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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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QuestionEASY

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

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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

Study other Related Concepts

Second Order Reaction Current Topic

Rate of Reaction
Rate Law
Order of Reaction
Zero Order Reaction
First Order Reaction
Second Order Reaction
nth Order Reaction
Methods of Determining Reaction Order
Molecularity of reaction
Pseudo First Order Reaction

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