequency f, fraction of collisions => activation energy p, fraction of collisions in proper orientation

k = Zpf

Collision Theory

04/08/23 47

At a given temperature, the fraction of molecular collisions, f, with energy greater than or equal to the activation energy, Ea, is related to activation energy by the expression:

An equation (Arrhenius) expressing the dependence of the rate constant, k, on temperature can be obtained by combining the relationship between the rate constant and fraction of collisions, f, that are >= to the “activation energy”, Ea

Collision Theory

04/08/23 48

-E RTf = e a

a

a

-E RT

-E RT

= Zpf

Let A = Zp = frequency factor

= Af

f = e

= Ae

k

k

k

Temperature dependence of reaction rate

k = Ae -Ea/RT

K = rate constant

A = frequency factor (pZ)Ea = activation energy (J)

R = gas constant (8.314 J/molK)

T = temperature (K) The negative exponential relationship between

temperature (T) and the rate constant k means that as the temperature increases, the negative exponent becomes smaller, so the value of k becomes larger, which means that the rate of the reaction increases

Higher T large k increased rate

Arrhenius Equation

04/08/23 49

The activation energy (Ea) can be calculated from the Arrhenius equation by taking the natural logarithm of both sides and rearranging the equation into a “straight line (y=b+mx) form

Arrhenius Equation

04/08/23 50

a

a

-E / RT

-E / RT

a

= Ae

ln = ln A + ln e

E 1ln = ln A -

R T

y = b + mx

k

k

k

A plot of ln k (y) vs. 1/T (x) gives a straight line whose slope (m) is -Ea/R and whose y intercept is Ln A (b)

Ea can be determined graphically from a series of k values at different temperatures Determine the slope from the plot Use slope formula = -Ea/R

Arrhenius Equation

04/08/23 51

Alternate Approach - Compute Ea mathematically if the rate constants at two temperatures are known

a a2 1

2 1

a a a a2 1

2 1 2 1

a a2 1 2

1 2 1 1 2 2 1

E E1 1ln k = ln A - ln k = ln A -

R T R T

E E E E1 1 1 1ln k - ln k = ln A - - ln A - = - +

R T R T R T R T

E Ek T T1 1 1 1ln = - - = - × - × = -

k R T T R T T T T

a 1 2

1 2

2 1 2a

1 1 2

E T - T

R T T

k T TE = -R × ln ×

k T - T

“ln A” term

drops out

Find the Activation Energy (Ea) for the decomposition of Hydrogen Iodide (HI)

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

The rate constants are:

9.51x10-9 L/mols at 500oK

1.10x10-5 L/mols at 600oK

Practice Problem

04/08/23 52

2aE = 1.76x10 kJ / mol

2 1 2a

1 1 2

k T TE = -R ×ln ×

k T - T

-5 o o

a -9 o o

1.10 x 10 L / mol •s 500 K ×600 KE = - 8.314J / mol • K ×ln ×

9.51 x 10 L / mol • s 500 K - 600 K

3 3aE = - 8.314J / mol • K ×ln 1.156677 x 10 × -3.00 x 10

3aE = - 8.314J / mol • K ×(7.053307)× -3.00 x 10

5a 3

1kJE = 1.76 x 10 J / mol ×

1 x 10 J

Rate – Affects of Temperature

04/08/23 53

REACTANTS

PRODUCTS

ACTIVATED STATEC

ollis

ion

Ene

rgy

Col

lisio

n E

nerg

y

Ea (forward)

Ea (reverse)

Molecules must collide with sufficient energy to reach “activation” status

Minimum collision energy is “energy of activation, Ea”

The forward reaction is exothermic because the reactants have more energy than the products.

Collision Theory: Proper Orientation (p)

04/08/23 54

Transition-state theory explains the reaction in terms of the collision of two high energy species – activated complexes An activated complex (transition state) is an

unstable grouping of atoms that can break up to form products.

A simple analogy would be the collision of three billiard balls on a billiard table.

Suppose two balls are coated with a slightly stick adhesive.

