# NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

**NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics-** Students appearing in the Class 12 board exam must check these NCERT Class 12 Chemistry Chapter 4 solutions. NCERT solutions for Class 12 Chemistry chapter 4 Chemical Kinetics covers all the questions from NCERT books for Class 12 Chemistry.

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In NCERT Class 12 Chemistry solutions chapter 4, there are questions and solutions of some important topics like average and instantaneous rate of a reaction, factors affecting the rate of reaction, integrated rate equations for zero and first-order reactions, etc. Read further to know all the Chemical Kinetics Class 12 exercise solutions.

**Also Read,**

**Check NCERT Class 12 solutions for other subjects.**

** Rate of reaction- ** It is defined as the rate of change in concentration of reactant or product. Unit of rate is .

Reactants, R Products, P

nA+mB pC+qD

## NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics -** **

** Answer ** :

We know that,

The average rate of reaction =

=

=

=

In seconds we need to divide it by 60. So,

=

= 6.67

** Question ** ** 4.2 ** In a reaction, P, the concentration of A decreases from to in 10 minutes. Calculate the rate during this interval?

** Answer ** :

According to the formula of an average rate

= ** (final concentration - initial conc.)/time interval **

=

=

=

=

** Question ** ** 4.3 ** For a reaction, ; the rate law is given by, . What is the order of the reaction?

** Answer ** :

Order of reaction = Sum of power of concentration of the reactant in the rate law expressions

So, here the power of A = 0.5

and power of B = 2

order of reaction = 2+0.5 =2.5

** Answer ** :

The order of a reaction means the sum of the power of concentration of the reactant in rate law expression.

So rate law expression for the second-order reaction is here R = rate

if the concentration is increased to 3 times means

new rate law expression = = = 9R

the rate of formation of Y becomes ** 9 times faster ** than before

** Question ** ** 4.5 ** A first order reaction has a rate constant . How long will of his reactant take to reduce to ?

** Answer ** :

Given data,

initial conc. = 5g

final conc. = 3g

rate const. for first-order =

We know that for the first-order reaction,

[log(5/3)= 0.2219]

= 444.38 sec (approx)

** Answer ** :

We know that t(half ) for the first-order reaction is

and we have given the value of half time

thus,

= 0.01155 /min

OR = 1.1925

** Alternative method **

we can also ** ** solve this problem by using the first-order reaction equation.

put

** Question ** ** 4.7 ** What will be the effect of temperature on rate constant ?

** Answer ** :

The rate constant of the reaction is nearly doubled on rising in 10-degree temperature.

Arrhenius equation depicts the relation between temperature and rates constant.

A= Arrhenius factor

Ea = Activation energy

R = gas constant

T = temperature

** Question ** ** 4.8 ** The rate of the chemical reaction doubles for an increase of in absolute temperature from Calculate .

** Answer ** :

Given data

(initial temperature) = 298K and (final temperature)= 308K

And we know that rate of reaction is nearly doubled when temperature rise 10-degree

So, and R = 8.314 J/mol/K

now,

On putting the value of given data we get,

Activation energy ( ) =

=52.9 KJ/mol(approx)

** Question ** ** 4.9 ** The activation energy for the reaction is at . Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

** Answer ** :

We have

Activation energy = 209.5KJ/mol

temperature= 581K

R = 8.314J/mol/K

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as

taking log both sides we get

= 18.832

x = antilog(18.832)

= 1.471

** NCERT Solutions for class 12 chemistry chapter 4 **

** Question 4.1(i) ** From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

** Answer ** :

Given pieces of information

Rate =

so the order of the reaction is 2

The dimension of k =

** Question 4.1(ii) ** From the rate expression for the following reactions determine their order of reaction and the dimensions of the rate constants.

** Answer ** :

Given rate =

therefore the order of the reaction is 2

Dimension of k =

** Question 4.1(iii) ** From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

** Answer ** :

Given

therefore the order of the reaction is 3/2

and the dimension of k

** Question 4.1(iv) ** From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

** Answer ** :

so the order of the reaction is 1

and the dimension of k =

** Question ** ** 4.2 ** For the reaction:

the rate = with . Calculate the initial rate of the reaction when . Calculate the rate of reaction after is reduced to .

