{"id":866325,"date":"2025-08-07T11:22:28","date_gmt":"2025-08-07T05:52:28","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=866325"},"modified":"2025-08-07T11:22:28","modified_gmt":"2025-08-07T05:52:28","slug":"ncert-solutions-class-11-chemistry-part-1-chapter-5-thermodynamics-free-pdf","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-1-chapter-5-thermodynamics-free-pdf\/","title":{"rendered":"NCERT Solutions Class 11 Chemistry (Part-1) Chapter 5: Thermodynamics (Free PDF)"},"content":{"rendered":"\n

In this unit, you have learned about the basic principles and laws relevant to thermodynamics. Below, we have presented some in-text exercises taken from the NCERT Class 11 Chemistry (Part I), Chapter 5: Thermodynamics, which will help you reinforce your understanding of key concepts, apply theoretical knowledge, and prepare effectively for exams.<\/p>\n\n\n\n\n\n\n

Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n

Chapter 1<\/a><\/strong><\/td>Chapter 2<\/a><\/strong><\/td>Chapter 3<\/a><\/strong><\/td>Chapter 4<\/a><\/strong><\/td>Chapter <\/strong>6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

NCERT Solutions Class 11 Chemistry (Part-1) Chapter 5: Thermodynamics<\/h2>\n\n\n\n

Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part-1) Chapter 5: Thermodynamics.<\/p>\n\n\n\n

Exercises<\/h2>\n\n\n\n
    \n
  1. 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<\/li>\n\n\n\n
  2. For the process to occur under adiabatic conditions, the correct condition is:
    (i) \u2206T = 0
    (ii) \u2206p = 0
    (iii) q = 0
    (iv) w = 0<\/li>\n\n\n\n
  3. The enthalpies of all elements in their standard states are:
    (i) unity
    (ii) zero
    (iii) < 0
    (iv) different for each element<\/li>\n\n\n\n
  4. \u2206U\u2070 of combustion of methane is \u2013X kJ mol\u20131. The value of \u2206H\u2070 is:
    (i) = \u2206U\u2070
    (ii) > \u2206U\u2070
    (iii) < \u2206U\u2070
    (iv) = 0<\/li>\n\n\n\n
  5. The enthalpies of combustion of methane, graphite, and dihydrogen at 298 K are \u2013890.3 kJ mol\u20131, \u2013393.5 kJ mol\u20131, and \u2013285.8 kJ mol\u20131, respectively. Enthalpy of formation of CH\u2084(g) will be:
    (i) \u201374.8 kJ mol\u20131
    (ii) \u201352.27 kJ mol\u20131
    (iii) +74.8 kJ mol\u20131
    (iv) +52.26 kJ mol\u20131<\/li>\n\n\n\n
  6. A reaction, A + B \u2192 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
    (iv) possible at any temperature<\/li>\n\n\n\n
  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?<\/li>\n\n\n\n
  8. The reaction of cyanamide, NH\u2082CN(s), with dioxygen was carried out in a bomb calorimeter, and \u2206U was found to be \u2013742.7 kJ mol\u20131 at 298 K. Calculate the enthalpy change for the reaction at 298 K.
    NH\u2082CN(s) + 3\/2 O\u2082(g) \u2192 N\u2082(g) + CO\u2082(g) + H\u2082O(l)<\/li>\n\n\n\n
  9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35\u00b0C to 55\u00b0C. Molar heat capacity of Al is 24 J mol\u20131 K\u20131.<\/li>\n\n\n\n
  10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0\u00b0C to ice at \u201310.0\u00b0C.
    \u2206fusH = 6.03 kJ mol\u20131 at 0\u00b0C
    Cp[H\u2082O(l)] = 75.3 J mol\u20131 K\u20131
    Cp[H\u2082O(s)] = 36.8 J mol\u20131 K\u20131<\/li>\n\n\n\n
  11. Enthalpy of combustion of carbon to CO\u2082 is \u2013393.5 kJ mol\u20131. Calculate the heat released upon formation of 35.2 g of CO\u2082 from carbon and dioxygen gas.<\/li>\n\n\n\n
  12. Enthalpies of formation of CO(g), CO\u2082(g), N\u2082O(g), and N\u2082O\u2084(g) are \u2013110, \u2013393, 81, and 9.7 kJ mol\u20131, respectively. Find the value of \u2206rH for the reaction:
    N\u2082O\u2084(g) + 3CO(g) \u2192 N\u2082O(g) + 3CO\u2082(g)<\/li>\n\n\n\n
  13. Given:
    N\u2082(g) + 3H\u2082(g) \u2192 2NH\u2083(g); \u2206rH\u2070 = \u201392.4 kJ mol\u20131
    What is the standard enthalpy of formation of NH\u2083(g)?<\/li>\n\n\n\n
  14. Calculate the standard enthalpy of formation of CH\u2083OH(l) from the following data:
    CH\u2083OH(l) + 3\/2 O\u2082(g) \u2192 CO\u2082(g) + 2H\u2082O(l); \u2206rH\u2070 = \u2013726 kJ mol\u20131
    C(graphite) + O\u2082(g) \u2192 CO\u2082(g); \u2206cH\u2070 = \u2013393 kJ mol\u20131
    H\u2082(g) + 1\/2 O\u2082(g) \u2192 H\u2082O(l); \u2206fH\u2070 = \u2013286 kJ mol\u20131<\/li>\n\n\n\n
  15. Calculate the enthalpy change for the process:
    CCl\u2084(g) \u2192 C(g) + 4Cl(g)
    and calculate the bond enthalpy of C\u2013Cl in CCl\u2084(g).
    Given:
    \u2206vapH\u2070(CCl\u2084) = 30.5 kJ mol\u20131
    \u2206fH\u2070(CCl\u2084) = \u2013135.5 kJ mol\u20131
    \u2206aH\u2070(C) = 715.0 kJ mol\u20131
    \u2206aH\u2070(Cl\u2082) = 242 kJ mol\u20131<\/li>\n\n\n\n
  16. For an isolated system, \u2206U = 0. What will be \u2206S?<\/li>\n\n\n\n
  17. For the reaction at 298 K:
    2A + B \u2192 C
    \u2206H = 400 kJ mol\u20131 and \u2206S = 0.2 kJ K\u20131 mol\u20131
    At what temperature will the reaction become spontaneous, considering \u2206H and \u2206S to be constant over the temperature range?<\/li>\n\n\n\n
  18. For the reaction:
    2Cl(g) \u2192 Cl\u2082(g), what are the signs of \u2206H and \u2206S?<\/li>\n\n\n\n
  19. For the reaction:
    2A(g) + B(g) \u2192 2D(g)
    \u2206U\u2070 = \u201310.5 kJ and \u2206S\u2070 = \u201344.1 J K\u20131
    Calculate \u2206G\u2070 for the reaction, and predict whether the reaction may occur spontaneously.<\/li>\n\n\n\n
  20. The equilibrium constant for a reaction is 10. What will be the value of \u2206G\u2070?
    (R = 8.314 J K\u20131 mol\u20131, T = 300 K)<\/li>\n\n\n\n
  21. Comment on the thermodynamic stability of NO(g), given:
    1\/2 N\u2082(g) + 1\/2 O\u2082(g) \u2192 NO(g); \u2206rH\u2070 = +90 kJ mol\u20131
    NO(g) + 1\/2 O\u2082(g) \u2192 NO\u2082(g); \u2206rH\u2070 = \u201374 kJ mol\u20131<\/li>\n\n\n\n
  22. Calculate the entropy change in surroundings when 1.00 mol of H\u2082O(l) is formed under standard conditions.
    \u2206fH\u2070 = \u2013286 kJ mol\u20131<\/li>\n<\/ol>\n\n\n\n

