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

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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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.

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.

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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

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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Hello,

Eligibility for Assam NEET state quota depends on specific criteria set by the state. Based on your situation:

  • Domicile Requirement: You must fulfill Assam’s domicile/residency criteria. Typically, a minimum of 20 years' parental residency might help, but rules vary.
  • Schooling: Completing schooling till Class 12 in Assam may strengthen your case.
  • Parental Residency: If your parents have proof of living in Assam for 20 years, it may support your application.
  • No Property Ownership: Not owning property in Assam doesn't disqualify you.

Check the latest Assam NEET guidelines for detailed eligibility criteria and required documents.

Hope it helps !

Yes, if your parents have been living in Assam for the past 20 years, you are likely eligible to claim the state quota in NEET, as this generally qualifies you for a domicile certificate in Assam, which is required to access the state quota seats in medical colleges there.


Domicile certificate requirements


You will need to obtain a domicile certificate from the Assam government to avail of the state quota.


Eligibility criteria:


Check the specific eligibility criteria set by the Assam government for claiming domicile, which may include details about your parents' residency duration and your own educational background within the state.


By claiming the state quota, you will have access to a larger pool of seats within Assam medical colleges compared to the All India Quota.


All the best for your future endeavours..

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I'm so sorry to hear that you're feeling depressed and lonely. It's completely normal to feel overwhelmed, especially during this critical phase of your academic journey.


Firstly, please know that you're not alone in this feeling. Many students, especially those preparing for competitive exams like NEET, experience similar emotions.


Here are some suggestions that might help you cope with these feelings:


Reach Out to Someone

1. *Talk to a family member*: Share your feelings with a parent, sibling, or relative. They might be able to offer valuable advice or just listen to you.

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3. *Mentor or teacher*: Talk to a trusted teacher or mentor who can offer guidance and encouragement.


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1. *Join a study group*: Collaborate with classmates or online groups to study for NEET. This can help you connect with others who share similar goals.

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1. *Exercise regularly*: Physical activity can help reduce stress and anxiety. Engage in sports, yoga, or any exercise that you enjoy.

2. *Mindfulness and meditation*: Practice mindfulness techniques, such as meditation or deep breathing, to calm your mind.

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1. *Consult a counselor*: If your feelings of depression and loneliness persist, consider seeking help from a counselor.They can provide you with personalized guidance and support.


Remember, it's essential to prioritize your mental health during this challenging time. Don't hesitate to reach out for help when you need it.


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Here are some resources and suggestions to help your friend find concise and crisp notes for NEET preparation:

Free NEET Notes (PDFs and Study Materials)

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    • Offers free videos and downloadable notes on Biology, Physics, and Chemistry.
  2. AIIMS/NEET Preparation Websites

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    • Websites like Studyrankers (https://www.studyrankers.com) and Exam Fear (https://www.examfear.com) provide notes for NCERT chapters and concepts essential for NEET.
  6. NCERT Summary Notes

    • Download NCERT summary PDFs for all subjects. These cover most NEET-relevant topics in a concise format. Available on myCBSEguide (https://mycbseguide.com) .
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Encourage your friend to actively use these materials while maintaining a consistent study schedule!


Sure! Here's a concise list of NEET marks ranges for different courses:

  1. MBBS : 650-720
  2. BDS : 600-680
  3. BAMS : 450-600
  4. BHMS : 400-550
  5. BVSc & AH : 450-580
  6. B.Sc Nursing : 350-500
  7. BPT : 300-450
  8. B.Sc Allied Health Sciences : 250-400
  9. BUMS : 350-500
  10. Private MBBS Colleges : 350-500
  11. Other Diploma Courses : 200-350

Let me know if you need more information on any specific course!

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Column II ( Their location)

Parotids

I

Below tongue

Sub-maxillary / sub-mandibular

Ii

Lower jaw

Sub-linguals

Iii

Cheek

Option: 1

a(i), b(ii) , c(iii)

 


Option: 2

a(ii), b(i), c(iii)

 


Option: 3

a(i), b(iii), c(ii)


Option: 4

a(iii), b(ii), c(i)


Ethyl \; ester \xrightarrow[(excess)]{CH_{3}MgBr} P

the product 'P' will be ,

Option: 1


Option: 2


Option: 3

\left ( C_{2}H_{5} \right )_{3} - C- OH


Option: 4


 

    

           

 Valve name                            

             

Function

    I   Aortic valve     A

Prevents blood from going backward from the pulmonary artery to the right ventricle.

    II   Mitral valve     B

 Prevent blood from flowing backward from the right ventricle to the right atrium.

    III   Pulmonic valve     C

 Prevents backward flow from the aorta into the left ventricle.

    IV   Tricuspid valve     D

 Prevent backward flow from the left ventricle to the left atrium.

 

Option: 1

I – A , II – B, III – C, IV – D


Option: 2

 I – B , II – C , III – A , IV – D


Option: 3

 I – C , II – D , III – A , IV – B


Option: 4

 I – D , II – A , III – B , IV – C 

 

 


Column A Column B
A

a) Organisation of cellular contents and further cell growth.  

B

b) Leads to formation of two daughter cells.

C

c) Cell grows physically and increase volume proteins,organells.

D

d)  synthesis and replication of DNA.

Match the correct option as per the process shown in the diagram. 

 

 

 

Option: 1

1-b,2-a,3-d,4-c
 


Option: 2

1-c,2-b,3-a,4-d


Option: 3

1-a,2-d,3-c,4-b

 


Option: 4

1-c,2-d,3-a,4-b


0.014 Kg of N2 gas at 27 0C is kept in a closed vessel. How much heat is required to double the rms speed of the N2 molecules?

Option: 1

3000 cal


Option: 2

2250 cal


Option: 3

2500 cal


Option: 4

3500 cal


0.16 g of dibasic acid required 25 ml of decinormal NaOH solution for complete neutralisation. The modecular weight of the acid will be

Option: 1

32


Option: 2

64


Option: 3

128


Option: 4

256


0.5 F of electricity is passed through 500 mL of copper sulphate solution. The amount of copper (in g) which can be deposited will be:

Option: 1

31.75


Option: 2

15.8


Option: 3

47.4


Option: 4

63.5


0.5 g of an organic substance was kjeldahlised and the ammonia released was neutralised by 100 ml 0.1 M HCl. Percentage of nitrogen in the compound is

Option: 1

14


Option: 2

42


Option: 3

28


Option: 4

72


0xone is

Option: 1

\mathrm{KO}_{2}


Option: 2

\mathrm{Na}_{2} \mathrm{O}_{2}


Option: 3

\mathrm{Li}_{2} \mathrm{O}


Option: 4

\mathrm{CaO}


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(2) Diacetylmorphine

(3) Possessing a white color

(4) Devoid of any odor

(5) Crystal compound with a bitter taste

(6) Obtained by extracting from the latex of the poppy plant

The above statements/information are correct for:

Option: 1

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Option: 2

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Option: 4

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