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Do you know why heat is absorbed in some chemical reactions while others release heat, what makes hot tea cool down when left on the table and why ice melts faster in cold water as compared to hot water? The answer to all these questions lies in Thermodynamics, a branch of Chemistry that deals with energy changes during Physical and Chemical processes. NCERT Solutions for Class 11 Chemistry Chapter 5 Thermodynamics helps students to develop strong understanding of concepts like internal energy, enthalpy, entropy, spontaneity, and the laws of Thermodynamics.
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NCERT Class 11 Chemistry Solutions of Chapter 5 Thermodynamics are designed by our subject experts in a systematic way to help you understand complex topics. These NCERT solutions include step by step answers , HOTS questions and approaches to solve questions to strengthen conceptual clarity and prepare you for exams. NCERT Solutions for Class 11 serves as an important resource for mastering Chapter 5 Thermodynamics and helps you to enhance performance in board exams as well as in the competitive exams like JEE Advanced, NEET, JEE Mains, etc.
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To get NCERT Solutions Class 11 Chemistry Thermodynamics PDF, click below on the download PDF icon. In this PDF, you will get detailed solutions to all the questions that are given in the NCERT textbook.
Get detailed and accurate NCERT solutions for Class 11 Chemistry Thermodynamics covering all exercise questions with step-by-step answers. These solutions help you understand important concepts like enthalpy, internal energy, entropy, and Gibbs free energy.
Answer :
A state function refers to the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of the path.
Like-
So, the correct option is (ii)
Question 5.2 For the process to occur under adiabatic conditions, the correct condition is:
(i)
(ii)
(iii) q = 0
(iv) w = 0
Answer :
A system is said to be under adiabatic conditions if there is zero exchange of heat between the system and its surroundings. Hence, under adiabatic conditions,
So, the correct option is (iii)
Answer :
The enthalpies of all elements in their standard states are zero. The standard state of an element refers to its most stable form under standard conditions, typically at a pressure of 1 bar and a temperature of 298 K.
So, the correct option is (ii)
Question 5.4
(i)
(ii)
(iii)
(iv) = 0
Answer :
Equation of combustion of methane-
Since,
from the equation
So the correct option is (iii).
(i) –74.8
(ii) –52.27
(iii) +74.8
(iv) +52.26
Answer :
So, the required equation is to get the formation of
Thus,
= [-393.5 + 2(-285.8) + 890.3]
= -74.8 kJ/mol
Therefore, the enthalpy of formation of methane is -74.8 kJ/mol. So, the correct option is (i)
Answer :
For the reaction to be feasible, the
According to the question,
Overall the
Therefore, the reaction is possible at any temperature. So, the correct option is (iv)
Answer :
The first law of Thermodynamics states that,
where,
q = heat and W = work
Given that,
q = +701 J (heat is absorbed)
W = - 394 (work is done by the system)
Substituting the value in the equation of the first law we get,
So, the change in internal energy for the process is 307 J
Answer :
Given information,
T = 298 K
R = 8.314
= (2 - 1.5)
= 0.5 moles
The enthalpy change for the reaction is expressed as;
where,
By putting the values we get,
Answer :
We know that
m = mass of the substance
c = heat capacity
By putting all these values we get,
= 1066.7 J
= 1.07 kJ
Question 5.10Calculate the enthalpy change on freezing of 1.0 mol of water at
Answer :
Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from
(ii) from
(iii) from
So, the total enthalpy change
Hence, the total enthalpy change in the transformation process is -7.151 kJ/mol
Question 5.11 Enthalpy of combustion of carbon to
Answer :
Formation of carbon dioxide from carbon and dioxygen reaction is-
1 mole of
So, the heat released in the formation of 44g of
therefore, In 35.5 g of
314.8 KJ/mol heat is released when 35.2 g of CO2 is formed from carbon and dioxygen.
Question 5.12 Enthalpies of formation of
Answer :
Given,
Enthalpies of formation of
We know that the
For the above reaction,
Substituting the given values we get,
Thus the value of
Question 5.13 Given
Answer :
the standard enthalpy of formation of any compound is the required change in enthalpy of formation of 1 mole of a substance in its standard form from its constituent elements.
Thus we can re-write the reaction as;
Therefore, standard enthalpy of formation of ammonia is =
= 1/2 (-92.4)
= -46.2 kJ/mol
Question 5.14 Calculate the standard enthalpy of the formation of CH3OH(l) from the following data:
Answer :
for the formation of
This can be obtained by the following expressions-
required eq = eq (ii) + 2 (eq iii) - eq (i)
Therefore,
Question 5.15 Calculate the enthalpy change for the process
Answer :
We have the following chemical reaction equations-
The enthalpy change for the process
And the bond enthalpy of C-Cl bond in
Question 5.16 For an isolated system,
Answer :
Since
Means change in internal energy
Question 5.17 For the reaction at 298 K,
2A + B
Answer :
From the equation,
Suppose the reaction is at equilibrium, So the change in temperature is given as;
To reaction should be spontaneous,
Question 5.18 For the reaction,
Answer :
The given reaction represents the formation of a chlorine molecule from its atom. Bond formation taking place, therefore the energy is released during this. So,
Two moles of an atom have more randomness than one mole of chlorine. So, spontaneity is decreased. Hence
Question 5.19 For the reaction
Calculate
Answer :
Given reaction is
We know that,
and
here
and
Substituting the given values in equations-
Hence the reaction will not occur spontaneously because the
Question 5.20 The equilibrium constant for a reaction is 10. What will be the value of
Answer :
Given values are,
and equilibrium constant = 10
It is known that,
Hence the value of
Question 5.21 Comment on the thermodynamic stability of NO(g), given
Answer :
The formation of
On the other hand,
Question 5.22 Calculate the entropy change in surroundings when 1.00 mol of
Answer :
-286 kJ/mol heat is evolved when 1 mol of the water molecule is formed. It means the same amount of energy is absorbed by the surrounding also.