** Answer ** :

The initial rate of reaction =

substitute the given values of [A], [B] and k,

rate =

=8

When [A] is reduced from 0.1 mol/L to 0.06 mol/L

So, conc. of A reacted = 0.1-0.06 = 0.04 mol/L

and conc. of B reacted = 1/2(0.04) = 0.02mol/L

conc. of B left = (0.2-0.02) = 0.18 mol/L

Now, the rate of the reaction is (R) =

=

** Question 4.3 ** The decomposition of on platinum surface is zero order reaction. What are the rates of production of and if ?

** Answer ** :

The decomposition of on the platinum surface reaction

therefore,

Rate =

For zero order reaction ** rate = k **

therefore,

So

and the rate of production of dihydrogen = 3 (2.5 )

= 7.5

** Answer ** :

Given that

So, the unit of rate is ** bar/min ** .( )

And thus the unit of k = unit of rate

** Question 4.5 ** Mention the factors that affect the rate of a chemical reaction.

** Answer ** :

The following factors that affect the rate of reaction-

- the concentration of reactants
- temperature, and
- presence of catalyst

** Question ** ** 4.6(i) ** A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is doubled

** Answer ** :

Let assume the concentration of reactant be x

So, rate of reaction,

Now, if the concentration of reactant is doubled then . So the rate of reaction would be

Hence we can say that the rate of reaction increased by 4 times.

** Question ** ** 4.6(ii) ** A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half ?

** Answer ** :

Let assume the concentration of reactant be x

So, rate of reaction, R =

Now, if the concentration of reactant is doubled then . So the rate of reaction would be

Hence we can say that the rate of reaction reduced to 1/4 times.

** Answer ** :

The rate constant is nearly double when there is a 10-degree rise in temperature in a chemical reaction.

effect of temperature on rate constant be represented quantitatively by Arrhenius equation,

where k is rate constant

A is Arrhenius factor

R is gas constant

T is temperature and

is the activation energy

** Question ** ** 4.8 ** In pseudo first order hydrolysis of ester in water, the following results were obtained:

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

** Answer ** :

The average rate of reaction between the time 30 s to 60 s is expressed as-

** Question ** ** 4.9(i) ** A reaction is first order in A and second order in B.

(i)Write the differential rate equation.

** Answer ** :

the reaction is first order in A and second order in B. it means the power of A is one and power of B is 2

The differential rate equation will be-

** Question ** ** 4.9(ii) ** A reaction is first order in A and second order in B.

(ii) How is the rate affected on increasing the concentration of B three times?

** Answer ** :

If the concentration of [B] is increased by 3 times, then

Therefore, the rate of reaction will increase 9 times.

** Question ** ** 4.9(iii) ** A reaction is first order in A and second order in B.

(iii) How is the rate affected when the concentrations of both A and B are doubled?

** Answer ** :

If the concentration of [A] and[B] is increased by 2 times, then

Therefore, the rate of reaction will increase 8 times.

What is the order of the reaction with respect to A and B?

** Answer ** :

we know that

rate law ( ) =

As per data

these are the equation 1, 2 and 3 respectively

Now, divide eq.1 by equation2, we get

from here we calculate that ** y = 0 **

Again, divide eq. 2 by Eq. 3, we get

Since y =0 also substitute the value of y

So,

=

=

taking log both side we get,

= 1.496

= approx 1.5

Hence the order of reaction w.r.t A is 1.5 and w.r.t B is 0(zero)

** Question ** ** 4.11 ** The following results have been obtained during the kinetic studies of the reaction:

2A + B C + D

Determine the rate law and the rate constant for the reaction .

** Answer ** :

Let assume the rate of reaction wrt A is and wrt B is . So, the rate of reaction is expressed as-

Rate =

According to given data,

these are the equation 1, 2, 3 and 4 respectively

Now, divide the equation(iv) by (i) we get,

from here we calculate that

Again, divide equation (iii) by (ii)

from here we can calculate the ** value of y is 2 **

Thus, the rate law is now,

So,

Hence the rate constant of the reaction is

Experiment | [A]/molL ^{-1} | [B]/molL^{-1} | Initial rate/ mol L^{-1} min^{-1} |

