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

The rate law for the reaction A + BP is:

\frac{d[A]}{dt}=-k[A][B]\; \; \; \; \; \; \; \; 11

Unlike the 2nd order reaction of the type AP with a rate law of \frac{d[A]}{dt}=-k[A]^2, eq11 cannot be integrated unless we express [B] in terms of [A], or [A] and [B] in terms of a third variable x. Since A and B react in a ratio of 1:1, the remaining concentrations of A and B at any time of the reaction are:

[A]=[A_0]-x\; \; \; and\; \; \; [B]=[B_0]-x

where [A0] and [B0] are the initial concentrations of A and B respectively.

Differentiating both sides of [A] = [A0] – x with respect to time, we have


Therefore, we can rewrite eq11 as

\frac{dx}{dt}=k\left ( a-x \right )\left ( b-x \right )\; \; \; \; \; \; \; \; 12

where a = [A0] and b = [B0]

Integrating eq12 using the partial fraction expression of \frac{1}{(a-x)(b-x)}=\frac{1}{b-a}\left ( \frac{1}{a-x}-\frac{1}{b-x} \right ),

\int_{0}^{x}\frac{1}{b-a}\left ( \frac{1}{a-x}-\frac{1}{b-x} \right )dx=k\int_{0}^{t}dt

After some algebra, we have,


which is equivalent to



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