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.
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.
(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.
For the process to occur under adiabatic conditions, the correct condition is
(a) ΔT = 0
(b) Δp = 0
(c) g = 0
(c): Adiabatic system does not exchange heat with the surroundings.
The enthalpies of all elements in their standard states are
(d) different for each element.
(b) : By convention, the standard enthalpy of formation of every element in its standard state is zero.
ΔU° of combustion of methane is – X kJ mol-1. The value of ΔH° is
(a) = ΔU°
(c) : CH4(g)+2O2 → CO2(g)+2H2Ol
Δn = 1 – 3 = -2. ‘
ΔH° = ΔU° + ΔnRT = -X- 2RT
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
(a) -74.8 kJ mol-1
(b) -52.27 kJ mol-1
(c) +74.8 kJ mol-1
(d) +52.26 kJ mol-1
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
(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.
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?
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
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+CO2(g)+H2Ol
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.
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
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol-1 at 0°C.
Cp[H2O(l)] = 75.3 J mol-1 K-1,
Cp[H2O(s)] = 36.8 J mol-1 K-1
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.
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
Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are -110, – 393, 81 and 9.7 kJ mol -1 respectively. Find the value of Δrf for the reaction
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
Given: N2(g) + 3H2(g) → 2NH3(g); ΔrH° = -92.4 kJ mol-1 What is the standard enthalpy of formation of NH3 gas?
N2(g) + 3H2(g) → 2NH3(g) ; ΔrH° = -92.4 kJ mol-1
∴ Standard enthalpy of formation of NH3(g)
Calculate the standard enthalpy of formation of CH3OHl from the following data :
Calculate the enthalpy change for the process
CCl4(g) → C(g) + 4Cl(g)
and calculate bond enthalpy of C – Cl in CCl4(g).
ΔvapH°(CCl4) = 30.5 kJ mol-1,
ΔfH°(CCl4) = -135.5 kJ mol-1,
ΔaH°(C) = 715.0 kJ mol-1, where ΔaH° is enthalpy of atomisation ΔaH°(Cl2) = 242 kJ mol-1
For an isolated system, ΔU = 0, what will be ΔS?
When energy factor has no role to play, for the process to be spontaneous ΔS must be +ve i.e., ΔS > 0.
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.
According to Gibbs Helmholtz equation, ΔG = ΔH – TΔS.
For the reaction, 2Cl(g) → Cl2(g), what are the signs of ΔH and ΔS?
ΔH is negative because bond energy is released and ΔS is negative because there is less randomness among the molecules than among the atoms.
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.
ΔH° = ΔU° + Δn(g)RT
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.
We know that ΔG° = – 2.303RT logK
= – 2.303 x 8.314 x 300 x log10
= – 5744.14 J/mol
Comment on the thermodynamic stability of NO(g), given
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.
Calculate the entropy change in surroundings when 1.00 mol of H2Ol is formed under standard conditions.
Δ(r)H° = – 286 kJ mol-1.
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