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Solutions

Overview

Step-by-step NCERT solutions for Thermodynamics (Chapter 5, NCERT Class 11 Chemistry) — every question and answer worked out in full, not just the final result. You can also read the Thermodynamics textbook chapter.

Solved

What these solutions cover

All 22 questions in Thermodynamics are solved in the PDF. Here's what's inside, exercise by exercise:

Exercises

  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.
  2. For the process to occur under adiabatic conditions, the correct condition is:
    • (i) DeltaT = 0
    • (ii) Deltap = 0
    • (iii) q = 0
    • (iv) w = 0
  3. The enthalpies of all elements in their standard states are:
    • (i) unity
    • (ii) zero
    • (iii) < 0
    • (iv) different for each element
  4. DeltaU of combustion of methane is -X kJ mol⁻¹. The value of DeltaH is
    • (i) = DeltaU
    • (ii) > DeltaU
    • (iii) < DeltaU
    • (iv) = 0
  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.
  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
  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?
  8. The reaction of cyanamide, NH₂CN(s), with dioxygen was carried out in a bomb calorimeter, and DeltaU 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)
  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.
  10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. DeltafusH = 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
  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.
  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 DeltarH for the reaction: N₂O₄(g) + 3CO(g) -> N₂O(g) + 3CO₂(g)
  13. Given N₂(g) + 3H₂(g) -> 2NH₃(g); DeltarH = -92.4 kJ mol⁻¹ What is the standard enthalpy of formation of NH₃ gas?
  14. Calculate the standard enthalpy of formation of CH₃OH(l) from the following data: CH₃OH(l) + 3/2 O₂(g) -> CO₂(g) + 2H₂O(l) ; DeltarH = -726 kJ mol⁻¹ C(graphite) + O₂(g) -> CO₂(g) ; DeltacH = -393 kJ mol⁻¹ H₂(g) + 1/2 O₂(g) -> H₂O(l); DeltafH = -286 kJ mol⁻¹.
  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). DeltavapH(CCl₄) = 30.5 kJ mol⁻¹ DeltafH(CCl₄) = -135.5 kJ mol⁻¹ DeltaaH(C) = 715.0 kJ mol⁻¹ (enthalpy of atomisation) DeltaaH(Cl₂) = 242 kJ mol⁻¹
  16. For an isolated system, DeltaU = 0, what will be DeltaS?
  17. For the reaction at 298 K, 2A + B -> C DeltaH = 400 kJ mol⁻¹ and DeltaS = 0.2 kJ K⁻¹ mol⁻¹ At what temperature will the reaction become spontaneous considering DeltaH and DeltaS to be constant over the temperature range.
  18. For the reaction, 2 Cl(g) -> Cl₂(g), what are the signs of DeltaH and DeltaS?
  19. For the reaction 2 A(g) + B(g) -> 2D(g) DeltaU = -10.5 kJ and DeltaS = -44.1 J K⁻¹. Calculate DeltaG for the reaction, and predict whether the reaction may occur spontaneously.
  20. The equilibrium constant for a reaction is 10. What will be the value of DeltaG? R = 8.314 J K⁻¹ mol⁻¹, T = 300 K.
  21. Comment on the thermodynamic stability of NO(g), given 1/2 N₂(g) + 1/2 O₂(g) -> NO(g); DeltarH = 90 kJ mol⁻¹ NO(g) + 1/2 O₂(g) -> NO₂(g); DeltarH = -74 kJ mol⁻¹
  22. Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. DeltafH = -286 kJ mol⁻¹.
Sample

Worked examples

3 questions from Thermodynamics, solved step by step.

Exercises

For the process to occur under adiabatic conditions, the correct condition is: (i) DeltaT = 0 (ii) Deltap = 0 (iii) q = 0 (iv) w = 0

Adiabatic process: a process in which no heat is exchanged between the system and its surroundings.

By definition, in an adiabatic process, q = 0.

DeltaT = 0 defines an isothermal (not adiabatic) process.

Deltap = 0 defines an isobaric process.

w = 0 is the condition for a process done at constant volume with no work — this does not define adiabatic.

Answer: (iii) q = 0

Exercises

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

By convention, the standard enthalpy of formation of all elements in their most stable standard states (at 298 K, 1 bar) is defined as zero.

This is the reference baseline from which enthalpies of all compounds are measured.

For example: DeltafH[O₂(g)] = 0, DeltafH[C(graphite)] = 0, DeltafH[H₂(g)] = 0.

Answer: (ii) zero

Exercises

DeltaU of combustion of methane is -X kJ mol⁻¹. The value of DeltaH is (i) = DeltaU (ii) > DeltaU (iii) < DeltaU (iv) = 0

Combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

Deltaₙ_gas = moles of gaseous products - moles of gaseous reactants

Deltaₙ_gas = 1 - (1 + 2) = -2

Relation: DeltaH = DeltaU + Deltaₙ_gas * R * T

DeltaH = -X + (-2) * R * T = -X - 2RT

Since 2RT is positive, DeltaH is more negative than DeltaU.

Therefore DeltaH < DeltaU (i.e., DeltaH is a more negative number than -X).

Answer: (iii) < DeltaU

All 22 questions in Thermodynamics are solved step by step in the Solutions PDF above.

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