Practice CBSE Class 11 Chemistry Thermodynamics MCQs Set D provided below. The MCQ Questions for Class 11 Chapter 5 Thermodynamics Chemistry with answers and follow the latest CBSE/ NCERT and KVS patterns. Refer to more Chapter-wise MCQs for CBSE Class 11 Chemistry and also download more latest study material for all subjects
MCQ for Class 11 Chemistry Chapter 5 Thermodynamics
Class 11 Chemistry students should review the 50 questions and answers to strengthen understanding of core concepts in Chapter 5 Thermodynamics
Chapter 5 Thermodynamics MCQ Questions Class 11 Chemistry with Answers
Question. The succeeding operations that enable this transformation of states are
(a) heating, cooling, heating, cooling
(b) cooling, heating, cooling, heating
(c) heating, cooling, cooling, heating
(d) cooling, heating, heating, cooling
Answer: c
Question. A gas expands from a volume of 1m3 to a volume of 2m3 against an external pressure of 105 Nm-2 . The work done by the gas will be
(a) 102 kJ
(b) 102 J
(c) 103 J
(d) 105 kJ
Answer: a
Question. A gas undergoes a thermodynamical process but the volume of the gas does not change. What is the nature of this process ?
(a) Isothermal
(b) Adiabatic
(c) Isobaric
(d) Isochoric
Answer: d
Question. An ideal gas expands in volume from 1 x 10-3 m3 to 1 x 10-2 m3 at 300 K against a constant pressure of 1 x 105 Nm-2 . The work done is
(a) – 9001
(b) – 900kJ
(c) 270kJ
(d) 900kJ
Answer: a
Question. An adiabatic process occurs in
(a) open system
(b) closed system
(c) isolated system
(d) in all the given system
Answer: c
Question. Which one of the following is an exothermic reaction?
(a) N2 (g) + O2 (g )+ 180. 8kJ → 2NO(g)
(b) N2 (g)+ 3H2 (g) – 92kJ → 2NH3 (g)
(c) C(g) + H2O(g) → CO(g) + H2 (g) – 131.1 kJ
(d) C(graphite)+2S(s) → CS2(l) – 91.9 kJ
Answer: b
Question. At 25°C, for the combustion of 1 mole of liquid benzene, the heat ofreaction at constant pressure is given by C6H6 (/) +15/2 O2 (g) → 6CO2 (g)+ 3H2O (/),
(ΔH = 780980cal)
Calculate the heat of reaction at constant volume
(a) 780.086 kcal
(b) – 780.086 kcal
(c) – 390.043 kcal
(d) 390.043 kcal
Answer: a
Question. A piston filled with 0.04 mole of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0°C. As it does so, it absorbs 208 J of heat. The values of q and W for the process will be (R = 8.314 J/mol K, In 7.5 = 2.01)
(a) q = + 208 J, W = – 208 J
(b) q = – 208 J, W = – 208 J
(c) q = – 208 J, W = + 208 J
(d) q = + 208 J, W = + 208 J
Answer: a
Question. IfC (s)+ O2(g) → CO2 (g);ΔH = r and l CO (g)+ 1/2 O2 → CO2 (g); ΔH =s, then the heat of 2 formation of CO is
(a) r+s
(b) r – s
(c) s – r
(d) rs
Answer: b
Question. C(s) + O2 (g) → CO2(g); ΔH = –94kcal
2CO(g) + O2 → 2CO2 (g); ΔH = –135.2 kcal
The heat of formation of CO (g) is
(a) – 26.4 kcal
(b) 41.2 kcal
(c) 26.4 kcal
(d) 229.2 kcal
Answer: a
Question. Internal energy is sum of
(a) kinetic energy and potential energy
(b) all types of energy of the system
(c) energy of internal system
(d) None of the above
Answer: b
Question. At 27° C one mole of an ideal gas is compressed isothermally and reversible from a pressure of 2 attn to 10 attn. The value of ΔE and q are ( R = 2 cal)
(a) 0, –965.84cal
(b) –965.84cal, –865.58cal
(c) + 865.58cal, –865.58cal
(d) +965.84cal, + 865.58cal
Answer: a
Question. Hess’s law is based on
(a) law of conservation of mass
(b) law of conservation of energy
(c) first law ofthem1odynamics
(d) None of the above
Answer: b
Question. The internal energy changes when a system goes from state A to B is 40 kJ/mol. If the system goes from A to B by a reversible path and returns to state A by an i.tTeversible path, what would be the net change in internal energy ?
