NCERT Solutions | Class 11 Chemistry Chapter 6

NCERT Solutions | Class 11 Chemistry Chapter 6 | Thermodynamics 

CBSE Solutions | Chemistry Class 11

Check the below NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics Pdf free download. NCERT Solutions Class 11 Chemistry  were prepared based on the latest exam pattern. We have Provided Thermodynamics Class 11 Chemistry NCERT Solutions to help students understand the concept very well.

NCERT | Class 11 Chemistry

NCERT Solutions Class 11 Chemistry

Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	11th
Subject:	Chemistry
Chapter:	6
Chapters Name:	Thermodynamics
Medium:	English

Thermodynamics | Class 11 Chemistry | NCERT Books Solutions

You can refer to MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Exercises

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 1.

Choose the correct answer. A thermodynamic state function is a quantity
(a) used to determine heat changes
(b) whose value is independent of path
(c) used to determine pressure volume work
(d) whose value depends on temperature only.

Solution.

(b): State function is a property of the system whose value depends only upon the state of the system and is independent of the path or the manner by which the state is reached.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 2.

For the process to occur under adiabatic conditions, the correct condition is
(a) ΔT = 0
(b) Δp = 0
(c) g = 0
(d) w=0

Solution.

(c): Adiabatic system does not exchange heat with the surroundings.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 3.

The enthalpies of all elements in their standard states are
(a) unity
(b) zero
(c) <0
(d) different for each element.

Solution.

(b) : By convention, the standard enthalpy of formation of every element in its standard state is zero.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 4.

ΔU° of combustion of methane is – X kJ mol^-1. The value of ΔH° is
(a) = ΔU°
(b) >ΔU°
(c) <ΔU°
(d) =0

Solution.

(c) : CH_4(g)+2O₂ → CO_{2(g)+2H2Ol
Δn = 1 – 3 = -2. ‘
ΔH° = ΔU° + ΔnRT = -X- 2RT}

NCERT Solutions for Class 11 Chemistry Chapter 6, 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 CH_4(g) will be
(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 6.

A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(a) possible at high temperature
(b) possible only at low temperature
(c) not possible at any temperature
(d) possible at any temperature

Solution.

(d): A + B → C + D + q, ΔS = +ve
Here, ΔH = -ve
ΔG = ΔH – TΔS
For reaction to be spontaneous, ΔG should be -ve. As ΔH = -ve and ΔS is +ve, ΔG will be -ve at any temperature.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 7.

In a process, 701 J of heat is absorbed by a system and 394 S of work is done by the system. What is the change in internal energy for the process?

Solution.

Heat absorbed by the system (q) = 701 J
Work done by the system (w) = -394 J
According to first law of thermodynamics,
ΔU = q + w = 701 + (-394) = 701 – 394 = 307 J

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 8.

The reaction of cyanamide, NH₂CN_(g), with dioxygen was carried out in a bomb calorimeter, and AU was found to be -742.7 kJ mol 1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN_(g) + \(\frac { 3 }{ 2 } { O }_{ 2(g) }\) → N₂+CO_2(g)+H₂O_l

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 9.

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.

Solution.

Mass of Al = 60 g
Rise in temperature, ΔT = 55 – 35 = 20°C
Molar heat capacity of Al = 24 J mo^-1 K^-1
Specific heat capacity of Al = \(\frac { 24 }{ 27 } J{ g }^{ -1 }{ K }^{ -1 }\)
∴ Energy required = m x c x ΔT
= \(60\times \frac { 24 }{ 27 } \times 20=\frac { 28800 }{ 27 } =1066.67\quad J\)
= 1.068kJ or 1.07kJ

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 10.

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ_fusH = 6.03 kJ mol^-1 at 0°C.
C_p[H₂O_(l)] = 75.3 J mol^-1 K^-1,
C_p[H₂O_(s)] = 36.8 J mol^-1 K^-1

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 11.

Enthalpy of combustion of carbon to CO₂ is -393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.

Solution.

C + O₂ → CO₂; ΔT = -393.5 kJ
∵ When 44 g of COz is formed from carbon and dioxygen gas, heat released = 393.5 kj
∴ When 35.2 g of CO₂ is formed from carbon and dioxygen gas, heat released
= \(\frac { 393.5\times 35.2 }{ 44 } =\frac { 138551.2 }{ 44 } =314.8kJ\)
Thus, ΔH = -314.8 kJ

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 12.