We know that,
Entropy changes for the surrounding is = q(surr)/ temp.
= 286000/ 298
= 959.73 J/mol/K
Enhance your conceptual understanding with HOTS questions from Class 11 Chemistry Chapter 5 Thermodynamics. These questions help you think deeper and prepare better for CBSE exams and entrance tests like NEET and JEE.
Question: At standard conditions, if the change in the enthalpy for the following reaction is
Given that bond energy of
1) 736
2) 518
3) 259
4) 368
Answer:
From the Bond energy concept, we have
Hence, the correct answer is option (4).
Question: The correct statement amongst the following is :
(1) The term 'standard state' implies that the temperature is
(2) The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K
(3)
(4)
Answer:
Option (1)
It asserts that the term "standard state" implies that the temperature is
In fact, "standard state" typically specifies a standard pressure (1 bar) and a chosen temperature-commonly 298 K for tabulated thermodynamic data-not necessarily
Thus, Option 1 is incorrect.
Option (2)
It claims that the standard state for a pure gas is defined at 1 bar and 273 K .
In modern thermochemistry, the standard state generally means the pure substance at 1 bar, but the standard temperature most commonly used is
Thus, Option 2 is incorrect.
Option (3)
It claims that
However, the standard state of oxygen is diatomic
Hence, Option 3 is incorrect.
Option (4)
The standard enthalpy of formation of any element in its most stable form is zero, regardless of temperature (though typically reported at 298 K ).
Thus, option (4) is correct
Hence, the correct answer is option (4).
Question: The hydration energies of
(1)
(2)
(3)
(4)
Answer:
Hence, the correct answer is option (3).
The basic understanding of how to solve questions from the Thermodynamics Class 11 NCERT is essential for building a strong foundation in physical chemistry. The approaches given below help students in developing the right techniques to solve questions.
1. Study the key concepts:
Learning about the basic concepts like system, surrounding, internal energy, work, heat, enthalpy and Laws of Thermodynamics is very important for solving questions of NCERT Class 11 Chemistry Chapter 5 Thermodynamics.
2). Identify the data and Read questions carefully
Before solving any question, first give it a thorough reading and note down all the information given in the question. Identify given data like pressure, volume, temperature, etc. and what we need to find.
3). Use the appropriate formula based on the concept like:
Using the correct formula is very important for Solving NCERT Class 11 Chemistry Chapter 5 Thermodynamics questions
(i). Internal Energy Change
(ii). First Law of Thermodynamics
(iii). Work Done in Expansion/Compression
(iv). Enthalpy and Enthalpy Change
(v). tandard Enthalpy Change of Reaction
(vi). Gibbs Free Energy
4). Convert units into standard form
Be careful while dealing with units and convert them if necessary to ensure consistency like temperature in Kelvin or volume in litres, etc.
5). Practice questions
Practice again and again, as it will help in mastering thermodynamics numerically.
6). Solve the NCERT textbook and NCERT exemplar questions.
Chapter 5 Thermodynamics Questions from the NCERT books are asked directly in the boards and other competitive exams. Do previous year questions from NEET and JEE to get used to question patterns.
All the topics and subtopics included in the NCERT Class 11 Chemistry Chapter 5 Thermodynamics are listed below:
5.1 Thermodynamics Terms
5.1.1 The System and the Surroundings
5.1.2 Types of the System
5.1.3 The State of the System
5.1.4 The Internal Energy as a State Function
5.2 Applications
5.2.1 Work
5.2.2 Enthalpy, H
5.3 Measurement of
5.4 Enthalpy Change,
5.5 Enthalpies for Different Types of Reactions
5.6 Spontaneity
5.7 Gibbs Energy Change and Equilibrium
Beyond the NCERT, students should focus on important concepts, formulas, different types of processes, and laws of thermodynamics according to the table given below:
Class 11 NCERT chapter-wise solutions are given below:
NCERT Subject-wise solutions are given below:
The NCERT books and syllabus links for class 11 are given below:
NCERT Books Class 11 Chemistry |
NCERT Syllabus Class 11 Chemistry |
NCERT Books Class 11 |
NCERT Syllabus Class 11 |
Thermodynamics is the study of energy transfer, particularly the relationships and conversions between heat, work, and chemical reactions.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics help students prepare well for the annual examinations. NCERT Solutions for Class 11 will help students to solve the problems given in the textbook. They give detailed and stepwise explanations of the problems given in the exercises of NCERT Solutions for Class 11.
Enthalpy is a thermodynamic quantity that represents the total heat content of a system. It is defined as the internal energy plus the product of pressure and volume. In thermodynamics, enthalpy changes indicate whether a reaction is exothermic or endothermic.
Gibbs free energy is a crucial thermodynamic potential that indicates the spontaneity of a process. It combines enthalpy and entropy into one value (G = H - TS), where T is temperature.
Entropy is a measure of the randomness or disorderness in a system. As described by the Second Law of Thermodynamics, It is an important concept in thermodynamics because it is related to the natural tendency of energy to disperse over time.
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