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# Second Order Reaction - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

## Quick Facts

• 16 Questions around this concept.

## Solve by difficulty

In a second-order reaction. How does the rate change when the concentration of a reactant is doubled?

## Concepts Covered - 1

Second Order Kinetics

Consider the reaction

$\mathrm{A\: \rightarrow \: P}$

For second-order reaction, the rate law is given as follows:

$\mathrm{Rate\: =\: k[A]^{2}\: =\: -\frac{dA}{dt}}$

• Integrated Rate Law: Since it is a second-order reaction, thus:
$\mathrm{\frac{-dA}{dt}\: =\: k[A]^{2}}$
Integrating both sides, we get:
$\mathrm{\int_{A_{o}}^{A}\frac{dA}{A^{2}}\: =\: -\int_{0}^{t}kdt}$

where Ao is the initial concentration of A at time t = 0
A is the remaining concentration of A after time 't'

$\mathrm{\Rightarrow\left [ \frac{-1}{A} \right ]^{A}_{A_{o}}\: =\: -k(t)^{t}_{0}}$

$\mathrm{\Rightarrow\left [ \frac{1}{A} \right ]^{A}_{A_{o}}\: =\: kt}$

$\mathrm{\Rightarrow\: \frac{1}{A}\: -\: \frac{1}{A_{o}}\: =\: kt\quad\quad\quad\quad\quad\quad...............(i)}$

$\mathrm{Thus, \mathbf{\frac{1}{[A]}\: =\: kt\: +\: \frac{1}{[A_{o}]}}}$

• Half-life(t1/2): We know that the half-life for a reaction is the time when the concentration of the reactant(A) is half of its initial value.
Thus, at time t = t1/2, A = Ao/2
From equation (i) we have:

$\mathrm{t\: =\: \frac{1}{k}\left [ \frac{1}{A}\: -\: \frac{1}{A_{o}} \right ]}$

$\mathrm{\Rightarrow t_{1/2}\: =\: \frac{1}{k}\left [ \frac{1}{A_{o}/2}\: -\: \frac{1}{A_{o}} \right ]}$

$\mathrm{Thus,\: \mathbf{t_{1/2}\: =\: \frac{1}{kA_{o}}}}$
This is the half-life for the second-order reaction.

## Study it with Videos

Second Order Kinetics

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