1. Chemistry 1011 Slot 51 Chemistry 1011 TOPIC Thermochemistry TEXT REFERENCE Masterton and Hurley Chapter 8

2. Chemistry 1011 Slot 52 8.5 Enthalpies of Formation YOU ARE EXPECTED TO BE ABLE TO: Relate  H to enthalpies of formation.

3. Chemistry 1011 Slot 53 Enthalpies of Formation Heats of Formation, or Enthalpies of Formation, are used to provide a concise collection of thermochemical data For consistency, enthalpy of formation data are recorded for reactions that take place under standard conditions Standard conditions are –Constant pressure of 1 atmosphere –Fixed temperature of 25 o C

4. Chemistry 1011 Slot 54 Standard Molar Enthalpy of Formation The enthalpy change when one mole of a compound is formed from its elements at 1 atm and 25 o C The elements must be in their stable states at this pressure and temperature 1 / 2 N 2(g) + 3 / 2 H 2(g)  NH 3(g) ;  H o f =  46.1kJ

5. Chemistry 1011 Slot 55 Notes About  H o f Enthalpies of formation are listed in most data books and are shown in Table 8.3 of the text (page 221) Most enthalpies of formation are negative. This means that the formation of a compound from its elements is exothermic There are no entries in a table of enthalpies of formation for elemental species such as Br 2, O 2, N 2, etc.  H o f for an element is zero

6. Chemistry 1011 Slot 56 Using Heats of Formation Enthalpies of formation can be used to determine the Standard Enthalpy Change for a reaction,  H o One way is to apply Hess’ Law: Use  H o f for CaCO 3, CaO and CO 2 to determine the standard heat of reaction for CaCO 3(s)  CaO (s) + CO 2(g)

7. Chemistry 1011 Slot 57 1.Ca (s) + C (s) + 3 / 2 O 2(g)  CaCO 3(s) ;  H = -1206.9kJ 2.Ca (s) + 1 / 2 O 2(g)  CaO (s) ;  H = -635.1kJ 3.C (s) + O 2(g)  CO 2(g) ;  H = -393.5kJ –Reverse equation 1 –Add equation 2 –Add equation 3 CaCO 3(s)  Ca (s) + C (s) + 3 / 2 O 2(g) ;  H = +1206.9kJ Ca (s) + 1 / 2 O 2(g)  CaO (s) ;  H = -635.1kJ C (s) + O 2(g)  CO 2(g) ;  H = -393.5kJ CaCO 3(s)  CaO (s) + CO 2(g) ;  H = +178.3kJ

8. Chemistry 1011 Slot 58 An Alternative Process The standard enthalpy change for a reaction is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants  H o =    f  products  –     f  reactants Coefficients of reactants and products in the balanced equation for the reaction must be taken into account

9. Chemistry 1011 Slot 59 Textbook Example 8.8 Calculate  H o for the combustion of one mole of propane, C 3 H 8 C 3 H 8(g) + 5 O 2(g)  3CO 2(g) + 4H 2 O (l)  H o = [3  H o f CO 2(g) + 4  H o f H 2 O (l) ] -  H o f  C 3 H 8(g) + 5  H o f O 2(g) ]  H o = [3 x -393.5 + 4 x -285.8] – [1 x –103.8 + 5 x 0]  H o = -2219.9kJ

10. Chemistry 1011 Slot 510 Enthalpies of Formation of Ions in Solution It is possible to construct a table to show the heats of formation of ions in solution However, all ionic solution processes involve both a + and a – ion HCl (g)  H + (aq) + Cl  (aq) To get around this problem,  H o f for the H + (aq) ion is set at zero

11. Chemistry 1011 Slot 511 Determining Enthalpies of Formation of Ions in Solution When HCl (g) is added to water HCl (g)  H + (aq) + Cl  (aq) ;  H o = -74.9kJ Using the summation method: -74.9kJ = [  H o f H + (aq) +  H o f Cl  (aq) ] -  H o f  HCl (g) ] -74.9kJ = 0 +  H o f Cl  (aq) + 92.3kJ (from Table 8.3)  H o f Cl  (aq) = -74.9kJ - 92.3kJ = -167.2kJ

12. Chemistry 1011 Slot 512 Using Enthalpies of Formation of Ions in Solution Example 8.10 Calculate  H o for the reaction: 2H + (aq) + CO 3 2  (aq)  CO 2(g) + H 2 O (l) Use the summation method