1442 Test 2 Chapters 16 & 17 Kinetics and Keq
website:
http://dipowell1.home.mindspring.com
( last item after powell is the
number one)
1.
[ratio M]
a
[ratio M]
b
= ratio of rates
Using experiment 1 & 2
:
[.1M/.1M]
a
[.2M/.3M]
b
= 0.004 M/s/0.0135 M/s
Since the A concentrations are equal we can cancel out that factor and:
[.2M/.3M]
b
= 0.004M/s
/
0.0135 M/s
To solve algebracially :
b log 0.667
=
log .296
b
=
3
If you don’t understand this process, go to the website home page and under
Chemistry Help Topics choose Rate Law: Kinetics.
Using Exp. 2&4 :
[.1M/.3M]
b
[.3M/.4M]
a
= 0.0135 M/s / .288 M/s
Since b did not stay constant in this problem, we must now substitute
the power of 3 for b, so the second factor becomes : [.3M/.4M]
3
[.1M/.3M]
a
[.3M/.4M]
3
= [0.0135 M/s
/
0.288 M/s]
[0.33]
a
[0.75 ]
3
=
0.0469
0.33
a
= .111
a = 2
( if you need help solving this see Rate Law: Kinetics)
Answer:
Rate = k[A]
2
[B]
3
2.
For an elementary reaction ( occurs in a single event) just use the coefficients in
the balanced equation for the powers.
So, the rate law is:
R = k[NO
2
]
2
No calculations are required if the reaction is elementary
3.
ln [A]
0
/ [A]
t
= kt
ln [1 atm]/ [0.803 atm]
t
= k (10,000s)
0.219 =
10000s k
k = 2.2 X 10
5
s
1
4.
1/ [A]
t
 1/ [A]
0
= kt
1/[A]
t
= kt  1/[A]
0
1/[A]
t
= (0.02 L mol
1
s
1
)(120s) + 1/[0.2]
[A]
t
=
0.14 M
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5.
To prove first order and second order plot the following:
ln[A]
0
/[A]
t
= kt
Plot the two things that change:
ln[A]
t
vs t
To prove second order:
1/[A]
t
1/[A]
0
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 Fall '08
 ROGERS
 Chemistry, Kinetics, 6 m, 2 k, 4 K

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