Physical Chemistry Third Edition

12.5 Chain Reactions 557

Once bromine atoms are formed in the initiation step, the reaction can proceed

without further initiation. Thechain lengthγis defined as the average number of

times the cycle of the two propagation steps is repeated for each initiation step. It is

possible to have a chain length as large as 10^6. In this reaction, the initiation step gives

two Br atoms, and each of these gives two molecules of HBr per cycle, so that the

number of molecules of product for each initiation step is equal to 4 times the chain

length.

To obtain the rate law we apply the steady-state approximation. For a three-step

mechanism we must write three differential equations. We choose the time derivatives

of [H 2 ] and the concentrations of the two chain carriers [H] and [Br]. We choose

[H 2 ] instead of [Br 2 ] or [HBr] because H 2 occurs in only one step of the mech-

anism and will give a simpler differential equation. The simultaneous differential

equations are

−

d[H 2 ]

dt

k 2 [Br][H 2 ]−k′ 2 [HBr][H] (12.5-4a)

d[Br]

dt

 2 k 1 [Br 2 ]−k 2 [Br][H 2 ]+k 3 [H][Br 2 ] (12.5-4b)

+k′ 2 [HBr][H]− 2 k′ 1 [Br]^2 ≈ 0

d[H]

dt

k 2 [Br][H 2 ]−k′ 2 [HBr][H]−k 3 [H][Br 2 ]≈ 0 (12.5-4c)

We have applied the steady-state approximation and set the time derivatives of the

concentration of the chain carriers H and Br equal to zero. To solve the algebraic

versions of Eqs. (12.5-4b) and (12.5-4c), we add Eqs. (12.5-4b) and (12.5-4c) to give

k 1 [Br 2 ]−k 1 ′[Br]^2  0 (12.5-5a)

which is the same as

[Br]

(

k 1

k′ 1

) 1 / 2

[Br 2 ]^1 /^2 (12.5-5b)

Equation (12.5-5b) is the same equation that would result from assuming that step 1

is at equilibrium. The relation of Eq. (12.5-5b) is substituted into Eq. (12.5-4b) or

Eq. (12.5-4c) to give (after several steps of algebra)

[H]

k 2

(

k 1 /k′ 1

) 1 / 2

[H 2 ][Br 2 ]^1 /^2

k 3 [Br 2 ]+k′ 2 [HBr]

(12.5-6)

We now simplify Eq. (12.5-4a) by noticing that the first two terms in Eq. (12.5-4c) are

the same as the two terms on the right-hand side of Eq. (12.5-4a), so that

−

d[H 2 ]

dt

k 3 [H][Br 2 ]

When Eq. (12.5-6) is substituted into this equation, we have

−

d[H 2 ]

dt



k 2

(

k 1 /k 1 ′

) 1 / 2

[H 2 ][Br 2 ]^1 /^2

1 +

k′ 2 [HBr]

k 3 [Br 2 ]

(12.5-7)