1. The specific heat of gold is 0.128 J/g  °C. How much heat would be needed to warm 250.0 g of gold from 25°C to 100°C? Example 3:

2. Heat of Fusion (H f ) Fusion means melting/freezing amount of energy needed to melt/freeze 1g of a substance Different for every substance – look on reference tables Q = mH f

3. Heat of Vaporization(H v ) Vaporization means boiling/condensing amount of energy needed to boil/condense 1g of a substance Different for every substance – look on reference tables Q = mH v

4. Examples: Calculate the mass of water that can be frozen by releasing 49370 J. Calculate the heat required to boil 8.65 g of alcohol (H v = 855 J/g). Calculate the heat needed to raise the temperature of 100. g of water from 25  C to 63  C.

5. Phase Change Diagram The flat points represent a phase change – temperature does not change while a phase change is occurring even though heat is being added. Diagonal points represent the 3 phases

6. Ice to Steam Handout

7. Enthalpy Thermodynamics

8. Enthalpy (H) Measures 2 things in a chemical reaction:  Energy change  Amount of work Most chemical reactions happen at constant pressure (atmospheric pressure)—open container

9. Enthalpy (ΔH) 2 types of chemical reactions:  Exothermic—heat released to the surroundings, getting rid of heat, -ΔΗ  Endothermic—heat absorbed from surroundings, bringing heat in, +ΔΗ **Enthalpy of reaction— amount of heat from a chemical reaction which is given off or absorbed, units = kJ/mol

10. Enthalpy of Reaction  H = H final - H initial H initial = reactants H final = products If H final > H initial then  H is positive and the process is ENDOTHERMIC If H final < H initial then  H is negative and the process is EXOTHERMIC

11. Enthalpy of Reaction H final < H initial and  H is negative

12. Enthalpy of Reaction H final > H initial and  H is positive

13. Water Formation 2H 2 + O 2 2H 2 0 ΔΗ = -967.28 kJ/mol

14. More Enthalpy The reverse of a chemical reaction will have an EQUAL but OPPOSITE enthalpy change HgO  Hg + ½ O 2 ΔH = + 90.83 kJ Hg + ½ O 2  HgO ΔH = - 90.83 kJ SOOO-----total ΔH = 0

16. Stoichiometry Returns

17. Example 1: Calculate the ΔH for the following reaction when 12.8 grams of hydrogen gas combine with excess chlorine gas to produce hydrochloric acid. H 2 + Cl 2  2HCl ΔH = -184.6 kJ/mol

18. Methods for determining ΔH  Calorimetry  Application of Hess’ Law  Enthalpies of Formation

19. Hess’ Law Enthalpy change for a chemical reaction is the same whether it occurs in multiple steps or one step ΔH rxn = ΣΔH A+B+C (sum of ΔH for each step) Break a chemical reaction down into multiple steps Add the enthalpies (ΔH) of the steps for the enthalpy for the overall chemical reaction

20. Guidelines for using Hess’ Law Use data and combine each step to give total reaction Chemical compounds not in the final reaction should cancel Reactions CAN be reversed but remember to reverse the SIGN on ΔH

21. Calculate  H for S (s) + 3/2O 2(g)  SO 3(g) knowing that S (s) + O 2(g)  SO 2(g)  H 1 = -296.8 kJ SO 2(g) + 1/2O 2(g)  SO 3(g)  H 2 = -98.9 kJ The two equations add up to give the desired equation, so  H net =  H 1 +  H 2 = -395.7 kJ USING ENTHALPY

22. Example 3: H 2 O (l)  H 2 O (g) ΔH° = ? Based on the following: H 2 + ½ O 2  H 2 O (l) ΔH° = -285.83 kJ/mol H 2 + ½ O 2  H 2 O (g) ΔH° = -241.82 kJ/mol

23. Example 4: NO (g) + ½ O 2  NO 2 (g) ΔH° = ? Based on the following: ½ N 2(g) + ½ O 2  NO (g) ΔH° = + 90.29 kJ ½ N 2(g) + O 2  NO 2 (g) ΔH° = +33.2 kJ

24. Homework Enthalpy Worksheet

25. Methods for determining ΔH  Calorimetry  Application of Hess’ Law  Enthalpies of Formation

26. Enthalpy of Formation (ΔH f °) Enthalpy for the reaction forming 1 mole of a chemical compound from its elements in a thermodynamically stable state. A chemical compound is formed from its basic elements present at a standard state (25°C, 1 atm) Enthalpy change for this reaction = ΔH f ° ΔH f °= 0 for ALL elements in their standard/stable state.

27. Enthalpy of Formation cont.  H rxn = H final – H initial Really, ΔH f (products) - ΔH f (reactants) Calculate ΔH rxn based on enthalpy of formation (ΔH f ) aA + bB  cC + dD ΔH° =[c (ΔH f °) C + d(ΔH f °) D ] - [a (ΔH f °) A + b (ΔH f °) B ]

28. Calculate the ΔH° for the reaction 2Mg (s) + O 2(g)  2MgO (s)  H o f for MgO = -601.6 kJ/mol Recall that  H o f for elements in their standard state = 0 kJ/mol  H rxn = ∑(  H o f products)(moles of products) – ∑  H o f reactants)(moles of reactants) = (-601.6kJ/mol)(2) – [(0kJ/mol)(2) + (1)(0kJ/ mol)] = - 1203.2kJ/mol

29. Enthalpy of Formation Values for #6 and 7 Chemical CompoundEnthalpy of Formation (ΔH f °) (kJ/mol) CO-110.5 CO 2 -393.5 Fe 2 O 3 -824.2 Al 2 O 3 -1676