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Ncert Solutions for Class 11 Chemistry Chapter 6

Ncert Solutions For Class 11 Chemistry Chapter 6 Download PDF

Class 11 chemistry NCERT solutions chapter 6 : Adda247 provides NCERT solutions for class 11 chemistry chapter 6. The NCERT Solutions provided here will enhance the concepts of the students, as well as suggest alternative methods to solve particular problems to the teachers.

These NCERT Solutions Class 11 chemistry chapter 6 are presented in a very simple language so that you can understand the basic of chemistry with ease. These NCERT Solutions class 11 Chemistry cover chapters 1 to 14 with all important questions and answers explained in a detailed way.

Students can download the Class 111 Chemistry NCERT Solutions, which they want to study with the comfort of their house.

The solutions available are in depth and simplest way. Thus will help the students beyond examination marks. This will help them develop a core understanding of the subject. Because this subject demands to understand rather than just memorizing solutions of Class 11 Chemistry. Here below we are providing you with the overview of all the chemistry Class 11 Chapters that are there in the NCERT textbook.

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Benefits of Solutions of NCERT class 11 Chemistry:

  • NCERT Solutions for Class 11 is helpful to solve questions from other reference books too.
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NCERT Solutions for Class 11 Chemistry Chapter – 6: Thermodynamics

Chemistry is the basic of things which we see around in our environment and is known as the “central science”. As chemistry is a mandatory subject for the students, it requires more focus from the exam perspective. The solutions provided here are equipped with all the basic details with questions which might appear in exam. Furthermore, these NCERT Solutions can also be downloaded in a PDF format.

The questions in NCERT Solution Chemistry Class 11 covered in this exercise assist students in gaining insight into the chapter, so that they can have in-depth knowledge about the topic and excel in their upcoming exams.

Thermodynamics, science of the relationship between heat, work, temperature, and energy. In broad terms, thermodynamics deals with the transfer of energy from one place to another and from one form to another.

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In particular, it describes how thermal energy is converted to and from other forms of energy and how it affects matter.

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Importance of thermodynamics

Thermodynamics is a very important branch of both physics and chemistry. It deals with the study of energy, the conversion of energy between different forms and the ability of energy to do work.

Use of thermodynamics in every-day life

Thermodynamics is used in everyday life all around us. One small example of thermodynamics in daily life is cooling down hot tea with ice cubes. … This heat transfer is the defined by the 2nd law of Thermodynamics when a system evolves towards thermodynamic equilibrium or the state of maximum entropy.

Basic Principles of Thermodynamics. A thermodynamical system is an arbitrarily but suitable chosen region of the space where certain phenomena are investigated. The system is enclosed by its surroundings. The system and its surroundings may be in equilibrium or may be interactions between them.


The first law of thermodynamics states that energy can either be created or destroyed, only altered in a form. In analyzing an open system using the first law of thermodynamics, the energy into the system is equal to the energy leaving the system.


The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.


One of the most important things we can do with heat is to use it to do work for us. A heat engine does exactly this—it makes use of the properties of thermodynamics to transform heat into work. Gasoline and diesel engines, jet engines, and steam turbines that generate electricity are all examples of heat engines.

Chemistry is a branch of Science which deals with molecules, atoms, states of matter, thermodynamics elements etc. Students have to pay attention to each of these concepts and learn them completely. Many students do not identify the crucial concepts in each chapter. So following the NCERT Solutions available at Adda247 will help you attain remarkable grades in the final exam. Furthermore, the NCERT Solutions provided on this page can be downloaded as a PDF on this page for free.


Subtopics included in NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

  1. Thermodynamics terms
  • The System and the Surroundings
  • Types of Thermodynamics
  • State of the System
  • Internal Energy as a State Function
  1. Applications
  • Work
  • Enthalpy, H
  1. Measurement of U and H Calorimetry
  2. Enthalpy Change and Reaction Enthalpy
  3. Enthalpies for Different Types of Reactions
  4. Spontaneity
  5. Gibbs Energy Change and Equilibrium
Ncert Solutions For Class 11 Chemistry Chapter 6_60.1
Ncert Solutions for Class 11 Chemistry Chapter 6

Important questions of NCERT Solutions of Chemistry Class 11 Chapter 6


Question :1 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are,  -890.3 kJ MIL) -1 -393.5 kJ Molly) -1, and -285.8 kJ mom) -1 respectively. Enthalpy of formation of CH4(g) will be

