second order reaction

Question

For a second-order reaction with equal initial concentrations of reactants, the half-life is:

RankGuru · Accepted Answer

For a reaction $2A 	o Products$ following **Second-Order Kinetics**, the integrated rate law is:
$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$
At half-life ($t = t_{1/2}$), $[A]_t = \frac{[A]_0}{2}$. Substituting this into the equation:
$\frac{1}{[A]_0/2} = \frac{1}{[A]_0} + k t_{1/2}$
$\frac{2}{[A]_0} - \frac{1}{[A]_0} = k t_{1/2}$
$t_{1/2} = \frac{1}{k [A]_0}$
- Thus, the half-life is **inversely proportional** to the initial concentration. This distinguishes it from first-order reactions where $t_{1/2}$ is independent of concentration.

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second order reaction

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