We’ll take a third ball covered with an extremely sticky adhesive and collide it with our joined pair.

Transition-State Theory

04/08/23 55

Transition-state theory (cont’) At the instant of impact, when all three

spheres are joined, we have an unstable transition-state complex

The “incoming” billiard ball would likely stick to one of the joined spheres and provide sufficient energy to dislodge the other, resulting in a new “pairing”

If we repeated this scenario several times, some collisions would be successful and others (because of either insufficient energy or improper orientation) would not be successful.

We could compare the energy we provided to the billiard balls to the activation energy, Ea

Transition-State Theory

04/08/23 56

Reaction of Methyl Bromide & OH-

Reaction is exothermic – reactants are higher in energy than products

Forward activation energy Ea(fwd) is less than reverse Ea(rev)

Difference in activation energies is “Heat of Reaction”

Hrxn = Ea(fwd) - Ea(rev)

Transition State Theory

04/08/23 57

Note the partial elongated C-O and C-Br bonds and the trigonal bipyramidal shape of the transition state

Exothermic Reaction Pathway

04/08/23 58

Transition State

Hrxn = Ea(fwd) - Ea(rev)

Ea(fwd) < Ea(rev)

Endothermic Reaction Pathway

04/08/23 59

Ea(fwd) > Ea(rev)

Steps in the overall reaction that detail how reactants change into products Reaction Mechanism – set of elementary

reactions that leads to overall chemical equation

Reaction Intermediate – species produced during a chemical reaction that do not appear in chemical equation

Elementary Reactions – single molecular event resulting in a reaction

Molecularity – number of molecules on the reactant side of elementary reaction

Rate Determining Step (RDS) – slowest step in the reaction mechanism

This is the reaction used to construct the rate law; it is not necessarily the overall reaction

Reaction Mechanisms

04/08/23 60

Proposed Overall Reaction

2 NO2(g) + 2 H2(g) 2 H2O(g) + N2(g)

A mechanism in 3 elementary reactions:

2 NO2 N2O2 (slow) (RDS)

H2 + N2O2 H2O + N2O (fast)

H2 + N2O H2O + N2 (fast)

The Overall Reaction from Elementary Reactions:

2 NO2(g) + 2 H2(g) 2 H2O(g) + N2(g)

N2O2 and N2O are reaction intermediates

Develop “Rate Law” from the “Rate Determining Step” (RDS) Rate law = Rate = k[NO2]2

Note: The reaction order in an elementary reaction comes from the stoichiometric coefficient, i.e., 2

Adding together the reactions in the mechanism provides the overall chemical equation

Reaction Mechanisms

04/08/23 61

Elementary Reactions (Steps) – The individual steps, which together make up a proposed reaction mechanism

Each elementary reaction describes a single molecular event, such as one particle decomposing or two particles colliding and combining.

Reaction Mechanisms

04/08/23 62

An elementary step is characterized by its “Molecularity – the number of reactant particles involved in the step

2O3(g) 3O2(g)

Proposed mechanism – 2 steps1st step – Unimolecular reaction

(decomposition)O3(g) O2(g) + O(g)

2nd step – Bimolecular reaction (2 particles react)

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

Reaction Mechanisms

04/08/23 63

Rate law for an elementary reaction can be deduced directly from molecularity of reaction (w/o experimentation)

An elementary reaction occurs in one step Its rate must be proportional to the product of the

reactant concentrations The stoichiometric coefficients are used as the

reaction orders in the rate law for an elementary step

The above statement holds only for an elementary reaction

In an overall reaction the reaction orders must be determined experimentally

Rate Law & Reaction Mechanisms

04/08/23 64

Steps in determining rate law from reaction mechanism Identify the rate determining step (RDS) of the

mechanism Write out the preliminary rate law from RDS Remove expressions for intermediates

algebraically Substitute into preliminary rate law to obtain

final rate law expression

Rate Law & Reaction Mechanisms

04/08/23 65

Practice Problem

04/08/23 66

PLAN:

SOLUTION:

The following two reactions are proposed as elementary steps in the mechanism of an overall reaction:

(1) NO2Cl(g) NO2(g) + Cl(g)

(2) NO2Cl(g) + Cl(g) NO2(g) + Cl2(g)

(a) Write the overall balanced equation(b) Determine the molecularity of each step(c) What are the reaction intermediates(d) Write the rate law for each step

(a) The overall equation is the sum of the steps

(b) Molecularity is the sum of the reactant particles in the step.

rate2 = k2[NO2Cl][Cl]

Step(1) is unimolecular.Step(2) is bimolecular.