Enthalpies of formation of CO_(g), CO_2(g), N₂O_(g) and N₂O_4(g) are -110, – 393, 81 and 9.7 kJ mol ^-1 respectively. Find the value of Δ_rf for the reaction
N₂O_4(g) + 3CO_(g) → N₂O_(g) + 3CO_2(g)

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 13.

Given: N_2(g) + 3H_2(g) → 2NH_3(g); Δ_rH° = -92.4 kJ mol^-1 What is the standard enthalpy of formation of NH₃ gas?

Solution.

N_2(g) + 3H_2(g) → 2NH_3(g) ; Δ_rH° = -92.4 kJ mol^-1
∴ Standard enthalpy of formation of NH_3(g)
= \(\frac { -92.4 }{ 2 } =\quad -46.2\quad kJ/mol\)

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 14.

Calculate the standard enthalpy of formation of CH₃OH_l from the following data :

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 15.

Calculate the enthalpy change for the process
CCl_4(g) → C_(g) + 4Cl_(g)
and calculate bond enthalpy of C – Cl in CCl_4(g).
Δ_vapH°(CCl₄) = 30.5 kJ mol^-1,
Δ_fH°(CCl₄) = -135.5 kJ mol^-1,
Δ_aH°(C) = 715.0 kJ mol^-1, where Δ_aH° is enthalpy of atomisation Δ_aH°(Cl₂) = 242 kJ mol^-1

Solution.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 16.

For an isolated system, ΔU = 0, what will be ΔS?

Solution.

When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 17.

For the reaction at 298 K, 2A + B → C,
ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ K^-1 mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.

Solution.

According to Gibbs Helmholtz equation, ΔG = ΔH – TΔS.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 18.

For the reaction, 2Cl_(g) → Cl_2(g), what are the signs of ΔH and ΔS?

Solution.

ΔH is negative because bond energy is released and ΔS is negative because there is less randomness among the molecules than among the atoms.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 19.

For the reaction
2A_(g) + B_(g) → 2D_(g), ΔU° = -10.5 kJ and ΔS° = – 44.1 JK^-1.
Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.

Solution.

ΔH° = ΔU° + Δn_(g)RT

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 20.

The equilibrium constant for a reaction is 10. What will be the value of ΔG° ?
R = 8.314J K^-1 mol^-1, T= 300 K.

Solution.

We know that ΔG° = – 2.303RT logK
= – 2.303 x 8.314 x 300 x log10
= – 5744.14 J/mol

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 21.

Comment on the thermodynamic stability of NO_(g), given

Solution.

Since Δ_(r)H° is +ve, i.e., enthalpy of formation of NO is positive, therefore NO is unstable. But Δ_(r)H° is negative for the formation of NO₂. So, NO₂ is stable.

NCERT Solutions for Class 11 Chemistry Chapter 6, Question 22.

Calculate the entropy change in surroundings when 1.00 mol of H₂O_l is formed under standard conditions.
Δ_(r)H° = – 286 kJ mol^-1.

Solution.

NCERT Class 11 Chemistry

Class 11 Chemistry Chapters | Chemistry Class 11 Chapter 6

Chapterwise NCERT Solutions for Class 11 Chemistry

NCERT Solutions For Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry
NCERT Solutions For Class 11 Chemistry Chapter 2 Structure of The Atom
NCERT Solutions For Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties
NCERT Solutions For Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure
NCERT Solutions For Class 11 Chemistry Chapter 5 States of Matter
NCERT Solutions For Class 11 Chemistry Chapter 6 Thermodynamics
NCERT Solutions For Class 11 Chemistry Chapter 7 Equilibrium
NCERT Solutions For Class 11 Chemistry Chapter 8 Redox Reactions
NCERT Solutions For Class 11 Chemistry Chapter 9 Hydrogen
NCERT Solutions For Class 11 Chemistry Chapter 10 The sBlock Elements
NCERT Solutions For Class 11 Chemistry Chapter 11 The pBlock Elements
NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques
NCERT Solutions For Class 11 Chemistry Chapter 13 Hydrocarbons
NCERT Solutions For Class 11 Chemistry Chapter 14 Environmental Chemistry

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