Class 11 Chemistry NCERT Solutions 2023 for Chapter 6 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics: If you are a student of class 11 you have reached the right platform. The NCERT Solutions for class 11 Chemistry in Hindi provided by us are designed in a simple, straightforward language, which is easy to memorize.

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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

NCERT Solutions Class 11 Chemistry Chapter 6 PDF which will help students understand topics such as Thermodynamics Terms, The system and surroundings, Types of the System, The state of the system, The internal Energy as a state function, application, Work, Enthalpy, Measurement of Du and Dh, Calorimetry, Enthalpy change, Drh of a reaction- Reaction Enthalpy, Enthalpies for different types of reactions, Spontaneity, Gibbs Energy Change, and Equilibrium.

Get 100% accurate CBSE, NCERT Solutions for Class 11 Chemistry Chapter 6 ( Thermodynamics) solved by expert chemistry teachers. We provide solutions for questions given in the class 11 Chemistry textbook as per CBSE Board guidelines from the latest NCERT book for class 11 chemistry.

Furthermore, the NCERT 11th Class Chemistry Chapter 6 Solutions pdf provided on this page can be downloaded for free by clicking the download button provided below.

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

 

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What will you learn in CBSE Class 11 Chemistry Chapter 6 Thermodynamics?

The sixth chapter in the NCERT class 11 textbook is “Thermodynamics”. Thermodynamics is a branch of science that deals with the relationship between heat and other forms of energy. A part of the universe where observations are made is called a system. The surrounding of a system is the part of the universe that does not contain the system.

Based on the exchange of energy and matter, thermodynamic systems can be classified into 3 types: a closed system, an open system, and an isolated system. In a closed system, only energy can be exchanged with the surroundings. In an open system energy as well as matter can be exchanged with the surroundings. In an isolated system, both energy and matter cannot be exchanged with the surroundings.

NCERT Class 11 Chemistry Chapter 6, Thermodynamics is from chemistry Part 1 and belongs to Unit 6. Unit 6 along with Unit 4, Unit 5 and Unit 6 and Unit 7 holds a weightage of 21 marks. In this chapter, the students will be introduced to Thermodynamics, Its meaning, application, and uses. This chapter will help students build a foundation for Grade 12 and further competitive exams.

Subtopics included in Class 11 Chemistry Chapter 6 Chemical Thermodynamics

• Thermodynamic Terms
  • The System and the Surroundings
  • Types of Thermodynamic Systems
  • State of the System
  • Internal Energy as a State Function
• Applications
  • Work
  • Enthalpy, H
• Measurement Of ΔU And ΔH: Calorimetry
• Enthalpy Change and Reaction Enthalpy
• Enthalpies For Different Types Of Reactions
• Spontaneity
• Gibbs Energy Change And Equilibrium

We cover all exercises in Chapter 6 – all 22 questions with solutions.

Other than given exercises students are encouraged to practice all the solved examples given in the book to clear their concepts on Thermodynamics.

Access NCERT Solutions for Class 11 Chemistry Chapter 6

Question 1. Choose the correct answer:
A thermodynamic state junction is a quantity
(i) used to determine heat changes
(ii) whose value is independent of the path
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only.

Answer: (ii) whose value is independent of the path

Question 2. For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T= 0 (ii) ∆p = 0
(iii) q = 0 (iv) w = 0

Ans. (iii) q = 0

Question 3. The enthalpies of all elements in their standard states are: ‘
(i) unity (ii) zero
(iii) < 0 (iv) different for each element

Answer:  (ii) zero

Question 4.

Answer:

Question 5. The enthalpy of combustion of methane, graphite, and dihydrogen at 298 K are -890.3 KJ mol^-1, – 393.5 KJ mol^-1, and – 285.8 KJ mol^-1 respectively. Enthalpy of formation of CHJg) will be
(i) – 74.8 KJ mol^-1  (ii) – 52.27 KJ mol^-1
(iii) + 74.8 KJ mol^-1 (iv) + 52.26 KJ mol^-1

Answer:  As per the available data :

Question 6. A reaction, A + B—>C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature (ii) possible only at low temperature
(iii) not possible at any temperature (iv) possible at any temperature

Answer:  (iv) possible at any temperature

Question 7. In a process, 701 ] 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?

Answer:  Heat absorbed by the system, q = 701 J Work done by the system = – 394 J Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

Question 8. The reaction of cyanamide, NH₂CN(s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be -742,7 KJ^-1 mol^-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.NH₂CN (S) + 3/202(g) —–>N₂(g) + CO₂(g) + H₂0(Z)

Answer:  ∆U = – 742.7 KJ^-1 mol^-1 ; ∆^ng = 2 – 3/2 = + 1/2 mol.
R = 8.314 x 10-3KJ^-1 mol^-1 ; T = 298 K
According to the relation,∆H = ∆U+∆^ng RT
∆H = (- 742.7 kj) + (1/2 mol) x (8.314 x10^-3 KJ^-1 mol^-1 ) x (298 K)
= – 742.7 kj + 1.239 kj = – 741.5 kj.

