Question

NCERT Solutions Class 11 Chemistry Chapter 6 Thermodynamics is a great way to learn the solutions of the given chapter. Our NCERT Solutions are solved and prepared by the experts on the subject.

Our NCERT Solutions Class 11 have all the solutions students need. We have provided an in-depth analysis and solutions of every concept such as

• Thermodynamics – Thermodynamics is the study of the science of the relation between heat, work, temperature, and energy.
• Thermodynamics terms – Common thermodynamic terms are
  • Work – the energy required to move or lift a weight for a unit distance is called work.
  • Temperature – the quantifiable difference between warm and cold objects is termed temperature.
  • Specific heat – the amount of heat required to raise the temperature of any substance by 1 degree C is called specific heat.
  • Chemical energy – when two chemicals react they either give off heat (exothermic reaction) or required heat (endothermic reaction).
  • Electrical energy – the energy related to the flow of electrons through a conductor.
  • Energy – the ability to perform any task is called energy. There are various forms of energy: potential energy, kinetic energy, thermal energy, chemical energy, and nuclear energy.
  • Enthalpy – term of units of energy this combines internal energy with a pressure/volume or flow work term.
  • Entropy – the degree of randomization or disorder as the property of matter is called entropy
  • Heat – motion of energy due to the temperature difference.
  • Internal energy – the activity within the molecular structure typically observed with temperature measurement.
  • Kinetic energy – energy due to the motion of any object is termed kinetic energy.
  • Nuclear energy – the energy of atoms and their fundamental particles. There are two ways to gain energy from nuclear form. Nuclear fission and another one is nuclear fusion.
  • Potential energy – the energy possessed by any particle due to its position, height, or shape is called potential energy.
• Gibbs Energy Change and Equilibrium - The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for reversible reactions as well. If the system is at equilibrium then ΔG = 0.

Our NCERT Solutions Class 11 Chemistry is one of the best ways to prepare for the important exams like the CBSE board, state boards, and also for the tough competitive exams like JEE Mains, JEE Advance, NTSE, and Olympiad.

Answer 1

NCERT Solutions Class 11 Chemistry Chapter 6 Thermodynamics

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:

A thermodynamic state function is a quantity whose value is independent of a path. 

Functions like p, V, T etc. depend only on the state of a system and not on the path. 

Hence, alternative (ii) is correct.

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:

A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0. 

Therefore, alternative (iii) is correct.

3. The enthalpies of all elements in their standard states are: 

(i) unity 

(ii) zero 

(iii) < 0 

(iv) different for each element

Answer:

The enthalpy of all elements in their standard state is zero. 

Therefore, alternative (ii) is correct.

4. ∆U^θ of combustion of methane is – X kJ mol^–1 . The value of ∆H^θ is

(i) = ∆U^θ 

(ii) > ∆U^θ 

(iii) < ∆U^θ 

(iv) = 0

Answer:

Since ∆H^θ = ∆U^θ + ∆n_gRT and ∆U^θ = –X kJ mol^–1, 

∆H^θ = (–X) + ∆n_gRT. ⇒ ∆H^θ < ∆U^θ 

Therefore, alternative (iii) is correct.

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 CH_4(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:

According to the question,

Thus, the desired equation is the one that represents the formation of CH_4(g) i.e.,

\(\therefore \) Enthalpy of formation of CH4(g) = –74.8 kJ mol^–1 

Hence, alternative (i) is correct.

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:

For a reaction to be spontaneous, ∆G should be negative. 

∆G = ∆H – T∆S 

According to the question, for the given reaction, 

∆S = positive 

∆H = negative (since heat is evolved) 

⇒ ∆G = negative 

Therefore, the reaction is spontaneous at any temperature. 

Hence, alternative (iv) is correct.

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:

According to the first law of thermodynamics, 

∆U = q + W    (i)

Where, 

∆U = change in internal energy for a process q = heat 

W = work 

Given, 

q = + 701 J (Since heat is absorbed) 

W = –394 J (Since work is done by the system) 

Substituting the values in expression (i), we get 

∆U = 701 J + (–394 J) 

∆U = 307 J 

Hence, the change in internal energy for the given process is 307 J.

8. The reaction of cyanamide, NH₂CN_(s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be –742.7 kJ mol^–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.

Answer:

Enthalpy change for a reaction (∆H) is given by the expression, 

∆H = ∆U + ∆n_gRT 

Where, ∆U = change in internal energy 

∆n_g = change in number of moles 

For the given reaction, 

∆n_g = ∑n_g (products) – ∑n_g (reactants) 

= (2 – 2.5) moles 

∆n_g = –0.5 moles

And, ∆U = –742.7 kJ mol^–1 

T = 298 K 

R = 8.314 × 10^–3 kJ mol^–1 K^–1

Substituting the values in the expression of ∆H: 

∆H = (–742.7 kJ mol^–1) + (–0.5 mol) (298 K) (8.314 × 10^–3 kJ mol^–1 K^–1) 

= –742.7 – 1.2 

∆H = –743.9 kJ mol^–1

9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^–1 K^–1.