    Also Read: NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 14: Biodiversity and Conservation (Free PDF)<\/strong><\/a><\/strong><\/p>\n\n\n\n

    Solutions<\/h2>\n\n\n\n
      \n
    1. (ii) Whose value is independent of the path<\/li>\n<\/ol>\n\n\n\n

      State functions depend only on the state, not the path (Section 5.2).<\/p>\n\n\n\n

        \n
      1. (iii) q = 0<\/li>\n<\/ol>\n\n\n\n

        In adiabatic processes, there is no heat exchange (Section 5.4.2).<\/p>\n\n\n\n

          \n
        1. (ii) Zero<\/li>\n<\/ol>\n\n\n\n

          Standard enthalpy of elements in their standard states is taken as zero (Section 5.7.1).<\/p>\n\n\n\n

            \n
          1. (ii) > \u2206U\u2070<\/li>\n<\/ol>\n\n\n\n

            \u2206H = \u2206U + \u2206n_gasRT \u2192 Enthalpy is generally greater than internal energy for reactions involving an increase in gas moles (Section 5.6.1).<\/p>\n\n\n\n

              \n
            1. (i) \u201374.8 kJ mol\u20131<\/li>\n<\/ol>\n\n\n\n

              Derived using Hess\u2019s law.<\/p>\n\n\n\n

                \n
              1. (iv) Possible at any temperature<\/li>\n<\/ol>\n\n\n\n

                If \u2206S > 0 and q is released (exothermic), the reaction is spontaneous at all temperatures.<\/p>\n\n\n\n

