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NCERT Exemplar solutions for Chemistry [English] Class 11 chapter 6 - Thermodynamics [Latest edition]

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NCERT Exemplar solutions for Chemistry [English] Class 11 chapter 6 - Thermodynamics - Shaalaa.com
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Solutions for Chapter 6: Thermodynamics

Below listed, you can find solutions for Chapter 6 of CBSE, Karnataka Board PUC NCERT Exemplar for Chemistry [English] Class 11.


Multiple Choice Questions (Type - I)
Multiple Choice Questions (Type - I) [Pages 68 - 77]

NCERT Exemplar solutions for Chemistry [English] Class 11 6 Thermodynamics Multiple Choice Questions (Type - I) [Pages 68 - 77]

1Page 68

Thermodynamics is not concerned about ______.

  • Energy changes involved in a chemical reaction.

  • The extent to which a chemical reaction proceeds.

  • The rate at which a reaction proceeds.

  • The feasibility of a chemical reaction

2Page 68

Which of the following statements is correct?

  • The presence of reacting species in a covered beaker is an example of open system.

  • There is an exchange of energy as well as matter between the system and the surroundings in a closed system.

  • The presence of reactants in a closed vessel made up of copper is an example of a closed system.

  • The presence of reactants in a thermos flask or any other closed insulated vessel is an example of a closed system.

3Page 68

The state of a gas can be described by quoting the relationship between ______.

  • Pressure, Volume, Temperature

  • Temperature, Amount, Pressure

  • Amount, Volume, Temperature

  • Pressure, Volume, Temperature, Amount

4Page 68

The volume of gas is reduced to half from its original volume. The specific heat will be ______.

  • reduce to half

  • be doubled

  • remain constant

  • increase four times

5Page 69

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is ______.

  • \[\ce{2C4H10 (g) + 1302 (g) -> 8CO2 (g) + 10H2O (l) ∆_cH = - 2658.0 kJ mol^{-1}}\]

  • \[\ce{ C4H10 (g) + 13/2 O2 (g) -> 4CO2 (g) + 5H2O (g) ∆_CH = -1329.0 kJ mol^{-1}}\]

  • \[\ce{C4H10 (g) + 13/2 O2 (g) -> 4CO2 (g) + 5H2O (l) ∆_cH = - 2658.0 kJ mol^{-1}}\]

  • \[\ce{C4H10 (g) + 13/2 O2 (g) -> 4CO2 (g) + 5H2O (l) ∆_cH = + 2658.0 kJ mol^{-1}}\]

6Page 69

\[\ce{∆_f U^Θ}\] of formation of \[\ce{CH4 (g)}\] at certain temperature is – 393 kJ mol–1. The value of \[\ce{∆_f H^Θ}\] is ______.

  • zero

  • \[\ce{< ∆_f U^Θ}\]

  • \[\ce{> ∆_f U^Θ}\]

  • equal to \[\ce{∆_f U^Θ}\]

7Page 69

In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

  • q = 0, ∆T ≠ 0, w = 0

  • q ≠ 0, ∆T = 0, w = 0

  • q = 0, ∆T = 0, w = 0

  • q = 0, ∆T < 0, w ≠ 0

8Page 69

The pressure-volume work for an ideal gas can be calculated by using the expression w = `- int_(v_i)^(v_f) p_(ex)  dV`. The work can also be calculated from the pV– plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf. choose the correct option.

  • w (reversible) = w (irreversible)

  • w (reversible) < w (irreversible)

  • w (reversible) > w (irreversible)

  • w (reversible) = w (irreversible) + `P_(ex).∆V`

9Page 70

The entropy change can be calculated by using the expression ∆S = `q_(rev)/T`. When water freezes in a glass beaker, choose the correct statement amongst the following :

  • ∆S (system) decreases but ∆S (surroundings) remains the same.

  • ∆S (system) increases but ∆S (surroundings) decreases.