Question 9. Calculate the number of kj of heat necessary to raise the temperature of 60 g of aluminum from 35°C to 55°C. The molar heat capacity of Al is 24 J mol^-1 K^-1.

Answer: No. of moles of Al (m) = (60g)/(27 g mol^-1) = 2.22 mol
Molar heat capacity (C) = 24 J mol^-1 K^-1.
Rise in temperature (∆T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C x m x T = (24 J mol^-1 K^-1) x (2.22 mol) x (20 K)
= 1065.6 J = 1.067 kj

Question 10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. A, H = 6.03 KJ mot1 at 0°C. Cp [H20(l)J = 75.3 J mol^-1 K^-1; Cp [H20(s)J = 36.8 J mol^-1 K^-1.

Answer: The change may be represented as:

Question 11. The enthalpy of combustion of carbon to carbon dioxide is – 393.5 J mol^-1. Calculate the heat released upon formation of 35.2 g of C0₂ from carbon and oxygen gas.

Answer: The combustion equation is:
C(s) + 0₂ (g) —–> C0₂(g); AcH = – 393.5 KJ mol^-1
The heat released in the formation of 44g of C0₂ = 393.5 kj
Heat released in the formation of 35.2 g of C0₂=(393.5 KJ) x (35.2g)/(44g) = 314.8 kj

Question 12. Calculate the enthalpy of the reaction:
N₂0₄(g) + 3CO(g) ———->N₂0(g) + 3CO₂(g)
Given that;∆_f H^–CO(g) = – 110 kj mot^-1; ∆_fHC0₂(g) = – 393 kj mol^-1
∆_fHN₂0(g) = 81 kj mot^-1; ∆_fN₂O₄(g) = 9.7 kj mol^-1

Answer:  Enthalpy of reaction (∆_r,H) = [81 + 3 (- 393)] – [9.7 + 3 (- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kj mol^-1

Question 13. Given : N₂(g) + 3H₂(g) ————> 2NH₃(g); ∆_{r H^–} = -92.4 kj mot^-1 What is the standard enthalpy of formation of NH₃ gas?

Answer:  ∆H^– NH₃ (g) = – (92.4)/2 = – 46.2 kj mol^-1

Question 14. Calculate the standard enthalpy of the formation of CH₃OH. from the following data:
(i) CH₃OH(l) + 3/2 0₂ (g) ———-> CO₂ (g) + 2H₂0 (l); ∆_rH^– = – 726kj mol^-1
(ii) C(s) + 0₂(g) —————>C0₂ (g); ∆_cH^– = -393 kj mol^-1
(iii) H₂(g) + 1/20₂(g) —————->H₂0 (l); ∆_fH^– = -286 kj mol^-1

Answer:  The equation we aim at;
C(s) + 2H₂(g) + l/20₂(g) ———> CH₃OH (l);∆_fH^– = ±? … (iv)
Multiply eqn. (iii) by 2 and add to eqn. (ii)
C(s) + 2H₂(g) + 20₂(g) ————->C0₂(g) + 2H₂0(Z)
∆H = – (393 + 522) = – 965 kj moH Subtract eqn. (iv) from eqn. (i)
CH₃OH(Z) + 3/20₂(g) ————> C0₂(y) + 2H₂0(Z); ∆H = – 726 kj mol^-1
Subtract: C(s) + 2H₂(y) + l/20₂(g) ———-> CH₃OH(Z); ∆_fHe = – 239 kj mol^-1

Question 15.

Answer:

Question 16. For an isolated system ∆U = 0; what will be ∆S? 

Answer: The change in internal energy (∆U) for an isolated system is zero for it does not exchange any energy with the surroundings. However, entropy tends to increase in case of spontaneous reaction. Therefore, ∆S > 0 or positive.

Question 17. For a reaction at 298 K
2A + B————->C
∆H = 40Q kj mot1 and AS = 0.2 kj Kr^-1 mol^-1.
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

Answer: As per the Gibbs Helmholtz equation:
ΔG = Δ H- TΔ S For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS
T = (400 KJ mol^-1)/(0.2 KJ K^-1 mol^-1) = 2000 k
Thus, the reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.

Question 18. For the reaction; 2Cl(g) ———-> Cl₂(g); what will be the signs of ∆H and ∆S?

Answer: ∆H : negative (- ve) because energy is released in bond formation
∆S : negative (- ve) because entropy decreases when atoms combine to form molecules.

Question 19.

Answer:

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