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    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.

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    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.

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    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.

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    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

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    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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