  • ∆S (system) decreases but ∆S (surroundings) increases.

  • ∆S (system) decreases and ∆S (surroundings) also decreases.

10Page 70

On the basis of thermochemical equations (a), (b) and (c), find out which of the algebric relationships given in options (i) to (iv) is correct.

(a) \[\ce{C (graphite) + O2 (g) -> CO2 (g) ; ∆_rH = xkJ mol^{-1}}\]

(b) \[\ce{C (graphite) + 1/2 O2 (g) -> CO (g) ; ∆_rH = ykJ mol^{-1}}\]

(c) \[\ce{CO (g) + 1/2 O2 (g) -> CO2 (g) ; ∆_r H = zkJ mol^{-1}}\]

  • z = x + y

  • x = y – z

  • x = y + z

  • y = 2z – x

11Page 70

Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (i) to (iv) is correct?

(a) \[\ce{C (g) + 4 H (g) -> CH4 (g); ∆_rH = xkJ mol^{-1}}\]

(b) \[\ce{C (graphic) + 2H2 (g) -> CH4 (g); ∆_rH = ykJ mol^{-1}}\]

  • x = y

  • x = 2y

  • x > y

  • x < y

12Page 70

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound ______.

  • is always negative

  • is always positive

  • may be positive or negative

  • is never negative

13Page 71

Enthalpy of sublimation of a substance is equal to ______.

  • Enthalpy of fusion + Enthalpy of vapourisation

  • Enthalpy of fusion

  • Enthalpy of vapourisation

  • Twice the enthalpy of vapourisation

14Page 71

Which of the following is not correct?

  • ∆G is zero for a reversible reaction

  • ∆G is zero for a reversible reaction

  • ∆G is negative for a spontaneous reaction

  • ∆G is positive for a non-spontaneous reaction

15Page 71

Thermodynamics mainly deals with:

(i) interrelation of various forms of energy and their transformation from one form to another.

(ii) energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.

(iii) how and at what rate these energy transformations are carried out.

(iv) the system in equilibrium state or moving from one equilibrium state to another equilibrium state.

16Page 71

In an exothermic reaction, heat is evolved, and system loses heat to the surrounding. For such system:

(i) qp will be negative

(ii) ∆rH will be negative

(iii) qp will be positive

(iv) ∆rH will be positive

17Page 71

The spontaneity means, having the potential to proceed without the assistance of external agency. The processes which occur spontaneously are:

(i) Flow of heat from colder to warmer body.

(ii) Gas in a container contracting into one corner.

(iii) Gas expanding to fill the available volume.

(iv) Burning carbon in oxygen to give carbon dioxide.

18Page 71

For an ideal gas, the work of reversible expansion under isothermal condition can be calculated by using the expression w = `- nRT` In `V_f/V_i`. A sample containing 1.0 mol of an ideal gas is expanded isothermally and reversibly to ten times of its original volume, in two separate experiments. The expansion is carried out at 300 K and at 600 K respectively.

(i) Work done at 600 K is 20 times the work done at 300 K.

(ii) Work done at 300 K is twice the work done at 600 K.

(iii) Work done at 600 K is twice the work done at 300 K.

(iv) ∆U = 0 in both cases.

19Page 72

Consider the following reaction between zinc and oxygen and choose the correct options out of the options given below:

\[\ce{2 Zn (s) + O2 (g) -> 2 ZnO (s); ∆H = - 693.8 kJ mol^{-1}}\]

(i) The enthalpy of two moles of \[\ce{ZnO}\] is less than the total enthalpy of two moles of \[\ce{Zn}\] and one mole of oxygen by 693.8 kJ.

(ii) The enthalpy of two moles of \[\ce{ZnO}\] is more than the total enthalpy of two moles of \[\ce{Zn}\] and one mole of oxygen by 693.8 kJ.

(iii) 693.8 kJ mol–1 energy is evolved in the reaction.

(iv) 693.8 kJ mol–1 energy is absorbed in the reaction.

