Chapter 13 - Chemical Kinetics II

Integrated Rate Laws Reaction Rates and Temperature

Reaction Order - Graphical Picture

Reactant Concentration vs. Time

A → Products

A—->Products

Integrated Rate Laws

Zero Order Reactions

Rate = k[A]0 = k (constant rate reactions) [A] = -kt +[A]0 (integrated rate law)

graph of [A] vs t is straight line (slope = -k and y intercept = [A]0)

t½ = [A]0/2k

•  Rate = k[A]0 = k

 constant rate reactions

•  [A] = -kt + [A]0

•  graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0

•  t ½ = [A0]/2k

•  when Rate = M/sec, k = M/sec

[A]0

[A]

time

slope = - k

First Order Reactions

Rate = k[A] ln[A] = -kt +ln[A]0 (integrated rate law)

graph of ln[A] vs t is straight line (slope = -k and y intercept = ln[A]0)

t½ = ln2/k = (0.693)/k (constant half-life)

•  Rate = k[A]0 = k

 constant rate reactions

•  [A] = -kt + [A]0

•  graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0

•  t ½ = [A0]/2k

•  when Rate = M/sec, k = M/sec

[A]0

[A]

time

slope = - k ln[A]

ln[A]0

Second Order Reactions

Rate = k[A]2 1/[A] = -kt + 1/[A]0 (integrated rate law)

graph of 1/[A] vs t is straight line (slope = +k and y intercept = 1/[A]0)

t½ = 1/(k[A]0)

'me

slope  =  

+k

1/[A]

1/[A]

Half-life, t½

the length of time it takes for the concentration of the reactants to fall to one-half its initial value

depends on the order of the reaction

Zero Order ReactionsDetermining Rate Graphically

First Order Reactions

Second Order Reactions

Data for 2NO2 (g) → 2NO (g) + O2 (g)

ln[NO2] vs time 1/[NO2] vs time

Problems Using Integrated Rate Laws

1) N2O decomposes at 1050 K through 1st order kinetics:

2 N2O —> 2 N2 + O2 Rate = k[N2O], k = 3.4 sec-1

For an initial concentration of [N2O] of 0.20 M, what is the [N2O] after 100 msec.

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

[A]kt = ln

(3.4 sec-1)(100 x 10-3sec)= ln [A]0

[A]

e =(3.4 sec-1)(100 x 10-3sec) [A]0

[A] e = (.34) [A]0

[A]

1.40 = 0.20 M

x x = 0.14 M

tmin 0 200 400 600 800 1000

[N2O5] 0.015 0.0096 0.0062 0.004 0.0025 0.0016

ln[N2O5] -4.20 -4.65 -5.08 -5.52 -5.99 -6.43

0

0.00375

0.00750

0.01125

0.01500

0 250 500 750 1000

[N2O5] vs time

-7.00

-5.25

-3.50

-1.75

0

0 250 500 750 1000

ln[N2O5] vs time

2) N2O5 decomposes at 25 ºC through 1st order kinetics:

2 N2O5 —> 4 NO2 + O2

Given the following concentration vs time data, determine the rate constant for the reaction.

[N2O5]

ln[N2O5]

-( )-slope = -2.23

60,000 sec

slope = 3.7 x 10-5 sec-1

[A]0

slope = -k

3) Cyclopropane is converted to propene (a rearrangement) at 500ºC with first order kinetics with a rate constant of 6.7 x 10-4 sec-1.

C3H6 —> C3H6

How long does it take for the concentration of cyclopropane to drop to 1/2 of its original concentration ? to 1/4 ?

t ½ = 0.69/(6.7 x 10-4sec-1) = 1.0 x 103 sec

t = 2 x t ½= 2.0 x 103 sec1/4

4) N2O5 decomposes at 25 ºC through 1st order kinetics:

2 N2O5 —> 4 NO2 + O2 k = 5.2 x 10-3 sec-1 at 65º

How long will it take for a sample which is 20 mM N2O5 to decrease to 2 mM?