Answer:

From the expression of heat (q), 

q = m. c. ∆T 

Where, 

c = molar heat capacity 

m = mass of substance 

∆T = change in temperature 

Substituting the values in the expression of q:

q = 1066.7 J

q = 1.07 kJ

Answer 2

10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 

10.0°C. ∆_fusH = 6.03 kJ mol^–1 at 0°C. 

C_p[H₂O(l)] = 75.3 J mol^–1 K^–1 

C_p[H₂O(s)] = 36.8 J mol^–1 K^–1

Answer:

Total enthalpy change involved in the transformation is the sum of the following changes: 

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C. 

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.

= (75.3 J mol^–1 K^–1) (0 – 10)K + (–6.03 × 10³ J mol^–1) + (36.8 J mol^–1 K^–1) (–10 – 0)K 

= –753 J mol^–1 – 6030 J mol^–1 – 368 J mol^–1 

= –7151 J mol^–1 

= –7.151 kJ mol^–1

Hence, the enthalpy change involved in the transformation is –7.151 kJ mol^–1.

11. Enthalpy of combustion of carbon to CO₂ is –393.5 kJ mol^–1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.

Answer:

Formation of CO₂ from carbon and dioxygen gas can be represented as:

(1 mole = 44 g) 

Heat released on formation of 44 g CO₂ = –393.5 kJ mol^–1 

\(\therefore\) Heat released on formation of 35.2 g CO₂. 

= –314.8 kJ mol^–1

12. Enthalpies of formation of CO_(g), CO_2(g), N₂O_(g) and N₂O4_(g) are –110 kJ mol^–1 , – 393 kJ mol^–1 , 81 kJ mol^–1 and 9.7 kJ mol^–1 respectively. Find the value of ∆rH for the reaction:

N₂O_4(g) + 3CO_(g) N₂O_(g) + 3CO_2(g)

Answer:

∆rH for a reaction is defined as the difference between ∆_fH value of products and ∆_fH value of reactants.

For the given reaction, 

N₂O_4(g) + 3CO_(g) N₂O_(g) + 3CO_2(g)

Substituting the values of ∆fH for N₂O, CO₂, N₂O₄, and CO from the question, we get:

Hence, the value of ∆rH for the reaction is -777.7 kJ mol^-1.

13. Given

N_2(g) + 3H_2(g) \(\longrightarrow\) 2NH_3(g) ; ∆_rH^θ = –92.4 kJ mol^–1

What is the standard enthalpy of formation of NH₃ gas ?

Answer:

Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1 mole of a substance in its standard form from its constituent elements in their standard state. 

Re-writing the given equation for 1 mole of NH_3(g),

\(\therefore\) Standard enthalpy of formation of NH_3(g)

= 1/2 ∆rH^θ 

= 1/2(–92.4 kJ mol^–1) 

= –46.2 kJ mol^–1

14. Calculate the standard enthalpy of formation of CH₃OH_(l) from the following data:

CH₃OH(l) + \(\frac32\) O_(2g) \(\longrightarrow\)CO_2(g) + 2H₂O_(l) ; ∆rH^θ = –726 kJ mol^–1

C_(g) + O₂(g) \(\longrightarrow\)CO_2(g) ; ∆_cH^θ = –393 kJ mol^–1

H_2(g) + \(\frac12\)O_(2g) \(\longrightarrow\)H₂O_(l) ; ∆_fH^θ = –286 kJ mol^–1.

Answer:

The reaction that takes place during the formation of CH₃OH_(l) can be written as:

C_(s) + 2H₂O_(g) + \(\frac12\)O_2(g) CH₃OH_(l)    (1)

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:

Equation (ii) + 2 × equation (iii) – equation (i)

∆_fH^θ [CH₃OH_(l)] = ∆_cH^θ + 2∆_fH^θ [H₂O_(l)] – ∆_rH^θ 

= (–393 kJ mol^–1) + 2(–286 kJ mol^–1) – (–726 kJ mol^–1)

= (–393 – 572 + 726) kJ mol^–1 

\(\therefore\) ∆_fH^θ [CH₃OH_(l)] = –239 kJ mol^–1

Answer 3

15. Calculate the enthalpy change for the process 

CCl_4(g) → C_(g) + 4Cl_(g) 

and calculate bond enthalpy of C–Cl in CCl_4(g). 

∆_vapH^θ (CCl₄) = 30.5 kJ mol^–1. 

∆_fH^θ (CCl₄) = –135.5 kJ mol^–1. 

