Ch. 26: 1, 4, 9, 10 (a, c, d, e), 15, 16, 17, 31, 36, 57, 59, 65 a , 65 b, 91 a, 91 b. Exercises p. 1198
1. (E) In the following structural formulas, the hydrogen atoms are omitted for simplicity.
Remember that there are four bonds to each carbon atom. The missing bonds are C— H
bonds.
(a)
(b)
CH 3CH 2 CHBrCHBrCH 3
C
Br Br
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
(c)
C
C
C
4.
C
C
C
C
C
C
(E) In the following structural formulas, the hydrogen atoms are omitted for simplicity.
Remember that there are four bonds to each carbon atom. The missing bonds are C—H
bonds.
(a) (CH3)3CCH2CH(CH3)CH2CH2CH3
(b)
(CH3)2CHCH2C(CH3)2CH2Br
C
C
C
C
C
C
C
C
C
C
C
(c)
Cl
C
Cl
C
C
C
C
Cl3CCH2CH(CH3)CH2Cl
Cl
C
C
CH2Cl
C
C
C
C
Br
9.
10.
(E) Structural (skeletal) isomers differ from each other in the length of their carbon atom
chains and in the length of the side chains. The carbon skeleton differs between these
isomers. Positional isomers differ in the location or position where functional groups are
located attached to the carbon skeleton. Geometric isomers differ in whether two
substituents are on the same side of the molecule or on opposite sides of the molecule from
each other; usually they are on opposite sides or the same side of a double bond.
(a)
The structures are identical.
(b)
The two compounds are constitutional isomers.
(c)
The two compounds have no relationship. They have different molecular formulas.
(d)
The two compounds are constitutional isomers.
(e)
The two compounds are stereoisomers.
(M) (a) The two compounds are identical.
c. The two compounds are enatiomers.
d. The two compounds are constitutional isomers.
e. The two compounds are constitutional isomers. They differ in the mode of attachment
of bromine atoms. In the first one, bromine atoms are attached to the same carbon,
whereas in the second one they are attached to two adjacent carbon atoms.
33. (a)
15. (M)
(b)
Cl
2C
CO2H
H
Carbon 2 is chiral
NH2
2C
CO2H
H
Carbon 2 is chiral
(c)
There are no
chiral carbon
atoms in
compound (c)
16. 34.
(M) (a)
OH
CH3
C
C
(b)
(c)
C
CH3
H
H
OH
Cl
H
chiral carbon
There are no
chiral carbon
atoms in
compound (c)
chiral carbon
17.
(E) (a)
Br
CH3CHCH2CH3
2-Bromobutane
alkyl halide (bromide)
(b)
aldehyde
(c)
ketone
(d) HO
OH
2, 3 or 4-hydroxyphenol
phenol, hydroxyl group, phenyl group
(only 4-hydroxyphenol shown)
31.
(E) (a) The longest chain is eight carbons long, the two substituent groups are methyl
groups, and they are attached to the number 3 and number 5 carbon atom. This is
3,5-dimethyloctane.
(b)
The longest carbon chain is three carbons long, the two substituent groups are methyl
groups, and they are both attached to the number 2 carbon atom. This is 2,2dimethylpropane.
(c)