(b)

rate1 = k1[NO2Cl](d)

(c) Cl(g) is reaction intermediateNO2(g) + Cl (g)(1) NO2Cl(g)(a)

(2) NO2Cl(g) + Cl (g) NO2(g) + Cl2(g)

2NO2Cl(g) 2NO2(g) + Cl2(g)

Criteria required for proposed reaction mechanism The elementary steps must add up to the

overall balanced equation The number of reactants and products

in the elementary reactions must beconsistent with the overall reaction

The elementary steps must be physically reasonable – they should involve one reactant (unimolecular) or at most two reactant particles (bimolecular)

The mechanism must correlate with the “rate law” – The mechanism must support the experimental facts shown by the rate law, not the other way around

Correlating Rate Law & Mechanism

04/08/23 67

If a slow step precedes a fast step in a two-step mechanism, do the substances in the fast step appear in the rate law?

Ans: No, the overall rate law must contain reactants only (no intermediates) and is determined by the slow step

If the first step in a reaction mechanism is slow, the rate law for that step is the overall rate law

If a fast step precedes a slow step in a two-step mechanism, how is the fast step affected?

Ans: If the slow step is not the first one, the faster preceding step produces intermediates that accumulate before being consumed in the slow step

How is this effect used to determine the validity of the mechanism?

Ans: Substitution of the intermediates into the rate law for the slow step will produce the overall rate law.

Practice Problem

04/08/23 68

Reaction between nitrogen dioxide & chlorine gas

Overall reaction2NO2(g) + F2(g) 2NO2F(g)

Experimental Rate LawRate = k[NO2][F2] (1st order)

Mechanism (1) NO2(g) + F2(g) NO2F(g) + F(g) [slow,

rds] (2) NO2(g) + F(g) NO2F(g) [fast]

Overall: 2NO2(g) + F2(g) 2NO2F(g)

Criteria 1: Elementary steps add up to experimental

Criteria 2: Both steps “Bimolecular”

Mechanism with a Slow Initial Step

04/08/23 69Con’t on next Slide

Criteria 3:Experimental Rate Law: k[NO2][F2] Elementary Reaction Rate Laws

Rate1 = k1[NO2][F2] (from rds)Rate2 = k2[NO2][F]

Rate 1 (k1) from rds is same as overall kThe 2nd [NO2] term (in Rate2) does not appear in

the overall rate law

Mechanism with a Slow Initial Step

04/08/23 70

Each step in mechanism has its own transition stateProposed transition state is shown in step 1Reactants for 2nd step are the F atom intermediateand the 2nd molecule of NO2

First step is slower – Higher Ea

Overall reaction is exothermic - Hrxn < 0

Nitric oxide, NO, is believed to react with chlorine (Cl2) according to the following mechanism

NO + Cl2 NOCl2 (Fast, equilibrium)

NOCl2 + NO 2 NOCl (slow, RDS)

1. What is the overall chemical equation for the reaction?

2. Identify the reaction intermediates.

3. Propose a viable rate law from the mechanism.

Mechanism with a Fast Initial Step

04/08/23 71

2

2

2

1(fwd) 2 1(rev) 2

12

Intermediate Reactant - NOClRDS = NOCl + NO 2NOCl

The rate law cannot be directly determined from the RDSbecause of the presence of NOCl , an"Intermediate Reactant"

k [NO][CL ] = k [NOCl ]k

[NOCl ] =

(fwd)2

1(rev)

21(fwd)2 2

1(rev)