∆_aH^θ (C) = 715.0 kJ mol^–1 , where ∆aHθ is enthalpy of atomisation 

∆_aH^θ (Cl₂) = 242 kJ mol^-1

Answer:

The chemical equations implying to the given values of enthalpies are:

(i) CCl_4(l) \(\longrightarrow\) CCl_4(g)   ; ∆_vapH^θ = 30.5 kJ mol^–1

(ii) C_(s) \(\longrightarrow\) C_(g)   ; ∆_aH^θ = 715.0 kJ mol^–1

(iii) Cl_2(g)  \(\longrightarrow\) 2Cl_(g)   ; ∆_aH^θ  = 242 kJ mol^-1

(iv) C_(g) + 4Cl_(g) \(\longrightarrow\)  CCl_4(g)  ; ∆_fH^θ  = –135.5 kJ mol^–1

Enthalpy change for the given process CCl_4(g) → C_(g) + 4Cl_(g) can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

∆H = ∆_aH^θ (C) + 2∆_aH^θ (Cl₂) – ∆_vapH^θ – ∆_fH 

= (715.0 kJ mol^–1) + 2(242 kJ mol^–1) – (30.5 kJ mol^–1) – (–135.5 kJ mol^–1) 

\(\therefore\) ∆H = 1304 kJ mol^–1

Bond enthalpy of C–Cl bond in CCl_4(g)

= \(\frac{1304}{4}\) kJ mol^-1

= 326 kJ mol^-1

16. For an isolated system, ∆U = 0, what will be ∆S?

Answer:

∆S will be positive i.e., greater than zero 

Since ∆U = 0, ∆S will be positive and the reaction will be spontaneous.

17. For the reaction at 298 K, 

2A + 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:

From the expression, 

∆G = ∆H – T∆S 

Assuming the reaction at equilibrium, ∆T for the reaction would be:

T = 2000 K 

For the reaction to be spontaneous, ∆G must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

18. For the reaction, 

2Cl_(g) → Cl_2(g), what are the signs of ∆H and ∆S ?

Answer:

∆H and ∆S are negative 

The given reaction represents the formation of chlorine molecule from chlorine atoms. 

Here, bond formation is taking place. Therefore, energy is being released. 

Hence, ∆H is negative. 

Also, two moles of atoms have more randomness than one mole of a molecule. 

Since spontaneity is decreased, ∆S is negative for the given reaction.

19. For the reaction 

2A_(g) + B_(g) → 2D_(g) 

∆U^θ = –10.5 kJ and ∆S^θ= –44.1 JK^–1 . 

Calculate ∆G^θ for the reaction, and predict whether the reaction may occur spontaneously.

Answer​​​​​​​:

For the given reaction, 

2 A_(g) + B_(g) → 2D_(g) 

∆n_g = 2 – (3) 

= –1 mole

Substituting the value of ∆U^θ in the expression of ∆H:

∆H^θ = ∆U^θ + ∆n_gRT 

= (–10.5 kJ) – (–1) (8.314 × 10^–3 kJ K^–1 mol^–1) (298 K) 

= –10.5 kJ – 2.48 kJ 

∆H^θ = –12.98 kJ

Substituting the values of ∆H^θ and ∆S^θ in the expression of ∆G^θ :

∆G^θ = ∆H^θ – T∆S^θ 

= –12.98 kJ – (298 K) (–44.1 J K^–1) 

= –12.98 kJ + 13.14 kJ 

∆G^θ = + 0.16 kJ

Since ∆G^θ for the reaction is positive, the reaction will not occur spontaneously.

20. The equilibrium constant for a reaction is 10. What will be the value of ∆G^θ? R = 8.314 JK^–1 mol^–1, T = 300 K.

Answer​​​​​​​:

From the expression, 

∆G^θ = –2.303 RT logK_eq 

∆G^θ for the reaction,

= (2.303) (8.314 JK^–1 mol^–1) (300 K) log10 

= –5744.14 Jmol^–1 

= –5.744 kJ mol^–1

21. Comment on the thermodynamic stability of NO_(g), given

Answer​​​​​​​:

The positive value of ∆_rH indicates that heat is absorbed during the formation of NO_(g). 

This means that NO(g) has higher energy than the reactants (N₂ and O₂). 

Hence, NO_(g) is unstable. 

The negative value of ∆_rH indicates that heat is evolved during the formation of NO_2(g) from NO_(g) and O_2(g). The product, NO_2(g) is stabilized with minimum energy. 

Hence, unstable NO_(g) changes to unstable NO_2(g).

22. Calculate the entropy change in surroundings when 1.00 mol of H₂O_(l) is formed under standard conditions. ∆_fH^θ = –286 kJ mol^–1.

Answer​​​​​​​:

It is given that 286 kJ mol^–1 of heat is evolved on the formation of 1 mol of H₂O_(l). Thus, an equal amount of heat will be absorbed by the surroundings. q_surr = +286 kJ mol^–1

Entropy change (∆S_surr) for the surroundings = \(\frac{q_{surr}}{7}\)

\(=\frac{286 \,kJ\,mol^{-1} }{298 k}\)

\(\therefore\) ∆S_surr = 959.73 J mol^–1 K^–1