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Important Questions for CBSE Class 11 Chemistry Chapter 6 - Thermodynamics

Last updated date: 16th May 2024
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CBSE Class 11 Chemistry Chapter-6 Important Questions - Free PDF Download

Important Questions for Chemistry Class 11 Chapter 6 are now available in PDF format on Vedantu. These questions and answers are based on the new NCERT curriculum and prepared by Vedantu's highly qualified teachers. The Class 11 Chemistry Chapter 6 Important Questions cover all the topics given in the chapter. Students can download and refer to the PDF of these Thermodynamics Class 11 Important Questions from Vedantu for free of cost. Thermodynamics Class 11 Important Questions will help them to understand the type of questions asked in the competitive exams from this chapter. 

The main goal of these critical questions is to assist students in prioritizing topics when studying for exams. These crucial questions were developed in accordance with the most recent CBSE guidelines. Students may also refer to other study materials for Class 11 Chemistry Chapter 6 on Vedantu for better exam preparation.

Download CBSE Class 11 Chemistry Important Questions 2024-25 PDF

Also, check CBSE Class 11 Chemistry Important Questions for other chapters:

CBSE Class 11 Chemistry Important Questions


Chapter No

Chapter Name


Chapter 1

Some Basic Concepts of Chemistry


Chapter 2

Structure of Atom


Chapter 3

Classification of Elements and Periodicity in Properties


Chapter 4

Chemical Bonding and Molecular Structure


Chapter 5

States of Matter


Chapter 6



Chapter 7



Chapter 8

Redox Reactions


Chapter 9



Chapter 10

The s-Block Elements


Chapter 11

The p-Block Elements


Chapter 12

Organic Chemistry - Some Basic Principles and Techniques


Chapter 13



Chapter 14

Environmental Chemistry

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Competitive Exams after 12th Science

Study Important Questions for class 11 chemistry Chapter 6 – Thermodynamics

1 Marks Question

1. Define a system

Ans: A system in thermodynamics is a part of the universe within a specified boundary.

2. Define surroundings

Ans: The remaining of the universe which might be in a position to exchange energy and matter with the system is termed its surroundings.

3. State the first law of thermodynamics

Ans: The first law of thermodynamics states that energy can neither be created nor be destroyed but can be transformed from one form to another.

4. What kind of system is the coffee held in a cup?

Ans: Coffee held in a cup is an open system because it can exchange matter (water vapour and energy (heat) with the surroundings.

5. Give an example of an isolated system

Ans: Coffee held in a thermos flask is an example of an isolated system because it can neither exchange energy nor matter with the surroundings.

6. Name the different types of the system.

Ans: There are three types of system 

  1. Open system

  2. Closed system

  3. Isolated system

7. What will happen to internal energy if work is done by the system?

Ans: There is a decrease in internal energy of the system if work is done by the system.

8. From a thermodynamic point of view, to which system the animals and plants belong?

Ans: Open system

9. How may the state of the thermodynamic system be defined?

Ans: The state of the thermodynamic system can be defined by specifying values of state variables like temperature, pressure and volume.

10. Define enthalpy

Ans:  It is defined as the total heat content of the system.

11. Give the mathematical expression of enthalpy.

Ans: Mathematically $H = U + PV$ where $U$ is internal energy

12. When is enthalpy change $\Delta H$- 

(i) positive (ii) negative.


  1. In an endothermic reaction it absorbs heat from the surroundings so enthalpy change in positive 

  2. In an exothermic reaction heat is evolved so enthalpy change is negative.

13. Give the expression for

 (i) isothermal irreversible change, 

and isothermal reversible change

Ans: For isothermal reversible change 

$Q=\text{ }\!\!~\!\!\text{ }-w$

$Q={{p}_{ext}}({{V}_{f}}-{{V}_{i}})\text{ }\!\!~\!\!\text{ }$

For isothermal reversible change 

$Q=\text{ }\!\!~\!\!\text{ }-w$

$Q=2.303nRT\log \left( \frac{{{V}_{f}}}{{{V}_{i}}} \right)\text{ }\!\!~\!\!\text{ }$

14. Define Heat capacity

Ans: Specific heat /specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin).

