Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap27

27

Molecular reaction dynamics

Solutions to exercises
Discussion questions
E27.1(b)

A reaction in solution can be regarded as the outcome of two stages: one is the encounter of two
reactant species, which is followed by their reaction, the second stage, if they acquire their activation
energy. If the rate-determining step is the former, then the reaction is said to be diffusion-controlled.
If the rate-determining step is the latter, then the reaction is activation controlled. For a reaction of
the form A + B → P that obeys the second-order rate law ν = k2 [A][B], in the diffusion-controlled
regime,
k2 = 4πR ∗ DNA
where D is the sum of the diffusion coefficients of the two reactant species and R ∗ is the distance
at which reaction occurs. A further approximation is that each molecule obeys the Stokes–Einstein
relation and Stokes’ law, and then
k2 ≈

8RT
3η

where η is the viscosity of the medium. The result suggests that k2 is independent of the radii of the
reactants.
E27.2(b)

In the kinetic salt effect, the rate of a reaction in solution is changed by modification of the ionic
strength of the medium. If the reactant ions have the same sign of charge (as in cation/cation or
anion/anion reactions), then an increase in ionic strength increases the rate constant. If the reactant
ions have opposite signs (as in cation/anion reactions), then an increase in ionic strength decreases

the rate constant. In the former case, the effect can be traced to the denser ionic atmosphere (see the
Debye–Huckel theory) that forms round the newly formed and highly charged ion that constitutes
the activated complex and the stronger interaction of that ion with the atmosphere. In the latter case,
the ion corresponding to the activated complex has a lower charge than the reactants and hence it has
a more diffuse ionic atmosphere and interacts with it more weakly. In the limit of low ionic strength
the rate constant can be expected to follow the relation
log k = log k ◦ + 2AzA zB I 1/2

E27.3(b)

Refer to Figs 27.21 and 27.22 of the text. The first of these figures shows an attractive potential energy
surface, the second, a repulsive surface.
(a) Consider Fig. 27.21. If the original molecule is vibrationally excited, then a collision with an
incoming molecule takes the system along the floor of the potential energy valley (trajectory C).
This path is bottled up in the region of the reactants, and does not take the system to the saddle
point. If, however, the same amount of energy is present solely as translational kinetic energy, then
the system moves along a successful encounter trajectory C∗ and travels smoothly over the saddle
point into products. We can therefore conclude that reactions with attractive potential energy
surfaces proceed more efficiently if the energy is in relative translational motion. Moreover, the
potential surface shows that once past the saddle point the trajectory runs up the steep wall of the
product valley, and then rolls from side to side as it falls to the foot of the valley as the products
separate. In other words, the products emerge in a vibrationally excited state.


MOLECULAR REACTION DYNAMICS

439

(b) Now consider the repulsive surface (Fig. 27.22). On trajectory C the collisional energy is largely
in translation. As the reactants approach, the potential energy rises. Their path takes them up

the opposing face of the valley, and they are reflected back into the reactant region. This path
corresponds to an unsuccessful encounter, even though the energy is sufficient for reaction. On
a successful trajectory C∗ , some of the energy is in the vibration of the reactant molecule and
the motion causes the trajectory to weave from side to side up the valley as it approaches the
saddle point. This motion may be sufficient to tip the system round the corner to the saddle point
and then on to products. In this case, the product molecule is expected to be in an unexcited
vibrational state. Reactions with repulsive potential surfaces can therefore be expected to proceed
more efficiently if the excess is present as vibrations.

