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Home / Chemistry Ncert Class 11 Solutions Pdf Chapter 6

Chemistry Ncert Class 11 Solutions Pdf Chapter 6

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Chapter 6 Thermodynamics

Download NCERT Solutions for Class 11 Chemistry

(Link of Pdf file is given below at the end of the Questions List)

In this pdf file you can see answers of following Questions


NCERT Solutions Exercises Questions


Question 6. 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.


Question 6. 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


Question 6. 3 The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element


Question 6. 4 ΔU0of combustion of methane is – X kJ mol–1. The value of ΔH0 is
(i) = ΔU0
(ii) > ΔU0
(iii) < ΔU0
(iv) = 0


Question 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 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.


Question 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
(v) possible at any temperature


Question 6. 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?


Question 6. 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–1 at 298 K. Calculate enthalpy change for the reaction at 298 K. NH2CN(g) + 3 2 O2(g) → N2(g) + CO2(g) + H2O(l)


Question 6. 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.


Question 6. 10 Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. ΔfusH =


Question 6. 03 kJ mol–1 at 0°C. Cp [H2O(l)] = 75.3 J mol–1 K–1 Cp [H2O(s)] = 3Question 6. 8 J mol–1 K–1


Question 6. 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.


Question 6. 12 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 ΔrH for the reaction:
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)


Question 6. 13 Given N2(g) + 3H2(g) → 2NH3(g) ; ΔrH0 = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas?


Question 6. 14 Calculate the standard enthalpy of formation of CH3OH(l) from the following data: CH3OH (l) + 3 2 O2(g) → CO2(g) + 2H2O(l) ; ΔrH0 = –726 kJ mol–1 C(g) + O2(g) → CO2(g) ; ΔcH0 = –393 kJ mol–1 H2(g) + 1 2 O2(g) → H2O(l) ; Δf H0 = –286 kJ mol–1.


Question 6. 15 Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g). ΔvapH0(CCl4) = 30.5 kJ mol–1. ΔfH0 (CCl4) = –135.5 kJ mol–1. ΔaH0 (C) = 715.0 kJ mol–1 , where ΔaH0 is enthalpy of atomisation ΔaH0 (Cl2) = 242 kJ mol–1 Question 6. 16 For an isolated system, ΔU = 0, what will be ΔS ?


Question 6. 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.


Question 6. 18 For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ΔH and ΔS ?


Question 6. 19 For the reaction 2 A(g) + B(g) → 2D(g) ΔU 0 = –10.5 kJ and ΔS0 = –44.1 JK–1. Calculate ΔG0 for the reaction, and predict whether the reaction may occur spontaneously.


Question 6. 20 The equilibrium constant for a reaction is 10. What will be the value of ΔG0 ? R = 8.314 JK–1 mol–1, T = 300 K.


Question 6. 21 Comment on the thermodynamic stability of NO(g), given 1 2 N2(g) + 1 2 O2(g) → NO(g) ; ΔrH0 = 90 kJ mol–1 NO(g) + 1 2 O2(g) → NO2(g) : ΔrH0= –74 kJ mol–1

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Chemistry Ncert Class 11 Solutions Pdf Chapter 6

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