15. Define specific heat.

Ans: Specific heat /specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin).

16. Give the mathematical expression of heat capacity.

Ans: The mathematical expression of heat capacity is as follows

$q = mc\Delta T$


 $m = 1$

$q = C\Delta T$

17. Define reaction enthalpy.

Ans: The enthalpy change concerning a reaction is termed as reaction enthalpy.

18. Define standard enthalpy.

Ans: The standard enthalpy of reaction is defined as the enthalpy change for a reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.

19. The standard heat of formation of ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ is ${\text{824}}{\text{.2}}\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ . Calculate heat change for the reaction. $4{\text{Fe}}(s) + 3{{\text{O}}_{\text{2}}}(g) \to 2{\text{F}}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s)$ 


$\Rightarrow \Delta {{H}_{R}}=[2\times \Delta {{H}_{f}}^{o}(\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s))-[4\Delta {{H}_{f}}^{o}(\text{Fe}(s))+3\times \Delta {{H}_{f}}^{o}({{\text{O}}_{\text{2}}})]$ $\Rightarrow \Delta {{H}_{R}}=[2\times \Delta {{H}_{f}}^{o}(\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s))-[4\Delta {{H}_{f}}^{o}(\text{Fe(}s))+3\times \Delta {{H}_{f}}^{o}({{\text{O}}_{\text{2}}})]$$\Rightarrow \Delta {{H}_{R}}=[2\times \Delta {{H}_{f}}^{o}(\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}(s))-[4\Delta {{H}_{f}}^{o}(\text{Fe}(s))+3\times \Delta {{H}_{f}}^{o}({{\text{O}}_{\text{2}}})]$

20. Define spontaneous process.

Ans: A process which occurs without use of an external agent is termed as spontaneous process.

21. Define non-spontaneous processes.

Ans:  A process which occurs with the use of an external agent is termed as spontaneous process.

22. What is the sign of enthalpy of formation of a highly stable compound?

Ans: Negative

23. Predict the sign of ${{\Delta S}}$for the following reaction

${\text{CaC}}{{\text{O}}_{\text{3}}}(s) \to {\text{CaO}}(s) + {\text{C}}{{\text{O}}_{\text{2}}}(g)$

Ans: $\Delta S$ is positive

24. Two ideal gases under the same pressure and temperature are allowed to mix in an isolated system – what will be a sign of entropy change?

Ans: Entropy change is positive since degree of disorderness decreases on mixing.

2 Marks Question

1. Change in internal energy is a state function while work is not, why?

Ans: In a process, the change in internal energy depends upon the initial and final state of the system. Therefore, it is a state function. But work is dependent on the path followed. Therefore, it is not a state function.

2. With the help of first law of thermodynamics and ${\text{H = U + pV}}$ prove ${{\Delta H = }}{{\text{q}}_{\text{p}}}$

Ans: Enthalpy is defined as follows


$\Delta H=\Delta (U+pV)$

$\Delta H=\Delta U+\Delta (pV)$

$\Delta H=\Delta U+p\Delta V+V\Delta p$……(i)

From first law of thermodynamics,

$ \Delta U=q+w$ 

 $ =q-p\Delta V$

From equation (i) and (ii), we get

$\Delta H=q-p\Delta V+p\Delta V+V\Delta p$

$\Delta H=q+V\Delta p$

At constant pressure,

$\Delta p=0$

$V\Delta p=0$

$\Delta H={{q}_{p}}$ at constant pressure

Therefore, $\Delta H={{q}_{p}}$

3. Why is the difference between $\Delta H$ and $\Delta U$ not significant for solids or liquids?

Ans: The difference between $\Delta H$ and $\Delta U$ is not significant for solids or liquids because systems made up entirely of solids and/or liquids do not experience significant volume changes when heated, the difference between and is usually insignificant.

4. What is an extensive and intensive property?

Ans: Extensive property is defined as the property which depends on the quantity or size of the matter present in the system.

