NCERT Solutions: Thermodynamics

Q5.1. 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.

Ans: (ii)

Explanation: A thermodynamic state function depends only on the current state of the system (for example pressure, volume, temperature) and not on how the system reached that state. Quantities such as heat (q) and work (w) depend on the path taken and are not state functions. Examples of state functions are P, V and T; therefore option (ii) is correct.

Q5.2. For the process to occur under adiabatic conditions, the correct condition is:
(i) ΔT = 0
(ii) Δp = 0
(iii) q = 0
(iv) w = 0

Ans: (iii) q = 0

Explanation: An adiabatic process is one with no heat exchange between the system and surroundings. This means the heat transferred, q, is zero. Work may still be done by or on the system, and temperature or pressure can change; hence only q = 0 is the correct condition for an adiabatic process.

Q5.3. The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element

Ans: (ii) zero

Explanation: By definition, the standard enthalpy of formation of an element in its most stable form (standard state) is chosen as zero. This provides a reference for calculating enthalpies of formation of compounds. Therefore the enthalpy of all elements in their standard states is zero.

Q5.4. ΔU^θ of combustion of methane is -X kJ mol^-1. The value of  Δ H^θ is
(i) = ΔU ^θ
(ii) >ΔU ^θ
(iii) < ΔU^θ
(iv) = 0

Ans: (iii) < ΔU^θ

Explanation: The relation between enthalpy and internal energy is ΔH = ΔU + Δn_gRT, where Δn_g is the change in moles of gas. For methane combustion the number of gas moles in products is less than in reactants, so Δn_g is negative. Thus the term Δn_gRT is negative and ΔH is smaller (more negative) than ΔU.

Q5.5. 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
Ans: (i) -74.8 kJ mol^-1
Explanation: Use Hess's law relating combustion enthalpies to formation enthalpy. The combustion reactions are:
Q5.6. 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
Ans: (iv) possible at any temperature
Explanation: For a reaction to be spontaneous, ΔG should be negative.
ΔG = ΔH - TΔS
According to the question, for the given reaction,
ΔS = positive
ΔH = negative (since heat is evolved)
⇒ ΔG = negative
Therefore, the reaction is spontaneous at any temperature.
Hence, alternative (iv) is correct.

Q5.7. 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
Ans: According to the first law of thermodynamics,
ΔU = q + W (i)
Where,
ΔU = 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
ΔU = 701 J + (-394 J)
ΔU = 307 J
Hence, the change in internal energy for the given process is 307 J.

Q5.8. The reaction of cyanamide, NH₂CN_(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be -742.7 kJ mol^-1at 298 K. Calculate enthalpy change for the reaction at 298 K.Ans: Use ΔH = ΔU + Δn_gRT
From the reaction (given in the image), Δn_g = (moles of gaseous products) - (moles of gaseous reactants) = 2 - 1.5 = 0.5.
Substitute values: ΔU = -742.7 kJ mol^-1, R = 8.314 × 10^-3 kJ mol^-1 K^-1, T = 298 K.
ΔH = -742.7 + (0.5)(8.314 × 10^-3)(298) kJ mol^-1
= -742.7 + 1.24 ≈ -741.5 kJ mol^-1.
Therefore, ΔH ≈ -741.5 kJ mol^-1.
Q5.9. 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.
Ans: q = n · c · ΔT.
Moles of Al = 60.0 g / 27 g mol^-1 ≈ 60/27 = 2.222 mol.
c = 24 J mol^-1 K^-1, ΔT = 55 - 35 = 20 K.
q = 2.222 × 24 × 20 = 1066.7 J = 1.07 kJ.
Hence, 1.07 kJ of heat is required.
Q5.10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. Δ_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
Ans: 

Q5.11. 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.
Ans: Formation of CO₂ from carbon and dioxygen gas can be represented as:
C_(s) + O_2(g) → CO_2(g)   Δ_fH = -393.5 kJ mol^-1

Q5.12. Enthalpies of formation of CO_(g), CO_2(g), N₂O_(g) and N₂O_4(g) are -110 kJ mol^-1, - 393 kJ mol^-1, 81 kJ mol^-1and 9.7 kJ mol^-1 respectively. Find the value of Δ_rH for the reaction:
N₂O_4(g) + 3CO_(g) N₂O_(g) + 3CO_2(g)

Ans: Δ_rH for a reaction is defined as the difference between Δ_fH value of products and Δ_fH value of reactants.
Δ_rH = ∑Δ_fH (products) - ∑Δ_fH (reactants)

Q5.14. Calculate the standard enthalpy of formation of CH₃OH_(l) from the following data:

C_(g) + O_2(g) CO_2(g) ; Δ_cH^θ = -393 kJ mol^-1

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

The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) - equation (i)
Δ_fH^θ [CH₃OH_(l)] = Δ_cH^θ + 2Δ_fH^θ [H₂O_(l)] - Δ_rH^θ
= (-393 kJ mol^-1) + 2(-286 kJ mol^-1) - (-726 kJ mol^-1)
= (-393 - 572 + 726) kJ mol^-1
∴Δ_fH^θ [CH₃OH_(l)] = -239 kJ mol^-1

