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 kJ reactants reaction progress

7. 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 XX

8. Ex: When only 50.0 kJ of energy is available for the reaction of bromine and chlorine to form monobromine monochloride how many grams of the product are formed? Br2 + Cl2 + 29.4 kJ 2 BrCl

9. On Day 1 we looked at thermochemical equations like the one below: (REVIEW For the equation below: the reaction is ____ thermic, the heat _______ (produced or absorbed) was ________, and the  H of the reaction is ______) C6H12O6 + 6O2 6CO2 + 6H2O + 2803 kJ Some of you inquisitively wondered where I got the number for the heat term. The heat term can be found several ways. You could do the reaction in a calorimeter and determine the heat given off from the temperature rise of the water (we’ll learn that on day 3). Another way to find the heat term without the laboratory is to use Hess’s Law.

10. Hess’s Law: -Hess’s law states that if you add two or more equations to get a third equation (like adding equations in algebra) you can add the  (s) of the original equations to find the  H of the final equation -This is a great way to find the  of an unknown reaction -To do this we need to know two things: oIf you reverse a reaction you change the sign of  oIf you multiply or divide an equation by a whole number you do the same to the  EX: Find the  of the following reaction: 2Al + Fe2O3 2Fe + Al2O3 Given: 4Al + 3O2 2Al2O3  -3352.0 kJ 2Fe + 1 1/2O2 Fe2O3  = -822.2 kJ 822.2 kJ -853.8 kJ -1676.0 kJ

11. -We have found most of the  fo (heat of formation) of many, many compounds (pg. 316) -We can use these to calculate the  of most of our reactions using a more useful form of Hess's Law:  rxn =  n(  fo products) -  m(  fo reactants) - I know this looks bad but in use it is really quite elegant and useful EX: What is the heat of the reaction for the decomposition of hydrogen peroxide?

12. -We have found most of the  fo (heat of formation) of many, many compounds (pg. 316) -We can use these to calculate the  of most of our reactions using a more useful form of Hess's Law:  rxn =  n(  fo products) -  m(  fo reactants) - I know this looks bad but in use it is really quite elegant and useful EX: What is the heat of the reaction for the decomposition of hydrogen peroxide? Summation (add up) coefficients in front of the substance in the balanced equation look up on a  Hf table (316) O2(g) You try the next one!!!!

13. Ex: 15.0 kJ of heat is applied to 10.0 g of calcium carbonate. Heat causes calcium carbonate to decompose to calcium oxide and carbon dioxide gas. How many L of carbon dioxide gas at STP will be formed? Balance!CaCO3 CaO + CO2 Look up the  Hf:-1207.0 -->-635.1 + -393.5 (all in kJ/mol) Do Hess' Law:  Hrxn =  n(Hfo products) -  m(Hfo reactants) = [1(-635.1) + 1(-393.5)] - [1(-1207.0)] = (-1029.0) - (-1207.0) = +178.0 kJ/mol Thermochemical equation is: CaCO3 + 178kJCaO + CO2 Do the stoichiometry: 15.0 kJ 1 mol CO2 22.4L of CO2 gas 1 178 kJ 1 mol CO2 gas 10.0 g CaCO3 1 mol CaCO3 1 mol CO2 22.4 L of CO2 gas 1 100 g CaCO3 1 mol CaCO3 1 mol CO2 x x x x x = = 1.89 L 2.24 L Answer is 1.89 L

14. Heat with no change of state: -Heat Capacity -Specific heat (C)q = m C  T (  T = Tf - Ti) -Calorimetry oEX: How much heat is absorbed by 875 mL of water as it is heated from 25o to 75o C 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?

15. Heat with no change of state: -Heat Capacity the amount of heat needed to increase the temperature 1 degree celcius -Specific heat (C)the amount of heat needed to raise 1 g of a substance 1 degreeq = 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?

16. Heat with no change of state: -Heat Capacity the amount of heat needed to increase the temperature 1 degree celcius -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)

18. 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

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

20. 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?

21. 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

23. 