PHYSICAL CHEMISTRY IN BRIEF

CHAP. 9: CHEMICAL KINETICS [CONTENTS] 319

Solution

We take the logarithm of equation (9.164) and substitute forTi,ki

lnk 1 = ln

(

RT 1 cst

NAh

)

+

∆S

R

−

∆H

RT 1

,

lnk 2 = ln

(

RT 2 cst

NAh

)

+

∆S

R

−

∆H

RT 2

The solution of this set of two equations for two unknowns is

∆H#=R

T 1 T 2

T 2 −T 1

ln

T 1 k 2

T 2 k 1

= 8. 314

556 · 781

781 − 556

ln

556 · 3. 954 · 10 −^2

781 · 3. 517 · 10 −^7

= 181 160J mol−^1 ,

∆S# =

∆H

T 1

+Rln

k 1 NAh

RT 1 cst

=

=

181 160

556

+ 8.314 ln

3. 517 · 10 −^7 · 6. 022 · 1023 · 6. 626 · 10 −^34

8. 314 · 556 · 1

=

= − 47. 814 J mol−^1 K−^1.

9.6.5 General relation for temperature dependence of the rate constant

The temperature dependence of the rate constant may be written in the form

k=a Tbe−c/T. (9.165)

The relations discussed in the preceding sections are special cases of this dependence. Forb= 0
we obtain the Arrhenius equation (9.159), wherea=Aandc=E∗/R. Forb= 1/2 we obtain
the relation (9.162) from the collision theory, wherea=Aandc=B. Forb= 1 we obtain the

relation (9.164) from the absolute-rates theory, wherea= (cst)(n−1)NRAhexp(∆S#/R) andc=

∆H#/R.