© University of South Carolina Board of Trustees Trial[NO] M [H 2 ] M Initial Rate, M/s...

© University of South Carolina Board of Trustees

Trial [NO]M

[H2]M

InitialRate, M/s

1 0.057 0.130 4.50x10-3

2 0.057 0.260 9.00x10-3

3 0.11 0.130 1.80x10-2

Student Problem

Write the rate law for the reaction given the following data

2NO + 2H2 N2 + 2H2O


Chapt. 13

Sec. 13.3Properties of

Common Rate Laws


Common Rate Laws

A products

a) First-orderRate = k [A]

b) Second-orderRate = k [A]2

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


1st-Order Rate Law

Differential Form

Rate = k [A](t)


1st-Order Rate Law

Differential Form

Rate = k [A](t)

Integral Forms

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


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

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


Concentration at a Later Time

C12H22O11 + H2O ® C6H12O6 + C6H12O6 sucrose glucose fructose

This reaction is 1st order with a rate constant of 6.2 x10-5 s-1. If the initial sucrose concentration is 0.40 M, what is the concentration after 2 hrs (7200 s)?


Time to Reach a Concentration

C12H22O11 + H2O ® C6H12O6 + C6H12O6 sucrose glucose fructose

This reaction is 1st order with a rate constant of 6.2 x10-5 s-1. If the initial sucrose concentration is 0.40 M, at what time does the concentration fall to 0.30 M?


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


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



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



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.


[A] vs t Data Rate Law

Method of Initial Rates (Sec. 13.2)

Trial and Error with Common Laws (Sec. 13.3)


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


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


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


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



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



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


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


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 Trial[NO] M [H 2 ] M Initial Rate, M/s...

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