• 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

MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers

November 23, 2020 by Prasanna Leave a Comment

Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Thermodynamics Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.

Thermodynamics Class 11 MCQs Questions with Answers

Question 1.
Third law of thermodynamics provides a method to evaluate which property?
(a) Absolute Energy
(b) Absolute Enthalpy
(c) Absolute Entropy
(d) Absolute Free Energy

Answer

Answer: (c) Absolute Entropy
Explanation:
The Third Law of Thermodynamics is concerned with the limiting behavior of systems as the temperature approaches absolute zero. Most thermodynamics calculations use only entropy differences, so the zero point of the entropy scale is often not important. However, the Third Law tells us about the completeness as it describes the condition of zero entropy.


Question 2.
One mole of which of the following has the highest entropy?
(a) Liquid Nitrogen
(b) Hydrogen Gas
(c) Mercury
(d) Diamond

Answer

Answer: (b) Hydrogen Gas
Explanation:
The measure of randomness of a substance is called entropy. Greater the randomness of molecules of a substance greater is the entropy. Here hydrogen gas has more entropy as it shows more randomness/disorderliness due to less molar mass than all the given substances and also in the gas phase.


Question 3.
The enthalpy of vaporisation of a substance is 8400 J mol-1 and its boiling point is –173°C. The entropy change for vaporisation is :
(a) 84 J mol-1K-1
(b) 21 J mol-1K-1
(c) 49 J mol-1K-1
(d) 12 J mol-1K-1

Answer

Answer: (a) 84 J mol-1K-1
Explanation:
∆S = (qrev)/ (T)
= (\(\frac {8400}{100}\))
= 84 J mol-1K-1


Question 4.
The species which by definition has ZERO standard molar enthalpy of formation at 298 K is
(a) Br2(g)
(b) Cl2(g)
(c) H2O(g)
(d) CH4(g)

Answer

Answer: (b) Cl2(g)
Explanation:
This is possible only for elements, chlorine is a gas at this temperature, but bromine is a liquid, so it is possible only for chlorine.


Question 5.
In a reversible process the system absorbs 600 kJ heat and performs 250 kJ work on the surroundings. What is the increase in the internal energy of the system?
(a) 850 kJ
(b) 600 kJ
(c) 350 kJ
(d) 250 kJ

Answer

Answer: (c) 350 kJ
Explanation:
∆E = q + w
= (600 – 250)
∆E = 350 J


Question 6.
Which of the following is true for the reaction? H2O (l) ↔ H2O (g) at 100° C and 1 atm pressure
(a) ∆S = 0
(b) ∆H = T ∆S
(c) ∆H = ∆U
(d) ∆H = 0

Answer

Answer: (a) ∆S = 0
Explanation:
Equilibrium
Therefore, ∆ G = 0 = ∆ H – T ∆ S
Or T∆ S = ∆H


Question 7.
Calculate the heat required to make 6.4 Kg CaC2 from CaO(s) and C(s) from the reaction: CaO(s) + 3 C(s) → CaC2(s) + CO (g) given that ∆f Ho (CaC2) = -14.2 kcal. ∆f H° (CO) = -26.4 kcal.
(a) 5624 kca
(b) 1.11 × 104 kcal
(c) 86.24 × 10³
(d) 1100 kcal

Answer

Answer: (b) 1.11 × 104 kcal
Explanation:
n = (Mass)/ (Molecular weight)
= (6.4 × 10³)/ (64)
= 100
For 1 mole of CaC2
∆ H = ∆Hf (CaC) + Hf (CO) – Hf (CaO)
= -14.2 – 26.4 + 151.6 = 111.1 kcal
For 100 moles, ∆H = 1.11 × 104 Kcal


Question 8.
In a system where ∆E = -51.0 kJ, a piston expanded against a pext of 1.2 atm giving a change in volume of 32.0 L. What was the change in heat of this system?
(a) -36 kJ
(b) -13 kJ
(c) -47 kJ
(d) 24 kJ

Answer

Answer: (c) -47 kJ
Explanation:
w = -1.2 (32) × 101.3
= – 3.89 KJ
= -4 (approx.)
= ∆ E = – 51.0 KJ
Therefore, E = q + w
– 51 = q – 4
Therefore, q = – 47 KJ


