Solutions Class 11 Chemistry Chapter 6 Thermodynamics

Topics and Subtopics in Solutions for ncert Class 11 Chemistry Chapter 6 Thermodynamics:

Section Name	Topic Name
6	Thermodynamics
6.1	Thermodynamic Terms
6.2	Applications
6.3	Measurement of ∆U and ∆H: Calorimetry
6.4	Enthalpy Change, ∆rH  of a Reaction – Reaction Enthalpy
6.5	Enthalpies for Different Types of Reactions
6.6	Spontaneity
6.7	Gibbs Energy Change and Equilibrium

NCERT Solutions -Thermodynamics Laws : Class 11 Chemistry

Question 1. Choose the correct answer:
A thermodynamic state junction 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

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

(i) unity (ii) zero
(iii) < 0 (iv) different for each element

Answer: 

(ii) zero

6.4.

Answer:

6.5. 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 CHJg) 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: 

As per the available data :

6.6. 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

6.7. In a process, 701 ] 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 = – 394 J Change in internal energy (∆U) = q + w = 701 – 394 = 307 J.

6.8. The reaction of cyanamide,NH₂CN(s) with dioxygen was carried out in a bomb calorimeter and ∆U was found to be -742,7 KJ^-1  mol^-1 at 298 K. Calculate the enthalpy change for the reaction at 298 K.NH₂CN (S) + 3/202(g) —–>N₂(g) + CO₂(g) + H₂0(Z)

Answer: 

∆U = – 742.7 KJ^-1  mol^-1 ; ∆^ng = 2 – 3/2 = + 1/2 mol.
R = 8.314 x 10-3KJ^-1  mol^-1 ; T = 298 K
According to the relation,∆H = ∆U+∆^ngRT
∆H = (- 742.7 kj) + (1/2 mol) x (8.314 x10^-3 KJ^-1  mol^-1 ) x (298 K)
= – 742.7 kj + 1.239 kj = – 741.5 kj.

6.9. Calculate the number of kj of heat necessary to raise the temperature of 60 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1 K^-1.

Answer: 

No. of moles of Al (m) = (60g)/(27 g mol^-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 x m x T = (24 J mol^-1 K^-1) x (2.22 mol) x (20 K)
= 1065.6 J = 1.067 kj

6.10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. A, H = 6.03 KJ mot1 at 0°C. Cp [H20(l)J = 75.3 J mol^-1 K^-1; Cp [H20(s)J = 36.8 J mol^-1 K^-1.

Answer: 

The change may be represented as:

6.11. Enthalpy of combustion of carbon to carbon dioxide is – 393.5 J mol^{-1   .}Calculate the heat released upon formation of 35.2 g of C0₂ from carbon and oxygen gas.

Answer: 

The combustion equation is:
C(s) + 0₂ (g) —–> C0₂(g); AcH = – 393.5 KJ mol^-1
Heat released in the formation of 44g of C0₂ = 393.5 kj
Heat released in the formation of 35.2 g of C0₂=(393.5 KJ) x (35.2g)/(44g) = 314.8 kj

6.12. Calculate the enthalpy of the reaction:
N₂0₄(g) + 3CO(g) ———->N₂0(g) + 3CO₂(g)
Given that;∆_fH^–CO(g) = – 110 kj mot^-1; ∆_fHC0₂(g) = – 393 kj mol^-1
∆_fHN₂0(g) = 81 kj mot^-1; ∆_fN₂O₄(g) = 9.7 kj mol^-1

Answer: 

Enthalpy of reaction (∆_r,H) = [81 + 3 (- 393)] – [9.7 + 3 (- 110)]
= [81 – 1179] – [9.7 – 330] = – 778 kj mol^-1

6.13. Given : N₂(g) + 3H₂(g) ————> 2NH₃(g); ∆_{r H^–} = -92.4 kj mot^-1 What is the standard enthalpy of formation of NH₃ gas?

Answer: 

∆H^– NH₃ (g) = – (92.4)/2 = – 46.2 kj mol^-1

6.14. Calculate the standard enthalpy of formation of CH₃OH. from the following data:
(i) CH₃OH(l) + 3/2 0₂ (g) ———-> CO₂ (g) + 2H₂0 (l); ∆_rH^– = – 726kj mol^-1
(ii) C(s) + 0₂(g) —————>C0₂ (g); ∆_cH^– = -393 kj mol^-1
(iii) H₂(g) + 1/20₂(g) —————->H₂0 (l); ∆_fH^– = -286 kj mol^-1

Answer: 

The equation we aim at;
C(s) + 2H₂(g) + l/20₂(g) ———> CH₃OH (l);∆_fH^– = ±? … (iv)
Multiply eqn. (iii) by 2 and add to eqn. (ii)
C(s) + 2H₂(g) + 20₂(g) ————->C0₂(g) + 2H₂0(Z)
∆H = – (393 + 522) = – 965 kj moH Subtract eqn. (iv) from eqn. (i)
CH₃OH(Z) + 3/20₂(g) ————> C0₂(y) + 2H₂0(Z); ∆H = – 726 kj mol^-1
Subtract: C(s) + 2H₂(y) + l/20₂(g) ———-> CH₃OH(Z); ∆_fHe = – 239 kj mol^-1

6.15.

Answer:

6.16.For an isolated system∆U = 0; what will be ∆S? 

Answer: 

Change in internal energy (∆U) for an isolated system is zero for it does not exchange any energy with the surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ∆S > 0 or positive.

6.17. For a reaction at 298 K
2A + B————->C
∆H = 40Q kj mot1 and AS = 0.2 kj Kr^-1 mol^-1.
At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range?

Answer:  

As per the Gibbs Helmholtz equation:
ΔG = Δ H- TΔ S For ΔG=0 ; ΔH=TΔS or T=ΔH/ΔS
T = (400 KJ mol^-1)/(0.2 KJ K^-1 mol^-1) = 2000 k
Thus, reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.

6.18. 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.

6.19.

Answer:

6.20.

Answer:

6.21.

Answer:

6.22.

Answer: