k[0.706]
3
[.1]
1
;
k
=
5.967
∗
10
−
5
0.706
3
0.1
1
=
1.69
∗
10
−
3
Rate II=
3.458
∗
10
−
5
==
k[.706]
3
[0.05]
1
;
k
=
3.458
∗
10
−
5
0.706
3
0.1
1
=
9.83
∗
10
−
4
Rate III=
1.109
∗
10
−
5
==
k[.353]
3
[0.1]
1
;
k
=
1.109
∗
10
−
5
0.353
3
0.1
1
=
2.52
∗
10
−
3
Rate IV=
0.112
∗
10
−
3
==
k[0.706]
3
[0.1]
1
;
k
=
0.112
∗
10
−
3
0.706
3
0.1
1
=
3.18
∗
10
−
3
30
o
C
POSTLAB QUESTIONS:
1)
Calculate the rate constant and write the rate law expression for the catalyzed
decomposition of hydrogen peroxide. Explain how you determined the order of the
reaction in H
2
O
2
and KI.
The Rate expression can be expressed as:
Rate=k[H
2
O
2
]
x
[I

]
y
Substituting the experimental data for I, and II, we have
Rate I=
5.967
∗
10
−
5
=
k[0.706]
x
[.1]
y
Rate II=
3.458
∗
10
−
5
=
k[0.706]
x
[.05]
y
Reactions I and II were chosen because the concentration of H
2
O
2
is held constant while the
concentration of I

is varied.
Divide Rate I by Rate II
Rate I=
5.967
∗
10
−
5
=
¿
k[0.706]
x
[.1]
y
Rate II=
3.458
∗
10
−
5
=
¿
k[0.706]
x
[.05]
y
RateI
RateII
=
1.726
=
2
y
; ylog
(
2
)
=
log
(
1.726
)
; y
=
log
(
1.726
)
log
(
2
)
=
0.787
≅
1
Substituting the experimental data for I, and III, we have
Rate I=
5.967
∗
10
−
5
=
k[0.706]
x
[.1]
y
Rate III=
1.109
∗
10
−
5
=
k[0.353]
x
[.1]
y
Divide Rate I by Rate I1
Rate I=
5.967
∗
10
−
5
=
k[0.706]
x
[.1]
y
Rate III=
1.109
∗
10
−
5
=
k[0.353]
x
[.1]
y
RateI
RateII I
=
5.38
=
2
x
; xlog
(
2
)
=
log
(
5.38
)
; x
=
log
(
5.38
)
log
(
2
)
=
2.69
≅
3
The rate equation is: Rate=k[H
2
O
2
]
3
[I

]
1
The rate is third order in H
2
O
2 and
first order I

. The
reaction is forth order overall.
2)
The following mechanism has been proposed for this reaction:
H
2
O
2
+ I
–
→ IO
–
H
2
O
H
2
O
2
+ IO
–
→ I
–
+ H
2
O + O
2
If this mechanism is correct, which step must be the ratedetermining step? Explain.
Answer:
No it is not valid for observed rate law, because here in mechanism order with
respect to H
2
O
2
to 1 but observed order is 3.
3)
Use the Arrhenius equation (shown below) to determine the
activation energy,
E
a
, for this reaction.
ln
k
1
k
2
=
−
E
a
R
(
1
T
2
−
1
T
1
)
k
1
=
1.69
∗
10
−
3
T
1
=21.8
R= 8.314
k
2
=
3.18
∗
10
−
3
T
2
= 30.0
ln
1.69
∗
10
−
3
3.18
∗
10
−
3
=
−
E
a
8.314
(
1
30.0
−
1
21.8
)
(−
0.144
∗
8.314
)
0.013
=
E
a
E
a
=−
92.09
KJ/mol
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 Spring '14
 BOBBYBURKES
 Chemistry