(a) 40 kJ
(b) > 40 kJ
(c) < 40 kJ
(d) Zero
Answer: d
Question. Mark out the enthalpy of formation of carbon monoxide (CO)
Given, C(s) + 1/2 O2 (g) → CO(g),
ΔH = – 39.3 kJ/rnol
CO(g) + 1/2 O2 (g) → CO2 (g), ΔH = –282.8 kJ/ mol
(a) 110.5 kJ/mol
(b) 676.1 kJ/mol
(c) 282.8 kJ/mol
(d) 300.0 kJ/mol
Answer: a
Question. In an adiabatic process
(a) p.ΔV = 0
(b) q =+W
(c) ΔE = q
(d) q = 0
Answer: d
Question. A plot of In k against 1/T (abscissa) is expected to be a straight line with intercept on coordinate axis equal to
(a) ΔS°/2.303R
(b) ΔS°/RE
(c) -ΔS°/R
(d) R x ΔS°
Answer: b
Question. A container of 1.0 L capacity filled with 1.0 mole of ideal gas is connected to an evacuated vessel of 9.0 L. Calculate change in entropy. (R = l.987 cal)
(a) 0.188cal. K-1 mol-1
(b) 0.4576cal.K-1 mol-1
(c) 4.576cal. K-1 mol-1
(d) 4.366cal. K-1 mol-1
Answer: d
Question. If t.Gfor a reaction is negative, you infer that the change is
(a) spontaneous
(b) non-spontaneous
(c) reversible
(d) irrevrsible
(e) equilibrium
Answer: a
Question. Using the data provided, calculate the multiple bond energy (kJ mol-1 )of a C = Cbond C2H2 . The energy is
(take the bond energy of a C—H bond as 350 kJ mol-1)
2C (s)+ H2 (g) → C2H2 (g); ΔH = 225 kJ mol-1
2C(s) → 2C(g); ΔH = 1410 kJ mol-1
H2 (g) → 2H(g); ΔH= 330 kJ mol-1
(a) 1165
(b) 837
(c) 865
(d) 815
Answer: d
Question. The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows
2/3 Al2O3 ➔ 4/3 Al +O2 , ΔrG° = + 966 kJ mol-1
The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least
(a) 4.5 V
(b) 3.0 V
(c) 2.5 V
(d) 5.0 V
Answer: c
Question. Identify the correct statement from the following in a chemical reaction.
(a) The entropy always increases
(b) The change in entropy along with suitable change in enthalpy decides the rate of reaction
(c) The enthalpy always decreases
(d) Both the enthalpy and the entropy remain constant
Answer: d
Question. For an isolated system, ΔU = 0, then
(a) ΔS = 0
(b) ΔS <0
(c) ΔS > 0
(d) The values of ΔS cannot be predicted
Answer: c
Question. For the reaction at 298 K
A(g) + B(g) ⇌ C(g) + D(g)
ΔH° = – 29.8 kcal, ΔS° = – 0.100 kcal K-1
What is the value of ΔG° ?
(a) l
(b) 0
(c) 2
(d) 4
Answer: b
Question. During which phenomenon does entropy decrease ?
(a) Crystallisation of sucrose from its solution
(b) Melting of ice
(c) Sublimation of camphor
(d) Rusting of iron
Answer: a
Question. Among the following for spontaneity of chemical reaction there should be
(a) decrease in entropy and increase in free energy
(b) decrease in entropy and free energy both
(c) increase in entropy and decrease in free energy
(d) increase in entropy and free energy both
Answer: c
Question. The incorrect expression an1ong the following is t.G
(a) ΔGsystem / ΔSsystem= – T
(b) In 1sotherrmal process, Wreversible = – nRT ln Vf/vi
(c) In K = ΔG° – TΔS / RT
(d) K = e –ΔG°/RT
Answer: c
Question. Which of the following reactions define ΔG°f ?
(a) C(diamond) + O2 (g) → CO2 (g)
(b) 1/2 H2(g)+ 1/2 F2(g) → HF(g)
(c) H4P2O7 + H2O → 2H3PO4
(d) SO2 (g) + 1/2 O2 (g) → SO3 (g)
Answer: b
Question. If ΔG is negative, the reaction will be
(a) at equilibrium
(b) not possible
(c) Both (a) and (b)
(d) possible
Answer: d
Question. Which of the following processes is associated with decrease in entropy ?
(a) Vaporisation ofa mole of water into steam at its b.p.
(b) Dissociation of a mole of common salt in water at 300K
(c) Mixing of two partially miscible liquids
(d) Crystallisation of a salt from its saturated solution
Answer: d
Question. For a particular reversible reaction at temperature T, /ΔH and ΔS were found to be both +ve. lfTe is the temperature at equilibrium, the reaction would be spontaneous when
(a) Te > T
(b) T > Te
(c) Te is 5 times T
(d) T = Te
Answer: b
Question. A gas expands isothermally against a constant external pressure of 1 atm from a volume of 10 dm3 to a volwne of 20 dm3 . It absorbs 800 J of thermal energy from its surroundings. The ΔU is
(a) –3121
(b) +123J
(c) –213 J
(d) + 231 J
Answer: c
Question. An ideal gas is allowed to expand both reversibly and ineversibly in an isolated system. If Ti is the mitial temperatw·e and Tf is the final temperature, which of the following statements is correct ?
(a) (Tf)irrev > (Tf)rev
(b) Tf > Ti for reversible process but Tf = Ti for irreversible process
(c) (Tf)rev = (Tf) irrev
(d) Tf = Ti for both reversible and irreversible processes
Answer: a
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Important Practice Resources for Class 11 Chemistry
MCQs for Chapter 5 Thermodynamics Chemistry Class 11
Students can use these MCQs for Chapter 5 Thermodynamics to quickly test their knowledge of the chapter. These multiple-choice questions have been designed as per the latest syllabus for Class 11 Chemistry released by CBSE. Our expert teachers suggest that you should practice daily and solving these objective questions of Chapter 5 Thermodynamics to understand the important concepts and better marks in your school tests.
Chapter 5 Thermodynamics NCERT Based Objective Questions
Our expert teachers have designed these Chemistry MCQs based on the official NCERT book for Class 11. We have identified all questions from the most important topics that are always asked in exams. After solving these, please compare your choices with our provided answers. For better understanding of Chapter 5 Thermodynamics, you should also refer to our NCERT solutions for Class 11 Chemistry created by our team.
Online Practice and Revision for Chapter 5 Thermodynamics Chemistry
To prepare for your exams you should also take the Class 11 Chemistry MCQ Test for this chapter on our website. This will help you improve your speed and accuracy and its also free for you. Regular revision of these Chemistry topics will make you an expert in all important chapters of your course.
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