                  \n
                1. \u2206U = q \u2013 w = 701 J \u2013 394 J = +307 J<\/li>\n\n\n\n
                2. \u2206H = \u2206U + \u2206n_gasRT
                  \u2206n = (1 + 1 + 1) \u2013 (0 + 1.5) = 1.5
                  \u2206H = \u2013742.7 kJ + (1.5 \u00d7 8.314 \u00d7 298) \/ 1000 = \u2013742.7 + 3.72 = \u2013738.98 kJ (Approx.)<\/li>\n\n\n\n
                3. q = n \u00d7 C \u00d7 \u2206T
                  Molar mass of Al = 27 g\/mol
                  n = 60\/27 = 2.22 mol
                  q = 2.22 \u00d7 24 \u00d7 (55 \u2013 35) = 1065.6 J = 1.07 kJ<\/li>\n\n\n\n
                4. Heat removal from 10\u00b0C to 0\u00b0C = 75.3 \u00d7 10 = 753 J
                  Freezing = \u20136.03 kJ
                  From 0\u00b0C to \u201310\u00b0C = 36.8 \u00d7 10 = 368 J
                  Total = \u20136.03 \u2013 0.753 \u2013 0.368 = \u20137.151 kJ<\/li>\n\n\n\n
                5. 35.2 g CO\u2082 = 0.8 mol
                  q = 0.8 \u00d7 (\u2013393.5) = \u2013314.8 kJ<\/li>\n\n\n\n
                6. \u2206rH = [\u2206fH(N\u2082O) + 3 \u00d7 \u2206fH(CO\u2082)] \u2013 [\u2206fH(N\u2082O\u2084) + 3 \u00d7 \u2206fH(CO)]
                  = [81 + (3 \u00d7 \u2013393)] \u2013 [9.7 + 3 \u00d7 (\u2013110)]
                  = [\u20131098] \u2013 [\u2013320.7] = \u2013777.3 kJ<\/li>\n\n\n\n
                7. \u2206rH = [2 \u00d7 \u2206fH(NH\u2083)] \u2013 [0] = \u201392.4
                  \u2192 \u2206fH(NH\u2083) = \u201392.4 \/ 2 = \u201346.2 kJ\/mol<\/li>\n\n\n\n
                8. \u2206fH(CH\u2083OH) = \u2206rH \u2013 [\u2206fH(CO\u2082) + 2 \u00d7 \u2206fH(H\u2082O)]
                  = \u2013726 \u2013 [\u2013393 + 2(\u2013286)] = \u2013726 \u2013 (\u2013965) = +239 kJ
                  But this is formation \u2192 correct method gives \u2013239 kJ\/mol<\/li>\n\n\n\n
                9. Total bond dissociation = \u2206vapH + \u2206aH(C) + 2 \u00d7 \u2206aH(Cl\u2082) \u2013 \u2206fH(CCl\u2084)
                  = 30.5 + 715 + 2\u00d7242 + 135.5 = 1365 kJ\/mol
                  Since 4 C\u2013Cl bonds: bond enthalpy = 1365 \/ 4 = 341.25 kJ\/mol<\/li>\n\n\n\n
                10. For an isolated system, q = 0, w = 0 \u2192 \u2206U = 0
                  \u2206S can increase, remain constant, or never decrease (Second Law). So, \u2206S \u2265 0<\/li>\n\n\n\n
                11. Spontaneous when \u2206G < 0: \u2206G = \u2206H \u2013 T\u2206S
                  0 = 400 \u2013 T(0.2) \u2192 T = 2000 K<\/li>\n\n\n\n
                12. 2Cl(g) \u2192 Cl\u2082(g): Bond formed \u2192 \u2206H < 0
                  Randomness decreases \u2192 \u2206S < 0
                  \u2206H = negative, \u2206S = negative<\/li>\n\n\n\n
                13. \u2206G = \u2206H \u2013 T\u2206S
                  \u2206H = \u2206U (no \u2206n_gas given) = \u201310.5 kJ
                  \u2206S = \u201344.1 J = \u20130.0441 kJ
                  \u2206G = \u201310.5 \u2013 (298)(\u20130.0441) = \u201310.5 + 13.14 = +2.64 kJ \u2192 Not spontaneous<\/li>\n\n\n\n
                14. \u2206G = \u2013RT ln K = \u2013(8.314)(300)ln(10)
                  = \u2013(8.314)(300)(2.303) = \u20135744 J = \u20135.74 kJ<\/li>\n\n\n\n
                15. NO formation is endothermic, so not thermodynamically stable.
                  Converts to NO\u2082, releasing energy \u2192 NO is less stable.<\/li>\n\n\n\n
                16. \u2206S_surr = \u2013\u2206H\/T = \u2013(\u2013286,000 J)\/298 = +959.73 J K\u207b\u00b9<\/li>\n<\/ol>\n\n\n\n

                  Also Read: NCERT Solutions Class 11 Geography Fundamentals of Physical Geography Chapter 14: Biodiversity and Conservation (Free PDF)<\/strong><\/a><\/strong><\/strong><\/p>\n\n\n\n

                  Download <\/strong>NCERT Solutions Class 11 Chemistry (Part-1) Chapter 5: Thermodynamics<\/h2>\n\n\n\n
                  Download PDF of NCERT Solutions Class 11 Chemistry (Part-1) Chapter 5: Thermodynamics<\/a><\/strong><\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

                  Download the Solutions of Other Chapters of Class 11 Chemistry<\/strong><\/p>\n\n\n\n

                  Chapter 1<\/a><\/strong><\/td>Chapter 2<\/a><\/strong><\/td>Chapter 3<\/a><\/strong><\/td>Chapter 4<\/a><\/strong><\/td>Chapter <\/strong>6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

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                  For more topics, follow LeverageEdu NCERT Study Material<\/strong><\/a> today! <\/p>\n","protected":false},"excerpt":{"rendered":"In this unit, you have learned about the basic principles and laws relevant to thermodynamics. 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