Therefore a rate law can be created:
rate = k (acetone)x⋅(I 2 )y⋅(H+)z
It is known that the amount of iodine does not affect the rate of the reaction. Therefore, the rate
order of I 2 is 0. Investigating the reaction in terms of the change in the concentration of iodine
over time gives the relationship
rate tI
∆∆ 2
=- ^h
The negative sign cancels the negative value of the change in I 2 which is due to the disappear-
ance of I 2 , making the rate a positive value. Since the rate of the overall reaction does not de-
pend on the iodine, the reaction rate can be studied by making the iodine the limiting reactant
present in excess acetone and H+, concentrations high enough so that their concentrations do
not change significantly through the course of the reaction. One simply then measures the time
required for the I 2 to be consumed by varying the concentrations of H+and acetone — easily
determined since I 2 is yellow in solution.
Scenario: A student prepared a data table for the results that he got in doing this experiment:
Mixture Acetone H+ (I 2 ) 0 Time (sec) Temp. °C rate = (I 2 ) 0 ⋅sec–1
I 0.80 M 0.20 M 0.0010 M 240 25 °C 4.2 × 10 –6
II 1.60 M 0.20 M 0.0010 M 120 25 °C 8.3 × 10 –6
III 0.80 M 0.40 M 0.0010 M 120 25 °C 8.3 × 10 –6
IV 0.80 M 0.20 M 0.00050 M 120 25 °C 4.2 × 10 –6Analysis:
- Determine the rate order for each reactant.
rate = k ⋅(acetone)x⋅((I 2 ) 0 )y⋅(H+)z
rate I = 4.2 × 10 –6= k ⋅(0.80)x⋅(0.0010)y⋅(0.20)z
rate II = 8.3 × 10 –6= k ⋅(1.60)x⋅(0.0010)y⋅(0.20)z
Ratio of Rate II to Rate I:....
...
.
.
rate Irate II
kk
0 0010 0 200 0010 0 20
42 1083 10 20
080160
yzyz
xx
66
##
::::::
===- =- ^^^
^^^
hhhhhh= log 2 = x ⋅log 2; x = 1
rate III = 8.3 × 10 –6= k ⋅(0.80)x⋅(0.0010)y⋅(0.40)z
rate IV = 4.2 × 10 –6= k ⋅(0.80)x⋅(0.00050)y⋅(0.20)z
Ratio of rate III to rate IV:...
...
...
rate IVrate III
kk
42 1083 10 20
0 80 0 00050 0 200 80 0 00050 0 40
xyzxyz
66
##
::
::::
===-- ^
^
^^^^
hh
hhhh= log 2 = z ⋅log 2; z = 1
Laboratory Experiments