20Page 72

18.0 g of water completely vapourises at 100°C and 1 bar pressure and the enthalpy change in the process is 40.79 kJ mol–1. What will be the enthalpy change for vapourising two moles of water under the same conditions? What is the standard enthalphy of vapourisation for water?

21Page 72

One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?

22Page 72

Standard molar enthalpy of formation, ∆fHΘ is just a special case of enthalpy of reaction, ∆rHΘ. Is the ∆rHΘ for the following reaction same as ∆fHΘ? Give reason for your answer.

\[\ce{CaO(s) + CO2(g) -> CaCO3(s); ∆_fH^Θ = - 178.3 kJ mol^{-1}}\]

23Page 72

The value of ∆fHΘ for \[\ce{NH3}\] is – 91.8 kJ mol–1. Calculate enthalpy change for the following reaction:

\[\ce{2NH3(g) -> N2(g) + 3H2(g)}\]

24Page 72

Enthalpy is an extensive property. In general, if enthalpy of an overall reaction A→B along one route is ∆rH and ∆rH1, ∆rH2, ∆rH3 ..... represent enthalpies of intermediate reactions leading to product B. What will be the relation between ∆rH for overall reaction and ∆rH1, ∆rH2 ..... etc. for intermediate reactions.

25Page 72

The enthalpy of atomisation for the reaction \[\ce{CH4(g) -> C(g) + 4H(g)}\] is 1665 kJ mol–1. What is the bond energy of \[\ce{C – H}\] bond?

26Page 73

Use the following data to calculate \[\ce{∆_{lattice}H^Θ}\] for \[\ce{NaBr}\].

\[\ce{∆_{sub}H^Θ}\] for sodium metal = 108.4 kJ mol–1

Ionization enthalpy of sodium = 496 kJ mol–1

Electron gain enthalpy of bromine = – 325 kJ mol–1

Bond dissociation enthalpy of bromine = 192 kJ mol–1

\[\ce{∆_fH^Θ}\] for \[\ce{NaBr (s)}\] = – 325 kJ mol–1 

27Page 73

Given that ∆H = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?

28Page 73

Heat has randomising influence on a system and temperature is the measure of average chaotic motion of particles in the system. Write the mathematical relation which relates these three parameters.

29Page 73

Increase in enthalpy of the surroundings is equal to decrease in enthalpy of the system. Will the temperature of system and surroundings be the same when they are in thermal equilibrium?

30Page 73

At 298 K. Kp for the reaction \[\ce{N2 O4 (g) ⇌ 2NO2 (g)}\] is 0.98. Predict whether the reaction is spontaneous or not.

31Page 73

A sample of 1.0 mol of a monoatomic ideal gas is taken through a cyclic process of expansion and compression as shown in figure 6.1. What will be the value of ∆H for the cycle as a whole?

32Page 73

The standard molar entropy of \[\ce{H2O (l)}\] is 70 JK–1 mol–1. Will the standard molar entropy of \[\ce{H2O (s)}\] be more, or less than 70 JK–1 mol–1?

33Page 73

Identify the state and path functions out of the following:

  1. Enthalpy
  2. Entropy
  3. Heat
  4. Temperature
  5. Work
  6. Free energy
34Page 73

The molar enthalpy of vapourisation of acetone is less than that of water. Why?

35Page 73

Which quantity out of ∆rG and ∆rGΘ will be zero at equilibrium?

36Page 73

Predict the change in internal energy for an isolated system at constant volume.

37Page 74

Although heat is a path function but heat absorbed by the system under certain specific conditions is independent of path. What are those conditions? Explain.

38Page 74

Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?

39Page 74

Heat capacity (Cp) is an extensive property but specific heat (c) is an intensive property. What will be the relation between Cp and c for 1 mol of water?

40Page 74

The difference between CP and CV can be derived using the empirical relation H = U + pV. Calculate the difference between CP and CV for 10 moles of an ideal gas.