[NO][Cl ]k

Substituting in RDS gives :k

Rate = [NO][Cl ][NO] = k NO Clk

2 2

2

2

NO + Cl NOCl

NOCl + NO 2NOCl

2NO + Cl 2NOCl

It is often necessary to “Speed up” a reaction in order to make it useful and in the case of industry, profitable

Approaches More energy (heat) – could be expensive!! Catalyst – Stoichiometrically small amount of a

substance that increases the rate of a reaction; it is involved in the reaction, but ultimately is not consumed

Catalysis – Speeding Up Reaction

04/08/23 72

Catalyst: Causes lower “activation energy”, (Ea) Lower activation energy is provided by a

change in the reaction mechanism Makes Rate constant larger Promotes higher reaction rate Speeds up forward & reverse reactions Does not improve yield – just makes it faster

Catalysis – Speeding Up Reaction

04/08/23 73

Homegeneous Catalysts Exist in “Solution” with the reactant mixture All homogenous catalysts are gases, liquids, or

“soluble” solids

Homogeneous Catalysts

04/08/23 74

Mechanism for the catalyzed hydrolysis of an organic ester.

R C

O

O

R'

H

H

O H R C

O

O

R'

H

H

O H

slow, rate-determining

step

R C

O

OHR'

H+

O Hall fast

R C

O

O

R'

H

R C

O

O

R'

H

R C

O

O

R'

H

resonance hybrid

resonance formsH+ , the catalyst, is a proton supplied by a strong acid

Step 1: Catalytic H+ ion bonds to electron rich oxygen.

Step 2: Slow, rate determining step.

The increased positive charge on the Carbon attracts the partially negative oxygen of the water more strongly, increasing the fraction of effective collisions, speeding up this rate determining step

H+ + R CO

OR'

R CO

OR'

Hfast

Step 2

Step 1

Steps 3-6

Heterogeneous Catalysts Speeds up a reaction that occurs in a separate

phase Ex. A solid interacting with gaseous or liquid

reactants The solid would have extremely large surface

area for contact If the rate-determining step occurs on the

surface of the catalyst, many reactions are zero order, because once the surface area is covered by the reactant, increasing the concentration has no effect on the rate

Heterogeneous Catalysts

04/08/23 75

Hydrogenation of Ethylene (Ethene) to Ethane catalyzed by Nickel (Ni), Palladium (Pd), or Platinum (Pt)

H2C=CH2(g) + H2(g) H3C – CH3

Finely divided Group 8B metals catalyze by adsorbing the reactants onto their surface

H2 lands and splits into separate H atoms chemically bound to solid catalyst’s metal atoms

H – H + 1catM(s) 2catM – H Then C2H4 absorbs and reacts with two H atoms,

one at a time, to form H3C–CH3

The H-H split is the rate determining step providing a lower energy of activation

Heterogeneous Catalysts

04/08/23 76

Ni, Pd, Pt

77

Effect of A CatalystComparison of Activation Energies in the Uncatalyzed and

Catalyzed Decompositions of Ozone

Catalyst: provides alternative mechanism for a reaction that has a lower activation energy

04/08/23

Ethyl Chloride, CH3CH2Cl2, used to produce tetraethyllead gasoline additive, decomposes, when heated, to give ethylene and hydrogen chloride. The reaction is first order. In an experiment, the initial concentration of ethyl chloride was 0.00100 M. After heating at 500 C for 155 s, this was reduced to 0.00067 M. What was the concentration of ethyl chloride after a total of 256 s?

Practice Problem

04/08/23 78t [A] = 0.00052 M

o

t

[A]ln = kt (1st order reaction)

[A]

o

t

[A] 0.00100Mln ln ln 1.492537[A] 0.4004780.00067M = = = = = 0.0026 / st 155s 155 155

k

Determine concentration after t = 256 secondst 0ln[A] - ln[A] = - tk

t 0ln[A] = ln[A] - tk

tln[A] = ln(0.00100 M) - 0.0026 / s × 256 s

tln[A] = - 6.90776 - 0.6656 = - 7.57336

The rate of a reaction increases by a factor of 2.4 when the temperature is increased from 275 K to 300 K. What is the activation energy of the reaction?

