Access the latest CBSE Class 11 Chemistry Thermodynamics Worksheet Set B. We have provided free printable Class 11 Chemistry worksheets in PDF format, specifically designed for Chapter 5 Thermodynamics. These practice sets are prepared by expert teachers following the 2025-26 syllabus and exam patterns issued by CBSE, NCERT, and KVS.
Chapter 5 Thermodynamics Chemistry Practice Worksheet for Class 11
Students should use these Class 11 Chemistry chapter-wise worksheets for daily practice to improve their conceptual understanding. This detailed test papers include important questions and solutions for Chapter 5 Thermodynamics, to help you prepare for school tests and final examination. Regular practice of these Class 11 Chemistry questions will help improve your problem-solving speed and exam accuracy for the 2026 session.
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Question. Enthalpy change for the reaction,
4H(g) → 2H2(g) is –869.6 kJ
The dissociation energy of H – H bond is
(a) – 434.8 kJ
(b) – 869.6 kJ
(c) + 434.8 kJ
(d) + 217.4 kJ
Answer. C
Question. Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 kJ mol–1 respectively. Enthalpy of formation of HCl is
(a) –93 kJ mol–1
(b) 245 kJ mol–1
(c) 93 kJ mol–1
(d) –245 kJ mol–1
Answer. A
Question. Consider the following reactions :
(i) H+ (aq) + OH–(aq) = H2O(l), ΔH = –X1 kJ mol–1
(ii) H2(g) + 1/2O2(g) = H2O(l), ΔH = –X2 kJ mol–1
(iii) CO2(g) + H2(g) = CO(g) + H2O(l),ΔH = –X3 kJ mol–1
(iv) C2H2(g) + 5/2O2(g) = 2CO2(g) + H2O(l),ΔH = +X4 kJ mol–1
Enthalpy of formation of H2O(l) is
(a) +X3 kJ mol–1
(b) –X4 kJ mol–1
(c) +X1 kJ mol–1
(d) –X2 kJ mol–1.
Answer. D
Question. Given that bond energies of H – H and Cl – Cl are 430 kJ mol–1 and 240 kJ mol–1 respectively and DHf for HCl is –90 kJ mol–1, bond enthalpy of HCl is
(a) 380 kJ mol–1
(b) 425 kJ mol–1
(c) 245 kJ mol–1
(d) 290 kJ mol–1
Answer. B
Question. The absolute enthalpy of neutralisation of the reaction :
MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) will be
(a) –57.33 kJ mol–1
(b) greater than –57.33 kJ mol–1
(c) less than –57.33 kJ mol–1
(d) 57.33 kJ mol–1
Answer. C
Question. If the bond energies of H–H, Br–Br, and H–Br are 433, 192 and 364 kJ mol–1 respectively, the ΔH° for the reaction H2(g) + Br2(g) → 2HBr(g) is
(a) –261 kJ
(b) +103 kJ
(c) +261 kJ
(d) –103 kJ
Answer. D
Question. For the reaction, 2Cl(g) → Cl2(g), the correct option is
(a) ΔrH > 0 and ΔrS > 0
(b) ΔrH > 0 and ΔrS < 0
(c) ΔrH < 0 and ΔrS > 0
(d) ΔrH < 0 and ΔrS < 0
Answer. D
Question. In which case change in entropy is negative?
(a) 2H(g) → H2(g)
(b) Evaporation of water
(c) Expansion of a gas at constant temperature
(d) Sublimation of solid to gas
Answer. A
Question. For a given reaction, DH = 35.5 kJ mol–1 and DS = 83.6 J K–1 mol–1. The reaction is spontaneous at (Assume that ΔH and ΔS do not vary with temperature.)
