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First Order Reaction - Practice Questions & MCQ

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

Quick Facts

  • First Order Reaction, Half Life of First Order Reaction, Graphs of First Order Kinetics are considered the most difficult concepts.

  • 74 Questions around this concept.

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t_{1/4}  can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its initial value. If the rate constant for a first order reaction is k, the  t_{1/4}  can be written as

The rate equation for the reaction 2A+B\rightarrow C is found to be : rate =k\left [ A \right ]\left [ B \right ]. The correct statement in relation to this reaction is that the

The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be :

Units of the rate constant of first and zero-order reactions in terms of molarity M unit are respectively.

Select the reaction which is not follows the first order Kinetics:


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If we start a reaction from 100 mol, having the rate constant is \mathrm{0.0693 \mathrm{~min}^{-1}}. What will be the rate of the reaction offer 20 min.


Which of the following reactions is first order reaction

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Which of the following correctly represents the first order reaction.

Which expression is not correctly represent the first order reaction?

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Consider a reaction \mathrm{A \rightarrow B+C} follow the first order kinetics with rate constant \mathrm{( k ) \: 0.02 \mathrm{~s}^{-1}}. What percentage of A will be left after after 5 half-lives?

Concepts Covered - 4

First Order Reaction

The rate of the reaction is proportional to the first power.


The chemical reaction occurs as follows:

R    \rightarrow        P

a                 0

a-x             x

We have,


\frac{-dx}{dt}=K(a-x)  [differentiate rate law]

ln \:[\frac{a}{a-x}]=kt \:(Integrated\: rate\: law)

\mathrm{k\: =\: \frac{1}{t}\, ln\left [ \frac{a}{a-x} \right ]}

Unit of k=sec^{-1}

Other Forms of Rate Law

We know that the first-order equation is given as follows:

\mathrm{log_{10}A\: =\: log_{10}A_{o}\: -\: \frac{kt}{2.303}}

But there are other forms of rate law also available that we use for different purposes. These forms are mentioned below:

  • Use to solve numericals:

    \mathrm{log_{10}A\: =\: log_{10}A_{o}\: -\: \frac{kt}{2.303}}

    \mathrm{\Rightarrow log_{10}\left [ \frac{A_{o}}{A} \right]\: =\: \frac{kt}{2.303}}

    \mathrm{Thus, t\: =\: \frac{2.303}{k}\, log_{10}\left [ \frac{A_{o}}{A} \right]}
  • Exponential form:

    \mathrm{log_{e}A\: -\: log_{e}A_{o}\: -kt}
    \\\mathrm{\Rightarrow log\frac{A}{A_{o}}\: =\: -kt}\\\\\mathrm{\Rightarrow \frac{A}{A_{o}}\: =\: e^{-kt}}\\\\\mathrm{Thus,\: A\: =\: A_{o}e^{-kt}}
    This equation is also known as exponential form.
Half Life of First Order Reaction

The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t1/2.
For a zero order reaction, rate constant is given as:
\text { At } t=t_{1 / 2}, \quad[\mathrm{R}]=\frac{1}{2}[\mathrm{R}]_{0}
The rate constant at t1/2 becomes:

k=\frac{[\mathrm{R}]_{0}-1 / 2[\mathrm{R}]_{0}}{t_{1 / 2}}

t_{1 / 2}=\frac{[\mathrm{R}]_{0}}{2 \mathrm{k}}
It is clear that t1/2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.
For the first order reaction,

k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}

\text { at } t_{1 / 2} \quad[\mathrm{R}]=\frac{[\mathrm{R}]_{0}}{2}
So, the above equation becomes

k=\frac{2.303}{t_{1 / 2}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]_{0} / 2}

\text { or } \quad t_{1 / 2}=\frac{2.303}{k} \log 2

t_{1 / 2}=\frac{0.693}{k}


Graphs of First Order Kinetics

Study it with Videos

First Order Reaction
Other Forms of Rate Law
Half Life of First Order Reaction

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