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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.

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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

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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

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

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

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

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

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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? 

 

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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?

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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}}

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

\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:
k=\frac{[\mathrm{R}]_{0}-[\mathrm{R}]}{t}
\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

Study other Related Concepts

First 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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