Outline: 2/5/07 Seminar Report - in front Extra Seminar – Wednesday @ 4pm Pick up Quiz #3 – from me Exam 1 – one week from Friday…

Outline Chapter 15 - Kinetics (cont’d): - integrated rate law calcs

Quiz #3 Average = 7.9Average = 7.9

Quiz #3

0

5

10

15

20

25

0 to 2 2 to 4 4 to 6 6 to 8 8 to 10

Summary: Can build a rate law from observed data:

Rate = k [A]m [B]n

m, n depend only on the chemical reaction under consideration….

Can use integrated rate laws to predict rates, concentrations at various times, etc.

15-3

15-5

There are two forms to know:

First order: ln[A] = ln[A]o k t

Second order: 1/[A] = 1/[A]o k t

(initial rates)

Note something weird about k:

The definition of rate constant varies from reaction to reaction…. e.g. Rate = k [A] [B] or Rate = k [C]

Therefore the units of k vary from reaction to reaction…. e.g. k is in M-1 sec-1 or k is in sec-1 or k is in atm-1 sec-1

The integrated forms of the rate laws are important:

Can predict [concentration] as a function of time!

Can predict rate as a function of time too!

e.g. CAPA-7 Question 10 “Popcorn kernels pop unimolecularly….”

29 kernels pop in 10 seconds when 400 total kernels are present. After 100 have popped, how many will pop over the next 10 seconds?

= first-order reaction

ln(A) ln(A0) = kt 29 kernels pop in 10 seconds when 400 total

kernels are present.

A0 = 400 (initial number of kernels) t = 10 sec A = 371 (number of kernels at time t) k = 0.00753 sec-1

1st ORDER: ln(A) ln(A0) = kt

and k = 0.00753 sec-1

After 100 have popped, how many will pop over the next 10 seconds?

A0 = 300 (initial number of kernels)

t = 10 sec k = 0.00753 sec1

A = 278.25 A = 21.75

Try Worksheet #5….

Work with somebody nearby and complete “Version 1”…

On Worksheet #5….answer (a):

25%

25%

25%

25% 1.1. Rate = k Rate = k ppSO2Cl2SO2Cl2

2.2. Rate = k Rate = k ppSO2Cl2SO2Cl222

3.3. Rate = k Rate = k ppSO2Cl2SO2Cl233

4.4. Rate = kRate = k

11 22 33 44 55

Try Worksheet #5….

First Order

y = -0.1675x + 1.6013

R2 = 0.9999

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

0 5 10 15 20

time (h)

ln (

p)

Try Worksheet #5….

2nd Order

y = 0.1695x - 0.0713

R2 = 0.9131

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

3.00

3.50

0 5 10 15 20

Time (h)

1/P

On Worksheet #5….answer (b):

11 22 33 44 55

25%

25%

25%

25% 1.1. k = 5.95 /hrk = 5.95 /hr

2.2. k = k = 5.95 atm/hr5.95 atm/hr

3.3. k = 0.168 /hrk = 0.168 /hr

4.4. k = k = 0.168 atm/hr0.168 atm/hr

5.5. None of the aboveNone of the above

Worksheet #5….

Version #1

-1.5

-1

-0.5

00.5

1

1.5

2

0 5 10 15 20

time (hrs)

ln P

(SO

2C

l2)

Rate = k (pSO2Cl2)

k = 1/5.95 = 0.168 hr

SO2Cl2 SO2 + Cl2

Practice! CAPA problems (sets 7-8) textbook examples (15.25-15.34)

How do you tell which integrated rate equation to use if you aren’t told?

How do we actually use “kinetics” for anything useful?

Like determining mechanisms? A + B C + D Rate = k [A]n[B]m

Keep one reagent constant (in excess), vary the other reagent….

O3 + isoprene prod Rate = k [O3]n[iso]m

the rate becomes:Rate = k’ [iso]m

where k’ = k[O3]n

Exp # 1: Let [O3]>>[iso]

[O3] = 5.4104 M [iso] = 2.5 10 M

O3 [isoprene]time (E-4 M) (E-6 M)

0 5.4 2.50.215 5.4 1.110.495 5.4 0.3050.955 5.4 0.037

“Isolation experiment”

Plot the data…. ln[iso] = ln[iso]o k t

1/[iso] = 1/[iso]o k t

Only one will be truly linear….Rate = k’ [iso]1

m=1, k’ = 4.4 s1

1st Order

R2 = 0.9988

-4-3-2-1012

0 0.5 1

Time (s)

ln([

iso

])

2nd Order

R2 = 0.8335

-10

0

10

20

30

0 0.5 1

Time (s)

1/[

iso

]

Perform the exp. again and graph data as 1st order to find k”

To find n, you change [O3] but keep [O3]>>[iso]

“Isolation experiment”

Exp # 2: Let [O3]>>[iso]

[O3] = 2.7104 M [iso] = 2.5106 M [O3]II = 1/2 [O3]I

k” = 2.2 s1

“Isolation experiment”

nnn

n

n

k

k

k

k

k

k)0.2(

7.2

4.5

]O[

]O[

]O[

]O[

''

'

II3

I3

II3

I3

n

k

k)0.2(

''

'

)0.2(ln''

'ln nk

k

1)0.2ln(

)0.2ln(

)0.2(ln

s2.2

s4.4ln

)0.2(ln

''

'ln 1

1

k

k

n

n

k

k)0.2(ln

''

'ln

That is all there is to it!

“Isolation experiment”

Rate = k [O3]1[iso]1

k’ = k[O3] so k = k’/[O3] = 8.110+3 M1s1

You’re about to do this…. Chem 114: CV lab!

Variation on a theme: Half-life Radioactive decay = 1st order decay Integrated rate law: ln(At/Ao) = kt

half-life is just another way to define the rate constant k.

In the case that At = 0.5 Ao

(half is left) t = t1/2

therefore: ln(0.5) = kt1/2