Physical Chemistry Third Edition

534 12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms

EXAMPLE12.6

For the gas-phase reaction

H 2 +I 2 −→2HI

at 373.15 K, the rate constant is equal to 8. 74 × 10 −^15 L mol−^1 s−^1. At 473.15 K it is equal

to 9. 53 × 10 −^10 L mol−^1 s−^1. Find the value of the activation energy and of the preexpo-

nential factor.

Solution

For any two temperaturesT 1 andT 2 , Eq. (12.3-2) gives

Ea

Rln

(

k(T 2 )

k(T 1 )

)

1

T 1

−

1

T 2

(12.3-3)

Substitution of the values gives

Ea

(8.3145 J K mol−^1 )ln

(

9. 53 × 10 −^10

8. 74 × 10 −^15

)

1

373 .15 K

−

1

473 .15 K

 1. 70 × 105 J mol−^1

AkeEa/RT

(8. 74 × 10 −^15 L mol−^1 s−^1 ) exp

⎛

⎝^1.^70 ×^10

(^5) J mol−^1
(
8 .3145 J K mol−^1
)
( 373 .15 K)
⎞
⎠
 5. 47 × 109 L mol−^1 s−^1
Exercise 12.6
A common rule of thumb is that the rate of a reaction doubles if the temperature is raised
by 10◦C.
a.Find the value of the activation energy if a rate constant doubles in value between 20◦C
and 30◦C.
b.Find the value of the activation energy if a rate constant doubles in value between 90◦C and
100 ◦C.
c.A common definition of the activation energy of a reaction is
EaRT^2
(
dln(k)
dT
)
(common definition ofEa) (12.3-4)
Show that ifkis given by Eq. (12.3-2), Eq. (12.3-4) gives the sameEaas in Eq. (12.3-2) ifAis
temperature-independent.