I | 0.1 | 0.1 | 2*10 ^{-2} |

II | - | 0.2 | 4*10 ^{-2} |

III | 0.4 | 0.4 | - |

IV | 0.2 | 2*10 ^{-2} |

** Answer ** :

The given reaction is first order wrt A and zero order in wrt B. So, the rate of reaction can be expressed as;

Rate = k[A]

from exp 1,

** k = 0.2 per min. **

from experiment 2nd,

** [A] ** =

from experiment 3rd,

from the experiment 4th,

from here ** [A] = 0.1 mol/L **

** Question ** ** 4.13 (1) ** Calculate the half-life of a first order reaction from their rate constants given below:

** Answer ** :

We know that,

half-life ( ) for first-order reaction =

=

** Question ** ** 4.13 (2) ** Calculate the half-life of a first order reaction from their rate constants given below:

** Answer ** :

the half-life for the first-order reaction is expressed as ;

= 0.693/2

= 0.35 min (approx)

** Question ** ** 4.13 (3) ** Calculate the half-life of a first order reaction from their rate constants given below:

** Answer ** :

The half-life for the first-order reaction is

= 0.693/4

= 0.173 year (approximately)

** Answer ** :

Given ,

half-life of radioactive decay = 5730 years

So,

per year

we know that, for first-order reaction,

= 1845 years (approximately)

Thus, the age of the sample is 1845 years

** Question ** ** 4.15 (1) ** The experimental data for decomposition of

in gas phase at 318K are given below:

Plot against t.

** Answer ** :

On increasing time, the concentration of gradually decreasing exponentially.

** Question ** ** 4.15 (2) ** The experimental data for decomposition of in gas phase at 318K are given below:

Find the half-life period for the reaction.

** Answer ** :

The half-life of the reaction is-

The time corresponding to the mol/ L = 81.5 mol /L is the half-life of the reaction. From the graph, the answer should be in the range of 1400 s to 1500 s.

** Question ** ** 4.15 (3) ** The experimental data for decomposition of in gas phase at 318K are given below:

** Answer ** :

0 | 1.63 | -1.79 |

400 | 1.36 | -1.87 |

800 | 1.14 | -1.94 |

1200 | 0.93 | -2.03 |

1600 | 0.78 | -2.11 |

2000 | 0.64 | -2.19 |

2400 | 0.53 | -2.28 |

2800 | 0.43 | -2.37 |

3200 | 0.35 | -2.46 |

** Question ** ** 4.15 (4) ** The experimental data for decomposition of in gas phase at 318K are given below:

** Answer ** :

Here, the reaction is in first order reaction because its log graph is linear.

Thus rate law can be expessed as

** Question ** ** 4.15 (5) ** The experimental data for decomposition of in gas phase at 318K are given below:

** Answer ** :

From the log graph,

the slope of the graph is =

= -k/2.303 ..(from log equation)

On comparing both the equation we get,

** Question ** ** 4.15 (6) ** The experimental data for decomposition of in gas phase at 318K are given below:

Calculate the half-life period from k and compare it with (ii).

** Answer ** :

The half life produce =

=

** Question ** ** 4.15(7) ** The rate constant for a first order reaction is . How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

** Answer ** :

We know that,

for first order reaction,

(nearly)

Hence the time required is

** Answer ** :

Given,

half life = 21.8 years

= 0.693/21.8

and,

by putting the value we get,

taking antilog on both sides,

[R] = antilog(-0.1071)

= 0.781

Thus 0.781 of will remain after given 10 years of time.

Again,

Thus 0.2278 of will remain after 60 years.

** Answer ** :

case 1-

for 99% complition,

CASE- II

for 90% complition,

Hence proved.

** Question ** ** 4.19 ** A first order reaction takes 40 min for 30% decomposition. Calculate

** Answer ** :

For the first-order reaction,

(30% already decomposed and remaining is 70%)

therefore half life = 0.693/k

=

= 77.7 (approx)

** Question 4.20 ** For the decomposition of azoisopropane to hexane and nitrogen at 543K, the following data are obtained.

Calculate the rate constant.

** Answer ** :

Decomposition is represented by equation-

After t time, the total pressure =

So,

thus,

for first order reaction,

now putting the values of pressures,

** when t =360sec **

** when t = 270sec **

So,

** Question ** ** 4.21 ** The following data were obtained during the first order thermal decomposition of at a constant volume.

Calculate the rate of the reaction when total pressure is 0.65 atm.