  1. -74.8 kJ mol)-1
  2. -52.27 kJ mol) -1
  • 8 kJ mol)-1
  1. 27 kJ mol)-1


  • -74.8 kJ mol-1

CH4(g) + 2O2(g) à CO2(g) + 2H2O(g)

H = -890.3 kJ mol-1

2.)         C(g) + O2(g) à CO2(g)

H = – 393.5 kJ mol-1

3.)          2H2(g) + O2(g) à 2H2O(g)

H = -285.8 kJ mol-1

C(s) + 2H2(g) à CH4(g)

fHCH4 = cHc + 2fHH2fHCO2

= [-393.5 + 2(-285.8) + (-890.3)] kJ mol-1

= -74.8 kJ mol-1


Question :2 A reaction, A + B à C + D + q is found to have a positive entropy change. The reaction will be

  1. Possible at high temperature
  2. Possible only at low temperature
  3. Not possible at any temperature
  4. Possible at any temperature


Possible at any temperature.

G should be –ve, for spontaneous reaction to occur

G = H +  TS

As per given in equation,

H is –ve

S is positive

Therefore, G is negative

So, the reaction will be possible at any temperature.


Question :3 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?


As per thermodynamics 1st law,

U = q + W(i);

U internal energy = heat

W = work done

W = -594 J (work done by system)

q = 801 J (+ve as heat is absorbed)


U = 801 + (-594)

U = 207 J


Question :4 The reaction of cyanide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, U was found to be -742.7 kJ mol)-1 at 298 K.  Calculate enthalpy change for the reaction at 298 K.

NH2CN(g) + 3/2O2(g) à N2(g) + CO2(g) + H2O(l)


H = is given by,

H = U + ngRT ……(1)

ng = change in number of moles

U = change in internal energy


T = 298K

U = -742.7 kJ mol-1

R = 8.314 x 10-3 kJ mol-1 K-1

Now, from (1)

H = (-742.7 kJ mol-1) + (0.5mol)(298K)(8.314 x 10-3kJmol-1K-1)

= -742.7 + 1.2

= -741.5 kJmol-1


Question :5 Calculate the number of kJ of heat necessary to raise the temperature of 60.0g aluminium from 35 degree C to 55 degree C.  Molar heat capacity of Al is 24 J mol)-1 K)-1.


Expression of heat,

Q = mCPT ………..(a)

T = change in temperature

c = molar heat capacity

From (a)

q = (60 / 27 mol)(24mol-1K-1)(20K)

q = 1066.67 J = 1.067 KJ.


Question :6 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 degree C to ice at -10.0 degree C. fusH = 6.03 kJ mol)-1 at 0 degree C.

Cp (H2O(l)) = 75.3 J mol-1 K-1

Cp (H2O(s)) = 36.8 J mol-1 K-1


Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.


(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.


(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at -10°C.


Total ΔH = Cp [H2OCI] ΔT  + ΔHfreezing + Cp [H2O(s)] ΔH

= (75.3 J mol-1 K-1) (0 – 10)K + (-6.03 × 103 J mol-1) + (36.8 J mol-1 K-1) (-10 – 0)K

= -753 J mol-1 – 6030 J mol-1 – 368 J mol-1

= -7151 J mol-1

= -7.151 kJ mol-1

Hence, the enthalpy change involved in the transformation is -7.151 kJ mol-1.


Question :7 Enthaply of formation of CO2 is -393.5 kJ mol)-1. Calculate the heart released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.


Formation of carbon dioxide from di-oxygen and carbon gas is given as:

C(s) + O2(g) à CO2(g): fH = -393.5 kJ mol-1

1 mole CO2 = 44g

Heat released during formation of 44 g CO2 = -393.5 kJ mol-1

Therefore, heat released during formation of 35.2 g of CO2 can be calculated as

= -393.5kJ mol-1 x 35.2 g / 44g

= -314.8 kJ mol-1


Question :8 Enthaplies of formation of CO(g), N2O(g), CO2(g) and N2O4(g) are -110, -393, 81 and 9.7 kJ mol-1 respectively. Find the value of H for the reaction :

N2O4(g) + 3CO(g) -> N2O(g) + 3CO2(g)


rH for any reaction is defined as the difference between fH value of product and fH value of reactants”

rH = fH(products) –   fH (reactants)

Now, for

N2O4(g) + 3CO(g) à N2O(g) + 3CO2(g)

rH = [fH+ (3fH(CO2)) – (fH(N2O4) + 3fH(CO))]

Now, substituting the given values in the above equation, we get:

rH = [{81kJ/mol + 3(-393) kJ / mol} – {9.7 kJ / mol + 3(-110) kJ / mol}]

rH = -777.7 kJ / mol.