Numerical exercises
E27.4(b)

The collision frequency is
z

=

so z =
=

21/2 σ v p
kT

where σ = π d 2 = 4π r 2 and v =

8RT 1/2
πM

21/2 p
8RT 1/2

16pNA r 2 π 1/2
(4πr 2 )
=
kT
πM
(RT M)1/2
16 × (100 × 103 Pa) × (6.022 × 1023 mol−1 ) × (180 × 10−12 m)2 × (π )1/2
[(8.3145 J K−1 mol−1 ) × (298 K) × (28.01 × 10−3 kg mol−1 )]1/2

= 6.64 × 109 s−1
The collision density is
1
zp
(6.64 × 109 s−1 ) × (100 × 103 Pa)
= 8.07 × 1034 m−3 s−1
ZAA = zN/V =
=
2
2kT
2(1.381 × 10−23 J K−1 ) × (298 K)
Raising the temperature at constant volume means raising the pressure in proportion to the temperature
√
ZAA ∝ T
so the per cent increase in z and ZAA due to a 10 K increase in temperature is 1.6 per cent , same as
Exercise 27.4(a).
E27.5(b)

The appropriate fraction is given by
f = exp

−Ea
RT

The values in question are
(a) (i) f = exp
(ii) f = exp
(b) (i) f = exp
(ii) f = exp

−15 × 103 J mol−1
(8.3145 J K−1 mol−1 ) × (300 K)
−15 × 103 J mol−1
(8.3145 J K−1 mol−1 ) × (800 K)
−150 × 103 J mol−1
(8.3145 J K−1 mol−1 ) × (300 K)
−150 × 103 J mol−1
(8.3145 J K−1 mol−1 ) × (800 K)

= 2.4 × 10−3
= 0.10
= 7.7 × 10−27
= 1.6 × 10−10


INSTRUCTOR’S MANUAL

440

E27.6(b)

−Ea
at the new temperature and
RT
compare it to that at the old temperature. An approximate approach would be to note that f changes
−Ea
−Ea
from f0 = exp
to f = exp
, where x is the fractional increase in the
RT
RT (1 + x)
−Ea
−Ea
temperature. If x is small, the exponent changes from
to approximately
(1 − x) and f
RT
RT
−x
−Ea
−Ea
−Ea (1 − x)
−Ea
= f0 f0−x .
exp
changes from exp
to exp
= exp
RT
RT

RT
RT
Thus the new Boltzmann factor is the old one times a factor of f0−x . The factor of increase is

A straightforward approach would be to compute f = exp

(a) (i) f0−x = (2.4 × 10−3 )−10/300 = 1.2
(ii) f0−x = (0.10)−10/800 = 1.03

(b) (i) f0−x = (7.7 × 10−27 )−10/300 = 7.4
(ii) f0−x = (1.6 × 10−10 )−10/800 = 1.3
E27.7(b)

The reaction rate is given by
8kB T 1/2
NA exp(−Ea /RT )[D2 ][Br2 ]
πµ

v = Pσ

so, in the absence of any estimate of the reaction probability P , the rate constant is
k =σ

8kB T 1/2
NA exp(−Ea /RT )
πµ

= [0.30 × (10

−9

2

m) ] ×

8(1.381 × 10−23 J K−1 ) × (450 K)

1/2

π(3.930 u) × (1.66 × 10−27 kg u−1 )

×(6.022 × 1023 mol−1 ) exp

−200 × 103 J mol−1
(8.3145 J K−1 mol−1 ) × (450 K)

= 1.71 × 10−15 m3 mol−1 s−1 = 1.7 × 10−12 L mol−1 s−1
E27.8(b)

The rate constant is
kd = 4πR ∗ DNA
where D is the sum of two diffusion constants. So
kd = 4π(0.50 × 10−9 m) × (2 × 4.2 × 10−9 m2 s−1 ) × (6.022 × 1023 mol−1 )
= 3.2 × 107 m3 mol−1 s−1
In more common units, this is
kd = 3.2 × 1010 L mol−1 s−1

E27.9(b)

(a) A diffusion-controlled rate constant in decylbenzene is

kd =

8 × (8.3145 J K−1 mol−1 ) × (298 K)
8RT
= 1.97 × 106 m3 mol−1 s−1
=
3η
3 × (3.36 × 10−3 kg m−1 s−1 )


