© University of South Carolina Board of Trustees

1st-Order Rate Law

Differential Form

Rate = k [A](t)

Integral Forms

[A](t) = [A]0 e-kt

ln [A](t) = ln [A]0 - k t y = b +mx

© University of South Carolina Board of Trustees

Graphing a 1st -Order Reaction

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A]

0.00 5.001.00 3.352.00 2.253.00 1.514.00 1.015.00 0.686.00 0.457.00 0.308.00 0.209.00 0.14

10.00 0.09

© University of South Carolina Board of Trustees

Graphing a 1st -Order Reaction

Time0 2 4 6 8 10

ln [A

]

-3

-2

-1

0

1

2

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

© University of South Carolina Board of Trustees

Graphing a 1st -Order Reaction

Time0 2 4 6 8 10

ln [A

]

-3

-2

-1

0

1

2

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

straight line this is a first-order reaction.

© University of South Carolina Board of Trustees

[A] vs t Data Rate Law

Method of Initial Rates (Sec. 13.2)

Trial and Error with Common Laws (Sec. 13.3)

© University of South Carolina Board of Trustees

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

Time0 2 4 6 8 10

ln [A

]

-3

-2

-1

0

1

2[A]0

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

intercept = ln [A]0

© University of South Carolina Board of Trustees

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

Time0 2 4 6 8 10

ln [A

]

-3

-2

-1

0

1

2

x

y

[A]0

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

( 1.2) 0.8

7.0 2.0

0.4 k

-1

yslope

x

s s

s

© University of South Carolina Board of Trustees

1st-Order Rate Law

Differential Form

Rate = k [A](t)

Integral Forms

[A](t) = [A]0 e-kt

ln [A](t) = ln [A]0 - k t

Half-life

© University of South Carolina Board of Trustees

1st -Order Half-Life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6 Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

[A] [A]/2 t1/2

© University of South Carolina Board of Trustees

1st -Order Half-Life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6 Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

[A] [A]/2 t1/2

5 2.5 1.75 s

t1/2

© University of South Carolina Board of Trustees

1st -Order Half-Life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6 Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

[A] [A]/2 t1/2

5 2.5 1.75 s

3 1.5 1.72 s

t1/2

© University of South Carolina Board of Trustees

Half-life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6 Time [A] ln [A]

0.00 5.00 1.611.00 3.35 1.212.00 2.25 0.813.00 1.51 0.414.00 1.01 0.015.00 0.68 -0.396.00 0.45 -0.797.00 0.30 -1.198.00 0.20 -1.599.00 0.14 -1.99

10.00 0.09 -2.39

[A] [A]/2 t1/2

5 2.5 1.75 s

3 1.5 1.72 s

1 0.5 1.73 s

t1/2

© University of South Carolina Board of Trustees

1st-Order Rate Law

Differential Form

Rate = k [A](t)

Integral Forms

[A](t) = [A]0 e-kt

ln [A](t) = ln [A]0 - k t

Half-life

always the same

© University of South Carolina Board of Trustees

1st-Order Rate Law

Differential Form

Rate = k [A](t)

Integral Forms

[A](t) = [A]0 e-kt

ln [A](t) = ln [A]0 - k t

Half-life

t1/2 = ln 2 / k

always the same

© University of South Carolina Board of Trustees

Half-Life Problem I

k = 6.2 x10-5 s-1 for the reaction

C12H22O11 + H2O C6H12O6 + C6H12O6

Calculate the half-life.

© University of South Carolina Board of Trustees

1st-Order Rate Law

Differential Form

Rate = k [A](t)

Integral Forms

[A](t) = [A]0 e-kt

ln [A](t) = ln [A]0 - k t

Half-life

t1/2 = ln 2 / k

always the same

© University of South Carolina Board of Trustees

Common Rate Laws

A products

a) First-orderRate = k [A]

b) Second-orderRate = k [A]2

c) Zero-orderRate = k [A]0 = k

© University of South Carolina Board of Trustees

2nd-Order Rate Law

Differential Form

Rate = k [A](t)2

© University of South Carolina Board of Trustees

2nd-Order Rate Law

Differential Form

Rate = k [A](t)2

Integral Form

0

0

( )1

tkt

AA

A

© University of South Carolina Board of Trustees

Graphing a 2nd-Order Reaction

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A]

0.00 5.001.00 2.502.00 1.673.00 1.254.00 1.005.00 0.836.00 0.717.00 0.638.00 0.569.00 0.50

10.00 0.45

0

0

( )1

tkt

AA

A

© University of South Carolina Board of Trustees

Graphing a 2nd-Order Reaction

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6

not a straight line

not a first-order reaction

Time0 2 4 6 8 10

ln [A

]

