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Second Order Reaction - Practice Questions & MCQ

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

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  • 16 Questions around this concept.

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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.

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Second Order Kinetics

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