MOLECULAR REACTION DYNAMICS

441

(b) In concentrated sulfuric acid
kd =

8 × (8.3145 J K −1 mol−1 ) × (298 K)
8RT
= 2.4 × 105 m3 mol−1 s−1
=
3η
3 × (27 × 10−3 kg m−1 s−1 )

E27.10(b) The diffusion-controlled rate constant is
kd =

8RT
8 × (8.3145 J K−1 mol−1 ) × (298 K)
= 1.10 × 107 m3 mol−1 s−1

=
−1
−3
−1
3η
3 × (0.601 × 10 kg m s )

In more common units, kd = 1.10 × 1010 L mol−1 s−1
The recombination reaction has a rate of
v = kd [A][B]

with [A] = [B]

so the half-life is given by
t1/2 =

1
1
=
= 5.05 × 10−8 s
−1
−1
10
k[A]0
(1.10 × 10 L mol s ) × (1.8 × 10−3 mol L−1 )

E27.11(b) The reactive cross-section σ ∗ is related to the collision cross-section σ by
σ∗ = Pσ

so

P = σ ∗ /σ.

The collision cross-section σ is related to effective molecular diameters by
σ = πd 2

so

d = (σ/π )1/2

2
1/2
1/2 2
2
Now σAB = πdAB
= π 21 (dA + dB ) = 41 σAA + σBB

so P =

σ∗
1
4

1/2

1/2 2

σAA + σBB

8.7 × 10−22 m

= 1
= 2.22 × 10−3
1/2
1/2
−9
2
+ (0.40) ) × 10 m]
4 [((0.88)
E27.12(b) The diffusion-controlled rate constant is
kd =

8 × (8.3145 J K−1 mol−1 ) × (293 K)
8RT
= 5.12 × 106 m3 mol−1 s−1
=
3η
3 × (1.27 × 10−3 kg m−1 s−1 )

In more common units, kd = 5.12 × 109 L mol−1 s−1
The recombination reaction has a rate of
v = kd [A][B] = (5.12 × 109 L mol−1 s−1 ) × (0.200 mol L−1 ) × (0.150 mol L−1 )
= 1.54 × 108 mol L−1 s−1


INSTRUCTOR’S MANUAL

442

E27.13(b) The enthalpy of activation for a reaction in solution is
‡

H = Ea − RT = (8.3145 J K−1 mol−1 ) × (6134 K) − (8.3145 J K −1 mol−1 ) × (298 K)
= 4.852 × 104 J mol−1 = 48.52 kJ mol−1

The entropy of activation is
‡

S = R ln

B=

A
−1
B

kRT 2
hp −−

where B =

(1.381 × 10−23 J K−1 ) × (8.3145 J K −1 mol−1 ) × (298 K)2
(6.626 × 10−34 J s) × (1.00 × 105 Pa)

= 1.54 × 1011 m3 mol−1 s−1
so

‡

8.72 × 1012 L mol−1 s−1

S = (8.3145 J K−1 mol−1 ) × ln

(1000 L m−3 ) × (1.54 × 1011 m3 mol−1 s−1 )

−1

= −32.2 J K−1 mol−1
Comment. In this connection, the enthalpy of activation is often referred to as ‘energy’ of activation.
E27.14(b) The Gibbs energy of activation is related to the rate constant by
k2 = B exp

− ‡G
RT

where B =

kRT 2
hp −−

so

‡

G = −RT ln

k2
B

k2 = (6.45 × 1013 L mol−1 s−1 )e−{(5375 K)/(298 K)} = 9.47 × 105 L mol−1 s−1
= 947 m3 mol−1 s−1

Using the value of B computed in Exercise 27.13(b), we obtain
‡

G = −(8.3145 × 10−3 kJ K−1 mol−1 ) × (298 K) × ln

947 m3 mol−1 s−1
1.54 × 1011 m3 mol−1 s−1

= 46.8 kJ mol−1
E27.15(b) The entropy of activation for a bimolecular reaction in the gas phase is
‡