-3

-2

-1

0

1

2

Time [A] ln [A]

0.00 5.00 1.611.00 2.50 0.922.00 1.67 0.513.00 1.25 0.224.00 1.00 0.005.00 0.83 -0.186.00 0.71 -0.347.00 0.63 -0.478.00 0.56 -0.599.00 0.50 -0.69

10.00 0.45 -0.79

© University of South Carolina Board of Trustees

2nd-Order Rate Law

Differential Form

Rate = k [A](t)2

Integral Form

01 1

( )kt

t

A A

y = b + mx

0

0

( )1

tkt

AA

A

© University of South Carolina Board of Trustees

1/ [A] = 1/ [A]0 + k tTime [A] ln [A] 1/[A]

0.00 5.00 1.61 0.20 1.00 2.50 0.92 0.40 2.00 1.67 0.51 0.60 3.00 1.25 0.22 0.80 4.00 1.00 0.00 1.00 5.00 0.83 -0.18 1.20 6.00 0.71 -0.34 1.40 7.00 0.63 -0.47 1.59 8.00 0.56 -0.59 1.80 9.00 0.50 -0.69 2.00

10.00 0.45 -0.79 2.20

Time0 2 4 6 8 10

1/[A] (

M-1

)

0.0

0.5

1.0

1.5

2.0

2.5Time

0 2 4 6 8 10

[A]

0

1

2

3

4

5

6

straight line this is a second-order reaction

© University of South Carolina Board of Trustees

1/ [A] = 1/ [A]0 + k tTime [A] ln [A] 1/[A]

0.00 5.00 1.61 0.201.00 2.50 0.92 0.602.00 1.67 0.51 1.003.00 1.25 0.22 1.404.00 1.00 0.00 1.805.00 0.83 -0.18 2.206.00 0.71 -0.34 2.607.00 0.63 -0.47 3.008.00 0.56 -0.59 3.409.00 0.50 -0.69 3.80

10.00 0.45 -0.79 4.20Time

0 2 4 6 8 10

[A]

0

1

2

3

4

5

6

1/[A]0

intercept = 1/[A]0

Time0 2 4 6 8 10

1/[A] (

M-1

)

0.0

0.5

1.0

1.5

2.0

2.5

© University of South Carolina Board of Trustees

1/ [A] = 1/ [A]0 + k tTime [A] ln [A] 1/[A]

0.00 5.00 1.61 0.201.00 2.50 0.92 0.602.00 1.67 0.51 1.003.00 1.25 0.22 1.404.00 1.00 0.00 1.805.00 0.83 -0.18 2.206.00 0.71 -0.34 2.607.00 0.63 -0.47 3.008.00 0.56 -0.59 3.409.00 0.50 -0.69 3.80

10.00 0.45 -0.79 4.20

1 1

1

(3.0 ) 1.0

7.0 2.0

0.4

M M

M k

-1

yslope

x

s s

s

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6

y

x

1/[A]0

Time0 2 4 6 8 10

1/[A] (

M-1

)

0.0

0.5

1.0

1.5

2.0

2.5

© University of South Carolina Board of Trustees

2nd-Order Rate Law

Differential Form

Rate = k [A](t)2

Half-life

Integral Form

01 1

( )kt

t

A A

0

0

( )1

tkt

AA

A

© University of South Carolina Board of Trustees

2nd -Order Half-Life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A]

0.00 5.001.00 2.502.00 1.673.00 1.254.00 1.005.00 0.836.00 0.717.00 0.638.00 0.569.00 0.50

10.00 0.45

[A] [A]/2 t1/2

© University of South Carolina Board of Trustees

2nd -Order Half-Life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6Time [A]

0.00 5.001.00 2.502.00 1.673.00 1.254.00 1.005.00 0.836.00 0.717.00 0.638.00 0.569.00 0.50

10.00 0.45

1.0 s

[A] [A]/2 t1/2

5 2.5 1.0 s

© University of South Carolina Board of Trustees

2nd -Order Half-Life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6

4.7 s

Time [A]

0.00 5.001.00 2.502.00 1.673.00 1.254.00 1.005.00 0.836.00 0.717.00 0.638.00 0.569.00 0.50

10.00 0.45

[A] [A]/2 t1/2

5 2.5 1.0 s

1 0.5 4.7 s

© University of South Carolina Board of Trustees

2nd -Order Half-Life

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6

4.7 s

Time [A]

0.00 5.001.00 2.502.00 1.673.00 1.254.00 1.005.00 0.836.00 0.717.00 0.638.00 0.569.00 0.50