** Answer ** :

The thermal decomposition of is shown here;

After t time, the total pressure =

So,

thus,

for first order reaction,

now putting the values of pressures, ** when t = 100s **

when

= 0.65 - 0.5

= 0.15 atm

So,

= 0.5 - 0.15

= 0.35 atm

Thus, rate of reaction, when the total pressure is 0.65 atm

rate = k( )

=

= 7.8

** Question ** ** 4.22 ** The rate constant for the decomposition of N2O5 at various temperatures

is given below:

Draw a graph between ln k and 1/T and calculate the values of A and

. Predict the rate constant at 30° and 50°C.

** Answer ** :

From the above data,

T/ | 0 | 20 | 40 | 60 | 80 |

T/K | 273 | 293 | 313 | 333 | 353 |

( ) | 3.66 | 3.41 | 3.19 | 3.0 | 2.83 |

0.0787 | 1.70 | 25.7 | 178 | 2140 | |

-7.147 | -4.075 | -1.359 | -0.577 | 3.063 |

Slope of line =

According to Arrhenius equations,

Slope =

12.30 8.314

= 102.27

Again, ** When T = 30 +273 = 303 K ** and 1/T =0.0033K

** k = **

** When T = 50 + 273 = 323 K ** and 1/T = 3.1 K

** k = 0.607 per sec **

** Answer ** :

Given that,

k =

= 179.9 KJ/mol

T(temp) = 546K

According to Arrhenius equation,

taking log on both sides,

= (0.3835 - 5) + 17.2082

= 12.5917

Thus A = antilog (12.5917)

A = 3.9 per sec (approx)

** Answer ** :

Given that,

k =

t = 100 s

Here the unit of k is in per sec, it means it is a first-order reaction.

therefore,

Hence the concentration of rest test sample is 0.135 mol/L

** Answer ** :

For first order reaction,

given that half life = 3 hrs ( )

Therefore k = 0.693/half-life

= 0.231 per hour

Now,

= antilog (0.8024)

= 6.3445

(approx)

Therefore fraction of sample of sucrose remains after 8 hrs is 0.157

** Question ** ** 4.26 ** The decomposition of hydrocarbon follows the equation k=(4.51011s^{-1})e^{-28000K/T} . Calculate

** Answer ** :

The Arrhenius equation is given by

.................................(i)

given equation,

............................(ii)

by comparing equation (i) & (ii) we get,

A= 4.51011 per sec

Activation energy = 28000 (R = 8.314)

= 232.792 KJ/mol

** Question ** ** 4.27 ** The rate constant for the first order decomposition of is given by the following equation:

.

Calculate for this reaction and at what temperature will its half-period be 256 minutes?

** Answer ** :

The Arrhenius equation is given by

taking log on both sides,

....................(i)

given equation,

.....................(ii)

On comparing both equation we get,

activation energy

half life ( ) = 256 min

k = 0.693/256

With the help of equation (ii),

T =

= 669 (approx)

** Question ** ** 4.28 ** The decomposition of A into product has value of k as at 10°Cand energy of activation 60 kJ mol–1. At what temperature would k be ?

** Answer ** :

The decomposition of A into a product has a value of k as at 10°C and energy of activation 60 kJ mol–1.

K1 =

K2 =

= 60 kJ mol–1

K2 =

** Answer ** :

We know that,

for a first order reaction-

Case 1

At temp. = 298 K

= 0.1054/k

Case 2

At temp = 308 K

= 2.2877/k'

As per the question

K'/K = 2.7296

From Arrhenius equation,

= 76640.096 J /mol

=76.64 KJ/mol

k at 318 K

we have , T =318K

A=

Now

After putting the calue of given variable, we get

on takingantilog we get,

k = antilog(-1.9855)

= 1.034

** Answer ** :

From the Arrhenius equation,

...................................(i)

it is given that

T1= 293 K

T2 = 313 K

Putting all these values in equation (i) we get,

** Activation Energy = 52.86 KJ/mo ** l

This is the required activation energy

## Topics and Sub-topics of Chemical Kinetics Class 12 NCERT Chapter 4

- 4.1 Rate of a Chemical Reaction
- 4.2 Factors Influencing Rate of a Reaction
- 4.3 Integrated Rate Equations
- 4.4 Temperature Dependence of the Rate of a Reaction
- 4.5 Collision Theory of Chemical Reactions