S = R ln

B=

A
−2
B

where B =

kRT 2
hp −−

(1.381 × 10−23 J K−1 ) × (8.3145 J K −1 mol−1 ) × [(55 + 273) K]2
(6.626 × 10−34 J s) × (1.00 × 105 Pa)

= 1.86 × 1011 m3 mol−1 s−1
The rate constant is

k2 = A exp

−Ea
RT

so A = k2 exp

A = (0.23 m3 mol−1 s−1 ) × exp
= 1.8 × 107 m3 mol−1 s−1

Ea
RT
49.6 × 103 J mol−1

(8.3145 J K−1 mol−1 ) × (328 K)


MOLECULAR REACTION DYNAMICS

and

‡

443

1.8 × 107 m3 mol−1 s−1

S = (8.3145 J K−1 mol−1 ) × ln

1.86 × 1011 m3 mol−1 s−1

−2

= −93 J K−1 mol−1
E27.16(b) The entropy of activation for a bimolecular reaction in the gas phase is
‡

S = R ln

A
−2
B

kRT 2
hp −−

where B =

For the collision of structureless particles, the rate constant is
k2 = NA

8kT 1/2
− E0
σ exp
πµ
RT

so the prefactor is
A = NA

8kT 1/2
σ = 4NA
πµ

RT 1/2
σ
πM

where we have used the fact that µ = 21 m for identical particles and k/m = R/M. So
A = 4 × (6.022 × 10

23

mol

−1

)×

(8.3145 J K −1 mol−1 ) × (500 K)

1/2

π × (78 × 10−3 kg mol−1 )

× (0.68 × 10−18 m2 )

= 2.13 × 108 m3 mol−1 s−1
B =

(1.381 × 10−23 J K−1 ) × (8.3145 J K −1 mol−1 ) × (500 K)2
(6.626 × 10−34 J s) × (1.00 × 105 Pa)

= 4.33 × 1011 m3 mol−1 s−1
and

‡

S = (8.3145 J K−1 mol−1 ) × ln

2.13 × 108 m3 mol−1 s−1
4.33 × 1011 m3 mol−1 s−1

−2

= −80.0 J K−1 mol−1
E27.17(b) (a) The entropy of activation for a unimolecular gas-phase reaction is
‡

S = R ln

A
−1
B

where B = 1.54 × 1011 m3 mol−1 s−1 [See Exercise 27.17(a)]

so ‡ S = (8.3145 J K−1 mol−1 )× ln

2.3 × 1013 L mol−1 s−1

(1000 L m−3 ) × (1.54 × 1011 m3 mol−1 s−1 )

−1

= −24.1 J K−1 mol−1
(b) The enthalpy of activation is
‡

H = Ea − RT = 30.0 × 103 J mol−1 − (8.3145 J K −1 mol−1 ) × (298 K)
= 27.5 × 103 J mol−1 = 27.5 kJ mol−1

(c) The Gibbs energy of activation is
‡

G =

‡

H − T ‡ S = 27.5 kJ mol−1 − (298 K) × (−24.1 × 10−3 kJ K−1 mol−1 )

= 34.7 kJ mol−1


INSTRUCTOR’S MANUAL

444

E27.18(b) The dependence of a rate constant on ionic strength is given by
log k2 = log k2◦ + 2AzA zB I 1/2
At infinite dilution, I = 0 and k2 = k2◦ , so we must find

log k2◦ = log k2 − 2AzA zB I 1/2 = log(1.55) − 2 × (0.509) × (+1) × (+1) × (0.0241)1/2
= 0.323

and

k2◦ = 1.08 L2 mol−2 min−1

Solutions to problems
Solutions to numerical problems
P27.1

8kT
πµ

A = NA σ ∗
∗

[Section 27.1 and Exercise 27.16(a); µ = 21 m(CH3 )]

= (σ ) × (6.022 × 10

23

mol

−1

)×

(8) × (1.381 × 10−23 J K−1 ) × (298 K)

(π ) × (1/2) × (15.03 u) × (1.6605 × 10−27 kg/u)

