Chapter 13 - Solutions
1. a.
H
H
H
H
+
N
C
C
-
H
H
C
C
N
H = (dissociation energy of bonds broken) – (dissociation energy of bonds formed)
H = D(C-N) – D(C-C) = 305 kJ – 348 kJ = –43 kJ
H < 0, therefore, the product is more stable.
k 
E
b. ln  1  = a
R
 k2 
 1

T
 2

1 
T1 
1

 = 161000 J·mol
4 
8.3145 J·mol 1·K 1
 6.6 10


ln 
k1
 1
1 



 500 K 300 K 
k1 = 4.0  1015 s1
 [CH 3 NC] 
 = kt

[CH
NC]
3
0

ln 
ln (.75) = (4.0  1015 s1)t
t = 7.2  1013 sec
 [CH 3 NC] 
 = kt
c. ln 
[CH 3 NC]0 


ln (.75) = k(3600 s);
ln
 8.0 10 5

 6.6 10  4

T1 = 470 K




=
k = 8.0  105 s1
161000 J·mol 1
 1
1 




8.3145 J·mol 1·K 1  500 K T1 
d. The role of the argon in this reaction is to behave as a catalyst for the dimerization
process when the argon is in high concentration. The argon provides a surface for
the dimerization reaction to occur, but the argon is not consumed during the
process.
A suggested mechanism would be:
Step 1:
Step 2:
CH3CN(g) + Ar(g)
CH3CN*(g)  CH3NC(g)
CH3CN*(g)
+ Ar(g)
The rate law would be:
For step 1:
Rate of the forward = Rate of the reverse
kf [CH3CN] [Ar] = kr [CH3CN*] [Ar]
kf [CH3CN] = kr [CH3CN*]
For step 2:
Rate = k2 [CH3CN*]
k 
Solving for [CH3CN*] in step 1: [CH3CN*] =  f  [CH 3CN]
 kr 
Substituting into the rate equation for step 2:
k 
Rate = k2  f  [CH 3CN] ; the equation is first-order with respect to CH3CN.
 kr 
When the argon has a low concentration, the proposed mechanism is less likely.
Another type of mechanism, a Lindemann mechanism, where the reaction
proceeds via the reaction between two molecules of the same gas. Therefore, the
following mechanism may be proposed.
k1
Step 1:
CH3CN(g) + CH3CN(g)
CH3CN*(g)
+ CH3CN(g)
k-1
Step 2:
CH3CN*(g)  CH3NC(g)
k2
The rate law would be based on the rate at which CH3CN*(g) changes into
products. That rate can be represented by considering that the CH3CN*(g) is
produced and consumed by the reactions shown above. Therefore the rate of the
d[CH 3CN* ]
reaction, which is the rate of the change in CH3CN*(g),
, can be
dt
shown by the following equation:
d[CH 3CN* ]
= k1[CH3CN]2 - k-1[CH3CN*] [CH3CN] - k2[CH3CN*]
dt
It turns out that the concentrations of CH3CN*(g) can be assumed to build up and
decay away very slowly during the course of the reaction, and the concentrations
usually do not ever get very large. If this is the case (and we will assume that it is
the case in this course) we can say that the rate of change of the concentration of
CH3CN*(g) is essentially zero over most of the course of the reaction.
d[CH 3CN* ]
= k1[CH3CN]2 - k-1[CH3CN*] [CH3CN] - k2[CH3CN*] = 0
dt
k1[CH3CN]2 - k-1[CH3CN*] [CH3CN] - k2[CH3CN*] = 0
Solving for [CH3CN*] yields:
k1[CH 3CN]2
[CH3CN*] =
k 2  k1[CH 3CN]
Remembering that Rate = k2[CH3CN*], then multiply the equation above by k2.
 k [CH CN]2 

1
3
Rate = k2[CH3CN*] = k 2 
k 2  k1[CH 3CN] 


And this rate law does not indicate that CH3CN(g) exhibits first-order behavior.
3. For mechanism i: Rate = k[C12H12O11]
For mechanism ii: Rate = k’[C12H22O11][H2O]
Recall that water is the solvent in this reaction; therefore, the concentration of H2O is
a constant. Hence rate for mechanism ii is Rate = k[C12H22O11] where k = k’[H2O].
Therefore, one cannot distinguish the two mechanisms. Mechanism ii is said to be
“pseudo first order,” because the concentration of one reactant (H2O) does not
effectively change.
5. [X]t = [X]0 ekt
t1/2 = ln2/k = 65.0 min  (60 sec·min-1) = 3900 sec


 ln2 t
[X]t = [X]0 e 3900
similarly,
 
 ln2 t
[M]t = [M]0 e 1500
[X]0 = [M]0


4
4
4
[X] t
 e  1.78 10 t  4.62 10 t  e 2.84 10 t = 2.5
[M]t
t = 3226 seconds = 54 min
7.
Recall that the definition of Km is
k k
K m  1 2
k1
Thus, the Michaelis-Menten constant Km is the sum of the rate constants for the
breakdown of the enzyme-substrate complex (k-1+k2), divided by the rate constant for
the formation of the enzyme-substrate complex (k1). A large Km means that the rate of
breakdown of the enzyme-substrate complex is large, relative to the rate of formation
of the enzyme-substrate complex. Therefore, a high Michaelis-Menten constant
indicates an unstable enzyme-substrate complex.
9.
Potential Energy
transition states
B
C
A
intermediates
D
Reaction progress
a. The overall reaction is exothermic. The final product is lower in energy than the
initial reactant.
b. There are two intermediates: B and C.
c. There are three transition states; they are the mountain tops.
There are two intermediates: B and C.
d. The step from B to C is the rate-determining step because it has the largest Ea.
e. The step from C to D is the fastest step because it has the smallest Ea.