• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • NCERT Solutions
  • RD Sharma Solutions
    • RD Sharma Class 12 Solutions
    • RD Sharma Class 11 Solutions
    • RD Sharma Class 10 Solutions
    • RD Sharma Class 9 Solutions
    • RD Sharma Class 8 Solutions
  • RS Aggarwal Solutions
    • RS Aggarwal Solutions Class 10
    • RS Aggarwal Solutions Class 9
    • RS Aggarwal Solutions Class 8
    • RS Aggarwal Solutions Class 7
    • RS Aggarwal Solutions Class 6
  • CBSE Sample Papers
  • ML Aggarwal Solutions

Learn Insta

RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

  • Concise Mathematics Class 10 ICSE Solutions 2018
  • NCERT Solutions
  • Extra Questions
  • MCQ Questions
  • CBSE Notes

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

July 17, 2020 by Veerendra Leave a Comment

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics.

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.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 1

Answer:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 2

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:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 3

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; Cp[H2O(l)] = 75.3 J mol-1 K-1 ; Cp[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)
= Cp[H2O(l)] × ∆T + ∆Hfreezmg + Cp [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.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 4

Answer:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 5

Question 14.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 6
Answer:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 7

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:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 8

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)
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 9

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
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 10

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

Answer:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 12

Question 21.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 13

Answer:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 14

Question 22.
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 15

Answer:
NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics 16

We hope the NCERT Solutions for Class 11 Chemistry at Work Chapter 6 Thermodynamics, help you. If you have any query regarding NCERT Solutions for Class 11 Chemistry at Work Chapter 6 Thermodynamics, drop a comment below and we will get back to you at the earliest.

Filed Under: CBSE Class 11

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

  • Maths NCERT Solutions
  • Science NCERT Solutions
  • Social Science NCERT Solutions
  • English NCERT Solutions
  • Hindi NCERT Solutions
  • Physics NCERT Solutions
  • Chemistry NCERT Solutions
  • Biology NCERT Solutions
RS Aggarwal Solutions RD Sharma Solutions
RS Aggarwal Class 10 RD Sharma Class 10
RS Aggarwal Class 9 RD Sharma Class 9
RS Aggarwal Class 8 RD Sharma Class 8
RS Aggarwal Class 7 RD Sharma Class 11
RS Aggarwal Class 6 RD Sharma Class 12

Recent Posts

  • Class 12 History Important Questions Chapter 6 Bhakti-Sufi Traditions: Changes in Religious Beliefs and Devotional
  • Geography Class 12 Important Questions Chapter 3 Population Composition
  • Geography Class 12 Important Questions Chapter 2 The World Population: Distribution, Density and Growth
  • Class 12 History Important Questions Chapter 5 Through the Eyes of Travellers: Perceptions of Society
  • Geography Class 12 Important Questions Chapter 1 Human Geography: Nature and Scope
  • Class 12 History Important Questions Chapter 4 Thinkers, Beliefs and Buildings: Cultural Developments
  • 1 Mark Questions for History Class 12 Chapter 15 Framing the Constitution: The Beginning of a New Era
  • 1 Mark Questions for History Class 12 Chapter 14 Understanding Partition: Politics, Memories, Experiences
  • 1 Mark Questions for History Class 12 Chapter 13 Mahatma Gandhi and the Nationalist Movement: Civil Disobedience and Beyond
  • 1 Mark Questions for History Class 12 Chapter 12 Colonial Cities: Urbanisation, Planning and Architecture
DMCA.com Protection Status

Footer

NCERT Solutions for Class 12
NCERT Solutions for Class 11
NCERT Solutions for Class 10
NCERT Solutions for Class 9
NCERT Solutions for Class 8
NCERT Solutions for Class 7
NCERT Solutions for Class 6
ML Aggarwal Class 10 ICSE Solutions
Concise Mathematics Class 10 ICSE Solutions
RS Aggarwal Solutions
RD Sharma Solutions
ML Aggarwal Solutions
CBSE Sample Papers
English Summaries
English Grammar
Like us on Facebook Follow us on Twitter
Watch Youtube Videos Follow us on Google Plus
Follow us on Pinterest Follow us on Tumblr
Percentage Calculator

Copyright © 2021 Learn Insta