1/2

= (5.52 × 1026 ) × (σ ∗ mol−1 m s−1 )
σ∗ =

(a)

2.4 × 1010 mol−1 dm3 s−1
5.52 × 1026 mol−1 m s−1

=

2.4 × 107 mol−1 m3 s−1
5.52 × 1026 mol−1 m s−1

= 4.4 × 10−20 m2

(b) Take σ ≈ πd 2 and estimate d as 2 × bond length; therefore
σ = (π ) × (154 × 2 × 10−12 m)2 = 3.0 × 10−19 m2
Hence P =
P27.3

σ∗
4.35 × 10−20
= 0.15
=
σ
3.0 × 10−19

For radical recombination it has been found experimentally that Ea ≈ 0. The maximum rate of
recombination is obtained when P = 1 (or more), and then
k2 = A = σ ∗ NA

8kT 1/2
= 4σ ∗ NA
πµ

kT 1/2
πm

µ = 21 m

σ ∗ ≈ πd 2 = π × (308 × 10−12 m)2 = 3.0 × 10−19 m2
Hence
k2 = (4) × (3.0 × 10−19 m2 ) × (6.022 × 1023 mol−1 )
×

(1.381 × 10−23 J K−1 ) × (298 K)
(π ) × (15.03 u) × (1.6605 × 10−27 kg/u)

1/2

= 1.7 × 108 m3 mol−1 s−1 = 1.7 × 1011 M−1 s−1
The rate constant is for the rate law
v = k2 [CH3 ]2
Therefore

d[CH3 ]

= −2k2 [CH3 ]2
dt


MOLECULAR REACTION DYNAMICS

445

1
1
−
= 2k2 t
[CH3 ] [CH3 ]0
For 90 per cent recombination, [CH3 ] = 0.10 × [CH3 ]0 , which occurs when

and its solution is

2k2 t =

9
[CH3 ]0

or

t=

9
2k2 [CH3 ]0

The mole fractions of CH3 radicals in which 10 mol% of ethane is dissociated is

(2) × (0.10)
= 0.18
1 + 0.10
The initial partial pressure of CH3 radicals is thus
p0 = 0.18 p = 1.8 × 104 Pa
1.8 × 104 Pa
RT
9RT
(9) × (8.314 J K −1 mol−1 ) × (298 K)
Therefore t =
=
(2k2 ) × (1.8 × 104 Pa)
(1.7 × 108 m3 mol−1 s−1 ) × (3.6 × 104 Pa)
and [CH3 ]0 =

= 3.6 ns
P27.6

Figure 27.1 shows that log k is proportional to the ionic strength for neutral molecules.

)

0.19

k/(

0.17

0.15
0

0.05

0.15

0.10

Figure 27.1
From the graph, the intercept at I = 0 is −0.182, so
k ◦ = 0.658 L mol−1 min−1
Comment. In comparison to the effect of ionic strength on reactions in which two or more reactants
are ions, the effect when only one is an ion is slight, in rough qualitative agreement with eqn 27.69.
P27.7

σ∗
≈
σ

e2
4πε0 d(I − Eea )

2

[Example 27.2]

Taking σ = πd 2 gives
∗

σ ≈π

e2
4πε0 [I (M) − Eea (X2 )]

2

=

6.5 nm2
(I − Eea )/eV


INSTRUCTOR’S MANUAL

446

Thus, σ ∗ is predicted to increase as I − Eea decreases. The data let us construct the following table
σ ∗ /nm2
Na
K
Rb
Cs

Cl2
0.45
0.72
0.77
0.97

Br 2
0.42

0.68
0.72
0.90

I2
0.56
0.97
1.05
1.34

All values of σ ∗ in the table are smaller than the experimental ones, but they do show the correct
trends down the columns. The variation with Eea across the table is not so good, possibly because
the electron affinities used here are poor estimates.
Question. Can you find better values of electron affinities and do they improve the horizontal trends
in the table?
P27.10

A + A → A2

v = −1


 
‡

S = R ln 

A
kT
h



 + 2

× pRT
−−

[27.63]

