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Chemistry 1011

Chemistry 1011. TOPIC Thermochemistry TEXT REFERENCE Masterton and Hurley Chapter 8. 8.5 Enthalpies of Formation. YOU ARE EXPECTED TO BE ABLE TO: Relate D H to enthalpies of formation. Enthalpies of Formation.

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Chemistry 1011

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  1. Chemistry 1011 TOPIC Thermochemistry TEXT REFERENCE Masterton and Hurley Chapter 8 Chemistry 1011 Slot 5

  2. 8.5 Enthalpies of Formation YOU ARE EXPECTED TO BE ABLE TO: • Relate DH to enthalpies of formation. Chemistry 1011 Slot 5

  3. 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 25oC Chemistry 1011 Slot 5

  4. Standard Molar Enthalpy of Formation • The enthalpy change when one mole of a compound is formed from its elements at 1 atm and 25oC • The elements must be in their stable states at this pressure and temperature 1/2N2(g) + 3/2H2(g) NH3(g); DHof = -46.1kJ Chemistry 1011 Slot 5

  5. Notes About DHof • 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 Br2, O2, N2, etc. DHof for an element is zero Chemistry 1011 Slot 5

  6. Using Heats of Formation • Enthalpies of formation can be used to determine the Standard Enthalpy Change for a reaction, DHo • One way is to apply Hess’ Law: • Use DHof forCaCO3, CaO and CO2 to determine the standard heat of reaction for CaCO3(s)  CaO(s) + CO2(g) Chemistry 1011 Slot 5

  7. Ca(s) + C(s) + 3/2O2(g) CaCO3(s) ; DH = -1206.9kJ • Ca(s) + 1/2O2(g) CaO(s) ; DH = -635.1kJ • C(s) + O2(g) CO2(g) ; DH = -393.5kJ • Reverse equation 1 • Add equation 2 • Add equation 3 CaCO3(s)  Ca(s) + C(s) + 3/2O2(g) ; DH = +1206.9kJ Ca(s) + 1/2O2(g) CaO(s) ; DH = -635.1kJ C(s) + O2(g) CO2(g) ; DH = -393.5kJ CaCO3(s)  CaO(s) + CO2(g);DH = +178.3kJ Chemistry 1011 Slot 5

  8. 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 • DHo = SDHofproducts–SDHofreactants • Coefficients of reactants and products in the balanced equation for the reaction must be taken into account Chemistry 1011 Slot 5

  9. Textbook Example 8.8 • Calculate DHo for the combustion of one mole of propane, C3H8 C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) DHo = [3 DHof CO2(g) + 4 DHof H2O(l)] - [DHofC3H8(g) +5DHof O2(g)] DHo = [3 x -393.5 + 4 x -285.8] – [1 x –103.8 + 5 x 0] DHo = -2219.9kJ Chemistry 1011 Slot 5

  10. 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, DHof for theH+(aq) ion is set at zero Chemistry 1011 Slot 5

  11. Determining Enthalpies of Formation of Ions in Solution • When HCl(g) is added to water HCl(g) H+(aq) + Cl-(aq);DHo = -74.9kJ • Using the summation method: -74.9kJ = [DHof H+(aq) + DHofCl-(aq)] - [DHofHCl(g) ] -74.9kJ = 0 + DHofCl-(aq) + 92.3kJ (from Table 8.3) DHofCl-(aq) = -74.9kJ - 92.3kJ = -167.2kJ Chemistry 1011 Slot 5

  12. Using Enthalpies of Formation of Ions in Solution • Example 8.10 • Calculate DHo for the reaction: 2H+(aq) + CO32-(aq)  CO2(g) + H2O(l) • Use the summation method Chemistry 1011 Slot 5

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