NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

These Solutions are part of NCERT Solutions for Class 11 Chemistry. Here we have given NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics.

Question 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:
(ii) whose value is independent of path

Question 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:
(iii) q = 0

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

Question 4.

Answer:

Question 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 CH₄ (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:

Question 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:
(iv) possible at any temperature

Question 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:
Heat absorbed by the system, q = 701 J
Work done by the system = – 304 J
Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

Question 8.
The reaction of cyanamide, NH₂CN(s) with oxygen was affected in a bomb calorimeter and ∆U was found to be – 742.7 kJ mol^-1 of cyanamide at 298 K. Calculate the enthalpy change for the reaction at 298 K.
NH₂CN(y) + 3/2O₂(g) → N₂fe) + CO₂(g) + H₂O(l)
Answer:
∆U = – 742.7 kJ mol^-1 ; ∆^ng = 2 – \(\frac { 3 }{ 2 } \) = + \(\frac { 1 }{ 2 } \) mol.
R = 8.314 × 10^-3 kJ K^-1 mol^-1 ; T = 298 K
According to the relation, ∆H = ∆U + ∆^ng RT
AH = (- 742.7 kJ) + (\(\frac { 1 }{ 2 } \) mol) × (8.314 × 10^-3 kJ K^-1 mol^-1) × (298 K)
= -742.7 kJ + 1.239 kJ = -741.5 kJ.

Question 9.
Calculate the number of kJ necessary to raise the temperature of 60 g of aluminium from 35 to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.
Answer:
No. of moles of Al (m) = \(\frac { 60g }{ 27gmo{ l }^{ -1 } } \) = 2.22 mol
Molar heat capacity (C) = 24 J mol^-1 K^-1
Rise in temperature (∆T) = 55 – 35 = 20°C = 20 K
Heat evolved (q) = C × m × T = (24 J mol^-1 K^-1) × (2.22 mol) × (20 K)
= 1065.6 J = 1.067 kJ.

Question 10.
Calculate the enthalpy change on freezing of 1.0 mole 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)l = 36.8 J mol^-1 K^-1.
Answer:
Total change in enthalpy (AH) for the freezing process may be calculated as :
∆H = (1 male of water at 10°C → 1 mole of water at 0°C) + (1 mole of water at °C → 1 mole of ice at 0°C)
+ (1 mole of ice at °C → 1 mole of ice at – 10°C)
= C_p[H₂O(l)] × ∆T + ∆H_freezmg + C_p [H₂O(s)] × ∆T.
= (75.3 Jk^-1 mol^-1) (0 – 10 K) + (-6.03 kJ mol^-1) + (36.8Jk^-1 mol^-1) × (-10 K)
= (-753 J mol^-1) – (6.03 kJ mol^-1) – (368 J mol^-1)
= (-0.753 kJ mol^-1) – (6.03 kJ mol^-1) – (0.368 kJ mol^-1)
= – 7.151 kJ mol^-1

Question 11.
Enthalpy of combustion of carbon to carbon dioxide is – 393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and oxygen gas.
Answer:
The combustion equation is :
C(s) + O₂(g) → CO₂(g) ; ∆_cH = – 393.5 kJ mol^-1 (44 g)
(44g)
Heat released in the formation of 44g of CO₂ = 393.5 kJ
Heat released in the formation of 35.2 g of CO₂ = \(\frac { (393.5kJ)\times (35.2g) }{ (44g) } \) = 314.8 kJ.

Question 12.
Calculate the enthalpy of the reaction :
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Given that;
∆_fH CO(g) = – 110 kJ mol^-1 ;
∆_fH CO₂(g) = – 393 kJ mol^-1
∆_fH. N₂O(g) = 81 kJ mol^-1 ;
∆_fH N₂O₄(g) = – 9.7 kJ mol^-1.
Answer:
Enthalpy of reaction (∆_rH) = [81 + 3(- 393)] – [9.7 + 3(- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kJ mol^-1.

Question 13.

Answer:

Question 14.

Answer:

Question 15.
Calculate the enthalpy change for the process
CCl₄ (g) → C (g) + 4 Cl (g) and calculate bond enthalpy of C—Cl in CC14 (g)
Given : ∆_vap H° (CCl₄) = 30.5 kJ mol^-1 ; ∆_fH°(CCl₄) = – 135.5 kJ mol^-1
∆_aH° (C) = 715.0 kJ mol^-1 where ∆_a H° is enthalpy of atomisation
∆_aH° (Cl₂) = 242 kJ mol^-1.
Answer:

Question 16.
For an isolated system ∆U = 0 ; what will be ∆S ?
Answer:
Change in internal energy (∆U) for an isolated system is zero because it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ∆S > 0 or positive.

Question 17.
For a reaction at 298 K
2 A + 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:
According to Gibbs-Helmholtz equation :
∆G = ∆H – T∆S
For ∆G = 0; ∆H = T∆S or T : = \(\frac { \triangle H }{ \triangle s } \)

T = \(\frac { (400kJmo{ l }^{ -1 }) }{ (0.2kJ{ K }^{ -1 }mo{ l }^{ -1 }) } \)
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.

Question 18.
For the reaction ; 2Cl (g) → Cl₂(g) ; what will be the signs of ∆H and ∆S ?
Answer:
∆H : negative (-ve) because energy is released in bond formation
∆S : negative (-ve) because entropy decreases when atoms combine to form molecules.

Question 19.
For a reaction ; 2A (g) + B (g) → 2D(g)

Calculate ∆U₂₉₈ for the reaction and predict whether the reaction is spontaneous or not.
Answer:
Let the mass of H₂ in the mixture = 20 g
The mass of O₂ in the mixture will be = 80 g

Question 20.

Answer:

Question 21.

Answer:

Question 22.

Answer:

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