 

= −(8.3145 J K −1 mol−1 ) × ln 



4.07 × 105 M−1 s−1 10m3 L
3

(1.381×10−23 J K −1 )×(300 K)2 ×(8.3145 J K−1 mol−1 )
(6.626×10−34 J s)×(1.013×105 Pa)

 + 2

= (8.3145 J K−1 mol−1 ) × [ln(2.631 × 10−9 ) + 2]
‡
‡

S = −148 J K−1 mol−1

H = Ea − 2RT = 65.43 kJ mol−1 − 2 × (8.3145 J K −1 mol−1 ) × (300 K)
10−3 kJ
×
[27.60, 27.61]
J

‡

H = 60.44 kJ mol−1

‡
‡

H =

‡

U =

‡

U+

‡

H−

‡

(pV )

(pV ) =

‡

H−

vRT

= (60.44 kJ mol−1 ) − (−1) × (8.3145 J K −1 mol−1 ) × (300 K) ×

‡

‡

‡

10−3 kJ
J

U = 62.9 kJ mol−1

G=

‡

H − T ‡ S = 60.44 kJ mol−1 − (300 K) × (−148 J K −1 mol−1 ) ×

G = 104.8 kJ mol−1 [27.59]



10−3 kJ
J


MOLECULAR REACTION DYNAMICS

P27.12

447

(a) The multilinear hypothesis is
Ea = c1 Gb + c2 I + c3
where the constants c1 , c2 , and c3 are independent of temperature. The substitutions Ea =
−RT ln(k/A) and Gb = −RT ln(Kb ) (eqns 25.25 and 9.19) give
k
A

−RT ln

= −c1 RT ln(Kb ) + c2 I + c3

c3 may be eliminated by subtracting the analogous equation for the methylbenzene reference.
Assuming that the pre-exponential A values for the reference and members of the series are
comparable, the working equation becomes
k

−RT ln

ktoluene

pk

substituting
pk

Kb

= −c1 RT ln

= − log

= c1 p kb +

k
ktoluene

Kb,toluene
,

p Kb

+ c2 (I − Itoluene )

= − log

Kb
Kb,toluene

,

I = I − Itoluene gives

c2 I
RT ln(10)

The temperature dependence of
RT ln(10) p Kb =
RT ln(10) p Kb =

Gb =

(1)
pk

p Kb .

Hb − T S b

Hb − T

= (cT − 1)

depends upon

Sb
Hb

(assuming that

Sb = C

Hb )

(2)

Evaluating the above equation at T = T0 = 273.15 K gives
Hbo =

−RT0 ln(10) p Kbo
cT0 − 1

(3)

where
Hb (T0 ) =
Hbo and p Kb (T0 ) = p Kbo . Assuming that Hb is approxiHbo . Substitute equation (3) into (2)
mately independent of temperature gives
Hb =
and substitute the result into equation (1) to get
pk

=

a T0 (cT − 1) p Kbo
b I
+
T (cT0 − 1)
RT ln(10)

where the symbols c1 and c2 have been replaced with the symbols a and b.

(b) The activation parameters for the ring destruction of p-xylene are determined with a linear
regression analysis of the experimental data plotted as ln(k) versus 1/T (eqn 25.25 and Example
25.5). The regression first gives:
slope = −8.875 × 103 K
intercept = 25.53


INSTRUCTOR’S MANUAL

448

Ea = −R × slope = 73.8 kJ mol−1
A = eintercept L mol−1 s−1 = 1.223 × 1011 L mol−1 s−1
‡

H and ‡ S values for solution reactions may be calculated with the following equation (k is
the Boltzmann constant in these equations).
‡