Intensive property is defined as the property which depends on the quantity or size of matter present in the system.

5. Show that for an ideal gas ${C_P} - {C_V} = R$

Ans: At constant pressure, when gas is heated, heat is required for increase in the temperature of gas and for doing mechanical work against expansion

At constant volume, heat capacity is written as ${C_V}$ and at constant pressure it is written as ${C_P}$

At constant volume ${q_v} = {C_V}\Delta T$ which is equal to $\Delta U$

At constant pressure ${q_P} = {C_P}\Delta T$ which is equal to $\Delta H$

For one mole ideal gas $\Delta H = \Delta U + \Delta (PV)$

$\Delta H = \Delta U + \Delta (RT)$

$\Delta H = \Delta U + R\Delta T$

$\Delta H = \Delta U + R\Delta T$

On substituting values of ${{\Delta H,\Delta U}}$ , we have

${C_P}\Delta T = {C_V}\Delta T + R\Delta T$


 ${{C}_{P}}-{{C}_{V}}=R $

6. Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to $\dfrac{3}{2}R$.

Ans: For an ideal gas, the average kinetic energy per mole of the gas at any temperature is given by ${E_k} = \dfrac{3}{2}RT$

Hence increase in average kinetic energy of gas for ${{\text{1}}^{\text{o}}}{\text{C}}$rise in temperature is

$\Delta \overrightarrow {{E_k}}  = \dfrac{3}{2}R(T + 1) - \dfrac{3}{2}RT$

By definition,$\overrightarrow {{E_K}} $is the molar heat capacity of gas at constant volume ${C_V}$

$\therefore {C_V} = \dfrac{3}{2}R$

7. A $1.25$ g sample of octane (${C_{18}}{H_{18}}$) is burnt in excess of oxygen in a bomb calorimeter. The temperature of the calorimeter rises from $294.05$ to $300.78$${\text{K}}$. If heat capacity of the calorimeter is $8.93\;{\text{KJ}}\;{{\text{K}}^{{\text{ - 1}}}}$. find the heat transferred to calorimeter.

Ans: Mass of octane,

$M = 1.250\;{\text{g}}$

$M = 0.00125$

Heat capacity, $c = 8.93\;{\text{KJ}}\;{{\text{K}}^{{\text{ - 1}}}}$

Increase in temperature,$\Delta T = 300.78 - 294.05$

$\Delta T = 6.73\;K$

So, the heat transferred to calorimeter is

$mc\Delta T = 0.00125 \times 8.93 \times 6.73$

$ = 0.075\;{\text{KJ}}$

8. Calculate the heat of combustion of ethylene (gas) to from ${\text{C}}{{\text{O}}_{\text{2}}}$ (gas) and ${{\text{H}}_{\text{2}}}{\text{O}}$ (gas) at ${\text{298}}\;{\text{K}}$ and 1 atmospheric pressure. The heats of formation of ${\text{C}}{{\text{O}}_{\text{2}}}$, ${{\text{H}}_{\text{2}}}{\text{O}}$ and${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}$  are $ - 393.7$,$ - 241.8$ , $ + 52.3$${\text{kJ}}$ per mole respectively.

Ans: The combustion reaction of ethylene gas is as follows

${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\, + \,\,3{{\text{O}}_{\text{2}}} \to 2{\text{C}}{{\text{O}}_{\text{2}}}(g)\,\, + \,\,2{{\text{H}}_{\text{2}}}{\text{O}}(g)$

$\Delta {H_f}({\text{C}}{{\text{O}}_{\text{2}}}) =  - 393.7\;{\text{KJ}}$

$\Delta {H_f}({{\text{H}}_{\text{2}}}{\text{O}}) =  - 241.8\;{\text{KJ}}$

$\Delta {H_f}({{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}) =  + 52.7\;{\text{KJ}}$

