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Know About NEET First Order Reaction Topic, Questions And Solutions

Know About NEET First Order Reaction Topic, Questions And Solutions

Edited By Sumit Saini | Updated on Jun 06, 2022 12:13 PM IST | #NEET
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First Order Reaction

Chemical Kinetics is one of the important chapters of class 12 Chemistry. It generally deals with understanding the rate of chemical reactions. Rate of chemical reaction is the speed at which chemical reaction takes place. Reaction rate is influenced by many factors including temperature, pressure, catalyst etc.

Know About NEET First Order Reaction Topic, Questions And Solutions
Know About NEET First Order Reaction Topic, Questions And Solutions

Reactions are classified in various orders like zero order reaction, First order reaction, second order reaction among others on the basis of rate law followed by them.

A first-order reaction refers to the reaction in which the rate of reaction is proportional to the first power of concentration of the reactant at any point of time. Hence, higher the concentration of reactant, higher will be the rate of reaction. Rate of first order reaction decreases as it proceeds due to decrease in concentration of reactant.

Background wave

Hence in first order reaction, rate of reaction is directly proportional to the concentration of reactant as described in the equation below.

Reaction-

Reactant →Product

Rate =k[Reactant]

Below are some of the examples of First Order reaction

2N2O5 → 4NO2+O2

Rate = k[N2O5]

2H2O2 → 2H2O + O2

Rate = k[H2O2]

After analysing the last five years of NEET papers, it has been observed that, total 10 questions were asked from the Chemical Kinetics Chapter and out of those, four questions belong to First order reaction. This shows that First order reaction is the most important concept of Chemical Kinetics and most of the questions can be solved if Integral form of first order reaction is understood well. Questions asked from first order reactions are discussed in the article below.

Differential Rate Law for First Order reaction

Differential Rate Law for a First-Order Reaction reflects the change in concentration with respect to time which is shown by equation below:

Let us assume a first order reaction: A(Reactant)→B(Product)

Rate = -d[A]/dt = k[A]1 = k[A]

In the above equation,

  • ‘k’ represents the rate constant of the first-order reaction, the unit of the rate constant is s-1.

  • ‘[A]’ represents the concentration of the reactant.

  • d[A]/dt represents the change in the concentration of reactant ‘A’ with respect to time.

Integrated Rate Law for First-Order Reaction

Integrated Rate Law of first order reaction is used to calculate the value of rate of rate constant, concentration of reactant a given time, half life etc. Most of the numericals are solved by using Integrated rate law only. Differential rate law is used to obtain the Integral rate law of a First order reaction.

Differential rate law is: -d[A]/dt = k[A]

\Rightarrow d[A]/dt = -k[A]

Integrating both sides, we get

\begin{array}{l}\int_{[A]_0}^{[A]}\frac{d[A]}{[A]} = -\int_{t_0}^{t}kdt\end{array}

Where [A]o is concentration of reactant at time t0 s, and [A] is the concentration of reactant at time t. We get,

\Rightarrow \begin{array}{l}\int_{[A]_0}^{[A]}\frac{1}{[A]}d[A] = -\int_{t_0}^{t}kdt\end{array}

Considering t0 =0, and solving the equation, we get

\Rightarrow ln[A] – ln[A]0 = -kt

\Rightarrow ln[A] =ln[A]0 -kt

Raising each side to component e, we get

\Rightarrow \begin{array}{l}e^{ln[A]} = e^{(ln[A]_0-kt)} \end{array}

\Rightarrow \begin{array}{l}[A] = [A]_0 e^{-kt}\end{array}

Hence, \begin{array}{l}[A] = [A]_0 e^{-kt}\end{array} is the final expression in the integrated form for first order reaction. The above derivation is useful in solving many numerical problems.

Using the Integrated Rate law, Concentration v/s time Graph for First order reaction can be obtained which is shown below

Concentration-time


As the first-order reaction follows the equation ln[A] = ln[A]0 – kt which is similar to that of a straight line (y = mx + c) with slope -k. Hence, this equation can be plotted as-

1653634323010

From the above graph, it can be said that the graph for ln[A] v/s t for a first-order reaction is a straight line with slope -k.

Half-Life of a First-Order Reaction

The half-life of a chemical reaction refers to the time required by the reactant concentration to become half of its initial value. It is represented by t_\frac{1}{2}.

Hence, according to the first order reaction, \begin{array}{l}[A] = [A]_0 e^{-kt}\end{array}. Where [A] refers to the concentration of the reactant at any time ‘‘t’’ and [A]o represents the initial concentration of the reactant.

At half life, t = t1/2 , [A] =A0/2

Putting the value of [A] =A0/2 and t = t1/2 in the equation \begin{array}{l}[A] = [A]_0 e^{-kt}\end{array}, we get

\begin{array}{l}\frac{[A]_0}{2} = [A]_0 e^{-kt_{1/2}}\end{array}

\begin{array}{l}e^{-kt_{1/2}}\end{array}=\frac{1}{2}

Taking ln on both sides, we get,

\begin{array}{l}ln(\frac{1}{2}) = -kt_{1/2}\end{array}

\begin{array}{l}\ t_{1/2} = \frac{0.693}{k}\end{array}

Hence half life of the first order reaction is 0.693/k, where k is the rate constant.

Below discussed are some of the questions asked in the NEET exam from first order reaction.

