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S-9
ADVANCE STUDY ASSIGNMENT
1.
What is the concentration of KI at the point of mixing when 20.0 mL of 0.0400 M KI are mixed
with 20.0 mL of 0.0600 M (NH
4
)2S2O
8
and 10.0 mL of 7.00 x 10
-4
M Na2S2O3?
Answer: 0.0160 M.
2.
What would happen if the S2O3²
-
were accidentally left out of a run?
Answer: Reaction (5) will not occur. What would you see?
3.
How does the iodide ion concentration in this experiment change before the blue color appears?
Why?
Answer: No change.
4.
What are the two requirements that have to be satisfied in proposing a reaction mechanism?
Answer: See page T-6
5.
What is a rate-determining step?
Answer: The slowest step in reaction mechanism
6.
In the ozonation of 1-pentene the following data are reported at 27.0°C:
O
3(g)
+ C
5
H
10(l)
C
5
H
10
O
3(l)
[C
5
H
10
] mol/L
[O3] mol/L
[
]
C
H
t
5
10
mol
L
s
0.010
0.0028
2.19 x 10
-5
0.027
0.0028
1.60 x 10
-4
0.010
0.0045
3.50 x 10
-5
a.
What is the order of the reaction with respect to O3 and C
5
H
10
?
Answer: First order in O3 and second order in C
5
H
10
b.
Determine the rate law for this reaction
Answer: rate = k[O3][C
5
H
10
]²
c.
Calculate the rate constant for one of the runs.
Answer: k = 78.2 M
-2
s
-1
REAGENTS AND EQUIPMENT
0.04xx M KI (potassium iodide)
0.100 M disodium EDTA
0.06xx M (NH
4
)2S2O
8
(ammonium peroxydisulfate)
3% starch solution
7.xx x 10
-4
M Na2S2O3 (sodium thiosulfate)
automatic dispenser for Na2S2O3
0.040 M KCl (potassium chloride)
2 burets