Practice Problem

04/08/23 79

aE = 65 kJ / mol

2 1 2a

1 1 2

k T ×TE = -R ×ln ×

k T - T

o o

1a o o

1

2.4 k 275 K × 300 KE = - 8.314J / mol • K ln ×

k 275 K - 300 K

aE = 65,846.88J / mol

The rate constant of a reaction at 250 oC is 2.69 x 10-3 1/M-s (L/mols). Given the activation energy for the reaction is 250 kJ, what is the rate constant for the reaction at 100 oC, assuming activation energy is independent of temperature?

Practice Problem

04/08/23 80

2 1 2a

1 1 2

k T ×TE = -R ×ln ×

k T - T

a 1 22 1

2 1

E T - Tln = ln + ×

-R T ×Tk k

-102k = 2.52x10 L / mol •s

o o-3

2 o o

kJ 1000J250 (250 C + 273.15)K - (100 C + 273.15)Kmol kJlnk = ln(2.69x10 L / mol • s + ×

-8.314J / mol • K (250 C + 273.1)K ×(100 C + 273.15)K

o

2 o 2

250,000J / mol 150 Klnk = 1.002694 L / mol • s + ×

-8.314J / mol • K 195,213.4225 K

237,500,000

lnk = 1.002694L / mol • s -1,623,004.395

2lnk = 1.002694L / mol • s - 23.105298 = -22.102604

If the half-life of a first-order reaction is 25 min, how long will it take for 20% of the reactant to be consumed?

Practice Problem

04/08/23 81

t = 8.0 min

1/ 2ln2 0.693

t = =k k

1/ 2 1/ 2

0.693 0.693 0.693k = = =

t t 25min

k = 0.02772 / min

o

t

[A]ln = kt

[A]

t 0[A] at time, t = 0.8[A]

oo

0t

[A][A] 1.00lnln ln0.8[A] ln 1.25[A] 0.2231435510.80

t = = = =k 0.02772 0.02772 0.02772 0.02772

For a reaction with the rate law given as rate = k[A]2, [A] decreases from 0.10 to 0.036 M in 161 min. What is the half-life of the reaction?

Practice Problem

04/08/23 82

1/ 2t = 91min

2k[A] represents the rate of a second order reaction

t 0

1 1- = kt

A A

t 0

1 1 1 1- -A A 27.7778L / mol -10L / mol 17.7778L / mol0.036M 0.10Mk = = = =t 161min 161min 161min

k = 0.11L / mol • min

ndHalf - Life for 2 order reaction

1/ 20

1t =

k A

1/ 21

t =0.11 L / mol • min x 0.10 mol / L

(See slide # 41)

The decomposition of the herbicide atrazine in the atmosphere by sunlight is first order, with a rate constant of 1.1 x 10-3 1/s. Following field application by spaying it is found that the atmospheric concentration of atrazine is 2.5 x 10-6 ppm at mid-day. How long (in hours) will it take for the atmospheric concentration of atrazine to reach the air quality standard of 1.0 x 10-9 ppm?

Practice Problem

04/08/23 83

t = 2.0 hr

stRate Law for 1 order reaction

0

t

A ln = kt

A

-60

3-9t

-3

A 2.5x10 mol / Llnln

ln 2.5x10A 1.0x10 mol / L 7.82405t = = = =

1hrk 3.9600 / hr 3.96 / hr1.1x10 / s3600s

The indirect photolysis of the pesticide atrazine (Atr, C8H14ClN5) in air by hydroxyl radical (OH) is shown below

C8H14ClN5 + OH C8H13ClN5 + H2O

The reaction is second order and follows the rate law [Atr]/t = kph[Atr][OH]. The concentration of OH is at steady-state during daylight hours at ~1.0e-18 M, and kph is 5.0e15 1/M-s. How long (in min) will it take for Atr to decrease from 2.5e-15 M to 1.0e-18 M (the air quality criteria) following application to a golf course assuming pseudo-first-order kinetics during daylight?

