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Second Order Reaction - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • 11 Questions around this concept.

Solve by difficulty

The half-life of a second-order reaction is dependent on concentration of Reactant:-
 

For a second-order reaction, if the concentration of the reactant is tripled, the rate of the reaction becomes:.
 

 A reaction follows second-order kinetics with the rate concentration of the reactant is 0.1 mott', what will be its concentration of ter 100 seconds?
 

Concepts Covered - 1

Second Order Kinetics

Consider the reaction

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

For ssecond 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 tis 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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