Question 9.
A system absorb 10 kJ of heat at constant volume and its temperature rises from 270 C to 370 C. The value of ∆ U is
(a) 100 kJ
(b) 10 kJ
(c) 0 kJ
(d) 1 kJ

Answer

Answer: (b) 10 kJ
Explanation:
At constant volume w = 0
Therefore, ∆ U = q = 10 kJ


Question 10.
An ideal gas is taken around the cycle ABCA as shown in P-V diagram The next work done by the gas during the cycle is equal to:
MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers 1
(a) 12P1V1
(b) 6P1V1
(c) 5P1V1
(d) P1V1

Answer

Answer: (c) 5P1V1
Explanation:
Work done = Area under P-V graph = (\(\frac {1}{2}\)) (5P1) (2V1) = 5P1 V1


Question 11.
In which of the following process, a maximum increase in entropy is observed?
(a) Dissolution of Salt in Water
(b) Condensation of Water
(c) Sublimation of Naphthalene
(d) Melting of Ice

Answer

Answer: (c) Sublimation of Naphthalene
Explanation:
The order of entropy in solid, liquid and gas is gas > liquid > solid .Hence, in sublimation of naphthalene, maximum increase in entropy is observed.


Question 12.
The bond energy (in kcal mol-1) of a C-C single bond is approximately
(a) 1
(b) 10
(c) 83-85
(d) 1000

Answer

Answer: (c) 83-85
Explanation:
C–C bond 83–85 kcal/mol

It is the energy required to break the bond .It is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolysis reaction at 0 K (absolute zero)


Question 13.
Which thermodynamic function accounts automatically for enthalpy and entropy both?
(a) Helmholtz Free Energy (A)
(b) Internal Energy (E)
(c) Work Function
(d) Gibbs Free Energy

Answer

Answer: (d) Gibbs Free Energy
Explanation:
Gibbs free energy combines the effect of both enthalpy and entropy. The change in free energy (ΔG) is equal to the sum of the change of enthalpy (∆H) minus the product of the temperature and the change of entropy (∆S) of the system.
∆G = ∆H – T∆S
ΔG predicts the direction in which a chemical reaction will go under two conditions: (1) constant temperature and (2) constant pressure.
If ΔG is positive, then the reaction is not spontaneous (it requires the input of external energy to occur) and if it is negative, then it is spontaneous (occurs without the input of any external energy).


Question 14.
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283.0 kJ mol-1 respectively. The enthalpy of formation of carbon monoxide is:
(a) -676 kJ
(b) 110.5 kJ
(c) -110.5 kJ
(d) 676.5 kJ

Answer

Answer: (c) -110.5 kJ
Explanation:
C(s) + O2 à CO2 ∆H1 = -393.5
CO + (\(\frac {1}{2}\)) O2 à CO2 ∆H2 = -283.0
C(s) + (\(\frac {1}{2}\)) O2 à CO ∆H = ∆H1 – ∆H2
= -393.5 + 283 = -110.5 KJ


Question 15.
The amount of the heat released when 20 ml 0.5 M NaOH is mixed with 100 ml 0.1 M HCl is x kJ. The heat of neutralization is
(a) -100 × kJ/mol
(b) -50 × kJ/mol
(c) 100 × KJ/mol
(d) 50 × kJ/mol

Answer

Answer: (a) -100 × kJ/mol
Explanation:
Normality of NaOH = Molarity × acidity
= 0.5 × 1 = 0.5 N
Total heat q produced = x kJ
Heat of neutralisation
= [(q)/ (Volume of acid or base)] ×1000× (1/normality of acid or base)
= (\(\frac {x}{20}\)) × 1000 × (\(\frac {1}{0.5}\))
= 100 x kJmol-1
Since heat is liberated, heat of neutralisation = −100 x kJmol-1


Question 16.
Based on the first law of thermodynamics, which one of the following is correct?
(a) For an isothermal process, q = +w
(b) For an isochoric process, ΔU = -q
(c) For an adiabatic process, ΔU = -w
(d) For a cyclic process, q = -w

