1. Chemistry

2. Chemical thermodynamics-II

3. Session Objectives

4. Session objectives 1.Enthalpy 2.Various types of enthalpy of reactions 3.Heat capacities of gases 4.Adiabatic process 5.Hess’s law 6.Bond energy 7.Lattice energy 8.Limitation of first law of thermodynamics

5. Enthalpy Enthalpy is the total heat contents of the system at constant pressure. Enthalpy is shown by ‘H’. H = E + PV H 2 – H 1 = E 2 – E 1 + P(V 2 – V 1 ) Where H 1, E 1 and V 1 are the enthalpy, internal energy and volume respectively in initial state while H 2, E 2 and V 2 are the enthalpy, internal energy and volume respectively in final state. Enthalpy change at constant pressure

6. Enthalpy Enthalpy is a state function At constant pressure PV 1 = n 1 RT(for initial state) PV 2 = n 2 RT(for final state) P(V 2 – V 1 ) = RT(n 2 – n 1 ) Where  n g = n p –n r (gaseous moles only)

7. Enthalpy of formation It is the change in enthalpy when one mole of a compound is formed from its elements in their naturally occuring physical states.

8. Enthalpy of combustion It is the change in enthalpy when one mole of the substance undergoes complete combustion.

9. Application of heat of combustion Amount of heat produced per gram of a substance (food or fuel) is completely burnt. Calorific value Calorific value of CH 4 (g) = Hydrogen has the highest calorific value (150 kJ/g)

10. Enthalpy of neutralization Difference in energy is used to ionize weak base. Strong acid and strong base Ionization energy of NH 4 OH = What is the Ionization energy of NH 4 OH?

11. Enthalpy of solution Amount of heat evolved or absorbed per mole of the substance in excess of water,

12. Enthalpy of fusion One mole of solid substance changes to its liquid state at its melting point.

13. Enthalpy of vaporization One mole of the substance changes from liquid state to gaseous state at its boiling point. Enthalpy change per mole of a solid converts directly to its vapours Enthalpy of sublimation

14. Heat capacity Quantity of heat required to raise the temperature of the system by one degree. Heat capacity = Heat capacity at constant pressure (Since at constant pressure dq = dH)

15. Heat capacity (Since at constant volume dq = dE) The difference between C p and C v is equal to the work done by 1 mole of gas in expansion when heated through 1° C. Heat capacity at constant volume

16. Heat capacity Specific heat capacity is the heat required to raise the temperature of unit mass by one degree. m = Mass of the substance q = Heat required = Temperature difference c = Specific heat capacity. Specific heat capacity of water is 4.18 J/g K.

17. Adiabatic work For adiabatic process, q = 0 it means no heat is exchanged with the surrounding.

18. Reversible Adiabatic expansion Relations for reversible adiabatic expansion of an ideal gas

19. Irreversible Adiabatic expansion (i) For free expansion, w = 0. Since P ext = 0 (ii)For intermediate expansion,

20. Hess’s Law C q 1 q 2 A B q According to Hess’s law q = q 1 + q 2

21. Applications of Hess’s law Determination of heat changes of transformation of rhombic sulphur into monoclinic sulphur. Given Subtracting (ii) from (i), we get S (rhombic) – S (monoclinic)

22. Resonance energy Resonance energy = Enthalpy of formation calculated from bond energy – Experimental value of enthalpy

23. Question

24. Illustrative example Calculate resonance energy of benzene from the following data: (i) benzene = –358.5 kJ mol –1 (ii) Heat of atomization of carbon is 716.8 kJ mol -1 (iii) Bond energies of C–H, C–C, C = C and H–H bonds are 490, 340, 620 and 436.9 kJ mol –1 respectively.

25. Solution The required equation is Bond energies of reactants – Bond energies of products = {6 [ H C(s) c(g) ] + 3 ( H H–H ) – 3{[ H C–C ] + 3 ( H C =C ) + 6 ( H C–H ) =6716.8 + (3 436.9) – (3 340) – (3620) – (6490) kJ mol –1 Resonance energy = H f (obs.) – H f (cal) = – 358.5 – ( – 208.5) = – 150.0 kJ mol –1

26. Determination of lattice energy

27. Figure

28. Determination of bond energies or bond enthalpies Energy required to break the bond or energy released during the bond formation is called bond energy. The average of these two bond dissociation energies gives the value of bond energy of S — H. Bond energy of S — H bond

29. Limitations of first law 1.The first law of thermodynamics states that one form of energy disappears, an equivalent amount of another form of energy is produced. But it is salient about the extent to which such conversion can take place. 2.It does not tell about the direction of flow of heat. 3.It does not tell about spontaneity of reaction.

30. Class exercise

31. Class exercise 1 For the reaction for the reaction will be (a) – 781.80 Kcal(b) – 780.009 Kcal (c) – 780.75 Kcal(d) 780.05 Kcal

32. Solution = –780.9 – (–1.5 RT) = – 780.9 + 0.891 = – 780.009 Kcal Hence, the answer is (b).

33. Class exercise 2 When 1 gram of methane (CH 4 ) burns in O 2 the heat evolved (measured under standard conditions) is 13.3 kcal. What is the heat of combustion? (a) –13.3 k cals(b) +213 k cals (c) – 213 k cals(d) – 416 k cals

34. Solution 13.3 Kcal/gm evolved = – 213 Kcal/mol Hence, the answer is (c).

35. Class exercise 3 When 4.184 J of heat is transferred to 1 g of water at 20° C, its temperature rises to 21° C. The molar heat capacity at this temperature is (a) 18 JK –1 (b) (c) 75.4 JK –1 (d) 4.184 JK –1 18 4184 1. JK - C = (4.184) × 18 = 75.4 J/K Hence, the answer is (c). Solution:

36. Class exercise 4 When 0.532 g of benzene (B.P 80° C) is burnt in a constant volume system with an excess of oxygen, 22.3 kJ of heat is given out. for the combustion process is given by (a) – 21 kJ(b) – 1234.98 kJ (c) – 221 kJ(d) – 3273.26 kJ

37. Solution = – 3269.5 kJ Hence, the answer is (d).

38. Class exercise 5 Consider the reaction = – 98.3 kJ. If the enthalpy of formation of SO 3 (g) is – 395.4 kJ, then the enthalpy of formation of SO 2 (g) is (a) 297.1 kJ(b) 493.7 kJ (c) – 493.7 kJ(d) – 297.1 kJ Solution: Hence, the answer is (d).

39. Class exercise 6 Calculate the heat change for the following reaction: for CH 4, H 2 O and CO 2 are –17.89, –68.3 and–94.05 kcal/mole.

40. Solution = – 2 × 68.3 – 94.05 + 17.89 – 0 = – 212.76 kcal/mol

41. Class exercise 7 Calculate the heat of combustion of benzene from the following data:

42. Solution The required reaction is 3 × (4) + 6 × (3) – (2) = 3(– 68320) + 6(– 93050) – (1720 Cal) = – 774.980 kcal/mol

43. Class exercise 8 Calculate for the following reaction at 27 o C. Given = – 337 kcal R= 1.987 cal deg –1 mole –1

44. Solution  H =  E + P  V = – 335.8078 kcal/mol

45. Class exercise 9 Calculate the heat of combustion of acetic acid at 25 o C if the heat of formation of CH 3 COOH(l),CO 2 (g) and H 2 O(l) are –116.4, –94.0 and –68.3 kcal mole –1 respectively. = 2(– 94.0) + 2(– 68.3) – (–116.4) = – 208.2 Kcal/mol Solution:

46. Thank you