Q5.15. 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).
Δ_vapH^θ (CCl₄) = 30.5 kJ mol^-1.
Δ_fH^θ (CCl₄) = -135.5 kJ mol^-1.
Δ_aH^θ (C) = 715.0 kJ mol^-1, where Δ_aH^θ is enthalpy of atomisation
Δ_aH^θ (Cl₂) = 242 kJ mol^-1
Ans: The chemical equations implying to the given values of enthalpies are:
(i) CCl_4(l) → CCl_4(g) Δ_vapH^θ = 30.5 kJ mol^-1
(ii) C_(s) → C_(g) Δ_aH^θ = 715.0 kJ mol^-1
(iii) Cl_2(g) → 2Cl_(g)  Δ_aH^θ = 242 kJ mol^-1
(iv) C_(g) + 4Cl_(g) → CCl_4(g) Δ_fH = -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)
ΔH = Δ_aH^θ(C) + 2Δ_aH^θ (Cl₂) - Δ_vapH^θ - Δ_fH
= (715.0 kJ mol^-1) + 2(242 kJ mol^-1) - (30.5 kJ mol^-1) - (-135.5 kJ mol^-1)
∴ΔH = 1304 kJ mol^-1
Bond enthalpy of C-Cl bond in CCl_{4 (g)}

= 326 kJ mol^-1
Q5.16. For an isolated system, ΔU = 0, what will be ΔS?
Ans: For an isolated system there is no exchange of energy or matter with the surroundings, so ΔU = 0. According to the second law of thermodynamics the entropy change of an isolated system satisfies ΔS ≥ 0: it increases for any spontaneous (irreversible) process (ΔS > 0) and remains zero only for a reversible process (ΔS = 0).

Q5.17. For the reaction at 298 K,
2A + B → C
ΔH = 400 kJ mol^-1 and ΔS = 0.2 kJ K^-1 mol^-1
At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
Ans: At equilibrium ΔG = 0, so T = ΔH / ΔS.
ΔH = 400 kJ mol^-1, ΔS = 0.2 kJ K^-1 mol^-1.
T = 400 / 0.2 = 2000 K.
For temperatures greater than 2000 K, ΔG = ΔH - TΔS will be negative and the reaction will be spontaneous. For T < 2000 K the reaction is non-spontaneous.

Q5.18. For the reaction,
2Cl_(g) → Cl_2(g), what are the signs of ΔH and ΔS ?
Ans: ΔH and ΔS 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, ΔH is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS  is negative for the given reaction.

Q5.19. For the reaction
2A_(g) + B_(g) → 2D_(g)
ΔU^θ = -10.5 kJ and ΔS^θ= -44.1 JK^-1.
Calculate ΔG^θ for the reaction, and predict whether the reaction may occur spontaneously.
Ans: For the given reaction,
2 A_(g) + B_(g) → 2D_(g)
Δn_g = 2 - (3)
= -1 mole
Substituting the value of ΔU^θ in the expression of ΔH:
ΔH^θ = ΔU^θ + Δn_gRT
= (-10.5 kJ) - (-1) (8.314 × 10^-3 kJ K^-1 mol^-1) (298 K)
= -10.5 kJ - 2.48 kJ
ΔH^θ = -12.98 kJ
Substituting the values of ΔH^θ and ΔS^θ in the expression of ΔG^θ:
ΔG^θ = ΔH^θ - TΔS^θ
= -12.98 kJ - (298 K) (-44.1 J K^-1)
= -12.98 kJ + 13.14 kJ
ΔG^θ = + 0.16 kJ
Since ΔG^θ for the reaction is positive, the reaction will not occur spontaneously.

Q5.20. The equilibrium constant for a reaction is 10. What will be the value of ΔG^θ? R = 8.314 JK^-1 mol^-1, T = 300 K.
Ans: Use ΔG° = -2.303 RT log K.
ΔG° = -2.303 × 8.314 J K^-1 mol^-1 × 300 K × log(10)
= -2.303 × 8.314 × 300 × 1 = -5744.14 J mol^-1 = -5.744 kJ mol^-1.
Therefore ΔG° ≈ -5.744 kJ mol^-1.
Q5.21. Comment on the thermodynamic stability of NO_(g), givenAns: The positive Δ_rH for formation of NO(g) indicates that formation of NO from N₂ and O₂ is endothermic; NO has higher enthalpy than the reactants and is therefore thermodynamically less stable. The further reaction NO + 1/2 O₂ → NO₂ is exothermic (negative Δ_rH), indicating NO₂ is thermodynamically more stable (lower energy) than NO. Thus NO(g) is less stable and tends to be converted to the more stable NO₂ under suitable conditions.
Q5.22. Calculate the entropy change in surroundings when 1.00 mol of H₂O_(l) is formed under standard conditions. Δ_fH^θ = -286 kJ mol^-1.
Ans: Heat released to surroundings q_surr = -ΔfH° = +286 kJ = 286000 J (since the system releases 286 kJ, the surroundings gain this heat).
ΔS_surr = q_surr / T = 286000 J / 298 K = 959.73 J K^-1 mol^-1.
Therefore the entropy change of the surroundings is 959.73 J mol^-1 K^-1.