showed that doubling the concentration of acetone increased the rate of the
reaction. Varying the rate of [Br
2
] appeared to have no effect on the rate of
reaction as shown when comparing experiment one to experiment two.
D.
Calculating x:

-.004616
=
k
[.004]
x
[.8]
y
[.2]
z
←DATA FROM EXP. 1
-.0045
k [.008]
x
[.8]
y
[.2]
z
←
DATA FROM EXP. 2
1.02578 = .5
x
log 1.02578 = x log .5
x = log 1.02578 / log .5
x = -3
Calculating y:
-.0045
=
k
[.008]
x
[.8]
y
[.2]
z
←DATA FROM EXP. 2
-.001016
k
[.008]
x
[1.6]
y
[.2]
z
←DATA FROM EXP. 4
4.4 = .5
y
log 4.4 = y log .5
y = log 4.4 / log .5
y = -2
Calculating z:
-.0045
=
k
[.008]
x
[.8]
y
[.2]
z
←DATA FROM EXP. 2
-.00198
k
[.008]
x
[.8]
y
[.4]
z
←DATA FROM EXP. 3
.44 = 2
z
log .44 = z log 2
z = log .44 / log 2
z = -1
Rate Law from Calculations:
Rate = k[Br
2
]
-3
[Acetone]
-2
[H
+
]
Negative exponents in this case do not seem logical given previous knowledge. Assuming
that this was a mathematical or calculation error, the exponents are to be made positive.
E.

Experiment #
k value
1
-5.6 x 10
5
M
-5
s
-1
2
-6.9 x 10
4
M
-5
s
-1
3
-1.5 x 10
4
M
-5
s
-1
4
-3.9 x 10
3
M
-5
s
-1
Experiment One:
Given the rate law is: Rate = k[Br
2
]
3
[Acetone]
2
[H
+
]
-0.004616 = k [.004]
3
[.8]
2
[.2]
k = -563476.56 M
-5
s
-1
= -5.6 x 10
5
M
-5
s
-1
Experiment Two:
Given the rate law is: Rate = k[Br
2
]
3
[Acetone]
2
[H
+
]
-.0045 = k[.008]
3
[.8]
2
[.2]
k = -68664.55 M
-5
s
-1
= -6.9 x 10
4
M
-5
s
-1

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- Spring '16
- Christine Gaudinski
- Chemistry, Reaction, Stoichiometry, Chemical reaction, Exponentiation