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First Order Reaction - Practice Questions & MCQ

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

Quick Facts

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

  • 95 Questions around this concept.

Solve by difficulty

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.

In a reaction, \mathrm{X_2 + 2Y_2 \rightarrow 2XY2}, the X disappears at

Identify the correct order of reaction for which initial concentration vs half-time plot is  - 

A reaction that is of the first order with respect to reactant \mathrm{A}  has a rate constant \mathrm{6 \mathrm{~min}^{-1}} . If we start with  \mathrm{[\mathrm{A}]=0.3 \mathrm{~mol}~ \mathrm{l}^{-1}}, when would \mathrm{[\mathrm{A}]}  reach the value \mathrm{0.03 \mathrm{~mol}~ \mathrm{l}^{-1}} ?

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A first order reaction has a specific reaction rate of  \mathrm{0.3 \times10^{-3} s^{-1}}. How much time will it take for 20 g of the reactant to reduce to 10 g? 


When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is 


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A radioactive element following first-order kinetics reduces to half in two years. How much time would it take for the radioactive element to decay such that only (1/4) of its original concentration remains?

Concepts Covered - 4

First Order Reaction

First Order Reactions

The rate of the reaction is proportional to the first power of the concentration of the reactant

Let us consider a chemical reaction which occurs as follows:

\mathrm{A\ \ \ \longrightarrow \ \ \ B}

We have,
\mathrm{rate (r)=K[A]^{1}}


\Rightarrow \mathrm{\frac{d[A]}{[A]}=-k dt}

Integrating both sides and putting the limits

\Rightarrow \mathrm{\int_{[A]_0}^{[A]_t}\frac{d[A]}{[A]}=-k \int _0^tdt}

\Rightarrow \mathrm{ln\left(\frac{[A]_t}{[A]_0} \right )=-kt}

Simplifying the above expression we have, 

\Rightarrow \mathrm{ln\left(\frac{[A]_0}{[A]_t} \right )=kt}

In case we are dealing in terms of a and x (where a is the initial concentration of A and x is the amount of A dissociated at any time t)

\Rightarrow \mathrm{k= \frac{1}{t}\ ln\left(\frac{[A]_0}{[A]_t} \right )=\frac{1}{t}\ ln\left(\frac{a}{a-x} \right )}

Graphical Representation of first order reaction



Units of \mathrm{k=time^{-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 numerical:

    \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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