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.

Question 1.

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

Answer:

(ii) whose value is independent of 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

Answer:

(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 CH₄ (g) 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:

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 J 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 = – 304 J
Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

Question 8.

The reaction of cyanamide, NH₂CN(s) with oxygen was affected in a bomb calorimeter and ∆U was found to be – 742.7 kJ mol^-1 of cyanamide at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH₂CN(y) + 3/2O₂(g) → N₂fe) + CO₂(g) + H₂O(l)

Answer:

∆U = – 742.7 kJ mol^-1 ; ∆^ng = 2 – \(\frac { 3 }{ 2 } \) = + \(\frac { 1 }{ 2 } \) mol.
R = 8.314 × 10^-3 kJ K^-1 mol^-1 ; T = 298 K
According to the relation, ∆H = ∆U + ∆^ng RT
AH = (- 742.7 kJ) + (\(\frac { 1 }{ 2 } \) mol) × (8.314 × 10^-3 kJ K^-1 mol^-1) × (298 K)
= -742.7 kJ + 1.239 kJ = -741.5 kJ.

Question 9.

Calculate the number of kJ necessary to raise the temperature of 60 g of aluminium from 35 to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.

Answer:

No. of moles of Al (m) = \(\frac { 60g }{ 27gmo{ l }^{ -1 } } \) = 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 × m × T = (24 J mol^-1 K^-1) × (2.22 mol) × (20 K)
= 1065.6 J = 1.067 kJ.

Question 10.

Calculate the enthalpy change on freezing of 1.0 mole of water at – 10.0°C to ice at – 10.0°C. ∆_fusH = 6.03 kJ mol^-1 at 0°C; C_p[H₂O(l)] = 75.3 J mol^-1 K^-1 ; C_p[H₂O(s)l = 36.8 J mol^-1 K^-1.

Answer:

Total change in enthalpy (AH) for the freezing process may be calculated as :
∆H = (1 male of water at 10°C → 1 mole of water at 0°C) + (1 mole of water at °C → 1 mole of ice at 0°C)
+ (1 mole of ice at °C → 1 mole of ice at – 10°C)
= C_p[H₂O(l)] × ∆T + ∆H_freezmg + C_p [H₂O(s)] × ∆T.
= (75.3 Jk^-1 mol^-1) (0 – 10 K) + (-6.03 kJ mol^-1) + (36.8Jk^-1 mol^-1) × (-10 K)
= (-753 J mol^-1) – (6.03 kJ mol^-1) – (368 J mol^-1)
= (-0.753 kJ mol^-1) – (6.03 kJ mol^-1) – (0.368 kJ mol^-1)
= – 7.151 kJ mol^-1

Question 11.

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

Answer:

The combustion equation is :
C(s) + O₂(g) → CO₂(g) ; ∆_cH = – 393.5 kJ mol^-1 (44 g)
(44g)
Heat released in the formation of 44g of CO₂ = 393.5 kJ
Heat released in the formation of 35.2 g of CO₂ = \(\frac { (393.5kJ)\times (35.2g) }{ (44g) } \) = 314.8 kJ.

Question 12.

Calculate the enthalpy of the reaction :
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Given that;
∆_fH CO(g) = – 110 kJ mol^-1 ;
∆_fH CO₂(g) = – 393 kJ mol^-1
∆_fH. N₂O(g) = 81 kJ mol^-1 ;
∆_fH N₂O₄(g) = – 9.7 kJ mol^-1.

Answer:

Enthalpy of reaction (∆_rH) = [81 + 3(- 393)] – [9.7 + 3(- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kJ mol^-1.

Question 13.

Answer:

Question 14.

Answer:

Question 15.

Calculate the enthalpy change for the process
CCl₄ (g) → C (g) + 4 Cl (g) and calculate bond enthalpy of C—Cl in CC14 (g)
Given : ∆_vap H° (CCl₄) = 30.5 kJ mol^-1 ; ∆_fH°(CCl₄) = – 135.5 kJ mol^-1
∆_aH° (C) = 715.0 kJ mol^-1 where ∆_a H° is enthalpy of atomisation
∆_aH° (Cl₂) = 242 kJ mol^-1.

Answer:

Question 16.

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

Answer:

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

Question 17.

For a reaction at 298 K
2 A + B → C
∆H = 400 kJ mol^-1 and ∆S = 0.2 kJ K’1 mol^-1.
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range ?

Answer:

According to Gibbs-Helmholtz equation :
∆G = ∆H – T∆S
For ∆G = 0; ∆H = T∆S or T : = \(\frac { \triangle H }{ \triangle s } \)

T = \(\frac { (400kJmo{ l }^{ -1 }) }{ (0.2kJ{ K }^{ -1 }mo{ l }^{ -1 }) } \)
Thus, 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.

For a reaction ; 2A (g) + B (g) → 2D(g)

Calculate ∆U₂₉₈ for the reaction and predict whether the reaction is spontaneous or not.

Answer:

Let the mass of H₂ in the mixture = 20 g
The mass of O₂ in the mixture will be = 80 g

Question 20.

Answer:

Question 21.

Answer:

Question 22.

Answer:

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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