2. Energy! Energy: the capacity to do work or supplying heat Energy is detected only by its effects Energy can be stored within molecules etc as chemical potential energy We will most often deal with energy changes with heat (q) For systems at constant pressure (no P  V work) then q =  H (enthalpy)

3. Measuring Heat (q): units: Joule, calorie, Calorie (1 calorie = 4.184 J 1 Calorie = 1000 calories) Use these equivalencies to convert between heat units: An 80 Cal apple has how many J? kJ? 80 Cal 1000 cal 4.184 J 1 X 1 Cal X 1 cal = 334720 J 334720 J 1 kJ 1 X 1000 J = 334.720 kJ

4. Heat is very often involved in chemical reactions There are two possibilities of how heat is involved: 1) heat is absorbed by the reaction ENDOTHERMIC 2) heat is given off by the reaction EXOTHERMIC How much heat is given off or absorbed is measured with enthalpy (H) We are more interested in the change in enthalpy therefore  H  H = Hproducts - Hreactants

5. Exothermic reactions: combustion reactions are obvious examples of exothermic reactions C6H12O6 + 6O2 6CO2 + 6H2O + 2803 kJ Heat is given off Heat is a product in a chemical reaction H of the products must be less than the H of the reactants Therefore  H must be negative For this reaction  H = - 2803 kJ We can also make a Enthalpy diagram for this reaction Reactants  H = -2803 kJ Products Reaction progress

6. Endothermic reactions: -Br2 + Cl2 + 29.4 kJ 2 BrCl Heat is absorbed by the reaction Heat is a reactant in a chemical reaction H of the products is more than the H of the reactants Therefore  H must be positive For this reaction  H = 29.4 kJ This reactions enthalpy diagram looks like: Products H  H = 29.4 kJreactants reaction progress Reactants Products  H = 29.4 kJ

7. Heat with no change of state: -Specific heat (C)the amount of heat needed to raise 1 g of a substance 1 degree q = m C  T (  T = Tf - Ti) see chart pg. 296, therefore the C of water = 4.184 J -Calorimetry measurement of heat change oEX: How much heat is absorbed by 875 mL of water as it is heated from 25o to 75o C q = m C  T q = (875)(4.184)(50) q = 183050 J oEX: 10.0 g of an unknown metal is heated to 100.0o C. The hot metal is added to 50 mL of water at 25.0o C. The final temperature of the solution is 35.6o C. What is the specific heat capacity of the unknown metal? heat lost by metal = heat gained by water -(m C  T) = m C  T -(10)(x)(-64.4) = (50)(4.184)(10.6) x = 3.44 J/ g(C)

9. q=  Hfus (mol) q=  Hvap (mol) Things to know: (where might one write them?)  Hfus = 6.01 kJ/mol  Hvap = 40.7 kJ/mol Cice = 2.1 J/g. oC Cwater = 4.184 J/g. oC Csteam = 1.7 J/g. oC

10. q = m C  T q=  Hfus (mol) q=  Hvap (mol)

11. From -25 to 0: q = m C  T q= (10)(2.1)(25) q= 525 J Melt the ice: q=  Hfus (mol) q= (6.01)(.556) q= 3.34 kJ = 3340 J Heat the water to boiling: q = m C  T q= (10)(4.184)(100) q= 4184 J Boil the water: q=  H vap (mol) q= (40.7)(.556) q= 22.63 kJ = 22630 J Heat the steam to 130: q = m C  T q= (10)(1.7)(30) q= 510 J Add up the total heat: Total q= 525+3340+ 4184+22630+510 = 31189 J = 31.2 kJ EX: How much heat is required to heat 10.0 g of ice at -25.0 oC to steam at 130.0 oC?

12. From -25 to 0: q = m C  T q= (10)(2.1)(25) q= 525 J Melt the ice: q=  Hfus (mol) q= (6.01)(.556) q= 3.34 kJ = 3340 J Heat the water to boiling: q = m C  T q= (10)(4.184)(100) q= 4184 J Boil the water: q=  H vap (mol) q= (40.7)(.556) q= 22.63 kJ = 22630 J Heat the steam to 130: q = m C  T q= (10)(1.7)(30) q= 510 J Add up the total heat: Total q= 525+3340+ 4184+22630+510 = 31189 J = 31.2 kJ EX: How much heat is required to heat 10.0 g of ice at -25.0 oC to steam at 130.0 oC?  from ice at -25 to water at 0o 525 + 3340 = 3865 J  from ice at -25 to water at 100o 525 + 3340 + 4184 = 8049 J  from ice at -25 to steam at 100o 525 + 3340 + 4184 + 22630 = 30679 J

13. Exothermic reactions: H eat is given off by the reaction Heat is a product in a chemical reaction Heat term is on the right  H must be negative E ndothermic reactions : H eat is absorbed by the reaction Heat is a reactant in a chemical reaction Heat term is on the left  H must be positive REVIEW

15. Now we can use this information in stoichiometry type problems (aren’t you happy!) The heat term can be treated just like balanced coefficient on a substance in a mole ratio. Ex: When 12.5 g of glucose (C6H12O6) is burned how many kJ of energy are released? C6H12O6 + 6O2 6CO2 + 6H2O + 2803 kJ