H = Ea − RT

‡

and

S = R ln

hA
KekT

where the transmission coefficient K is assumed to equal 1. These equations may be deduced
by modification of Sections 27.4 and 27.5 concepts to the solution phase. Eqn 27.42 becomes
[C‡ ] = K ‡ [A][B]; eqn 27.44 becomes k2 = k ‡ K ‡ = kνK ‡ (eqn 27.45). Eqns 27.5 and 27.53
‡
‡
become k2 = Kν(kT K / hν) = KkT K / h.
‡
Eqn 27.58 becomes k2 = (KkT / h) e− G/RT and eqn 27.60 becomes
‡ S/R

k2 = (K kT / h) e

‡
e− H /RT . According to the last equation,

∂ ln k2
1
1
=
+
∂T
T
R
p

∂ ‡S
∂T

+
p

‡ S✟✟

1
1 ∂
✟
=
+
R ✟✟∂T
T
✟
=

p

‡H

1
−
RT
RT 2

∂ ‡H
∂T

p



∂ ✘

1 
T✘✘✘
+
−
✘✘
RT
∂T
RT 2 ✘





✘
‡ H✘✘

‡H

(eqn 4.16)

p

‡H
1
+
T
RT 2

Substitution into the formal definition of activation energy (eqn 25.26) Ea = RT 2 (∂ ln k2 /∂T )p ,
gives Ea = ‡ H + RT or ‡ H = Ea − RT . Subtitution of this condusion into the k2 equation

gives
k2 = (K RT / h)e
= (Ke kT / h)e

‡ S/R

‡
e− H /RT = (K kT / h)e

‡ S/R

e−(Ea −RT )/RT

‡ S/R

Substituting of k2 = Ae−Ea /RT (eqn 25.25) and solving for
‡

T /K
293.15
303.15
313.15
323.15
333.15
343.15

S = R ln

‡

S gives the final result.

hA
Ke kT

k/10−2 L mol−1 s−1
0.86
2.5
5.4
13
47
59

‡

H /k J mol−1
71.4
71.3
71.2
71.1
71.0
70.9

‡

S/J K−1 mol−1
−40.8
−41.1
−41.4
−41.6

−41.9
−42.1

Entropy decreases upon formation of the transition state.
(c) The 6 temperatures at which rate constants are measured may be indexed as i = 0, 1, 2, . . . , 5.
The 7 arenes studied may be indexed as j = 0, 1, 2, . . . , 6. p K, p Kbo , and I values may
be calculated for each arene at each temperature. Methybenzene (toluene is the reference arene).
The values for p kexp (T ) are calcluated with the arrhenius parameters. The constants a, b, and
c that appear in the equation deduced in part (a) are determined by systematically altering their
values so that the sum of the squares of errors (SSE) between p kexp and the fitted equation
p kfit is minimized.


MOLECULAR REACTION DYNAMICS

449

6

5
p kexp (Ti ) −

SSE(a, b, c) =
j =0

i=0

p kfit (a, b, c, Ti )

2

j

Mathematical software like mathcad’s given/minerr solve block easily perform the minimization.
We find a best fit when
a = 0.413

b = −0.192

c = 1.39 × 10−3 K −1

The goodness of the fit may be graphically evaluated by plotting the ratio p kfit / p kexp against
p Kexp for the 7 arenes at a temperature of choice. A good fit gives a ratio of 1 to within
experimental error. The following plot gives the ratio at 293.15 K.

∆p kfit
∆p kexp

1.05

1

0.95

0.9
–5

–4

–3

–2

–1
∆p kexp

0

1

2

3

Figure 27.2

p kfit volumes are within about ±3% of the experimental values. This is a good fit, which
confirms that the activation energy for arene distinction is multilinear in the basicity constant
and the ionization energy. This is also evidence for the proposed arene ring oxidation mechanism.

Solutions to theoretical problems
P27.14

Programs for numerical integration using, for example, Simpson’s rule are readily available for
personal computers and hand-held calculators. Simplify the form of eqn 27.40 by writing
z2 =

kx 2
,
4D

τ = kt,

j=

A
n0

π D 1/2
[J]∗
k

Then evaluate
j=

τ
0

1 1/2 −z2 /τ −τ
e
e dτ +
τ

for various values of k.