Answer

Answer: (d) For a cyclic process, q = -w
Explanation:
(1) ΔU = q + w. For an isochoric process, w = −PΔV = 0. Hence, ΔU = qv
(2) For an adiabatic process, q = 0. Hence, ΔU = w
(3 ) For an isothermal process, ΔU = 0 Hence, q = −w
(4) For a cyclic process , ΔU = 0. Hence, q = −w


Question 17.
A system absorb 10 kJ of heat at constant volume and its temperature rises from 270°C to 370°C. The value of ∆ U is
(a) 100 kJ
(b) 10 kJ
(c) 0 kJ
(d) 1 kJ

Answer

Answer: (b) 10 kJ
Explanation:
At constant volume w = 0
Therefore, ∆ U = q = 10 kJ


Question 18.
The temperature of the system decreases in an ______.
(a) Adiabatic Compression
(b) Isothermal Expansion
(c) Isothermal Compression
(d) Adiabatic Expansion

Answer

Answer: (d) Adiabatic Expansion
Explanation:
In adiabatic process heat is neither added nor removed from system. So the work done by the system (expansion) in adiabatic process will result in decrease of internal energy of that system (from first law).
As internal energy is directly proportional to the change in temperature there will be temperature drop in an adiabatic process.


Question 19.
Which of the following salts will have maximum cooling effect when 0.5 mole of the salt is dissolved in same amount of water. Integral heat of solution at 298 K is given for each salt?
(a) KNO3 (∆H = 35.4 kJ mol-1)
(b) NaCl (∆H = 5.35 kJ mol-1)
(c) HBr (∆H = -83.3 kJ mol-1)
(d) KOH ( ∆H = -55.6 kJ mol-1)

Answer

Answer: (a) KNO3 (∆H = 35.4 kJ mol-1)
Explanation:
More the heat absorbed, more will be the cooling effect. Hence, more the positive value of ∆ H, more the cooling effect.


Question 20.
Standard enthalpy of vapourisation ΔHvap for water at 100oC is 40.66 kJmol-1. The internal energy of vapourisation of water at 100°C (in kJmol-1) is (Assume water vapour to behave like an ideal gas)
(a) 43.76
(b) 40.66
(c) 37.56
(d) -43.76

Answer

Answer: (c) 37.56
Explanation:
For gaseous reactants and products, we have a relation between standard enthalpy of vaporization (ΔHvap) and standard internal energy (ΔE) as-
ΔHvap = ΔE+ Δng​ RT whereas,
Δng = n2​ − n1, i.e., difference between no. of moles of reactant and product.
For vaporization of water,
H2O (l) → H2O(g)
​Therefore, Δng = 1 − 0 = 1
Therefore, ΔHvap = ΔE+ Δng RT
⇒ ΔE = ΔHvap − Δng
​RT = 40.63 − (1×8.314×10-3 × 373) = 37.53 KJ/mol
Hence the value ΔE for this process will be 37.53KJ/mol.


We hope the given NCERT MCQ Questions for Class 11 Chemistry Chapter 6 Thermodynamics with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Chemistry Thermodynamics MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Filed Under: MCQ Questions

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

  • CBSE Important Extra Questions for Class 12 Economics Chapter Wise
  • MCQ Questions for Class 12 Geography Chapter 5 Primary Activities with Answers
  • MCQ Questions for Class 12 Geography Chapter 4 Human Development with Answers
  • MCQ Questions for Class 12 Geography Chapter 3 Population Composition with Answers
  • MCQ Questions for Class 12 Geography Chapter 2 The World Population: Distribution, Density and Growth with Answers
  • Geography Class 12 Important Questions Chapter 7 Mineral and Energy Resources
  • Class 12 History Important Questions Chapter 13 Mahatma Gandhi and the Nationalist Movement: Civil Disobedience and Beyond
  • MCQ Questions for Class 12 Geography Chapter 1 Human Geography: Nature and Scope with Answers
  • Geography Class 12 Important Questions Chapter 6 Water Resources
  • MCQ Questions for Class 12 History Chapter 15 Framing the Constitution: The Beginning of a New Era with Answers
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