(a) T > 425 K
(b) all temperatures
(c) T > 298 K
(d) T < 425 K
Answer. A
Question.The correct thermodynamic conditions for the spontaneous reaction at all temperatures is
(a) ΔH < 0 and ΔS > 0
(b) ΔH < 0 and ΔS < 0
(c) ΔH < 0 and ΔS = 0
(d) ΔH > 0 and ΔS < 0
Answer. A,C
Question. For the reaction, X2O4(l) → 2XO2(g) ΔU = 2.1 kcal, ΔS = 20 cal K–1 at 300 K
Hence, DG is
(a) 2.7 kcal
(b) – 2.7 kcal
(c) 9.3 kcal
(d) – 9.3 kcal
Answer. B
Question. A reaction having equal energies of activation for forward and reverse reactions has
(a) ΔH = 0
(b) ΔH = ΔG = ΔS = 0
(c) ΔS = 0
(d) ΔG = 0
Answer. A
Question. The enthalpy of fusion of water is 1.435 kcal/mol.The molar entropy change for the melting of ice at 0°C is
(a) 10.52 cal/(mol K)
(b) 21.04 cal/(mol K)
(c) 5.260 cal/(mol K)
(d) 0.526 cal/(mol K)
Answer. C
Question. If the enthalpy change for the transition of liquid water to steam is 30 kJ mol–1 at 27°C, the entropy change for the process would be
(a) 10 J mol–1 K–1
(b) 1.0 J mol–1 K–1
(c) 0.1 J mol–1 K–1
(d) 100 J mol–11 K–1
Answer. D
Question. Standard entropies of X2, Y2 and XY3 are 60, 40 and 50 J K–1 mol–1 respectively. For the reaction 1/2X2 + 3/2Y2 ⇔ XY3, ΔH = –30 kJ, to be at equilibrium, the temperature should be
(a) 750 K
(b) 1000 K
(c) 1250 K
(d) 500 K
Answer. A
Question. For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJ mol–1 and 108.8 J K–1 mol–1, respectively.
The temperature when Gibbs’ energy change (DG) for this transformation will be zero, is
(a) 273.4 K
(b) 393.4 K
(c) 373.4 K
(d) 293.4 K
Answer. C
Question. The values of DH and DS for the reaction, C(graphite) + CO2(g) → 2CO(g) are 170 kJ and 170 J K–1, respectively. This reaction will be spontaneous at
(a) 910 K
(b) 1110 K
(c) 510 K
(d) 710 K
Answer. B
Question. For the gas phase reaction,
PCl5(g) ⇔ PCl3(g) + Cl2(g)
which of the following conditions are correct?
(a) DH < 0 and DS < 0
(b) DH > 0 and DS < 0
(c) DH = 0 and DS < 0
(d) DH > 0 and DS > 0
Answer. D
Question. Identify the correct statement for change of Gibbs’ energy for a system (DGsystem) at constant temperature and pressure.
(a) If DGsystem < 0, the process is not spontaneous.
(b) If DGsystem > 0, the process is spontaneous.
(c) If DGsystem = 0, the system has attained equilibrium.
(d) If DGsystem = 0, the system is still moving in a particular direction.
Answer. C
Question. The enthalpy and entropy change for the reaction:
Br2(l) + Cl2(g) → 2BrCl(g)
are 30 kJ mol–1 and 105 J K–1 mol–1 respectively.
The temperature at which the reaction will be in equilibrium is
(a) 300 K
(b) 285.7 K
(c) 273 K
(d) 450 K
Answer. B
Question. Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?
(a) Exothermic and increasing disorder
(b) Exothermic and decreasing disorder
(c) Endothermic and increasing disorder
(d) Endothermic and decreasing disorder
Answer. A
Question. A reaction occurs spontaneously if
(a) TΔS < ΔH and both ΔH and ΔS are +ve
(b) TΔS > ΔH and ΔH is +ve and ΔS is –ve
(c) TΔS > ΔH and both ΔH and ΔS are +ve
(d) TΔS = ΔH and both ΔH and ΔS are +ve
Answer. C
Question. Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are –382.64 kJ mol–1 and –145.6 J mol–1, respectively.
Standard Gibbs’ energy change for the same reaction at 298 K is
(a) – 221.1 kJ mol–1
(b) – 339.3 kJ mol–1
(c) – 439.3 kJ mol–1
(d) – 523.2 kJ mol–1
Answer. B
Question. Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is
(a) ΔSsystem + ΔSsurroundings > 0
(b) ΔSsystem – ΔSsurroundings > 0
(c) ΔSsystem > 0 only
(d) ΔSsurroundings > 0 only.
Answer. A
Question. What is the entropy change (in J K–1 mol–1) when one mole of ice is converted into water at 0°C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol–1 at 0°C.)
(a) 20.13
(b) 2.013
(c) 2.198
(d) 21.98
Answer. D
Question. The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm–3, respectively. If the standard free energy difference (DG°) is equal to 1895 J mol–1, the pressure at which graphite will be transformed into diamond at 298 K is
(a) 9.92 × 108 Pa
(b) 9.92 × 107 Pa
(c) 9.92 × 106 Pa
(d) 9.92 × 105 Pa
Answer. A
Question. Unit of entropy is
(a) J K–1 mol–1
(b) J mol–1
(c) J–1K–1 mol–1
(d) J K mol–1
Answer. A
Question. 2 moles of ideal gas at 27°C temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change.
