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Thermodynamics is an important branch of Chemistry. It deals with energy, its transformation, and the laws governing the entire process of transformation of energy. Have you ever wondered why a hot cup of tea cools down after some time or why ice melts, so the only reason behind this kind of phenomenon is thermodynamics. NCERT Class 11 Chapter 5 plays a very important role in Chemistry as it is the basic chapter that builds the foundations of advanced chapters and it also helps to explain why changes occur around us.
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In Thermodynamics, there are 22 questions in the exercise. The NCERT solutions for class 11 Chemistry thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you in the preparation of the class 11 final examination as well as in competitive exams like JEE Mains, NEET, etc. If you are looking for an answer from any other chapter even from any other class then go with NCERT Solutions, there you will get all the answers of NCERT easily.
Also Read :
Thermodynamics Class 11 Chemistry NCERT Exemplar Solutions |
Thermodynamics Class 11 Chemistry Chapter 5 Notes |
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 :
In adiabatic conditions, the heat exchange between the system and the surroundings through its boundary is not allowed.
So, the correct option is (iii)
Answer :
The enthalpies of all elements in their standard states are Zero.
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 5.10 Calculate 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
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 (i) + 2 (eq iii) - eq (i)
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
So, the
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
Thermodynamics enables us to study energy changes quantitatively and to make useful predictions. For these, we divide the universe into the system and the surroundings. In thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes surroundings. After completing NCERT solutions for class 11 chemistry Thermodynamics you will be able to explain terms like system, surroundings, and boundary; explain three types of systems open, closed, and isolated systems; explain work, heat, and internal energy, etc. In this chapter, we will also discuss the first law of thermodynamics and second laws of thermodynamics and express them mathematically, internal energy, enthalpy, entropy, and Gibbs energy. You will find NCERT solutions for class 11 Chemistry Thermodynamics by scrolling down.
5.1 Thermodynamic Terms
5.2 Applications
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
Chapter 1 | |
Chapter-2 | |
Chapter-3 | |
Chapter-4 | |
Chapter-5 | Thermodynamics |
Chapter-6 | |
Chapter-7 | |
Chapter-8 | |
Chapter-9 |
Also Check NCERT Books and NCERT Syllabus here:
NCERT Books Class 11 Chemistry |
NCERT Syllabus Class 11 Chemistry |
NCERT Books Class 11 |
NCERT Syllabus Class 11 |
Some of the important topics covered in this chapter are:
official website of NCERT: http://www.ncert.nic.in/
Weightage of thermodynamics for CBSE board exam is of 8 marks
Internal Energy is the energy within the system while Enthalpy is the internal energy plus the energy required to push back the atmosphere. Enthalpy represents the amount of heat absorbed or released at constant pressure.
Refer to this link: https://school.careers360.com/ncert/ncert-solutions-class-11-chemistry
The first law of thermodynamics states that energy can neither be created nor be destroyed it can only transfer from one form to another form.
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