CHEMCA Free Chemistry Notes for JEE NEET Chemistry MadeEasy: Second Order Reactions

A reaction is said to be of the second order if the rate of the reaction depends on the concentration of two reactant molecules (or square of one).

1. Differential Rate Law

For a reaction $2A \rightarrow \text{Products}$:

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

For a reaction $A + B \rightarrow \text{Products}$:

$$ \text{Rate} = -\frac{d[A]}{dt} = k[A][B] $$

Note: We will focus on the simpler case where $[A]_0 = [B]_0$, which mathematically reduces to the single reactant case.

Rearranging the differential equation:

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

Integrating both sides from $[A]_0$ to $[A]_t$:

$$ \int_{[A]_0}^{[A]_t} -[A]^{-2} d[A] = \int_0^t k \, dt $$ $$ \left[ \frac{1}{[A]} \right]_{[A]_0}^{[A]_t} = kt $$

Final Equation:

$$ \frac{1}{[A]_t} - \frac{1}{[A]_0} = kt $$

Alternatively: $kt = \frac{x}{a(a-x)}$ where $a$ is initial conc. and $x$ is reacted amount.

The time required for the concentration to reduce to half its initial value ($[A]_t = [A]_0 / 2$).

$$ \frac{1}{[A]_0/2} - \frac{1}{[A]_0} = k t_{1/2} $$ $$ \frac{2}{[A]_0} - \frac{1}{[A]_0} = k t_{1/2} $$ $$ \frac{1}{[A]_0} = k t_{1/2} $$

$$ t_{1/2} = \frac{1}{k[A]_0} $$

Key Difference: Unlike 1st order reactions, the half-life of a 2nd order reaction depends inversely on the initial concentration.

Rearranging the integrated equation to $y = mx + c$ form:

$$ \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} $$

A straight line with positive slope confirms second order kinetics.

Decomposition of Nitrogen Dioxide:
$$ 2NO_2(g) \rightarrow 2NO(g) + O_2(g) $$
Saponification of Ester (Alkaline Hydrolysis):
$$ CH_3COOC_2H_5 + NaOH \rightarrow CH_3COONa + C_2H_5OH $$
(When concentrations of Ester and Base are equal)
Formation of HI:
$$ H_2(g) + I_2(g) \rightarrow 2HI(g) $$

Unit of Rate Constant ($k$):
Since Rate = $k[Concentration]^2$:
Unit = $M^{-1} s^{-1}$ or $L \, mol^{-1} \, s^{-1}$.