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Unformatted text preview: Chemistry 2000 Spring 2001 Section B Assignment 1
Solutions
1. 4 ¡ nH2 O 0 0036 ¡ ¦ ¥ ¥ (a) Suppose that we take 1 L of this solution. This liter of solution contains 1.697 mol of
acetone and weighs 0.9849 kg (from the concentration and density, respectively). The
acetone weighs ¡ £¤ ¡ © £ macetone ¡ ¡ ¤ ¢ 2. ¡ § ¡ §¡ ¨ §¦ §
¦
¡
¡¡ ¢¡
¡ ¤ ¢
¡ £¤
£
¢ ¡¡
¡ mH2 O XK 8 0g
0 051 mol
158 032 g mol
250 mL 0 998 23 g mL
250 g
250 g
13 9 mol
18 015 g mol
nK
0 051 mol
nH2 O nK
nMnO
13 9 0 051 0 051 mol ¡ nKMnO4 1 697 mol 58 079 g mol 98 56 g The water contained in our liter of solution therefore weighs ¡ ¡ 1 697 mol
0 8863 kg 98 56 g 886 3 g ¡¢ ¡ c
˜ ¡ ¡ The molality is therefore 984 9 ¡¡ mwater 1 915 mol kg (b) Consider again our liter of solution. We know how many moles of acetone it contains.
The number of moles of water can be determined from the mass: ¡
¡
¡ ¢
¡ ¡
¡§¡
¡ ¡ ¥
¡
¡ ¢¡ ¡ Xacetone 3. (a) ∆A
∆t 0 382 886 3 g
49 20 mol
18 015 g mol
1 697 mol
0 03334
1 697 49 20 mol ¡ nwater 0 400 mol L
0 32 s 1 0 056 mol L 1s 1 (b) ∆A
∆B
2
0 113 mol L 1s 1
∆t
∆t
Notes: I am using ∆’s instead of d ’s because these are average rates. However, it would
not be wrong to use derivative notation. It’s a matter of taste, and I’m not completely
consistent about it myself. Also note that I got the answer to this question using more
digits than are shown in my answer to question 3a. This is the correct procedure, even
if the two numbers are then not quite in the stoichiometric ratio of 2:1. This is ﬁne
since the last signiﬁcant ﬁgure shown in any calculation is supposed to be somewhat
uncertain. (That’s why we stop at that digit rather than some other.) ¡ ¡ ¢ 4. The rate law is assumed to be of a simple form, e.g. dx dt kxn . If we increase x by a factor
of 3, dx dt increases by a factor of 9 32 . Therefore, the order of the reaction with respect
to the reactant (n) is 2. ¢ (a) Experiments 1, 2 and 3 were performed at identical nitrogen monoxide concentrations.
The concentration of ozone was doubled from experiment 1 to experiment 2 and the
rate increased by a factor of 2. Comparing experiments 1 and 3, the ratio of the ozone
concentrations is 3 and the ratio of the rates is also 3. The rate is therefore proportional
to the ozone concentration.
Experiments 3, 4 and 5 were performed at identical ozone concentrations. The NO
concentration was doubled in experiment 4 compared to experiment 3 and the rate
doubled. Comparing experiments 3 and 5, we see that both the NO concentrations and
rates are in ratios of 1:3. Thus, the rate is also proportional to the NO concentration.
The rate law is therefore
d NO2
k NO O3
dt ¡ (b) I’ll use data from experiment 1, but we could in principle use any of the data points:
k ¤ £ ¡
¢¡
¡
¡ ¤
¢ ¡ £ ¤ ¢ ¡ ¡ £
¢ d NO2 dt
NO O3
66 0 10 6 mol L 1s 1
1 00 10 6 mol L 3 00 10 6 mol L
2 20 107 L mol 1s 1 Note that I got rid of the SI preﬁxes on the quantities appearing in this calculation to
avoid getting an answer in awkward units (22 0 L µmol 1s 1 or 22 0 L µM). Strange
units like these are sometimes useful, but it is generally considered bad style to have a
preﬁx in the denominator of units in the SI system. The preﬁxes should really only be
used as preﬁxes, i.e. at the beginning of a unit, whether it is a simple unit (g, mol, L,
etc.) or a compound unit (such as µmol L 1s 1 ). 5. 2 (c) ¡ ¡ £¤
¡£
k NO O3 3 1 1 30 10 6 ¡ £¤ ¢ 2 20 107 L mol 1 s
143 µmol L 1s 1 mol L 5 00 10 6 ¤¢ d NO2
dt mol L ...
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 Fall '06
 Roussel
 Chemistry

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