3rd order reaction of the type (A + B + C → P)

The rate law for the 3rd order reaction of the type A + B + C → P is:

$\frac{d[A]}{dt}=-[A][B][C]$

Using the same logic described in a previous article, we can rewrite the rate law as:

$\frac{dx}{dt}=k(a-x)(b-x)(c-x)$

where a = [A₀], b = [B₀] and c = [C₀].

Integrating the above equation throughout, we have

$\int_{0}^{x}\frac{dx}{(a-x)(b-x)(c-x)}=k\int_{0}^{t}dt$

Substituting the partial fraction expression $\small \frac{1}{(a-x)(b-x)(c-x)}=\frac{1}{(a-x)(b-a)(c-a)}+\frac{1}{(b-x)(a-b)(c-b)}+\frac{1}{(c-x)(a-c)(b-c)}$ in the above integral and after some algebra, we have,

$kt=\frac{1}{(b-a)(c-a)}ln\frac{a}{a-x}+\frac{1}{(a-b)(c-b)}ln\frac{b}{b-x}+\frac{1}{(a-c)(b-c)}ln\frac{c}{c-x}$

3rd order reaction of the type (A + B + C → P)

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