(R = 2 cal/mol K)
(a) 92.1
(b) 0
(c) 4
(d) 9.2
Answer. D
Question. PbO2 → PbO; ΔG298 < 0
SnO2 → SnO; ΔG298 > 0
Most probable oxidation state of Pb and Sn will be
(a) Pb4+, Sn4+
(b) Pb4+, Sn2+
(c) Pb2+, Sn2+
(d) Pb2+, Sn4+
Answer. D
Question. Cell reaction is spontaneous when
(a) ΔG° is negative
(b) ΔG° is positive
(c) ΔE°red is positive
(d) ΔE°red is negative.
Answer. A
Question. Identify the correct statement regarding entropy.
(a) At absolute zero of temperature, the entropy of all crystalline substances is taken to be zero.
(b) At absolute zero of temperature, the entropy of a perfectly crystalline substance is +ve.
(c) At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero.
(d) At 0°C, the entropy of a perfectly crystalline substance is taken to be zero.
Answer. C
Question. Following reaction occurring in an automobile
2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)
The sign of ΔH, ΔS and ΔG would be
(a) –, +, +
(b) +, +, –
(c) +, –, +
(d) –, +, –
Answer. D
Short Answer
Question. What are the necessary conventions regarding thermochemical equations?
Answer. The necessary conventions regarding thermochemical equations are:
1. The coefficients in a balanced thermochemical equation refer to the number of moles of reactants and products involved in the reaction.
2. The numerical value of Δr HƟ refers to the number of moles of substances specified by an equation.
Question. Distinguish between ionization energy and electron affinity?
Answer. 1. Ionization energy is the amount of energy needed by a gaseous atom in order to remove an electron from its outermost orbital whereas electron affinity is the amount of energy released when a neutral atom or molecule gains an electron from outside.
2. Ionization energy is used to describe electron removing whereas electron affinity is used to describe electron gaining.
3. Ionization energy describes the absorption of energy from outside whereas electron affinity describes the release of energy to the surrounding.
Question. What do you mean by the term standard enthalpy of formation?
Answer. The standard enthalpy of formation of a compound is the change of enthalpy during the formation of one mole of the substance from its constituent elements, with all substances in their standard states.
Question. Why Gibbs energy is free at equilibrium?
Answer. Gibbs free energy is a measure of how much potential a reaction has left to do a net something. So if the free energy is zero, then the reaction is at equilibrium, and no more work can be done. It may be easier to see this using an alternative form of the Gibbs free energy, such as ΔG=−TΔS.
Question. What do you mean by the term Born-Haber cycle?
Answer. The Born Haber cycle is an approach to analyse reaction energies. Born Haber cycle is mainly used to calculate the lattice energy. It also involves several steps or processes such as electron affinity, ionization energy, sublimation energy, the heat of formation and dissociation energy.
Long Answer
Question. Explain second and third law of thermodynamics?
Answer. Second law thermodynamics: The Second Law of Thermodynamics states that processes that involve the transfer or conversion of heat energy are irreversible. It is very important because it talks about entropy and as we have discussed, entropy dictates whether or not a process or a reaction is going to be spontaneous. The change in entropy delta S is equal to the heat transfer delta Q divided by the temperature T, for example: Foods like carbohydrates and fats, and liquid fuels like gasoline petrol for some of us, have highly concentrated potential energy stored in their chemical bonds. They have a lot of energy in a fairly small space and it is efficient for us to convert that concentrated energy into useful energy to keep our bodies and our machines going. Third law of thermodynamics: The third law of thermodynamics states that as the temperature approaches absolute zero in a system, the absolute entropy of the system approaches a constant value. It helps in the calculation of the absolute entropy of a substance at any temperature T. These determinations are based on the heat capacity measurements of the substance. For any solid, let S0 be the entropy at 0 K and S be the entropy at T K, for example: turning on a light would seem to produce energy; however, it is electrical energy that is converted.
Question. Difference between enthalpy and entropy?
Answer. 1. Enthalpy is the thermodynamic quantity equivalent to the total heat content of a system whereas entropy represents the unavailability of a system’s thermal energy for conversion into mechanical work.
2. Enthalpy relates to the first law of thermodynamics whereas entropy relates to the second law of thermodynamics.
3. In enthalpy it gives the heat transfer takes place in constant pressure whereas in entropy it gives the idea of the randomness of the system.