10.00 0.45

[A] [A]/2 t1/2

5 2.5 1.0 s

1 0.5 4.7 s

reaction gets ‘slower’ as it proceeds

© University of South Carolina Board of Trustees

2nd-Order Rate Law

Differential Form

Rate = k [A](t)2

Half-life

t1/2 = 1 / k[A]

varies - not so useful

Integral Form

01 1

( )kt

t

A A

0

0

( )1

tkt

AA

A

© University of South Carolina Board of Trustees

Example: 2nd-Order Reaction

The reaction

2NOCl 2NO + Cl2

obeys the rate law

Rate = (0.020 M-1s-1) [NOCl]2

Calculate the concentration of NOCl after 30 minutes, if the initial concentration was 0.050 M.

© University of South Carolina Board of Trustees

Common Rate Laws

A products

a) First-orderRate = k [A]

b) Second-orderRate = k [A]2

c) Zero-orderRate = k [A]0 = k

© University of South Carolina Board of Trustees

0th -Order Rate Law

Differential Form

Rate = k [A](t)0

Rate = k

© University of South Carolina Board of Trustees

0th -Order Rate Law

Differential Form

Rate = k [A](t)0

Rate = k y = b + mx

Integral Forms

[A](t) = [A]0 - kt

© University of South Carolina Board of Trustees

Graphing a 0th -Order ReactionTime [A]

0.00 5.001.00 4.202.00 3.403.00 2.604.00 1.805.00 1.006.00 0.207.00 0.008.00 0.009.00 0.00

10.00 0.00

Time0 2 4 6 8 10

[A]

0

1

2

3

4

5

6

straight line this is a 0th -order reaction.

© University of South Carolina Board of Trustees

[A](t)=[A]0 - ktTime [A]

0.00 5.001.00 4.202.00 3.403.00 2.604.00 1.805.00 1.006.00 0.207.00 0.008.00 0.009.00 0.00

10.00 0.00

Time0 2 4 6 8 10

[A] (M

)

0

1

2

3

4

5

6

y

x

(1.0 ) 4.2

5.0 1.0

0.8 /

M M

M k

yslope

x

s s

s

© University of South Carolina Board of Trustees

0th -Order Rate Law

Differential Form

Rate = k [A](t)0

Rate = k

Half-life

y = b + mx

Integral Forms

[A](t) = [A]0 - kt

© University of South Carolina Board of Trustees

0th -Order Rate Law

Differential Form

Rate = k [A](t)0

Rate = k

Half-life

varies - not so useful

Integral Forms

[A](t) = [A]0 - kt

© University of South Carolina Board of Trustees

Summary: Differential Rate Laws

Zero-Order

First-Order

Second-Order 2A

Ad

rate kdt

AA

drate k

dt

Adrate k

dt

Rate vs Concentration

© University of South Carolina Board of Trustees

Summary: Integral Rate Laws

Zero-Order [A](t) = [A]0 − kt

First-Order [A](t) = [A]0 e−kt

Second-Order

0

0

( )1

AA

At

kt

Concentration vs Time

© University of South Carolina Board of Trustees

Summary: Graphical Tests

Zero-Order [A](t) = [A]0 − k t

First-Order ln [A](t) = ln[A]0 − k t

Second-Order 01 1

( )A Ak t

t

y = b + m x

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Summary: Half-Life

Zero-Order varies

First-Order constant:

Second-Order varies

1/ 22ln

tk

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Half-Life Problem II

A sample of wood has 58% of the 14C (t1/2 = 5730 yr) originally present.  What is the age of the wood sample?

© University of South Carolina Board of Trustees

Chapt. 13Kinetics

Sec. 4Theory of the Rate Constant (k)

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Bimolecular Rate TheoryA + B products

Rate = k(T) [A][B] Experiment

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Bimolecular Rate TheoryA + B products

Rate = frequency of collisions

Rate = k(T) [A][B] Experiment

© University of South Carolina Board of Trustees

Bimolecular Rate TheoryA + B products

Rate = frequency of collisions (Z0[A][B])

Rate = pZ0 e- [A][B] Theory

Rate = k(T) [A][B] Experiment

© University of South Carolina Board of Trustees

Bimolecular Rate TheoryA + B products

Rate = frequency of collisions (Z0[A][B])

Rate = pZ0 e- [A][B] Theory

Rate = k(T) [A][B] Experiment

correct conc. dependencerates much too largeno temperature dependence

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k Changes with Temperature

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Activation Energy Diagram

DGThermodynamics

Kinetics

Reactants

Products

Transition State