41Page 74

If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.

42Page 74

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.

\[\ce{H2(g) + Br(g) -> 2HBr(g)}\]

Given that Bond energy of \[\ce{H2, Br2}\] and \[\ce{HBr}\] is 435 kJ mol–1, 192 kJ mol–1 and 368 kJ mol–1 respectively.

43Page 74

The enthalpy of vapourisation of \[\ce{CCl4}\] is 30.5 kJ mol–1. Calculate the heat required for the vapourisation of 284 g of \[\ce{CCl4}\] at constant pressure. (Molar mass of \[\ce{CCl4}\] = 154 g mol–1).

44Page 74

The enthalpy of reaction for the reaction: \[\ce{2H2 (g) + O2(g) -> 2H2O (l)}\] is ∆rHΘ = – 572 kJ mol–1. What will be standard enthalpy of formation of \[\ce{H2O (l)}\]?

45Page 74

What will be the work done on an ideal gas enclosed in a cylinder, when it is compressed by a constant external pressure, pext in a single step as shown in figure. Explain graphically.

46Page 74

How will you calculate work done on an ideal gas in a compression, when change in pressure is carried out in infinite steps?

47Page 74

Represent the potential energy/enthalpy change in the following processes graphically.

(a) Throwing a stone from the ground to roof.

(b) \[\ce{1/2 H2(g) + 1/2 Cl2 (g) ⇌ HCl (g) Δ_rH^Θ = - 92.32 kJ mol^{-1}}\]

In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?

48Page 75

Enthalpy diagram for a particular reaction is given in figure. Is it possible to decide spontaneity of a reaction from given diagram. Explain.

49Page 75

1.0 mol of a monoatomic ideal gas is expanded from state (1) to state (2) as shown in figure. Calculate the work done for the expansion of gas from state (1) to state (2) at 298 K.

50Page 75

An ideal gas is allowed to expand against a constant pressure of 2 bar from 10 L to 50 L in one step. Calculate the amount of work done by the gas. If the same expansion were carried out reversibly, will the work done be higher or lower than the earlier case? (Given that 1 L bar = 100 J)

51Page 75

Match the following :

A B
(i) Adiabatic process (a) Heat
(ii) Isolated system (b) At constant volume
(iii) Isothermal change (c) First law of thermodynamics
(iv) Path function (d) No exchange of energy and matter
(v) State function (e) No transfer of heat
(vi) ΔU = q (f) Constant temperature
(vii) Law of conservation of energy (g) Internal energy
(viii) Reversible process (h) Pext = o
(ix) Free expansion (i) At constant pressure
(x) ΔH = q (j) Infinitely slow process which proceeds through a series of equilibrium states.
(xi) Intensive property (k) Entropy
(xii) Extensive property (l) Pressure
  (m) Specific heat
52Page 76

Match the following processes with entropy change:

Reaction Entropy change
(i) A liquid vapourises (a) ∆S = 0
(ii) Reaction is non-spontaneous
at all temperatures and ∆H
is positive
(b) ∆S = positive
(iii) Reversible expansion of an ideal gas (c) ∆S = negative
53Page 76

Match the following parameters with description for spontaneity 

∆ (Parameters)
\[\ce{∆_rH^Θ, ∆_rS^Θ, ∆_rG^Θ}\]
Description
(i) \[\ce{+ – +}\] (a) Non-spontaneous at high temperature
(ii) \[\ce{– – + at high T}\] (b) Spontaneous at all temperatures
(iii) \[\ce{– + –}\] (c) Non-spontaneous at all temperatures
54Page 76

Match the following :

Column I Column II
(i) Entropy of vapourisation (a) decreases
(ii) K for spontaneous process (b) is always positive
(iii) Crystalline solid state (c) lowest entropy
(iv) ∆U in adiabatic expansion of ideal gas (d) `(∆H_(vap))/T_b`
55Page 76

Assertion (A): Combustion of all organic compounds is an exothermic reaction.