4. Enthalpy can use to measure the change in the energy of the system after the reaction whereas entropy can use to measure the degree of the disorder of the system after the reaction.
Question. Explain Hess’s law of constant heat summation?
Answer. Hess's law of constant heat summation states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function. Reaction enthalpy changes can be determined by calorimetry for many reactions. The values are usually stated for processes with the same initial and final temperatures and pressures, although the conditions can vary during the reaction. Hess' law can be used to determine the overall energy required for a chemical reaction, when it can be divided into synthetic steps that are individually easier to characterize. This affords the compilation of standard enthalpies of formation that may be used as a basis to design complex syntheses. Hess' Law of Constant Heat Summation is useful in the determination of enthalpies of the following:
1. Heats of formation of unstable intermediates like CO (g) and NO (g).
2. Heat changes in phase transitions and allotropic transitions.
3. Lattice energies of ionic substances by constructing Born-Haber cycles if the electron affinity to form the anion is known.
4. Electron affinities using a Born-Haber cycle with a theoretical lattice energy.
Question. Write short note on spontaneity in thermodynamics?
Answer. A spontaneous process is the time-evolution of a system in which it releases free energy and it moves to a lower, more thermodynamically stable energy state, for cases involving an isolated system where no energy is exchanged with the surroundings, spontaneous processes are characterized by an increase in entropy. Spontaneous reactions are accompanied by an increase in overall entropy or disorder. Criteria for spontaneity in terms of free energy change: (i) If ΔG is negative, the process is spontaneous. (ii) If ΔG is positive, the direct process is non-spontaneous. (iii) If ΔG is zero, the process is in equilibrium. Entropy is a universal criterion of spontaneity. Most of the chemical process takes place at constant temperature and pressure. During course of every spontaneous process, Gibb's function decreases. A spontaneous process is an irreversible process but we can actually reverse it by the application of some external agents. The entropy of any system is the amount of randomness in it. Total entropy change is the essential parameter which defines the spontaneity of any process. Since most of the chemical reactions fall under the category of a closed system and open system; we can say there is a change in enthalpy too along with the change in entropy. So, change in enthalpy also increases or decreases the randomness by affecting the molecular motions, entropy change alone cannot account for the spontaneity of such a process.
Question. What are the different types of reactions for enthalpies explain?
Answer. The different types of reactions for enthalpies are:
1. Standard enthalpy of combustion: The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one mole of the compound is burned completely in oxygen under the standard conditions and with everything in its standard state. Enthalpies of combustion can be used to compare which fuels or substances release the most energy when they are burned. They can be calculated using a bomb calorimeter.
2. Enthalpy of atomization: The enthalpy of atomization is the enthalpy change that accompanies the total separation of all atoms in a chemical substance.Enthalpy of atomisation (ΔatH) is enthalpy change involved when one mole of gaseous atoms is formed from a substance in elemental state. Enthalpy is the energy required to break one mole of bond to give separated atoms in gaseous state.
3. Bond enthalpy: Bond enthalpy is the enthalpy change when one mole of gaseous molecules each breaks a covalent bond to form two free radicals, averaged over a range of compounds.
4. Lattice enthalpy: Lattice enthalpy is simply the change in enthalpy associated with the formation of one mole of an ionic compound from its gaseous ions under standard conditions. The lattice energy is exothermic, i.e., the value of ΔHlattice is negative because it corresponds to the coalescing of infinitely separated gaseous ions in vacuum to form the ionic lattice. The lattice enthalpy is reported as a positive value.
5. Enthalpy of solution: The enthalpy of solution is the enthalpy change associated with the dissolution of a substance in a solvent at constant pressure resulting in infinite dilution. The enthalpy of solution is most often expressed in kJ/mol at constant temperature.
Important Practice Resources for Class 11 Chemistry
Chapter 5 Thermodynamics CBSE Class 11 Chemistry Worksheet
Students can use the Chapter 5 Thermodynamics practice sheet provided above to prepare for their upcoming school tests. This solved questions and answers follow the latest CBSE syllabus for Class 11 Chemistry. You can easily download the PDF format and solve these questions every day to improve your marks. Our expert teachers have made these from the most important topics that are always asked in your exams to help you get more marks in exams.
NCERT Based Questions and Solutions for Chapter 5 Thermodynamics
Our expert team has used the official NCERT book for Class 11 Chemistry to create this practice material for students. After solving the questions our teachers have also suggested to study the NCERT solutions which will help you to understand the best way to solve problems in Chemistry. You can get all this study material for free on studiestoday.com.
Extra Practice for Chemistry
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