Cliffs AP Chemistry, 3rd Edition

IV. Reaction Order with Respect to Iodide Ion

rate time s()

I

s

==∆^3510 # -^4 M

7 - A

. /.
s

50 10 mol L 1 42857 10

35

#^45

= #

• 
  
  
  
  • . /.
    s
  
  
  
  

50 10 mol L 9 61538 10

52

#^46

= #

• 
  
  
  
  • . /.
    s
  
  
  
  

50 10 mol L 8 06452 10

62

#^46

= #

• 
  
  
  
  • . /.
    s
  
  
  
  

mol L

71

50 10# -^4 =7 04225 10# - 6

. /.
s

mol L

140

50 10# -^4 =3 57143 10# -^6

V. Overall Reaction Order and the Rate Constant

1. Add all the partial orders to obtain the total reaction order.
   Total reaction order: 1+0+1 = 2


2. Average experimental rate constant using the experimental rate equation:
   rate = k[H 2 O 2 ]^1 [H 3 O+]^0 [I–]^1 for parts II, III, and IV.
   Part II. rate = k[H 2 O 2 ]^1
   (1) 1.06383 × 10 –5= k(0.04) k 1 06383 10. 004. 2 6595 10.
   #^54
   ==#

(2) 9.80392 × 10 –6= k(0.03) k 9 80392 10. 003. 3 2680 10.

#^64

==#

(3) 8.33333 × 10 –6= k(0.022) k 8 33333 10. 0 022. 3 7878 10.

#^64

==#

(4) 7.14286 × 10 –6= k(0.018) k 7 14286 10. 0 018. 3 9682 10.

#^64

==#

(5) 5.5556 × 10 –6= k(0.014) k 5 5556 10. 0 014. 3 9682 10.

#^64

==#

• 
  
  
  
  • Average k for Part II
    3.53 × 10 –4
  
  
  
  

Part III. rate = k[H 3 O+]^0

(1) 1.11111 × 10 –5= k(0.04) k 1 11111 10. 004. 2 7777 10.

#^54

==#

(2) 1.0000 × 10 –5= k(0.03) k 1 0000 10. 003. 3 3333 10.^4

#^5

==#

(3) 6.66667 × 10 –6= k(0.022) k 6 66667 10. 0 022. 3 0303 10.

#^64

==#

(4) 4.80769 × 10 –6= k(0.018) k 4 80769 10. 0 018. 2 6710 10.^4

#^6

==# -

(5) 3.84615 × 10 –6= k(0.014) k 3 84615 10. 0 014. 2 7472 10.

#^64

==#

• 
  
  
  
  • Average k for Part III
    2.91 × 10 –4
  
  
  
  

2.7777 × 10 –4

3.3333 × 10 –4

3.0303 × 10 –4

2.6710 × 10 –4

+2.7472 × 10 –4

0.00145595 ÷5 = 2.91 × 10 –4

2.6595 × 10 –4

3.2680 × 10 –4

3.7878 × 10 –4

3.9682 × 10 –4

+3.9682 × 10 –4

0.00176517 ÷5 = 3.53 × 10 –4

log (rate)

log [I-]

Graph

to find

y slope = order = y

Part III: AP Chemistry Laboratory Experiments