Careers360 Logo
ask-icon
share
    Know About NEET First Order Reaction Topic, Questions And Solutions

    Know About NEET First Order Reaction Topic, Questions And Solutions

    Sumit SainiUpdated on 06 Jun 2022, 12:13 PM IST
    Know About NEET First Order Reaction Topic, Questions And Solutions
    Ace First Order Reaction For NEET(Image: Shutterstock)

    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.

    Live | Jul 18, 2026 | 10:25 AM IST

    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.

    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.

    Confused About College Admissions?

    Get expert advice on college selection, admission chances, and career path in a personalized counselling session.

    Book a Counselling Slot
    Select Date
    Pick a Slot

    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.

    JSS University Mysore Allied Sciences 2026

    NAAC A+ Accredited| Ranked #21 in University Category by NIRF | Applications open for multiple UG & PG Programs

    Aakash iACST – NEET Repeater

    Register for iACST. Get instant Scholarship on NEET Repeater Courses.

    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

    Aakash iACST – NEET Repeater Programs
    Register for NEET Repeater programs. Get instant scholarship with iACST
    Register Now

    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

    Emversity Allied Health Programs

    Get Job Ready in Healthcare | Employability-Focused Programs

    Virohan Allied & Healthcare Programs

    Allied & Healthcare programs | 20+ Partner Universities & Institutes | 98% placement record

    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.

    NEET Syllabus: Subjects & Chapters
    Select your preferred subject to view the chapters
    Articles
    |
    Upcoming Medicine Exams
    Ongoing Dates
    SVNIRTAR CET Result Date

    9 Jul'26 - 30 Dec'26 (Online)

    Ongoing Dates
    Admit Card Date

    15 Jul'26 - 24 Jul'26 (Online)

    Certifications By Top Providers
    Management of Medical Emergencies in Dental Practice
    Via Tagore Dental College and Hospital, Chennai
    Online M.Sc Psychology
    Via Centre for Distance and Online Education, Andhra University
    School Counseling
    Via Avinashilingam Institute for Home Science and Higher Education for Women, Coimbatore
    Digital Forensic
    Via Dr Harisingh Gour Vishwavidyalaya, Sagar
    Counseling Psychology
    Via Savitribai Phule Pune University, Pune
    Explore Top Universities Across Globe
    University College London, London
    Gower Street, London, WC1E 6BT
    University of Essex, Colchester
    Wivenhoe Park Colchester CO4 3SQ
    University of Nottingham, Nottingham
    University Park, Nottingham NG7 2RD
    University of Wisconsin, Madison
    329 Union South 1308 W. Dayton Street Madison, WI 53715-1149
    University of Alberta, Edmonton
    116 St. and 85 Ave., Edmonton, Alberta, Canada T6G 2R3
    Keele University, Newcastle
    Staffordshire, UK, ST5 5BG

    Questions related to NEET

    On Question asked by student community

    Have a question related to NEET ?

    You can watch recorded NEET counselling webinars from several trusted sources:

    Medical Counselling Committee (MCC) – The official MCC website publishes counselling schedules, notices, information bulletins, and updates. While it does not regularly host recorded webinars, it is the most reliable source for official counselling information.

    Hey there,

    With 343 marks in NEET UG and Bihar domicile, getting an MBBS seat at Katihar Medical College through the mop-up or stray vacancy rounds is possible but not guaranteed. Admission depends on your NEET rank, category, counselling round, and seat availability. You should participate in all Bihar UGMAC

    Hey there,

    With 115 marks in Re-NEET 2026, getting a BAMS seat through the management quota is unlikely, as you must first qualify NEET and also meet the eligibility criteria prescribed by the counselling authority. If your score is above the official qualifying cutoff for your category, you may have

    Hello Dear Student,
    With 526 marks in the NEET 2026 exam, your expected All India Rank (AIR) will be roughly in the range of 26,000 to 39,000. For the OBC category specifically, your state rank in Bihar will likely hover between 4,000 and 6,000.

    You can check, find and access

    Hello Dear Student,
    No, NEET is not compulsory for BPT admission in the 2026–27 academic year. The National Commission for Allied and Healthcare Professions (NCAHP) issued an official notification deferring the mandatory NEET requirement for Bachelor of Physiotherapy (BPT) and Bachelor of Occupational Therapy (BOT) courses until the 2027–28 academic