**NCERT Chapter-wise Solutions for Class 12 Chemistry**

Chapter 1 | The Solid State |

Chapter 2 | Solutions |

Chapter 3 | Electrochemistry |

Chapter 4 | Chemical Kinetics |

Chapter 5 | Surface chemistry |

Chapter 6 | General Principles and Processes of isolation of elements |

Chapter 7 | The P-block elements |

Chapter 8 | The d and f block elements |

Chapter 9 | Coordination compounds |

Chapter 10 | Haloalkanes and Haloarenes |

Chapter 11 | Alcohols, Phenols, and Ethers |

Chapter 12 | Aldehydes, Ketones and Carboxylic Acids |

Chapter 13 | Amines |

Chapter 14 | Biomolecules |

Chapter 15 | Polymers |

Chapter 16 | Chemistry in Everyday life |

### More About Class 12 Chemistry Chapter 4 Chemical Kinetics NCERT solutions

A total of 5 marks of questions will be asked in the Class 12 CBSE board exam of Chemistry from this chapter.

In this chapter, there are 30 questions in the exercise and 9 questions that are related to topics studied.

To clear doubts of students, the Chemical Kinetics NCERT solutions are prepared in a comprehensive manner by subject experts.

This chapter is vital for both CBSE Board exam as well as for competitive exams like JEE Mains, VITEEE, BITSAT, etc. so, students must pay special attention to the concepts of this chapter.

The NCERT solutions provided here are completely free and you can also download them for offline use also if you want to prepare or any other subject or any other class.

By referring to the NCERT Solutions for Class 12 chemistry chapter 4 PDF download, students can understand all the important concepts and practice questions well enough before their examination.

**NCERT Solutions for Class 12 Subject wise**

- NCERT Solutions for Class 12 Biology
- NCERT solutions for class 12 Maths
- NCERT solutions for class 12 Chemistry
- NCERT Solutions for Class 12 Physics

### Benefits of NCERT solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics

First, the easy steps given in the NCERT Class 12 Chemistry solutions chapter 4 will help you to understand the chapter easily.

The revision will be so easy that you always remember the concepts and get very good marks in your class.

Homework will be easy now, all you need to do is check the detailed CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics and you are good to go.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get the answers with solutions that will help you score well in your exams.

**Also check NCERT Exemplar Class 12 Solutions**

- NCERT Exemplar Class 12th Maths Solutions
- NCERT Exemplar Class 12th Physics Solutions
- NCERT Exemplar Class 12th Chemistry Solutions
- NCERT Exemplar Class 12th Biology Solutions

### Also Check NCERT Books and NCERT Syllabus here:

- NCERT Books Class 12 Chemistry
- NCERT Syllabus Class 12 Chemistry
- NCERT Books Class 12
- NCERT Syllabus Class 12

## Frequently Asked Question (FAQs) - NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

**Question: **What is the weightage of NCERT class 12 Chemistry chapter 4 in JEE Mains?

**Answer: **

Weightage of NCERT class 12 Chemistry chapter 4 in JEE Mains is 4 marks.

**Question: **What is the weightage of NCERT class 12 Chemistry chapter 4 in NEET?

**Answer: **

The weightage of NCERT class 12 Chemistry chapter 4 in NEET is 3%. Solve more previous year NEET papers to get more problems

**Question: **What is the weightage of NCERT class 12 Chemistry chapter 4 in CBSE board exam?

**Answer: **

The weightage of NCERT class 12 Chemistry chapter 4 in CBSE board exam is 5 marks. To practice questions on the chapter refer to NCERT book exercise and NCERT exemplar problems.

**Question: **What are the important topics of this chapter?

**Answer: **

Important topics of this chemical kinetics are the rate of reaction, concept of collision theory, effect of temperature in activation energy, Arrhenius equation, concept of collision theory.

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## Questions related to CBSE Class 12th

### i couldnt clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

### i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

### if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

### I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

### I will have my boards in Feb 2023 (Class 12) and I still havent completed 1 chapter in any subject (Physics, chemistry and maths) , My 11th concepts are not that great but its okish, I wanted to know that If I do start studying from 15th of August Can I get above 85?

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.