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Unformatted text preview: Chemistry 2000B Spring 2002 Assignment 1 Solutions
(a) 1 L of solution contains 1.875 mol of sodium sulfate (Na2 SO4 ) and weighs 1.2106 kg.
The mass of sodium sulfate is
266 3 g £ ¡ ¡ 1 875 mol 142 043 g mol ¢ ¤ ¢ £ ¢ mNa2 SO4 ¢ 1. The mass of water is therefore
0 2663 kg 0 9443 kg ¢ ¢ ¥ 1 2106 ¢ mH2 O ¢ The molality is thus
1 986 mol kg
¤ 1 875 mol
0 9443 kg ¢ ¢ ¢ c
˜ ¢ (b) The solution contains sodium and sulfate ions and water. Taking, as before, 1 L of
solution, we have ¢ 3 750 mol
3 750 1 875 mol
© ¢ § ¢ © ¤ ¢ ¦ ©§ 4 0 06461 ¢ ¢ ¢ § § © ¨ XNa ¦ 944 3 g
52 42 mol
18 0152 g mol
nNa
nH2 O nNa
nSO2
52 42 ¢ nH2 O ¢ 3 750 mol
¢ 4 ¢ 2nSO2 ¦ nNa ¢ 1 875 mol
¢ 4 ¢ nSO2 2. 1 mol L 1 s 1 ¢ ¤ ¢ mol L 1 s 5
5 10 7 89 ¢ ¢ ¢ rate 1 42 10 6 mol L
18 0 10 3 s
1∆ B
3 94 10
2 ∆t ¢ ∆B
∆t
¨ 3. Comparing experiments 1 and 2, we see that cutting the concentration of ammonium ion
in half reduces the rate by the same factor. In experiments 2 and 3, the ratio of the NO2
concentrations is 1.5 and the ratio of the rates is 5 4 3 6 1 5. Thus, the reaction is ﬁrst
order in each of the two reactants:
¢ ¢ ¤ ¢ k NH4 NO2 ¢
v The rate constant can be recovered from any one of the experiments. For instance, using data
from experiment 1:
10
¢ £ ¤ ¢ ¡ £ 1 30 4 L mol 1s
¤ ¢ ¢ ¡ NH4 NO2 7 2 10 6 mol L 1 s 1
0 24 mol L 0 10 mol L v 1
¢
k (a)
Br2 g
NO g
NO g
Br2 g # # © ©
©
## NOBr2 g
2NO g
©
## ©
# ! # # Br2 g NOBr2 g
2NOBr g "
! #
$© ¨ NO g NOBr2 g
2NOBr g 2NOBr g NO g
NOBr2 g
4. © (b)
¥
%
k2 NO NOBr2 ¥ 2k2 NO NOBr2
¥ k1 NO Br2
k1 NO Br2 k2 NO NOBr2
k1 NO Br2 ¥
&
d NO
dt
d Br2
dt
d NOBr2
dt
d NOBr
dt
(c) If the ﬁrst step was rate determining, the rate law would be v k1 NO Br2 . Since this
doesn’t agree with the experimental rate law, the ﬁrst step is not rate determining. In
fact, using techniques you will learn in Chemistry 2710 (if you opt to take this course),
we can show that this rate law is consistent with the second step being rate determining.
(d) At the very least, we need to do a set of experiments in which we measure the rate at
ﬁxed [NO] varying Br2 , and another set at ﬁxed Br2 varying [NO]. For instance, we
might perform the following series of experiments:
Br2
1
1
1
2
4
Experiment [NO]
1
1
2
2
3
4
4
1
5
1 Note that the units of concentration don’t matter so I haven’t written any in. I’m essentially proposing to look at two doublings of each of the concentrations. It’s important
to study at least three different concentrations for each reactant because it is possible to
be fooled if we use the minimum amount of data.
If the experimental rate law is correct, then the measured rate in experiment 2 should
be four times the rate in experiment 1, and the rate in experiment 3 should be four
times the rate in experiment 2. Also, the rate in experiment 4 should be twice the rate
in experiment 1 and the rate in experiment 5 should be twice the rate in experiment 4.
(a) If it’s a ﬁrstorder reaction, we should get a straight line when we plot ln NOBr vs t :
2 5. 3 3.2 ln [NOBr] 3.4 3.6 3.8 4 4.2
0 10 20
t (s) 30 40 We do not get a straight line, therefore it is not a ﬁrstorder reaction.
If it’s a secondorder reaction, a plot of 1 NOBr vs t should be linear:
'¤ 65
60 1/[NOBr] (L/mol) 55
50
45
40
35
30
25
20
0 10 20
t (s) 30 40 This is a very good ﬁt. This is therefore a secondorder reaction. The rate constant is the
slope of this relationship. Because it was slightly easier than some of the alternatives,
I decided to get the slope by measuring the points at which the line enters and leaves
3 ¢
'¤ ¤ 0 806 L mol 1s
¢ 61 5 21 2 L mol
45
5s 1
¢ ¢ ¤ ¢ £ ¥ ¥ ¡ ¢ ¥
'¤ k 21 2 L mol. At
¤ the graph. At t
5 s, the line passes through the point 1 NOBr
t 45 s, 1 NOBr 61 5 L mol. The slope is therefore ¥ For comparison, the value obtained by linear regression is k 0 803 L mol 1s 1 .
Obviously, there are greater uncertainties in this determination than in most other calculations you might carry out in this class, so your answer might be a little different from
mine. I would think that, from a carefully drawn graph with a similarly carefully drawn
line through the data, you should be able to get the slope to within 0 02 L mol 1s 1
or so.
¢ ¢ ( (b) The equation relating concentration to time is 4 00 10 4 mol
¢ 0 0400 mol L and we want NOBr ¤ ¤ ¢ L. Thus 3082 s 0 1
0 0400 mol L ¢ ¤ ¢ ¥ 1
10 4 mol L ¤ 1
1
NOBr
NOBr 0
1
0 803 L mol 1s 1 4 00
0 ¢ ) ¥ 1
k ) t kt
© We have NOBr 0 1
NOBr 0 ¢ 1
NOBr ¢ I used the value of k obtained by linear regression, but I could just as easily have used
the other value. Obviously you would use whatever you calculated in the previous part
of this question. Note also that there’s no point worrying too much about signiﬁcant
ﬁgures in this problem since there are signiﬁcant uncertainties in the rate constant. © 1 4 NO g were the ratelimiting step. NOBr2 g 2NOBr g k NOBr 2 . This
(c) The rate law, as determined in the previous part of this question, is v
would be the rate law for the reaction if ...
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This note was uploaded on 03/03/2012 for the course CHEM 2000 taught by Professor Roussel during the Fall '06 term at Lethbridge College.
 Fall '06
 Roussel
 Chemistry

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