Chemical KineticsChapter 14

Reminders

• Assignment 1 due today (end of class)

• Assignment 2 up on ACME, due Jan. 29 (in class)

• Assignment 3 will be up Mon., Jan 29 and will be due Mon., Feb. 05

• Assignment 4 (Ch. 15) will not be due before Midterm 1, but Ch. 15 will be on the midterm

First-Order Reactions

14.3

A product

rate = -[A]t

rate = k [A]

k = rate[A]

= s-1M/sM=

[A]t = k [A]-

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

Average rate

Rate law

Differential rate law

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

Integrated rate law [A] = [A]0exp(-kt)

Integrated rate law(linear form)

Rate constant

First-Order Reactions

14.3

A product

rate = -[A]t

rate = k [A]

[A]t = k [A]-

Average rate

Rate law

Differential rate law

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

Integrated rate law [A] = [A]0exp(-kt)

Integrated rate law(linear form)

[A] is the concentration of A at any time t

[A]0 is the concentration of A at time t=0

The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?

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

kt = ln[A]0 – ln[A]

ln[A]0 – ln[A]k

= 66 s

[A]0 = 0.88 M

[A] = 0.14 M

t =ln

[A]0

[A]k

=ln

0.88 M0.14 M

2.8 x 10-2 s-1=

14.3

yxyx lnlnln

Recall:

First-Order Reactions

14.3

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

ln[A]0

[A]0/2k

=t½ln2k

=0.693

k=

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order? units of k (s-1)

tln

[A]0

[A]k

=

First-Order Reactions

14.3

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.

t½ = t when [A] = [A]0/2

What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?

t½ln2k

=0.693

5.7 x 10-4 s-1= = 1200 s = 20 minutes

How do you know decomposition is first order? units of k (s-1)

The half-life of a 1st-order reaction is independent of the initial concentration of the reactant.

A product

First-order reaction

# of half-lives [A] = [A]0/n

1

2

3

4

2

4

8

16

14.3

n= 2 for each half-lifeelapsed

21

41

21

21

x

81

41

21

x

161

81

21

x

1

14.3

Second-Order Reactions

14.3

A product rate = -[A]t

rate = k [A]2

k = rate[A]2 = 1/M•sM/s

M2=[A]t = k [A]2-

[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=0

1[A]

=1

[A]0+ kt

t½ = t when [A] = [A]0/2

t½ = 1k[A]0

Distinguishing 1st and 2nd order reactions

• Is the rate law 1st or 2nd order?– If (a) plot gives a straight line, then 1st order and rate = k[A]– If (b) plot gives a straight line, then 2nd order and rate = k[A]2

ln[A] = ln[A]0 - kt1

[A]=

1[A]0

+ kt

NO2(g) NO(g) + ½ O2(g) at 300oC

Zero-Order Reactions

14.3

A product rate = -[A]t

rate = k [A]0 = k

k = rate[A]0 = M/s

[A]t = k-

[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=0

t½ = t when [A] = [A]0/2

t½ = [A]0

2k

[A] = [A]0 - kt

Summary of the Kinetics of Zero-Order, First-Orderand Second-Order Reactions

Order Rate LawConcentration-Time

Equation Half-Life

0

1

2

rate = k

rate = k [A]

rate = k [A]2

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

1[A]

=1

[A]0+ kt

[A] = [A]0 - kt

t½ln2k

=

t½ = [A]0

2k

t½ = 1k[A]0

14.3

Temperature and reaction rates

• Rates of most chemical reactions increase as the temperature rises

– Dough rises faster at r.t. than when refrigerated

– Plants grow more rapidly in warm weather than in cold

Temperature and reaction rates

• This effect of temperature on reaction rate can be seen directly by observing a chemiluminescent reaction

Temperature affects rate of chemiluminescence reaction in CyalumeTM light sticks

• Left: light stick in hot water

• Right: light stick in cold water

• At the higher temperature, the reaction is initially faster and produces a brighter light

• Although the light stick glows more brightly initially, its luminescence also dies out more rapidly

Temperature Dependence of the Rate Constant

14.4

1st order reaction, ln[CH3NC]t = -kt + ln[CH3NC]0

[A] = [A]0exp(-kt) ln[A] = ln[A]0 - kt

Temperature Dependence of the Rate Constant

14.4

1st order reaction, ln[CH3NC]t = -kt + ln[CH3NC]0

• For a successful put, we need:– Enough energy to get over the hill– Directionality

• For a successful reaction, we need:– Enough energy to get over the activation barrier– Collisions between reacting molecules– Reacting molecules with the right orientation

A + B C + D

Exothermic Reaction Endothermic Reaction

The activation energy (Ea) is the minimum amount of energy required to initiate a chemical reaction.

14.4

Cl + NOCl NO + Cl2

Temperature Dependence of the Rate Constant

k = A • exp( -Ea/RT )

Ea is the activation energy (J/mol)

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature

A is the frequency factor

lnk = -Ea

R1T

+ lnA

(Arrhenius equation)

14.4

14.4

lnk = -Ea

R1T

+ lnA

14.5

Reaction Mechanisms

The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.

The sequence of elementary steps that leads to product formation is the reaction mechanism.

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

N2O2 is detected during the reaction!

Elementary step: NO + NO N2O2

Elementary step: N2O2 + O2 2NO2

Overall reaction: 2NO + O2 2NO2

+