[A]0

[A]kt = ln [A]0

[A] t = ln 1

k= ln 20mM

2 mM 1

5.2 x 10-3

= (192)ln(10) = 442 sec

5) HI decomposes at 700 K through 2nd order kinetics:

2 HI —> H2 + I2 k = 1.8 x 10-3 M-1sec-1

How long will it take for 1/2 of a 0.10 M sample to decompose ? for 3/4 ?.

1/(1.8 x 10-3 M-1sec-1)(0.010 M)t½ =

t½ = 5.6 x 104 sec

t = 1/(1.8 x 10-3 M-1sec-1)(0.005 M) 1/41/2 →

t = 11.2 x 104 sec 1/41/2 →

Total time for 3/4 to decompose = 1.7 x 105 sec

Reaction Rates and Temperature

Temperature: Changing the temperature changes the rate constant

in the rate law.

These factors are incorporated into rate constant k by the Arrhenius equation:

R = universal gas constant (in J/mol·K) T = temperature (in kelvin)

A is a collisional frequency factor. (Includes frequency of collisions and an orientation factor)

Ea = Activation energy, The minimum energy of molecular collisions required to break bonds in reactants, leading to

formation of products.

Collision Theory

In order for a chemical reaction to occur, reactant molecules must collide with sufficient energy and in the proper orientation.

Effective collisions involve both of these factors.

When two molecules have an effective collision, a temporary, high-energy (unstable) chemical species, the reactive

intermediate*, is formed.

*also referred to as an activated complex or transition state

Arrhenius equation:

Frequency Factor

Activation Energy

Exponential Factor

Reaction Energy Profile

High-energy transition state (“activated

complex”)

Activation energy (Ea )

Reaction Energy Profile

Partially broken and partially formed bonds

The amount of energy necessary to form the “activated complex”

The Reaction of CH3Br and OH-

Reaction Energy Diagram for the Reaction of CH3Br and OH-

The Arrhenius equation can be rearranged:

by

Graphical Determination of Ea (Arrhenius Plot)

Slope  =  –Ea/R  Intercept  =  ln  A  

According to the rearranged Arrhenius equation, the activation energy for a reaction can be determined by plotting k vs 1/T which yields a line whose slope is -Ea/R and whose intercept is lnA.

Mathematical Determination of Ea

Consider determination of k two different T’s:

By subtraction,

k = (A)(e- ) Ea RT

The Exponential Factor

A number between 0 and 1, representing the fraction of reactant molecules with sufficient energy to

make it over the energy barrier

Ea (J/mol)Temp -Ea/RT -Ea/RTe

10,000523 K -2.30

-11.5 0.00001

0.100

The Effect of Ea on the Fraction of Collisions with Sufficient Energy to Allow Reaction

k = (A)(e- ) Ea RT

523 K

523 K

20,000

50,000

-4.60 0.001

The higher the energy barrier, the fewer molecules that have sufficient energy to overcome the energy barrier.

The Effect of Ea on the Fraction of Collisions with Sufficient Energy to Allow Reaction

Ea (J/mol) Temp -Ea/RT -Ea/RTe

10,000

10,000

10,000

298 K

523 K

773 K

-4.04

-2.30

-1.56 0.211

0.100

0.0176

The Effect of Temperature on the Fraction of Collisions with Sufficient Energy to Allow Reaction

k = (A)(e- ) Ea RT

The extra kinetic energy of the reactants is converted into potential energy when molecules collide.

Increased temperature increases kinetic energy of molecules and molecular

collisions.

The Effect of Temperature on the Fraction of Collisions with Sufficient Energy to Allow Reaction

k = (A)(e- ) Ea RT

Frequency Factor

A number representing the number of reactant molecules which can approach each other with the correct orientation

and sufficient energy to make it over the energy barrier

A=(p)(z)Frequency Factor

p zOrientation

FactorCollision Frequency

Factorfor reactions of single atoms, p=1

for most molecules p<<1for a collision to overcome the energy

barrier, the reactants mus have sufficient kinetic energy so that the

collision leads to the formation of an “activated complex”

c

Effective Collisions-Kinetic Energy Effect