$\Delta {H_R} = \sum {\Delta {H_f}^o(p) - \sum {\Delta {H_f}^o(r)} } $

$\Delta {H_R} = [2 \times \Delta {H_f}^o({\text{C}}{{\text{O}}_{\text{2}}}) + 2 \times \Delta {H_f}^o({{\text{H}}_{\text{2}}}{\text{O}})] - [\Delta {H_f}^o({{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}) + 3 \times \Delta {H_f}^o({{\text{O}}_{\text{2}}})]$

$\Delta {H_R} = [2 \times  - 393.7 + 2 \times  - 241.8] - [523 + 0]$

Enthalpy of formation is zero for elementary change

$\Delta {H_R} =  - 1323.3\;{\text{KJ}}$

9. Give two examples of reactions which are driven by enthalpy change.

Ans: The examples of reactions which are driven by enthalpy change is as follows

For a highly exothermic process its enthalpy change is negative and large value but entropy change is negative is driven by enthalpy change

  1. ${{\text{H}}_{\text{2}}}(g) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}(g) \to {{\text{H}}_{\text{2}}}{\text{O}}(l)$

$\Delta {H_f}^o =  - 285.8\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

  1. ${{\text{N}}_{\text{2}}}(g) + 3{{\text{H}}_{\text{2}}}(g) \to 2{\text{N}}{{\text{H}}_{\text{3}}}(g)$

$\Delta H_f^o =  - 92\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

10. Will the heat released in the following two reactions be equal? Give reasons in support of your answer

  1. ${{\text{H}}_{\text{2}}}(g) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}(g) \to {{\text{H}}_{\text{2}}}{\text{O}}(g)$

  2. ${{\text{H}}_{\text{2}}}(g) + \dfrac{1}{2}{{\text{O}}_{\text{2}}}(g) \to {{\text{H}}_{\text{2}}}{\text{O}}(l)$

Ans: The heats released in the two reactions are not equal. The heat released in a reaction depends on the reactants, products and the physical states.

In reaction:

  1. Water is produced in the gaseous state whereas in 

  2. It is in liquid state.

Also, when water vapors condensed to form water latent heat of vaporization is released. Therefore, more heat is released in reaction (ii).

11. What is the relation between the enthalpy of reaction and bond enthalpy?

Ans: In a chemical reaction the breaking of bonds and formation of new bonds in products takes place. The heat of reaction is dependent on the values needed to break the bond formation. Thus

Heat of reaction = Heat required for breaking of bonds in reactants$ - $Heat required for breaking of bonds in products 

  1. $\Delta {H^o} = $ Bond energy is required to break the bonds - Bond energy required to form the bonds = Bond energy of reactants – Bond energy of products.

12. The reaction ${\text{C}}(graphite) + {{\text{O}}_{\text{2}}}(g) \to {\text{C}}{{\text{O}}_{\text{2}}}(g) + 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ represents the formation of ${\text{C}}{{\text{O}}_{\text{2}}}$ and also combustion of carbon. Write the $\Delta {H^o}$ values of the two processes.

Ans: The standard enthalpy of formation of ${\text{C}}{{\text{O}}_{\text{2}}}$ is $ - 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}$

$\Delta H_f^o({\text{C}}{{\text{O}}_{\text{2}}}(g)) =  - 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

The standard enthalpy of formation of combustion of carbon is ${\text{ - 393}}{\text{.5}}\;{\text{kJ}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

$\Delta {H_{comb}}({\text{C}}{{\text{O}}_{\text{2}}}(g)) = \; - 393.5\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

13. Explain how is enthalpy related to spontaneity of a reaction?

Ans: Most of the exothermic reactions are spontaneous due to an increase in energy.

Burning a substance is a spontaneous process.

${\text{C}}(s) + {{\text{O}}_{\text{2}}}(g) \to {\text{C}}{{\text{O}}_{\text{2}}}(g)$, $\Delta H = \; - 394\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$

Many spontaneous reactions start with the absorption of heat. Conversion of water into water vapour is an endothermic spontaneous reaction. Hence the change in enthalpy is not the only criteria for deciding the spontaneity of the reaction.

