CHEMISTRY 1AA3
March 29April 2, 2010
TUTORIAL PROBLEM SET 11 SOLUTIONS
________________________________________________________________________
Chemistry 1AA3
1. The decomposition of N
2
O
5
in the gas phase at room temperature was studied. Using the
following data verify that the rate law is first order in [N
2
O
5
] and calculate the rate constant.
Time (s)
[N
2
O
5
] (M)
0
0.100
50
0.0707
100
0.0500
200
0.0250
300
0.0125
400
0.00625
SOLUTION:
Plot
ln
[N
2
O
5
] versus time
this gives a straight line confirming that the reaction is first order in
N
2
O
5
.
Plot:
ln
[N
2
O
5
]
t
= 
k
t +
ln
[N
2
O
5
]
0
slope = 
k
= 6.93 x10
3
s
1
2. Using the data in the previous question, calculate [N
2
O
5
] at 150 s after the start of the reaction.
Since 150 s is halfway between 100 s and 200 s (two time points at which [N
2
O
5
] was
measured) is it possible to use the average of the two values to obtain the [N
2
O
5
] at 150s? If your
answer is no, explain why.
SOLUTION:
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View Full DocumentCHEMISTRY 1AA3
March 29April 2, 2010
TUTORIAL PROBLEM SET 11 SOLUTIONS
________________________________________________________________________
Chemistry 1AA3
It is not possible to use the average to determine the [N
2
O
5
] because it is the
ln
[N
2
O
5
] that is
proportional to time not simply [N
2
O
5
].
You must use the integrated rate equation for first order reactions:
ln([A]
t
) = 
k
t + ln([A]
0
).
Thus,
ln
[N
2
O
5
]
t
= 
k
t + ln[N
2
O
5
]
0
t= 150 s
k
= 6.93 x10
3
s
1
[N
2
O
5
]
0
= 0.100 M
[N
2
O
5
]
t
= exp(3.343) = 0.0353 M
Notice that this value is not halfway between 0.0500 and 0.0250 M.
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 Fall '10
 Tammie
 Linear Algebra, Algebra, Reaction, Rate equation, Tutorial Problem Set

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