Physical Chemistry Third Edition

548 12 Chemical Reaction Mechanisms I: Rate Laws and Mechanisms

Solution

We write equilibrium expressions for steps (1) and (2):

K 1 

k 1

k 1 ′



[N 2 O 4 ][H 2 O]

[H+][HNO 2 ][NO− 3 ]

(12.4-23)

K 2 

k 2

k 2 ′



[NO 2 ]^2

[N 2 O 4 ]

(12.4-24)

The rate differential equation for step (3) is

ratek 3 [NO 2 ][fer] (12.4-25)

where [fer] stands for the concentration of the hexacyano iron(III) ion, Fe(CN)^36 −, also known

as the ferricyanide ion. Equations (12.4-23) and (12.4-24) are solved simultaneously. From

Eq. (12.4-24)

[NO 2 ](K 2 [N 2 O 4 ])^1 /^2 (12.4-26)

From Eq. (12.4-23)

[N 2 O 4 ]

K 1 [H+][HNO 2 ][NO− 3 ]

[H 2 O]

(12.4-27)

Equation (12.4-27) is substituted into Eq. (12.4-26) and the resulting equation is substituted

into Eq. (12.4-25). The result is

rate

(

k 3 K 2 K 1 [H+][HNO 2 ][NO− 3 ]

[H 2 O]

) 1 / 2

[fer]

kapp[H+]^1 /^2 [HNO 2 ]^1 /^2 [NO− 3 ]^1 /^2 [fer] (12.4-28)

We have incorporated the concentration of H 2 O intokapp, the apparent rate constant, since

the concentration of water is nearly constant in dilute aqueous solutions. The reaction is

pseudo 5/2 order overall.

Exercise 12.12

Apply the steady-state approximation to the mechanism of Example 12.12. Omit the reverse

reaction in step 2.

EXAMPLE12.13

The following is a famous gas-phase reaction, with a mechanism proposed by Ogg.a

2N 2 O 5 −→4NO 2 +O 2 (12.4-29)

aR. A. Ogg, Jr.,J. Chem. Phys., 15 , 337, 613 (1947).