2. Enthalpy C 6 H 12 O 6 (s) + 6O 2 (g) --> 6CO 2 (g) + 6H 2 O(l) + 2803 kJ 2C 57 H 110 O 6 + 163O 2 (g) --> 114 CO 2 (g) + 110 H 2 O(l) + 75,520 kJ The heat lost or gained by a system undergoing a process at constant pressure is related to the change in ENTHALPY (  H) of the system  H = H final - H initial = q p Note:  H is a state function Most physical and chemical changes take place under the constant pressure of the Earth’s atmosphere.

3. For an exothermic process at constant pressure  H < 0 For an endothermic process at constant pressure  H > 0  H > 0  H < 0

4. Enthalpy, H is defined as H = E + PV For a system undergoing pressure-volume work, the change in enthalpy in the system is  H =  E +  PV If the pressure is a constant external pressure, P ext,  H =  E + P ext  V From the 1st law:  E = q + w For constant P:  H = q p + w + P ext  V Since w = - P ext  V  H = q p - P ext  V + P ext  V = q p

5. If volume is held constant, no pressure-volume work can be done on that system or by that system. work = - P ext  V = 0 at constant volume  E = q + w, For constant volume processes,  E = q v A constant pressure calorimeter measures  H A constant volume calorimeter (like a bomb calorimeter) measures  E.

6. Relationship between  H and  E C(s) + 1/2 O 2 (g) --> CO(g)  H = -110.5 kJ Determine the change in internal energy accompanying this reaction.  H =  E +  (PV)  E =  H -  (PV)  (PV) =  (nRT) = RT(  n g )  E =  H - RT(  n g ) or  H =  E + RT  n g For this reaction  n g = 0.5 mol; hence  E = -111.7 kJ For reactions that do not involve gases;  H ≈  E

7. Enthalpy of Physical Change Phase transitions involve change in energy Vaporization is endothermic; condensation exothermic. Since phase transitions typically take place at constant pressure, the heat transfer is the change in enthalpy.  H vap = H vapor - H liquid liquid --> gas  H fusion = H liquid - H solid solid --> liquid  H freezing = -  H fusion

9.  H forward = -  H reverse

10.  H sublimation =  H fusion +  H vaporization

11. Heating curve of water H 2 O(s) --> H 2 O(l)  H = 6.01 kJ H 2 O(l) --> H 2 O(g)  H = 40.7 kJ

12. Enthalpy of Chemical Change The enthalpy change for a chemical reaction is given by  H =  H(products) -  H(reactants) For example: 2H 2 (g) + O 2 (g) --> 2H 2 O(g)  H = - 483.6 kJ Thermochemical reaction

13. In a chemical reaction, the enthalpy change during the reaction indicates whether the reaction releases energy or consumes energy. If  H < 0, the reaction releases heat and is EXOTHERMIC If  H > 0, the reaction absorbs heat and is ENDOTHERMIC

14. The magnitude of  H for a reaction is directly proportional to the amount of reactants consumed by the reaction. CH 4 (g) + 2O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ Calculate the amount of heat that would be released when 4.50 g of CH 4 (g) is burned in an oxygen atmosphere at constant pressure. 4.50g CH 4 => 0.28 mole CH 4 => (0.28 mol CH 4 )(-890 kJ/mol CH 4 ) = -250 kJ

15. The enthalpy change for a reaction is equal in magnitude but opposite in sign to  H for the reverse reaction CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ CO 2 (g) + 2H 2 O(l) --> CH 4 (g) + 2 O 2 (g)  H = +890 kJ The enthalpy change for a reaction depends on the phase of the reactants and products. 1) CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ 2) CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2H 2 O(g)  H = -802 kJ 3) 2H 2 O(g) --> 2H 2 O(l)  H = -88kJ

16. Standard Reaction Enthalpies CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H = -890 kJ CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2H 2 O(g)  H = -802 kJ

17. Standard State: reactant and products in their pure forms at a pressure of 1 atm (or 1 bar). A solute in a liquid solution is in its standard state when its concentration is 1 mol L -1 In order to compare  H’s accompanying reactions need to define a standard state While temperature is not part of the definition of the standard state, standard reaction enthalpies are usually reported for a temperature of 298.15 K.

18. Standard Reaction Enthalpies (  H o ): enthalpy change accompanying a reaction when reactants in their standard states change to products in their standard states. CH 4 (g) + 2 O 2 (g) --> CO 2 (g) + 2H 2 O(l)  H o = -890 kJ

19. Hess’s Law If a reaction is carried out in a series of steps,  H for the reaction will be equal to the sum of the enthalpy change for the individual steps.

20. The enthalpy of combustion of C to CO 2 is -393.5 kJ/mol, and the enthalpy of combustion of CO to CO 2 is -283.0 kJ/mol CO. (1) C(s) + O 2 (g) --> CO 2 (g)  H = -393.5 kJ Calculate the enthalpy change of combustion of C to CO (2) CO(g) + O 2 (g) --> CO 2 (g)  H = -283.0 kJ 1 2 (3) C(s) + O 2 (g) --> CO (g)  H = ? 1 2

21. C(s) + O 2 (g) --> CO 2 (g)  H = -393.5 kJ CO 2 (g) --> CO(g) + O 2 (g)  H = 283.0 kJ 1 2 C(s) + O 2 (g) --> CO (g)  H = -110.5 kJ 1 2

22. Standard Enthalpy of Combustion Change in enthalpy per mole of a substance that is burned in a combustion reaction under standard conditions

23. Calculate the mass of propane that you would need to burn to obtain 350 kJ of heat which is just enough energy to heat I L of water from room temperature (20 o C) to boiling at sea level. Assume that all the heat generated is absorbed by the water. C 3 H 8 (g) + 5 O 2 (g) --> 3 CO 2 (g) + 4H 2 O(l)  H o = -2220 kJ Combustion of 1 mole of propane generates 2220 kJ of heat Mass or propane required = (350 kJ) (1 mole C 3 H 8 /2220kJ) (44.09 g/mol) = 6.95 g

24. If you were to use butane instead of propane in the previous example, how much butane (in grams) would you need? 2 C 4 H 10 (g) + 13 O 2 (g) --> 8 CO 2 (g) + 10 H 2 O(l)  H o = -5756 kJ Answer: 7.07 g If you were to use ethanol instead of propane in the previous example, how much ethanol (in grams) would you need? C 2 H 5 OH(l) + 3 O 2 (g) --> 2 CO 2 (g) + 3 H 2 O(l)  H o = -1368 kJ Answer: 11.8 g

25. Enthalpies of Formation The enthalpy of formation,  H f, or heat of formation, is defined as the change in enthalpy when one mole of a compound is formed from its stable elements. The standard enthalpy of formation (  H f o ) of a compound is defined as the enthalpy change for the reaction that forms 1 mole of compound from its elements, with all substances in their standard states. 2C(s) + 1/2 O 2 (g) + 3 H 2 (g) --> C 2 H 5 OH(l)  H f o = -277.69 kJ

26. Standard enthalpies of formation of substance can be used to determine standard reaction enthalpies.

27. The standard enthalpy of formation of the most stable form of an element under standard conditions is ZERO.  H f o for C(graphite), H 2 (g), O 2 (g) are zero The stoichiometry for formation reactions indicate the formation of 1 mole of the desired compound, hence enthalpies of formation are always listed as kJ/mol.