14. The $\Delta H$ and $\Delta S$ for $2{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}(s) \to 4{\text{Ag}}(s) + {{\text{O}}_{\text{2}}}(g)$ are given $ + \,61.17\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$and $ + \,132\;{\text{J}}{{\text{k}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ respectively. Above what temperature will the reaction be spontaneous?

Ans: From the above question we have $\Delta H$and $\Delta S$ for $2{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}(s) \to 4{\text{Ag}}(s) + {{\text{O}}_{\text{2}}}(g)$ are given as $ + \,61.17\;{\text{kJ}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ and $ + \,132\;{\text{J}}{{\text{k}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ respectively.

As per the Gibbs Helmholtz equation, $\Delta G = \Delta H - T\Delta S$

Here, $\Delta G$ is the change in Gibbs free energy, $\Delta H$is the change in the enthalpy, $T$ is the absolute temperature.

For the reaction $2{\text{A}}{{\text{g}}_{\text{2}}}{\text{O}}(s) \to 4{\text{Ag}}(s) + {{\text{O}}_{\text{2}}}(g)$

It is spontaneous when $\Delta G$is negative.

Here $\Delta H,\Delta S$is positive. So $\Delta G$is negative when $\Delta H < T\Delta S$

$T > \dfrac{{\Delta H}}{{\Delta S}}$

$T = \dfrac{{61170\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}}}{{132\;{\text{J}}{{\text{k}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}}}$

$T = 463.4K$

Therefore, the process is spontaneous above a temperature of ${\text{T}} = {\text{463}}{\text{.4K}}$.

3 Marks Question

1. Give the relationship between ${{\Delta H}}$ and ${{\Delta U}}$ for gases.

Ans: For an appreciable volume change,

Let ${V_A}$ represent the volume of gaseous reactants

Let ${V_B}$ represent the volume of gaseous products

Let ${n_A}$ be the moles of reactant

Let ${n_B}$ be the moles of product

At constant pressure and temperature

$p{V_A} = {n_A}RT$

$p{V_B} = {n_B}RT$

$p{V_B} - p{V_A} = ({n_B} - {n_A})RT$

$p\Delta V = \Delta {n_g}RT$

$\Delta {n_g} = {n_B} - {n_A}$

Now substituting the value of $p\Delta V$ , we get

$\Delta H = \Delta U + \Delta {n_g}RT$

Heat change at constant pressure,$\Delta H = {q_p}$

Heat change at constant temperature, $\Delta H = {q_V}$

For gaseous system

${q_P} = {q_V} + \Delta {n_g}RT$

2. It has been found that $221.4\;{\text{J}}$ is needed to heat $30\;{\text{g}}$ of ethanol from${\text{1}}{{\text{5}}^{\text{o}}}{\text{C}}$  to ${\text{1}}{{\text{8}}^{\text{o}}}{\text{C}}$. calculate (a) specific heat capacity, and (b) molar heat capacity of ethanol.


  1. From the question we have $221.4\;{\text{J}}$of energy is needed to heat $30\;{\text{g}}$ of ethanol from${\text{1}}{{\text{5}}^{\text{o}}}{\text{C}}$  to ${\text{1}}{{\text{8}}^{\text{o}}}{\text{C}}$.

Since, specific heat capacity is given by the formula,

$c = \dfrac{q}{{m\Delta T}}$

$q = 221.4\;{\text{J}}$

$m = 30\,{\text{g}}$

$C = \dfrac{{221.1{\text{J}}}}{{30{\text{g}}({{18}^ \circ }{\text{C}} - {{15}^ \circ }{\text{C}})}}$

$C = \dfrac{{221.4}}{{30 \times 3}}$

$C = 2.46\;{\text{J}}{{\text{g}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{c}}^{{\text{ - 1}}}}$

As ${{\text{1}}^{\text{o}}}{\text{C}}$is equal to ${\text{1K}}$ ,therefore specific heat capacity of ethanol is $C = 2.46\;{\text{J}}{{\text{g}}^{ - 1}}{\,^ \circ }{{\text{c}}^{ - 1}}$