Question 1. A first order reaction has a specific reaction rate of 10-2 sec-1. How much time will it take for 20 g of the reactant to reduce to 5 g? [NEET 2017]

  1. 238.6 sec

  2. 138.6 sec

  3. 346.5 sec

  4. 693.0 sec

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Ans. The rate of first order reaction is proportional to the first power of the concentration of the reaction.

Let us consider a reaction, R → P , wherein

R → P

At t=0 a 0

At t=t a-x x

rate[r]=K[R]^{1}

\frac{-d(a-x)}{dt}=K(a-x)

After solving this equation, we get

t= \frac{2.303}{k} log \frac{a}{a-x}

Now, putting the values

t= \frac{2.303}{10^{-2}} log \frac{20}{5}

t= 138.6 sec

Alternatively, this can be solved by the half life method. According to which after the first half life concentration becomes half and after the second half life, concentration becomes ¼ th of the initial concentration. Hence after two half life, the concentration becomes from 20g to 5g.

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

t_\frac{1}{2} \:\:of \:\:the \:\:reaction=\frac{0.693}{K}

\frac{0.693}{10^{-2}}sec=69.3\:sec

Now, 2t1/2= 69.3×2= 138.6 sec.

Hence option ‘’C’’ is correct.

Question 2. The correct difference between first and second-order reactions is that [NEET 2018]

  1. a first-order reaction can be catalysed; a second-order reaction cannot be catalysed

  2. the half-life of a first-order reaction does not depend on [A]o; the half-life of a second-order reaction does depend on [A]o

  3. the rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations

  4. the rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

Ans. For first Order reaction, rate is directly proportional to the first power of concentration.

For a reaction, A→B

Rate = -d[A]/dt = k[A] and half life is

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

Where unit of k=sec^{-1}.

Hence \left ( t_{\frac{1}{2}} \right ) 1^{st} order = Independent \; of \; concentration

For second order reaction, For a reaction, A→B

Rate = k[A]^2 and its half life is t_{1/2}= \frac{1}{k[A]_o}

\left ( t_{\frac{1}{2}} \right ) 2^{nd} order \alpha \frac{1}{Initial \;Concentration}

Hence, The half-life of a first-order reaction does not depend on [A]o; the half-life of a second-order reaction does depend on [A]o

Option B is correct.

Question 3. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99 per cent of the reaction is given by: [NEET 2019]

  1. t = 0.693/k

  2. t = 6.909/k

  3. t = 4.606/k

  4. t = 2.303/k

Ans. Let us consider a reaction, R → P , wherein

R → P

At t=0 a 0

At t=t a-x x

For the First Order reaction, Rate of reaction is proportional to the first power of the concentration of the reaction.

rate[r]=K[R]^{1}

\frac{-d(a-x)}{dt}=K(a-x)

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

Unit of k=sec^{-1}

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

t= \frac{2.303}{k}\log_{10}\frac{a}{a-x}

a = 100\%\;\;\;\;x=99\%

a - x= 1\%

t_{99.9%} = \frac{2.303}{k}\log_{10}\frac{100}{1} = \frac{2.303}{k}\log_{10}10^2

t_{99.9%} = \frac{2.303}{k}\times 2 = \frac{4.606}{k}

Hence option C is correct.

Question 4. The rate constant for a first order reaction is 4.606 \times 10^{-3}s^{-1} The time required to reduce 2 g of the reactant to 0.2g is: [NEET 2020]

  1. 1000s

  2. 100s

  3. 200s

  4. 500s

Ans. Given,

K=4.606 \times 10^{-3}s^{-1}

Hence the given reaction is of first order.

Thus, we have:

\mathrm{K\: t\: =\: ln\frac{A_{o}}{A}}

Hence it is from the option of keeping th

\mathrm{4.606\: x\: 10^{-3}\: t\: =\: 2.303log\left (\frac{2}{0.2} \right )}

Thus, t=0.5\times10^{3}\; \text {sec}

\text {t}=500\; \text {sec}

Therefore, Option(D) is correct.

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To get admission to a BAMS course without NEET, here are a few possible routes:

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Devanshi Goyal 01 April,2025
In the month may 2025, I will get my 12 result .. So I want government college for bsc in microbiology.so suggest some ..I had not given neet and cuet ..so without that suggest me ..

Dear

If you're looking to pursue a BSc in Microbiology from a government college without appearing for NEET or CUET, you can still explore many excellent options. Most government colleges admit students to BSc courses based on the marks obtained in your 12th-grade exams. Here are some of the top government colleges you can consider for BSc in Microbiology:

1. University of Delhi (DU)

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2. Banaras Hindu University (BHU)

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  • Admission Process: Admission is usually based on the entrance test, but if you’re looking for colleges with direct 12th marks-based admission, consider other options.
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3. University of Calcutta

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  • Admission Process: Based on merit (your 12th-grade marks).
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  • Research and Apply Early: Keep an eye on the official websites for admission dates and specific criteria.
  • Check Cutoffs: Colleges like DU and BHU release cutoff lists, which are based on your 12th-grade performance. Ensure you meet the required percentage.
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Devanshi Goyal 31 March,2025
Is samay neet ka phise kitna chal raha hai

Hello,

As of March 2025, the NEET application fees are:

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- SC/ST/PwD/Third Gender:1,000

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Fees are payable online via net banking, credit/debit cards, or UPI.

hope this helps you,

Thank you

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