The longest carbon chain is 7 carbon atoms long, there are two chloro groups
attached to carbon atom 3, and an ethyl group attached to carbon 5. This is 3,3dichloro-5-ethylheptane.
57.
(E) The four structural isomers are 1,2-dichloropropane, 1,3-dichloropropane, 2,2dichloropropane and 1,1-dichloropropane.
59.
(M)
65.
(M)
91.
(M) (a) alcohol, amine(secondary), arene
(b) C1= sp2, C2 = sp3, C3 = sp3, C4 = sp3, N = sp3
Ch 27, 7, 10(a, c, d), 17, 31. P. 1259
7. (E)
10. (M) (a) Rate=k[2-bromo-2-methylpentane]
(c) The rate will double.
(d) The rate will not change because it is independent on the concentration of ethanol.
17. (M) The reaction most likely proceeded via SN1 mechanism resulting in racemization of
the product (equal amounts of S and R isomers).
Ch 14: 12, 17, 49, 87, 101, 111, 115 . p. 646
12.
(M) From Expt. 1 and Expt. 2 we see that [B] remains fixed while [A] triples. As a result,
the initial rate increases from 4.2  103 M/min to 1.3  102 M/min, that is, the initial
reaction rate triples. Therefore, the reaction is first-order in [A]. Between Expt. 2 and Expt.
3, we see that [A] doubles, which would double the rate, and [B] doubles. As a
consequence, the initial rate goes from 1.3  102 M/min to 5.2  102 M/min, that is, the
rate quadruples. Since an additional doubling of the rate is due to the change in [B], the
reaction is first-order in [B]. Now we determine the value of the rate constant.
Rate
5.2  102 M / min
1
1
Rate = k A B
k=
=
= 5.8  103 L mol 1min 1
A B
3.00 M  3.00 M
c
h
The rate law is Rate = 5.8  103 L mol 1min 1 A B .
17.
1
1
(M)
(a) Since the half-life is 180 s, after 900 s five half-lives have elapsed, and the original
quantity of A has been cut in half five times.
bg
final quantity of A = 0.5  initial quantity of A = 0.03125  initial quantity of A
About 3.13% of the original quantity of A remains unreacted after 900 s.
or
More generally, we would calculate the value of the rate constant, k, using
ln 2 0.693
k=
=
= 0.00385 s 1 Now ln(% unreacted) = -kt = -0.00385 s-1×(900s)
t1/2
180 s
= -3.465
(% unreacted) = 0.0313 × 100% = 3.13% of the original quantity.
5
(b)
49.
Rate = k A = 0.00385 s1  0.50 M = 0.00193 M / s
(E)
(a)
There are two intermediates (B and C).
(b)
There are three transition states (peaks/maxima) in the energy diagram.
(c)
The fastest step has the smallest Ea, hence, step 3 is the fastest step in the reaction
with step 2 a close second.
(d)
Reactant A (step 1) has the highest Ea, and therefore the slowest smallest constant
(e)
Endothermic; energy is needed to go from A to B.
(f)
Exothermic; energy is released moving from A to D.
87.
(M)
ΔCCl 3
 rateformation  ratedisappearance  0
Δt
so rateformation  ratedecomposition
k 2 [Cl(g)][CHCl 3 ]  k 3[CCl 3 ][[Cl(g)] and, simplifying, [CCl 3 ] 
k2
[CHCl 3 ]
k3
k

