NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics- Thermodynamics is the branch of science which deals with the transfer of energy or changes in energy in physical or chemical processes. This is an important chapter for class 11th students because some concepts of this chapter will also be applied in the next chapter which is equilibrium and also some in the 12th class chapters like chemical kinetics etc. In CBSE NCERT solutions for class 11 chemistry chapter 6 Thermodynamics, there are 22 questions in the exercise. The solutions of NCERT class 11 chemistry chapter 6 thermodynamics are prepared in a very comprehensive manner. These NCERT solutions of class 11 can help you in the preparation of class 11 final examination as well as in the 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.
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 chapter 6 Thermodynamics you will be able to explain the terms like system, surroundings, and boundary; explain three types of system open, close 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, Gibbs energy. You will find NCERT solutions for class 11 chemistry chapter 6 Thermodynamics by scrolling down.
Important points of NCERT solutions for class 11 chemistry chapter 6 Thermodynamics-
1. System and Surroundings - in thermodynamics, the system is the part of the universe where observations are made and the remaining universe constitutes surroundings. The universe = The system+ The Surroundings.
2. The First Law of Thermodynamics- according to it under any circumstances, energy cannot be created and nor be destroyed, it can be converted from one form to another form with the interaction of work, heat and internal energy. Mathematically, this is represented as: , where w is the work done on or by the system, q is the heat exchanged between the system and its surroundings and ΔU is the total change in internal energy of the system.
3. The second law of thermodynamics- it states that entropy of the entire universe(total entropy of both the system and its surroundings), as an isolated system, will always increase over time.
Topics of NCERT Grade 11 Chemistry Chapter 6 Thermodynamics
6.1 Thermodynamic Terms
6.3 Measurement of ∆U and ∆H: Calorimetry
6.4 Enthalpy Change, ∆H of a Reaction – Reaction Enthalpy
6.5 Enthalpies for Different Types of Reactions
6.7 Gibbs Energy Change and Equilibrium
NCERT solutions for class 11 chemistry chapter 6 Thermodynamics- Exercise Questions
A state function refers to the equilibrium state of a system. A thermodynamic state function is a quantity whose value is independent of path.
Like- depends on the state of the system, not on the path.
So, the correct option is (ii)
In adiabatic condition, the heat exchange between the system and the surrounding through its boundary is not allowed.
So, the correct option is (iii) = 0
The enthalpies of all elements in their standard states are Zero.
So, the correct option is (ii)
Equation of combustion of methane-
from the equation
so the correct option is (iii).
................. -890.3 kJ/mol
So, the required equation is to get the formation of by combining these three equations-
= [-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)
For the reaction to be feasible the should be negative
According to question,
= positive unit
= negative (as heat is evolved in the reaction)
Overall the is negative.
Therefore the reaction is possible at any temperature. So, the correct option is (iv)
The first law of Thermodynamics state that,
where, = change in internal energy for the process
q = heat and W = work
q = +701 J (heat is absorbed)
W = - 394 (work is done by the system)
Substituting the value in equation of first law we get,
So, the change in internal energy for the process is 307 J
T = 298 K
R = 8.314
= (2 - 1.5)
= 0.5 moles
The enthalpy change for the reaction is expressed as;
where, = change in internal energy and
change in no. of moles
By putting the values we get,
We know that
m = mass of the substance
c = heat capacity
= change in temperature
By putting all these values we get,
= 1066.7 J
= 1.07 kJ
Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from to of 1 mol of water (water water )
(ii) from to of 1 mol of ice (water ice)
(iii) from to of 1 mole of ice (ice ice)
So, the total enthalpy change
Hence the total enthalpy change in the transformation process is 7.151 kJ/mol
Formation of carbon dioxide from carbon and dioxygen reaction is-
............. = -393.5 kJ/mol
1 mole of = 44 g
So, the heat released in the formation of 44g of = -393.5 kJ/mol
therefore, In 35.5 g of the amount of heat released =
Enthalpies of formation of are –110, – 393, 81 and 9.7
We know that the (product)(reactants)
For the above reaction,
Substituting the given values we get,
Thus the value of of the reaction is -777.7 kJ/mol
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
for the formation of the reaction is ,
This can be obtained by the following expressions-
required eq = eq (i) + 2 (eq iii) - eq (i)
We have the following chemical reactions equations-
The enthalpy change for the process by the above reaction is calculated as;
And the bond enthalpy of C-Cl bond in (g) = 1304/4 = 326 kJ/mol
So, the (positive) therefore the reaction will be feasible.
From the equation,
Suppose the reaction is at equilibrium, So the change in temperature is given as;
( at equilibrium is zero)
To reaction should be spontaneous, should be neagtive. So, that for the given reaction T should be greater than 2000 K
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, is negative.
Two moles of an atom have more randomness than the one mole of chlorine. So, spontaneity is decreased. Hence is negative.
Given reaction is
We know that,
substituting the given values in equations-
Hence the reaction will not occur spontaneously because the value is positive for this reaction.
Given values are,
and equilibrium constant = 10
It is known that,
Hence the value of is -5.744 kJ/mol
The formation of is unstable because is positive it means heat is absorbed during the reaction. So, (g) has higher energy than its reactants and .
On the other hand, is stable because is negative means heat is released during the formation of . It is stablised with minimum energy. Hence unstable changes to stable .
-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
NCERT solutions for class 11 chemistry
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