1 1/2 −z2 /τ −τ
e
e
τ

INSTRUCTOR’S MANUAL

450

P27.16

Ka =

[H+ ][A− ] 2
[H+ ][A− ]γ± 2
γ± ≈
[HA]γHA
[HA]

Therefore, [H+ ] =

[HA]Ka
[A− ]γ±2

and log[H+ ] = log Ka + log

[HA]
[HA]
− 2 log γ± = log Ka + log − + 2AI 1/2
−
[A ]
[A ]

Write v = k2 [H+ ][B]
then log v = log(k2 [B] + log[H+ ]

[HA]
= log(k2 [B]) + log − + 2AI 1/2 + log Ka
[A ]
[B][HA]Ka
= log v ◦ + 2AI 1/2 , v ◦ = k2
[A− ]
That is, the logarithm of the rate should depend linearly on the square root of the ionic strength,
log v ∝ I 1/2
P27.18

k1 =

kT
q‡
× e−β E [Problem 27.17]
h
q

q ‡ = qz‡V qy‡V qxR ≈
qR ≈

kT 2 R
q
hν ‡

(T /K)3/2
1.027
×
[Table 20.4, A = B = C] ≈ 80
σ

(B/cm−1 )3/2

q = qzV qyV qxV ≈

kT 3
hν

ν3
Therefore, k1 ≈ 80 × ‡2 e−β E0 ≈ 80 × 5.4 × 104 s−1 [Problem 27.15] = 4 × 106 s−1
ν
Consequently, D ≈ (80) × (2.7 × 10−15 m2 s−1 ) =
P27.20

2 × 10−13 m2 s−1 if ν ‡ = ν and

9 × 10−13 m2 s−1 if ν ‡ = 21 ν.
It follows that, since N s and l are the same for the two experiments,
σ (CH2 F2 )
ln 0.6
=
[Problem 27.17] = 5
σ (Ar)
ln 0.9
CH2 F2 is a polar molecule; Ar is not. CsCl is a polar ion pair and is scattered more strongly by the
polar CH2 F2 .

Solutions to applications
P27.22

Collision theory gives for a rate constant with no energy barrier

k = Pσ

8kT 1/2
NA
πµ

so P =

k
σ NA

π µ 1/2
8kT


MOLECULAR REACTION DYNAMICS

P =

k/(L mol−1 s−1 ) × (10−3 m3 L−1 )
(σ/nm2 ) × (10−9 m)2 × (6.022 × 1023 mol−1 )
×

=

451

1/2

π × (µ/u) × (1.66 × 10−27 kg)

8 × (1.381 × 10−23 J K−1 ) × (298 K)

(6.61 × 10−13 )k/(L mol−1 s−1 )
(σ/nm2 ) × (µ/u)1/2

The collision cross-section is
1/2

2
σAB = πdAB

where dAB =

1/2

+σ
σ
1
(dA + dB ) = A 1/2B
2
2π

1/2

so

σAB =

(σA

1/2

+ σ B )2
4

The collision cross-section for O2 is listed in the Data Section. We would not be far wrong if we took
that of the ethyl radical to equal that of ethene; similarly, we will take that of cyclohexyl to equal that
of benzene. For O2 with ethyl
(0.401/2 + 0.641/2 )2
nm2 = 0.51 nm2
4
(32.0 u) × (29.1 u)
m O me
=
= 15.2 u
µ=
mO + m e
(32.0 + 29.1) u
σ =

(6.61 × 10−13 ) × (4.7 × 109 )
= 1.6 × 10−3
(0.51) × (15.2)1/2
For O2 with cyclohexyl

so P =

(0.401/2 + 0.881/2 )2
nm2 = 0.62 nm2
4

mO m C
(32.0 u) × (77.1 u)
µ=
=
= 22.6 u
mO + m C
(32.0 + 77.1) u
σ =

so P =

(6.61 × 10−13 ) × (8.4 × 109 )
= 1.8 × 10−3
(0.62) × (22.6)1/2