Question :9 Given

N2(g) + 3H2(g) -> 2NH3(g);  rH = -92.4 kJ mol-1

What is the standard enthalpy of formation of NH3 gas?


“ Standard enthalpy of formation of a compound is the enthalpy that takes place during the formation of 1 mole of  a substance in its standard form, from its constituents elements in their standard atste”

Dividing the chemical equation given in the question by 2, we get

(0.5)N2(g) + (1.5)H2(g) à 2NH3(g)

Therefore, standard enthalpy for formation of ammonia gas

= 0.5 rH

= (0.5)(-92.4 kJ mol-1)

= -46.2 kJ / mol


Question :10 Calculate the standard enthaply of formation of CH3OH(l) from the following data:

CH3OH(l) + 3/2 O2(g) -> CO2(g) + 2H2O(l); H = -726kJ mol-1rH = -726 kJ mol-1

C(graphite) + O2(g) -> CO2(g);   cH = -393 kJ mol-1

H2(g) + ½ O2(g) -> H2O(l); fH = -286 kJ mol-1

C(s) + 2H2O(g) ½ O2(g) à CH3OH(l) ….(i)

CH3OH(l) can be obtained as follows,

fH[CH3OH(l)] = cH


= (-393 kJ / mol) + 2(-286 kJ / mol) – (-726 kJ / mol)

= (-393 – 572 + 726) kJ / mol

= -239 kJ / mol


Question :11 Calculate the enthalpy change for the process

CCl4(g) -> C(g) + 4Cl(g)

And calculate bond enthalpy of C-Cl in CCl4(g).

vaoH(CCl4) = 30.5 kj mol-1

fH(CCl4) = -135.5 kJ mol-1

aH(C) = 715.0 kJ mol-1 , where aHis the enthalpy of atomisation

aH(CCl2) = 242 kJ mol-1


The chemical equations implying to the given values of enthalpies are:

(i) CCl4(l)    →    CCL4(g)      ΔvapH0 = 30.5 kJ mol-1

(ii) C(s)    →   C(g)               ΔaH0 = 715.0 kJ mol-1

(iii) Cl2(g)  →   2Cl(g)          ΔaH0 = 242 kJ mol-1

(iv) C(g)  + 4Cl(g)  →  CCl4(g)  ΔfH = -135.5 kJ mol-1

Enthalpy change for the given process  C(g)  + 4Cl(g)  →  CCl4(g)   can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

ΔH = ΔaH0(C)  +  2ΔaH0 (Cl2) –  ΔvapH0 – ΔfH

= (715.0 kJ mol-1) + 2(242 kJ mol-1) – (30.5 kJ mol-1) – (-135.5 kJ mol-1)

∴ΔH = 1304 kJ mol-1

Bond enthalpy of C-Cl bond in CCl4(g) = 326 kJ mol-1


Frequently Asked questions on NCERT Solutions for Chemistry Class 11 Chapter 6

  1. What is thermodynamics according to NCERT Solutions for Class 11 Chemistry Chapter 6?

Ans. The branch of science which deals with the quantitative relationship between heat and other forms of energies is called thermodynamics. Thermal equilibrium If there is no flow of heat from one portion of the system to another, the system is said to be in thermal equilibrium.

The First law of thermodynamics is same as law of conservation of energy. … According to first law of thermodynamics:- The change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings.


  1. How NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics is helpful for exam preparation?

Ans. NCERT Solutions for Class 11 Chapter 6 Thermodynamics are created by the Adda247 expert faculty to help students in the preparation of their examinations. These expert faculty solve and provide the NCERT Solutions for class 11, which would help student to solve the problems comfortably. They give a detailed and stepwise explanations to the problems given in the exercise in the NCERT Solutions for Class 11. These solutions help students prepare for their upcoming board Exams by covering the whole Syllabus, in accordance with the NCERT guidelines.


  1. Is Adda247 providing answers for all questions present in NCERT Solutions for Class 11 Chemistry Chapter 6?

Ans. NCERT Solutions for Class 11 Chemistry Chapter 6 are useful for Students as it helps them to score well in the class exams. We, in our aim to help students, have devised detailed chapter-wise solutions for them to understand the concepts easily. We have followed the latest syllabus, while creating the NCERT Solutions and it is framed in accordance with the exam pattern of the CBSE Board. These solutions are designed by subject matter experts who have assembled model questions covering all the exercise questions from the textbook.

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