NCERT Solutions for Class 11 Chemistry Chapter 6: Thermodynamics

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Subtopics included in NCERT Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

  1. Thermodynamic Terms
    • The System and the Surroundings
    • Types of Thermodynamic Systems
    • State of the System
    • Internal Energy as a State Function
  2. Applications
    • Work
    • Enthalpy, H
  3. Measurement of ΔU and ΔH: Calorimetry
  4. Enthalpy Change and Reaction Enthalpy
  5. Enthalpies for Different Types of Reactions
  6. Spontaneity
  7. Gibbs Energy Change and Equilibrium

Access Answers to NCERT Solutions for Class 11 Chemistry Chapter 6

Q-1: Choose the correct answer. A thermodynamic state function is a quantity

(i) used to determine heat changes
(ii) whose value is independent of the path
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only

Ans:

(ii) whose value is independent of the path.

Reason:

Functions like pressure, volume and temperature depend on the state of the system only and not on the path.

 

Q-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

Ans:

(iii) q = 0

Reason:

For an adiabatic process, heat transfer is zero, i.e., q = 0.

 

Q-3: The enthalpies of all elements in their standard states are:

(i) Unity   (ii) Zero

(iii) < 0     (iv) Different for every element

Ans:

(ii) Zero

 

Q-4: ∆U0 of combustion of methane is – X kJ mol–1. The value of ∆H0 is

(i) =








Q-11 Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon the formation of 35.2 g of CO2 from carbon and dioxygen gas.

Ans:

Formation of carbon dioxide from di-oxygen and carbon gas is given as:

C(s) + O2(g) → CO2(g);


Q-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 ∆rH for the reaction:

N2O4(g) + 3CO(g) →  N2O(g) + 3 CO2(g)

Ans:

Q-13: Given N2 (g) + 3H2 (g) → 2NH3 (g) ; ∆rH0= –92.4 kJ mol–1
What is the standard enthalpy of the formation of NH3 gas?

Ans:

“Standard enthalpy of formation of a compound is the enthalpy that takes place during the formation of 1 mole of a substance in its standard form, from its constituent elements in their standard state.”

Dividing the chemical equation given in the question by 2, we get

(0.5)N2(g) + (1.5)H2(g) → 2NH3(g)

Therefore, standard enthalpy for formation of ammonia gas

= (0.5)


Q-15: Calculate the enthalpy change for the process

CCl4(g) → C(g) + 4Cl(g) and determine the value of bond enthalpy for C-Cl in CCl4(g).





Q-20: The equilibrium constant for a reaction is 10. What will be the value of ∆G0? R = 8.314 JK–1 mol–1, T = 300 K.

Ans:

Now,



Q-21: Comment on the thermodynamic stability of NO(g), given,

(1/2)N2(g) + (1/2)O2(g) → NO(g);

represents that during NO2(g) formation from O2(g) and NO(g),  heat is evolved. The obtained product,

 NO2(g), gets stabilised with minimum energy.

Thus, unstable NO(g) converts into stable  NO2(g).

 

Q-22: Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. ∆f H0= –286 kJ mol–1.

 Ans:









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