Page 375 - Elements of Chemical Reaction Engineering Ebook

346 Nonelementary Reaction Kinetics Chap. 7

Solution
From the linear plot we know that

5 = A +BC, =A +B(x) (E7-1.1)
I
where C, (X). Inverting yields
- - 1 (E7-1.2)
1-
I o A+B(X)
From rule 1 of Table 7-1, the denominator suggests that alcohol collides with the
active intermediate:
X + intermediate + deactivation products (E7-1.3)

The alcohol acts as what is called a scavenger to deactivate the active intermediate.
The fact that the addition of CC14 or CS2 increases the intensity of the luminescence,
1 (CSd (E7-1.4)

leads us to postulate (rule 3 of Table 7-1) that the active intermediate was probably
formed from CS2:
M+CS, ___) CS; +M (E7- 1.5)

where M is a third body (CS2, H20, etc.).
We also know that deactivation can occur by the reverse of Reaction (E7-1.5).
Combining this information, we have as our mechanism:
Activation: M + CS, kt > CS; + M (E7- 1.5)

Deactivation: M + CS; '' > CS2 + M (E7-1.6)
The mechanism
Deactivation: X, + CS; k3 > cs2 + x (E7-1.3)

Luminescence: cs; k4 > CS2 + hv (E7- I .7)

I = kd(Cs;) (E7-1.8)

Using the PSSH on CS; yields
res; = 0 = k,(CS,)(M) - kz(CS;)(M) - k3(X)(CS;) - k4(CS;)

Solving for CS; and substituting into Equation (E7-1.8) gives us

(E7-1.9)

In the absence of alcohol,

(E7- 1.10)