Reason (R): The enthalpies of all elements in their standard state are zero.

  • Both A and R are true and R is the correct explanation of A.

  • Both A and R are true but R is not the correct explanation of A.

  • A is true but R is false.

  • A is false but R is true.

56Page 77

Assertion (A): Spontaneous process is an irreversible process and may be reversed by some external agency.

Reason (R): Decrease in enthalpy is a contributory factor for spontaneity.

  • Both A and R are true and R is the correct explanation of A.

  • Both A and R are true but R is not the correct explanation of A.

  • A is true but R is false.

  • A is false but R is true.

57Page 77

Assertion (A): A liquid crystallises into a solid and is accompanied by decrease in entropy.

Reason (R): In crystals, molecules organise in an ordered manner.

  • Both A and R are true and R is the correct explanation of A.

  • Both A and R are true but R is not the correct explanation of A.

  • A is true but R is false.

  • A is false but R is true.

58Page 77

Derive the relationship between ∆H and ∆U for an ideal gas. Explain each term involved in the equation.

59Page 77

Extensive properties depend on the quantity of matter but intensive properties do not. Explain whether the following properties are extensive or intensive.

60Page 77

The lattice enthalpy of an ionic compound is the enthalpy when one mole of an ionic compound present in its gaseous state, dissociates into its ions. It is impossible to determine it directly by experiment. Suggest and explain an indirect method to measure lattice enthalpy of \[\ce{NaCl(s)}\].

61Page 77

\[\ce{∆G}\] is net energy available to do useful work and is thus a measure of “free energy”. Show mathematically that ∆G is a measure of free energy. Find the unit of \[\ce{∆G}\]. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?

62Page 77

Graphically show the total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from \[\ce{(p_i, V_i)}\] to \[\ce{(p_f , V_f )}\]. With the help of a pV plot compare the work done in the above case with that carried out against a constant external pressure \[\ce{p_f}\].

Solutions for 6: Thermodynamics

Multiple Choice Questions (Type - I)
NCERT Exemplar solutions for Chemistry [English] Class 11 chapter 6 - Thermodynamics - Shaalaa.com

NCERT Exemplar solutions for Chemistry [English] Class 11 chapter 6 - Thermodynamics

Shaalaa.com has the CBSE, Karnataka Board PUC Mathematics Chemistry [English] Class 11 CBSE, Karnataka Board PUC solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT Exemplar solutions for Mathematics Chemistry [English] Class 11 CBSE, Karnataka Board PUC 6 (Thermodynamics) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT Exemplar textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Chemistry [English] Class 11 chapter 6 Thermodynamics are Thermodynamic Terms, The State of the System, The Internal Energy as a State Function - Work, The Internal Energy as a State Function - Heat, The Internal Energy as a State Function - the General Case, Work, Calorimetric Measurement of ΔU and ΔH, Standard Enthalpy of Reactions, Enthalpy Changes During Phase Transformations, Standard Enthalpy of Formation, Thermochemical Equations, Hess’ Law of Constant Heat Summation, Is Decrease in Enthalpy a Criterion for Spontaneity, Entropy and Spontaneity, Gibbs Energy and Spontaneity, Enthalpy (H), Entropy and Second Law of Thermodynamics, Absolute Entropy and Third Law of Thermodynamics, Gibbs Energy Change and Equilibrium.

Using NCERT Exemplar Chemistry [English] Class 11 solutions Thermodynamics exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Exemplar Solutions are essential questions that can be asked in the final exam. Maximum CBSE, Karnataka Board PUC Chemistry [English] Class 11 students prefer NCERT Exemplar Textbook Solutions to score more in exams.

Get the free view of Chapter 6, Thermodynamics Chemistry [English] Class 11 additional questions for Mathematics Chemistry [English] Class 11 CBSE, Karnataka Board PUC, and you can use Shaalaa.com to keep it handy for your exam preparation.

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