a. 26 b. 1.8 c. 527 d. 4,218 e. 119

Practice Problem

04/08/23 84Solution on next Slide

Photolysis of Atrazine (con’t) The 2nd order rate law ([Atr]/t = kph[Atr][OH]) is stated

in terms of two reactants This would result in a different integrated form of the 2nd

order reaction, more complicated math Since the concentration of hydroxyl, [OH], is constant,

the rate law can be reduced to a pseudo 1st order reaction by combining the Kph & [OH] terms (both constants) into a new rate constant:

Practice Problem

04/08/23 85

15 -18 -3phk = k ×[OH] = 5.0 x 10 L / mol • s ×1.0 x 10 mol / L = 5.0 x 10 / s

nd stΔ[Atr]= k[Atr] (Restatement of 2 order rate law into pseudo 1 order)

Δt

o

t

[A]ln = kt (Integrated 1st order Rate Law))

[A]

-15o

3-18t

-3 -3 -3

[A] 2.5 x 10 mol / Lln ln[A] ln(2.5 x 10 ) 7.8240461.0 x 10 mol / Lt = = = =k 5.0 x 10 / s 5.0 x 10 / s 5.0 x 10 / s

3 1mint = 1.5648 x 10 s× = 26 min

60s

A convenient rule of thumb is that the rate of a reaction doubles for a 10 oC change in temperature.

What is the activation energy for a reaction whose rate doubles from 10.0 oC to 20.0 oC?

a. 47.8 kJ b. 19.5 kJ c. 24.3 kJ d. 10.1 kJ e. 69.2 kJ

Practice Problem

04/08/23 86

2 1 2a

1 1 2

T TE = - R × ln ×

T - T

k

k

o2 1At 20 C the rate constant k = 2 k

oo o o

1a oo o o1

1

10 C + 273.15 K × 20 C + 273.15) K2E = -8.314J / mol • K × ln ×

10 C + 273.15 K - 20 C + 273.15) K

k

k

4o 2

a o

8.30054 x 10 KE = -8.314J / mol • K × ln 2 ×

-10 K3o

aE = -8.314J / mol • K × 0.693147 × 8.30054 x 10 K

4a

kJE = 4.783455 x 10 J / mol × = 47.8kJ / mol

1000J

Rate Equations - Summary

04/08/23 87

Integrated Rate Laws

0

01/ 2t 0

0 t

Δ Arate = - = k[A] = k Zero Order Rate Reaction

ΔtA

Rate Law : A - A = - t Half - Life t = 2

Δ Arate = - = [A] First Order Reaction

Δt

Rate Law : ln[A] - ln[A] = t

kk

k

k

1/ 2

2

1/ 2t 0 0

ln2 0.693 Half - Life t = =

Δ Arate = - = [A] Second Order Reaction

Δt1 1 1

Rate Law : - = t Half - Life t = ln[A] ln[A] [A]

k k

k

kk

Rate Equations - Summary

04/08/23 88

An Overview of Zero-Order, First-Order, and Simple Second-Order Reactions

Zero Order First Order Second Order

Plot for straight line

Slope, y intercept

Half-life

Rate law rate = k rate = k[A] rate = k[A]2

Units for k mol/L*s 1/s L/mol*s

Integrated rate law in straight-line form

[A]t = -kt + [A]0 ln [A]t = -kt + ln [A]0 1/[A]t = kt + 1/[A]0

[A]t vs. t ln [A]t vs. t 1/[A]t = t

k, [A]0 -k, ln [A]0k, 1/[A]0

[A]0/2k ln 2/k 1/k[A]0

Activation Energy (Ea)k = Zpff = e -E

a/RT

k = Zpe -Ea/RT = Ae -Ea/RT

k, rate constantZ, collision frequencyf, fraction of collisions that are => activation energyp, fraction of collisions in proper orientationA = frequency factor (pZ)Ea = activation energy (J)R = gas constant (8.314 J/molK)T = temperature (K)

Rate Equations - Summary

04/08/23 89

2 1 2a

1 1 2

k T TE = -R ×ln ×

k T - T

Arrhenius Equation