Cliffs AP Chemistry, 3rd Edition

Ratio of rate IV to rate I:

.

.

.

..
.(.)

.(. )

rate I

rate IV

k

k

020

020

42 10

42 10 10

0 80 0 0010

0 80 0 00050

x y z

x y z

6

6

#

#

::

::

:

:

===-

• ^

^

^

^

h

h

h

h

= log 1 = y log 0.5; y = 0

As the concentration of the acetone doubles, the rate doubles; x = 1.

As the concentration of the H+doubles, the rate doubles; z = 1.

The concentration of the I 2 has no affect on the rate; y = 0.

2. Determine the rate specific constant, k, for the reaction.

()()

k

acetone H

= rate

+

Mixture 1 : k (. )(. )080 02042 10. 26 10.

#^65

==#

• 
  
  
  
  • 
    
  
  
  
  

Mixture 2 : k (. )(. )160 02083 10. 26 10.

#^65

==#

• 
  
  
  
  • 
    
  
  
  
  

Mixture 3 : k (. )(. )080 02083 10. 26 10.

#^65

==#

• 
  
  
  
  • 
    
  
  
  
  

Mixture 4 : k (. )(. )080 02042 10. 26 10.

#^65

==#

• 
  
  
  
  • 
    
  
  
  
  

Average: k = 2.6 × 10 –5

3. Given the following information, predict the time required for the reaction to reach
   completion:
   Acetone HCl I 2 H 2 O
   25 mL 25 mL 25 mL 25 mL
   3.2 M 4.0 M 0.020 M —

Total volume of mixture = 25 mL + 25 mL + 25 mL + 25 mL = 1.0 × 102 mL

Concentration of each after mixing:

Acetone:

..
.'.

L

L

moles

1 L sol n M

0 025 32

010

##^1 = 080

HCl:

..
.'.

L

L

moles

1 L sol n M

0 025 40

010

##^1 = 10

I 2 :

..
.'.

L

L

moles

1 L sol n M

0 025 0 020

010

##^1 =0 0050

rate = k ⋅(acetone)x⋅((I 2 )o)y⋅(H+)z

= 2.6 × 10 –5⋅(0.80)^1 ⋅(0.0050)^0 ⋅(1.0)^1 = 2.1 × 10 –5moles ⋅L–1⋅sec–1

Part III: AP Chemistry Laboratory Experiments