NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 exams must go through NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

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NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.

NCERT Exercises

Question 1.
Choose the correct answer. A thermodynamic state function is a quantity
(a) used to determine heat changes
(b) whose value is independent of path
(c) used to determine pressure volume work
(d) whose value depends on temperature only.
Solution.
(b): State function is a property of the system whose value depends only upon the state of the system and is independent of the path or the manner by which the state is reached.

Question 2.
For the process to occur under adiabatic conditions, the correct condition is
(a) ΔT = 0
(b) Δp = 0
(c) g = 0
(d) w=0
Solution.
(c): Adiabatic system does not exchange heat with the surroundings.

Question 3.
The enthalpies of all elements in their standard states are
(a) unity
(b) zero
(c) <0
(d) different for each element.
Solution.
(b) : By convention, the standard enthalpy of formation of every element in its standard state is zero.

Question 4.
ΔU° of combustion of methane is – X kJ mol^-1. The value of ΔH° is
(a) = ΔU°
(b) >ΔU°
(c) <ΔU°
(d) =0
Solution.
(c) : CH_4(g)+2O₂ → CO_{2(g)+2H2Ol
Δn = 1 – 3 = -2. ‘
ΔH° = ΔU° + ΔnRT = -X- 2RT}

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 CH_4(g) will be
(a) -74.8 kJ mol^-1
(b) -52.27 kJ mol^-1
(c) +74.8 kJ mol^-1
(d) +52.26 kJ mol^-1
Solution.

Question 6.
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(a) possible at high temperature
(b) possible only at low temperature
(c) not possible at any temperature
(d) possible at any temperature
Solution.
(d): A + B → C + D + q, ΔS = +ve
Here, ΔH = -ve
ΔG = ΔH – TΔS
For reaction to be spontaneous, ΔG should be -ve. As ΔH = -ve and ΔS is +ve, ΔG will be -ve at any temperature.

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 S of work is done by the system. What is the change in internal energy for the process?
Solution.
Heat absorbed by the system (q) = 701 J
Work done by the system (w) = -394 J
According to first law of thermodynamics,
ΔU = q + w = 701 + (-394) = 701 – 394 = 307 J

Question 8.
The reaction of cyanamide, NH₂CN_(g), with dioxygen was carried out in a bomb calorimeter, and AU was found to be -742.7 kJ mol 1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN_(g) + → N₂+CO_2(g)+H₂O_l
Solution.

Question 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.
Solution.
Mass of Al = 60 g
Rise in temperature, ΔT = 55 – 35 = 20°C
Molar heat capacity of Al = 24 J mo^-1 K^-1
Specific heat capacity of Al =
∴ Energy required = m x c x ΔT
=
= 1.068kJ or 1.07kJ

Question 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
Solution.

Question 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.
Solution.
C + O₂ → CO₂; ΔT = -393.5 kJ
∵ When 44 g of COz is formed from carbon and dioxygen gas, heat released = 393.5 kj
∴ When 35.2 g of CO₂ is formed from carbon and dioxygen gas, heat released
=
Thus, ΔH = -314.8 kJ

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

Question 13.
Given: N_2(g) + 3H_2(g) → 2NH_3(g); Δ_rH° = -92.4 kJ mol^-1 What is the standard enthalpy of formation of NH₃ gas?
Solution.
N_2(g) + 3H_2(g) → 2NH_3(g) ; Δ_rH° = -92.4 kJ mol^-1
∴ Standard enthalpy of formation of NH_3(g)
=

Question 14.
Calculate the standard enthalpy of formation of CH₃OH_l from the following data :

Solution.

Question 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°(CCl₄) = 30.5 kJ mol^-1,
Δ_fH°(CCl₄) = -135.5 kJ mol^-1,
Δ_aH°(C) = 715.0 kJ mol^-1, where Δ_aH° is enthalpy of atomisation Δ_aH°(Cl₂) = 242 kJ mol^-1
Solution.

Question 16.
For an isolated system, ΔU = 0, what will be ΔS?
Solution.
When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.

Question 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.
Solution.
According to Gibbs Helmholtz equation, ΔG = ΔH – TΔS.

Question 18.
For the reaction, 2Cl_(g) → Cl_2(g), what are the signs of ΔH and ΔS?
Solution.
ΔH is negative because bond energy is released and ΔS is negative because there is less randomness among the molecules than among the atoms.

Question 19.
For the reaction
2A_(g) + B_(g) → 2D_(g), ΔU° = -10.5 kJ and ΔS° = – 44.1 JK^-1.
Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.
Solution.
ΔH° = ΔU° + Δn_(g)RT

Question 20.
The equilibrium constant for a reaction is 10. What will be the value of ΔG° ?
R = 8.314J K^-1 mol^-1, T= 300 K.
Solution.
We know that ΔG° = – 2.303RT logK
= – 2.303 x 8.314 x 300 x log10
= – 5744.14 J/mol

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

Solution.
Since Δ_(r)H° is +ve, i.e., enthalpy of formation of NO is positive, therefore NO is unstable. But Δ_(r)H° is negative for the formation of NO₂. So, NO₂ is stable.

Question 22.
Calculate the entropy change in surroundings when 1.00 mol of H₂O_l is formed under standard conditions.
Δ_(r)H° = – 286 kJ mol^-1.
Solution.

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NCERT Solutions for Class 11 Chemistry