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

The rate law for the 3rd order reaction of the type A + 2BP 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 yields

\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 working out some algebra gives

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.

 

 

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