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

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


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


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,




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





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



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