This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chemistry 2000 Spring 2002 Midterm Examination
Solutions
(a) The concentrations in experiments 1 and 2 are in a 3:1 ratio, as are the rates. This is
therefore a ﬁrstorder reaction with rate law
¢
£¡ 2
¤ k CH3 N v ¦
§¥ 1. The rate constant can be determined from either experiment:
25 4 10
© ¦ © © ¦ ¦ ¥ 2 1 s 6 0 10 6 mol L 1s
2 4 10 2 mol L 1
¤ v
CH3 N ¦ k ¢
¨¡ (b)
∆ C2 H6
∆t
10 4 mol L
6 0 10 6 mol L 1s
¡ ¦ 17 s
¦ 1 ¡ ¥ © ¡ ¥ ¦ ∆t d C2 H6
dt
∆ C2 H6
v ¥ v
¢ ¤ t (min)
ln %bp 2. If it’s a ﬁrstorder reaction, a graph of ln percent vs t should be a straight line of slope
In this case, we have k. 2.21 5.16 13.26 22.84 40.52 60.41
4.55 4.42 4.17 3.78 3.14 2.30 ¢ ¤ My graph is shown in Fig. 1. The data ﬁt a straight line reasonably well, which conﬁrms
that the decomposition of bp is a ﬁrstorder process. The rate constant is found by taking the
slope of the line. The line passes through the points 0 4 65 and 70 1 98 (estimated from
the graph). The slope is therefore
¢ ¢ ¤ 1
0 0381 min ¦ ¦ ¦ ¦ ¦ 0 0381 min 1 .
¦ k, this means that k (a)
17 5 103 J mol
8 314 472 J K 1mol 1 243 15 K
¤ ¤ ¦ ¦ 1 ¦ ¢ 1 !
"¤ exp ¦ 106 L mol 1s 1 © © © ¢ ¦ 2 09 1010 L mol 1 s 12 ¢ Ea RT
k∞ e
k ¦ 3. 4 65 1 98
0 70 min ¦ Since the slope is ∆y
∆x ¤ slope 5
4.5 ln(percent) 4
3.5 # 3
2.5
2
1.5
0 10 20 30 40 50 60 70 t (min) ¢ Figure 1: Plot of ln %bp vs t from the data of question 2.
¤ ¡ d OH
k OH CF3 CH2 OCHF2
dt
(c) If the initial concentrations are equal, then by stoichiometry the two concentrations
d OH
remain equal for all time. Thus
k OH 2 , which is a simple secondorder
dt
reaction.
¡ ¥ ¡ ¥ ¥ (b) ¡ ¡ ¥ ¥ (d) According to the secondorder integrated rate law, ¦ 1
0 0405 mol L
! ¦ ¦ ¢ $ 0 140 K
¦ ¦ ¤ ¦ ¦ ¤ ¤ ¥ ¡
¦ © ¢ ¢ ¦ © ¦ ¦ ¡ ¥ 0 223 0 111
0 177
¤ aPCl3 aCl2
aPCl5 ¥ ¦ ¡ ¦ ¢ Q ¡ 4. ¥ t ! R 1
R0
0 004 05 mol L
1
1
6 L mol 1 s 1 0 004 05 mol L
2 09 10
1 06 10 4 s
¦ 11
kR
0 10 R 0 t This means that the amounts of products will have to increase to reach equilibrium, which
in turn means that the reaction is spontaneous as written.
2 10 4 mol L. Since Kw
© 13 ¦ ¡
¥ 5. Potassium hydroxide is a strong base. Accordingly, OH
aH aOH ,
¢ ¢ ¤& 10
© 22 11
¦ © ¦ © ¦ ¦ ¦ & % % pH Kw
2 9 10 15
aOH
1 3 10 4
log10 aH
10 65 ¦ ¤ % aH
¢ a2
PBr P 2
Br
. We need to know the equilibrium pressures of bromine atoms and
aBr2
PBr2 P
molecules. The number of moles of Br2 at equilibrium is 100 1 20% 98 8% of the initial number, i.e. 0 988 1 05 mol
1 0374 mol. The number of molecules dissociated is
0 0120 1 05 mol
0 0126 mol. The number of bromine atoms produced is twice as large
(by stoichiometry), so nBr 0 0252 mol. The volume of the ﬂask is 0 980 L 1000 L m3
9 80 10 4 m3 . The temperature is 1600 273 15 K 1873 K. The pressures of the reactant
and product are therefore
¤
('
)¤ ¦ ¤ ¢
¢ © © ¦ ¦ ¦ 1 © © ¢ ¦ ¦ ¦ 1 ¦ ¦ ¦ ¤ ¦ ¦ ¢ ¢ ¤ ¦ ¢ ¤ ¦ ¤ ¢ ¤ ¦ ¦ ¦ 0 ¦ ¢ ¤ 1 ¤ ¦ 1873 K ¢ ¤ ¦ ¢ 1873 K 1 ¤ ¤ nBr2 RT
V
1 0374 mol 8 314 472 J K 1mol
9 80 10 4 m3
1 65 107 Pa 163 atm
nBr RT
V
0 0252 mol 8 314 472 J K 1mol
9 80 10 4 m3
4 00 105 Pa 3 95 atm
3 952
9 60 10 2
163 ¢ ¢ ¦ © © ¦ ¦ ¦ K ¦ PBr ' PBr2 ¦ 6. K 7. In these problems, we only ever need to worry about the ﬁrst proton. By Le Chatelier’s
principle, we know that the acidic environment created by the rst proton will inhibit the
dissociation of the second proton. The relevant equilibrium is thus
4 0
2
3 ¢ ¦ % aC2 O4 H aH
aC2 O4 H2 ¤ ¤& ¢ Ka H aq
for which we have C2 O4 H aq C2 O4 H2 aq The Ka is quite large. It seems extremely likely that we will be able to ignore the autoionization of water. However, we probably should not treat this equilibrium as if little dissociation
occurred because of the size of the equilibrium constant. Suppose that the amount of oxalic
acid which dissociates, expressed in terms of activities, is x. Then
3 aH
0
x
x
% & aC2 O4 H2 aC2 O4 H
Initial
0.043
0
Change
x
x
Final
0 043 x
x
¦ The Ka expression becomes 10 3 ! ¦ ¢
¦ ¢ ¤ © ¦ 6
75 © © ¦ ¦ ¤ © ¦ © © % ¦ ¦ ¢ 0 ¦ ¦ 4 1 53 © 2 795 4 ¤ © ¦ ¦ © x ¦ ¦ 0 aH
pH 10 x2
0 043 x
x2 6 5 10 2x 2 795 10 3
1
6 5 10 2
6 5 10 2 2
2
2 96 10 2
2 96 10 2
log10 aH
log10 2 96 10 2
¦ % 65 2 ¦ ...
View
Full
Document
 Fall '06
 Roussel
 Reaction

Click to edit the document details