1. 1  H = H final - H initial If H final > H initial then  H is positive Process is ENDOTHERMIC If H final > H initial then  H is positive Process is ENDOTHERMIC If H final < H initial then  H is negative Process is EXOTHERMIC If H final < H initial then  H is negative Process is EXOTHERMICENTHALPYENTHALPY For systems at constant pressure, the heat content is the same as the Enthalpy, or H, of the system. Heat changes for reactions carried out at constant pressure are ths same as changes in enthalpy (∆H)

2. 2 Standard Enthalpy Values Most  H values are labeled  H o P = 1 atmosphere ( = 760 torr = 101.3 kPa) Concentration = 1 mol/L T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas o means measured under standard conditions

3. 3 Endo- and Exothermic Surroundings Heat q sys > 0 System ENDOTHERMICENDOTHERMIC Heat q sys < 0 q sys < 0 Surroundings System EXOTHERMIC

4. 4 But the reverse reaction, the decomposition of water : H 2 O(g) + 242 kJ ---> H 2 (g) + 1/2 O 2 (g) Endothermic reaction — heat is a “reactant”,  H = +242 kJ. This does not occur spontaneously. Consider the combustion of H 2 to form water.. H 2 (g) + 1/2 O 2 (g) ---> H 2 O(g)  242 kJ Exothermic reaction — heat is a “product”.  H = -242 kJ. This is spontaneous and proceeds readily once initiated. USING ENTHALPY BUT... Decomposition of water can be made to occur by coupling to another, spontaneous process...

5. 5 How can we make H 2 gas ? N. Lewis, American Scientist, Nov. 1995, page 534. Nov. 1995, page 534.

6. 6 liquid Making H 2 from liquid H 2 O involves two steps. H 2 O(liq) + 44 kJ  H 2 O(g) H 2 O(g) + 242 kJ  H 2 (g) + 1/2 O 2 (g) --------------------------------------------------- H 2 O(liq) + 286 kJ  H 2 (g) + 1/2 O 2 (g) This is an example of HESS’S LAW — If a reaction is the sum of 2 or more others, the net  H is the sum of the  H’s of the other rxns. If a reaction is the sum of 2 or more others, the net  H is the sum of the  H’s of the other rxns.

7. Heat and Changes of State Heat of combustion (∆H)= the heat of reaction for the complete burning of one mole of a substance Molar heat of fusion (∆H fus )= the heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature Molar heat of solidification (∆H solid )= heat lost when one mole of a liquid freezes to a solid at a constant temperature (equal to the negative heat of fusion) Molar heat of vaporization (∆H vap )= the heat absorbed by one mole of a substance in vaporizing from liquid to a gas Molar heat of condensation (∆H cond )= heat released by one mole of a vapor as it condenses

8. Example (Heat of combustion) The standard heat of combustion (∆H° rxn ) for glucose (C 6 H 12 O 6 ) is -2808 kJ/mol. If you eat and burn 70.g of glucose in one day, how much energy are you getting from the glucose? –Step one: convert g of glucose to moles 70. g glucose x 1 mol = 0.28 mol glucose 1 246 g –Step two: Use (∆H° rxn ) to find amount of kJ gained 0.28 mol glucose x 2808 kJ = 790 kJ gained (+ b/c gained not lost) 1 1 mol

9. Example (Other ∆H’s) You have a sample of H2O with a mass 23.0 grams at a temperature of -46.0 °C. ΔHfus= 6.01 kJ/mol ΔHvap= 40.7 kJ/mol How many kilojoules of heat energy are necessary to carry out each step? Heat the ice to 0 °C? –Which equation do you need? q = mcΔT q = (23g)(4.184 J/g °C)(0 °C - - 46 °C) q = 4427 J = 4.43 kJ Boil the water? –Which equation do you need? ΔH = mol x ΔHvap ΔH = (23.0g/1)(1 mol/18g)(40.7 kJ/mol) ΔH = 52 kJ

10. 10 Calc.  H rxn for S(s) + 3/2 O 2 (g) --> SO 3 (g) knowing that S(s) + O 2 (g) --> SO 2 (g)  H 1 = -320.5 kJ SO 2 (g) + 1/2 O 2 (g) --> SO 3 (g)  H 2 = -75.2 kJ Hess’s Law - a second example : The two rxns. add to give the desired rxn., S(s) + 3/2 O 2 (g) --> SO 3 (g) so  H rxn =  H 1 +  H 2 = -395.7 kJ  H 3 = -395.7 kJ

11. 11 - the enthalpy change when 1 mol of compound is formed from elements under standard conditions. By definition,  H o f = 0 for elements in their standard states.  H o f = standard molar enthalpy of formation H 2 (g) + 1/2 O 2 (g) --> H 2 O(g)  H o f = -241.8 kJ/mol H 2 (g) + 1/2 O 2 (g) --> H 2 O(g)  H o f = -241.8 kJ/mol

12. 12 Using Standard Enthalpy Values In general, when ALL enthalpies of formation are known,  H o rxn =   H o f (products) -  H o f (reactants) Calculate  H of reaction?