since rate  k 3[CCl 3 ][Cl(g)]  k3  2 [CHCl 3 ]  [Cl(g)]  k 2 [CHCl 3 ][Cl(g)]
k3

1/2
1/2
k

k

We know: [Cl(g)]   1 [Cl 2 (g)]  then rate overall = k 2[CHCl 3 ]  1 [Cl 2 (g)] 
 k -1

 k -1

1/2
1/2
k 
 4.8 103 
and the rate constant k will be: k  k2  1   (1.3 102 ) 
 0.015
3 
 3.6 10 
 k -1 
101. (E) The answer is (c). The rate constant k is only dependent on temperature, not on the
concentration of the reactants
111. (D) The overall stoichiometry of the reaction is determined by adding the two reactions
with each other: A + 2B
C+D
(a) Since I is made slowly but is used very quickly, its rate of formation is essentially zero.
The amount of I at any given time during the reaction can be expressed as follows:
d  I
dt
 0  k1  A  B  k 2  B I 
 I 
k1
A
k2
Using the above expression for [I], we can now determine the overall reaction rate law:
d  C
dt
 k 2  I B  k 2 
k1
 A   B  k1  AB
k2
(b) Adding the two reactions given, we still get the same overall stoichiometry as part (a).
However, with the given proposed reaction mechanisms, the rate law for the product(s) is
given as follows:
d  B2 
dt
 B2  
 k1  B  k 1  B2   k 2  A  B  0
2
k1  B 
2
k 1  k 2  A 
Therefore,
d  C
dt
 k 2  A  B2  
k 2 k1  A  B
2
k 1  k 2  A 
which does not agree with the observed reaction rate law.
115. (E) The answer is (c), remain the same. This is because for a zero-order reaction,
d[A]/dt = k[A]0 = k. Therefore, the reaction rate is independent of the concentration of the
reactant.
Ch. 15: 11, 43, 47, 61, 64, 104, 110 P.688
11.
(E) Add one-half of the reversed 1st reaction with the 2nd reaction to obtain the desired
reaction.
1
1
1
NO  g 
Kc =
2 N 2  g  + 2 O2  g 
2.11030
NO  g  + 12 Br2  g 
NOBr  g 
Kc = 1.4
net : 12 N 2  g  + 12 O 2  g  + 12 Br2  g 
NOBr  g 
Kc =
1.4
2.110
30
= 9.7 10 16
43.
(M)
We organize the solution around the balanced chemical equation.
Equation:
2 Cr 3+  aq  + Cd(s)
2 Cr 2+  aq   Cd 2   aq 
Initial :
1.00 M
0M
0M
Changes:
2 x M
2 x
x
Equil:
(1.00  2 x) M
2x
x
2
Cr 2+  Cd 2+ 
(2 x) 2 ( x)
Kc =
=
= 0.288
2
2
1.00  2 x 
Cr 3 
Via successive approximations , one obtains
x = 0.257 M
Therefore, at equilibrium, [Cd2+] = 0.257 M, [Cr2+] = 0.514 M and [Cr3+] = 0.486 M
Minimum mass of Cd(s) = 0.350L  0.257 M  112.41 g/mol = 10.1 g of Cd metal
47.
61.
(E)
K
aconitate
citrate
Q
4.0 105
 0.031
 0.00128
Since Q = K, the reaction is at equilibrium,
(E) If the total pressure of a mixture of gases at equilibrium is doubled by compression,
the equilibrium will shift to the side with fewer moles of gas to counteract the increase in
pressure. Thus, if the pressure of an equilibrium mixture of N 2(g), H2(g), and NH3(g) is
doubled, the reaction involving these three gases, i.e., N2(g) + 3 H2(g)
2 NH3(g), will
proceed in the forward direction to produce a new equilibrium mixture that contains
additional ammonia and less molecular nitrogen and molecular hydrogen. In other words,
P{N2(g)} will have decreased when equilibrium is re-established. It is important to note,
however, that the final equilibrium partial pressure for the N 2 will, nevertheless, be higher
than its original partial pressure prior to the doubling of the total pressure.
64.
(M) The equilibrium position for a reaction that is exothermic shifts to the left
(reactants are favored) when the temperature is raised. For one that is endothermic, it
shifts right (products are favored) when the temperature is raised.
(a) NO  g 
(b)
1
2
N2  g  + 12 O2  g 
SO3  g 
(c) N2 H4  g 
(d)
COCl2  g 
H o = 90.2 kJ shifts left, % dissociation 
SO2  g  + 12 O2  g 
N2  g  + 2 H2  g 
CO  g  + Cl2  g 
H o = +98.9 kJ shifts right, % dissociation 
H o = 95.4 kJ shifts left, % dissociation 
H o = +108.3 kJ shifts right, % dissociation 
104. (E) The answer is (b). At half the stoichiometric values, the equilibrium constant is K 1/2. If
the equation is reversed, it is K-1. Therefore, the K’ = K-1/2 = (1.8×10-6)-1/2 = 7.5×10-2.
110. (E) Since K >>1, there will be much more product than reactant