Chapter 2 Exercise 1: Given: ) (1) 21(‡

2‡1 GGGG ∆−∆=∆−∆ α

From Transition State Theory:

∆−

∆=

TRH

RS

hTkk

g

oi

g

oii

‡‡

expexp Where i = 1, 2 (2)

∆−

∆=

TRH

RS

hTkk

g

oi

g

oii

‡‡

exp (3)

And since , substitute into equation (3). ‡‡‡

ooo STHG ∆−∆=∆ ‡oG∆

∆−=

TRG

hTkk

g

oii

‡

exp (4)

Use equation (4) to find

1

2lnkk

.

∆−

∆−

=

TRG

hTk

TRG

hTk

kk

g

o

g

o

‡2

‡1

1

2

exp

exp

lnln (5)

∆+

∆−=

TRG

TRG

kk

g

o

g

o‡1

‡2

1

2 explnln (6)

Substitute equation (1) to simplify further.

∆−

∆=

TRG

TRG

kk

g

o

g

o 21

1

2ln α (7)

2.1

Since TRG

Kg

oii

∆−=ln , express equation (7) in terms of equilibrium constants.

( )

=−=

1

212

1

2

lnln

lnlnlnlnKK

KKkk

αα

Exercise 2:

)( OH)( propene)( propanol2

)( H)( acetone)( propanol2

2

2

2

1

EDA

CBAk

k

+⇒−

+⇒−

Express the selectivity to acetone in terms of the rates of formation of acetone (B) and propene (D).

DB

BB raterate

rateS

+= (1)

Therefore,

)1( BB

BD S

Srate

rate −= (2)

Use equation (2) to convert selectivity data into rate of propene formation. Write rate expressions for the formation of acetone and propene. (3) α

AB Ckrate 1=

(4) βAD Ckrate 2=

Use the Arrhenius equation to substitute for the rate constant (k) in equations (3) and (4).

ABg

CrateTRE

Ak lnlnlnln 111 α−=−= (5)

ADg

CrateTRE

Ak lnlnlnln 222 β−=−= (6)

Plot ln(rateB) and ln(rateD) vs. 1/T and calculate the slope to find the activation energy (E) for each reaction. This analysis was possible because CA was constant for each experiment.

2.2

y = -19665x + 19.596R2 = 0.9991

-15-14.5

-14-13.5

-13-12.5

-12

0.0016 0.00165 0.0017 0.00175 0.0018

1/T (K-1)

ln(rate B

)

y = -28316x + 32.47R2 = 0.9749

-20

-15

-10

-5

0

0.0016 0.00165 0.0017 0.00175 0.0018

1/T (K-1)

ln(rate D

)

From the graphs, E1 = 1.63 x 105 J mol-1 E2 = 2.35 x 105 J mol-1

Exercise 3:

H3C

H3C

+O

HH3C

H3C

O

H

DMB (A) 3,4-Dimethyl-∆3-tetrahydro-benzaldehyde (C)Acrolein (B)

Write the Arrhenius equation and take the natural log of both sides.

2.3

−=

TREAkg

exp (1)

TREAkg

−+= lnln (2)

Using the rate constants calculated in Exercise 7 of Chapter 1, plot ln(k) vs. 1/T to find

the slope

−

gRE

and the y-intercept )A(ln .

Temperature (K) Rate Constant (L mol-1 hr-1)

323 0.100 298 0.028 278 0.0033

y = -6782.8x + 18.856R2 = 0.9725

-6

-5

-4

-3

-2

-1

0

0.003 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037

1/T (K-1)

ln (k

)

From the graph, E = 56400 J mol-1 A = 1.55 x 108 L mol-1 hr-1 Exercise 4: Consider a molecule bounded to a surface that desorbs into the gas phase:

2.4

A A

A

The transition state may have more degrees of freedom than the adsorbed state. For example, the transition state may have translational degrees of freedom in the plane of the surface and/or additional degrees of rotational freedom (for polyatomic molecules). Thus

‡oS∆ can be greater than zero and the pre-exponential factor )/exp( ‡

go RShTk

∆ can exceed

1013 s-1 at normal temperatures. Additional discussion of adsorption and desorption of molecules on surfaces can be found in Chapter 5. Exercise 5: No solution will be given. Exercise 6: According to Example 2.3.1, the rate constant of the reaction A+B S is related to the charge of species in solution given by the following expression:

( ) IZZkk

BAo

=

log , where ko is the rate constant at infinite dilution. If the

charge on the hydrated electron (A) where zero, the rate constant would be independent of I , the ionic strength of the solution. If the charge on A were (-1), then variation of

concentration of ions (B) would yield a linear relationship between

okklog and

I with a slope of BA ZZ .

2.5