November 7, 2019December 14, 2022 by gary_mail@hotmail.com

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

The rate law for the reaction A + B → P is:

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

Unlike the 2nd order reaction of the type A → P 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

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

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,

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

which is equivalent to

$kt=\frac{1}{[B_0]-[A_0]}ln\frac{[A_0][B]}{[B_0][A]}$

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