3rd order reaction of the type (A + 2B → P)

The rate law for the 3rd order reaction of the type A + 2B → P is:

$\frac{d[A]}{dt}=-k[A][B]^2$

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

$\frac{dx}{dt}=k(a-x)(b-2x)^2$

where a = [A₀] and b = [B₀].

Integrating the above equation throughout, we have

$\int_{0}^{x}\frac{dx}{(a-x)(b-2x)^2}=k\int_{o}^{t}dt\; \; \; \; \; \; \; \; 15$

Substituting the partial fraction expression $\frac{1}{(a-x)(b-2x)^2}=\frac{2}{(2a-b)(b-2x)^2}+\frac{1}{(2a-b)^2(a-x)}-\frac{2}{(2a-b)^2(b-2x)}$ in eq15 and after some algebra, we have,

$kt=\frac{2x}{b(2a-b)(b-2x)}+\frac{1}{(2a-b)^2}ln\frac{a(b-2x)}{b(a-x)}$

Question

How to compute $\int_{0}^{x}\frac{2}{(2a-b)(b-2x)^2}dx$ ?

Answer

$\int_{0}^{x}\frac{2}{(2a-b)(b-2x)^2}dx=\int_{0}^{x}\frac{2b}{b(2a-b)(b-2x)^2}dx$

$=\int_{0}^{x}\frac{2[(b-2x)+2x]}{b(2a-b)(b-2x)^2}dx=\int_{0}^{x}\frac{2[(2a-b)(b-2x)+2x(2a-b)]}{b(2a-b)^2(b-2x)^2}dx$

$=\int_{0}^{x}\frac{2b(2a-b)(b-2x)-4bx(b-2a)}{b^2(2a-b)^2(b-2x)^2}dx=\frac{2x}{b(2a-b)(b-2x)}$

Note that the integrand in the 5th equality can be obtained by differentiating the last term using the quotient rule.

3rd order reaction of the type (A + 2B → P)

Question

Answer

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