  1. Molar heat capacity is product of specific heat and molar mass

${C_m} = 2.46 \times 46$

${C_m} = 113.2{\text{J}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{C}}^{{\text{ - 1}}}}$

Therefore, molar heat capacity of ethanol is ${C_m} = 113.2\,{\text{J}}\;{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{C}}^{{\text{ - 1}}}}$

The Following Topics Are Covered in Class 11 Chemistry Chapter 6 Important Questions

  • Thermodynamics state

  • Applications 

  • Measurement of ΔU & ΔH: Calorimetry

  • Enthalpy change, ΔH of a reaction

  • Enthalpies for different types of reactions

  • Spontaneity 

  • Gibbs energy change and Equilibrium

After going through all these topics, student will be able to understand the system (close, open & isolated system) and surroundings, internal energy, work, heat, the first law of thermodynamics, state function, enthalpy change, Hess’s law, the difference between extensive and intensive properties, spontaneous and nonspontaneous processes and Gibbs free energy. On Vedantu, there are some essential Class 11 Chemistry Chapter 6 Extra Questions that clarify the concepts of this chapter in an applicational approach.


In exams like IIT or NEET, thermodynamics is one of the most important topics because it has a weightage of around 3%. Most students find it difficult to understand the concepts of thermodynamics. Students can learn Class 11 Chemistry Chapter 6 Important Questions on Vedantu online as well as offline. Vedantu is one of the leading online learning websites, that enables students to learn the concepts with a better understanding with the help of free PDFs. Our team of experts has developed all of the study materials in accordance with the existing CBSE board syllabus. Students are advised to practice more and more questions and answers to excel in their examinations.

Important Related Links for CBSE Class 11 Chemistry


FAQs on Important Questions for CBSE Class 11 Chemistry Chapter 6 - Thermodynamics

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

The NCERT solutions at Vedantu are devised by experts to help the student answer any question that is asked in the examination without having any second thoughts. These solutions have numerous exercises pertaining to the topic that will make the contents easier for the student to understand. All these exercises have with them solved solutions so that all the doubts and confusions of the students are cleared. With the practice of these solutions, the student will definitely be able to score high marks in the examination. 

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

Thermodynamics is the branch of chemistry, the study of which deals with the relations between heat, work, temperature and energy. The laws under the field of thermodynamics describe the process of the change of energy in a system and whether this change has any useful effect on its surroundings. To know more students can download a free PDF of Important questions from the vedantu website (vedantu,com).

3. What are the main topics in thermodynamics?

The main topics that this chapter includes are;

  • Introduction to the concept of thermodynamics

  • Thermal equilibrium

  • Zeroth law of thermodynamics

  • Heat, internal energy and work

  • The first law of thermodynamics

  • Specific heat capacity

  • Thermodynamic state variables and equation of state

  • Thermodynamic processes

  • Heat engines

  • Refrigerators and heat pumps

  • The second law of thermodynamics

  • Reversible and irreversible processes

  • Carnot engine

To do well in the exams, it is avoidable that students make their own revision notes and refer to the NCERT solutions. With adequate practise the student will be able to retain all the important topics. All the study material for all topics can be accessed from the vedantu app.

4. What is the importance of thermodynamics?

Thermodynamics is a very important branch of study concerning both physics and chemistry, as it deals with the study of energy and the conversion of this energy along with the different forms and abilities of this energy to do work. In our day to day life, it helps in the understanding of the heating and cooling systems in the home and building. It also helps in the study of the engines used in the vehicle. The laws of thermodynamics put forward the process in which the energy gets converted into heat, this heat is then transferred and gets converted to perform useful work. For a more detailed explanation, visit Vedantu website (

5. How many questions are asked from thermodynamics in NEET?

There are a total of 12 marks that the topic of Heat and Thermodynamics entails in the question paper set for NEET.  These 12 marks are extremely important for the aspirants to get into the university or college of their choice. Thus, it becomes evident that a thorough reading of this chapter is not just important for scoring good marks in the class 11 examination, but it also goes on to prove its worth in other competitive exams that the student might sit for in the future.