1. READING for Thursday: 7.3 – 7.4 READING for Thursday: 7.3 – 7.4 HOMEWORK – DUE TUESDAY 10/20/15 HOMEWORK – DUE TUESDAY 10/20/15 HW-BW 7.1 (Bookwork) CH 7 #’s 5, 7-12 all, 14, 15, 20, 21, 24, 28-31 all, 34 HW-BW 7.1 (Bookwork) CH 7 #’s 5, 7-12 all, 14, 15, 20, 21, 24, 28-31 all, 34 HW-WS 12 (Worksheet) (from course website) HW-WS 12 (Worksheet) (from course website) HOMEWORK – DUE THURSDAY 10/22/15 HOMEWORK – DUE THURSDAY 10/22/15 HW-BW 7.2 (Bookwork) CH 7 #’s 39, 42, 48-52 all, 55-60 all, 64, 69, 71, 72, 78, 90 HW-BW 7.2 (Bookwork) CH 7 #’s 39, 42, 48-52 all, 55-60 all, 64, 69, 71, 72, 78, 90 HW-WS 13 (Worksheet) (from course website) HW-WS 13 (Worksheet) (from course website) Lab Lab Wednesday/Thursday – finish EXP 8 Wednesday/Thursday – finish EXP 8 Next Monday/Tuesday – EXP 9 Next Monday/Tuesday – EXP 9

2. More Enthalpy Diagram Draw and enthalpy diagram and calculate the  H rxn for: N 2 O (g) + 2 O 2(g)  N 2 O 5(s) Enthalpy  elements N 2(g) + 5/2 O 2(g) reactants N 2 O (g) + 2 O 2(g) 1) formation of reactants products N 2 O 5(g)  H = 1 mol (  82.05 kJ/mol) =  82.05 kJ 4) heat release in formation of products  43.1 kJ heat of formation of reactants  82.05 kJ  H =  43.1 kJ – (  82.05 kJ) =  125.2 kJ 2) formation of N 2 O 5(g)  H = 1 mol (  43.1 kJ/mol) =  43.1 kJ

3. Heat of Reaction for Heats of Formation Just like in the enthalpy diagrams, but with less diagram C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)

4. Heat of Reaction for Heats of Formation Just like in the enthalpy diagrams, but with less diagram N 2 O (g) + 2 O 2(g)  N 2 O 5(s)

5. Heat Capacity When a system absorbs heat, its temperature increases When a system absorbs heat, its temperature increases The increase in temperature is directly proportional to the amount of heat absorbed The increase in temperature is directly proportional to the amount of heat absorbed The proportionality constant is called the heat capacity, C The proportionality constant is called the heat capacity, C units of C are J/°C or J/K units of C are J/°C or J/K q = C x  T

6. The heat capacity of an object depends on its amount of matter The heat capacity of an object depends on its amount of matter usually measured by its mass usually measured by its mass 200 g of water requires twice as much heat to raise its temperature by 1°C as does 100 g of water 200 g of water requires twice as much heat to raise its temperature by 1°C as does 100 g of water The heat capacity of an object depends on the type of material The heat capacity of an object depends on the type of material 1000 J of heat energy will raise the temperature of 100 g of silver by 42 °C, 100 g of sand 12 °C, but only raise the temperature of 100 g of water by 2.4 °C 1000 J of heat energy will raise the temperature of 100 g of silver by 42 °C, 100 g of sand 12 °C, but only raise the temperature of 100 g of water by 2.4 °C Heat Capacity

7. The specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C The specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C units are J/(g∙°C) units are J/(g∙°C) The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C The molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C Specific Heat Capacity

8. heat (J or cal) mass (g) specific heat ( or ) change in temp. ( o C or K)  T = T final - T initial Specific Heat Capacity

9. The amount of energy (J or cal) that is required to raise 1 g of a substance by 1 degree Celsius. Specific heat for water =or Specific heat for silver = or Specific Heat Capacity

10. A SMALLER specific heat means a LARGER temperature change given the same amount of energy. Specific Heat Temperature change when 51000 calories are added to 1000.0 grams of each: ~900 o C

11. 75.0 mL of ethanol at 25.0 o C is poured into a perfectly insulating coffee cup calorimeter containing 50.0 g of water at 33.4 o C. What is the final temperature of the ethanol? C (water) = 4.184 J/g o C, C (ethanol) = 2.42 J/g o C, d (ethanol) = 0.789g/mL

12. Specific Heat How many joules are needed to raise the temperature of 500.0 grams of water by 23.0 o C? joules 500.0 grams23.0 o C raise

13. Specific Heat How many joules are needed to raise the temperature of 500.0 grams of water by 23.0 o C?

14. Specific Heat 3.718 kJ is removed from a 51.73 gram piece of copper which was originally at 391.0 o C. What is the final temperature? (S.H. = 0.385 J/g o C) removed3.718 kJ51.73 gram 0.385 J/g o C 391.0 o C

15. Specific Heat 3.718 kJ is removed from a 51.73 gram piece of copper which was originally at 391.0 o C. What is the final temperature? (S.H. = 0.385 J/g o C)

16. Specific Heat When 67.89 joules was added to a piece of gold, its temperature rose from 23.0 o C to 89.7 o C. What was the mass of the gold? (S.H. = 0.0305 cal/g o C)

17. Specific Heat When 67.89 joules was added to a piece of gold, its temperature rose from 23.0 o C to 89.7 o C. What was the mass of the gold? (S.H. = 0.0305 cal/g o C)

18. Enthalpy (  H) C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)  H rxn = –2202.0 kJ If it takes 7.20x10 3 kJ to perfectly cook a steak, how many grams of propane are necessary (assuming perfect heat transfer)? What mass of carbon dioxide is produced when enough propane is burned to release 7.20x10 3 kJ?

19. Enthalpy (  H) C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g)  H rxn = –2202.0 kJ What mass of propane would need to be burned to raise 2500 grams of water from 25.0 o C to 100.0 o C? (MM propane = 44.11 g/mol) 1 kJ absorbed by the water = 1 kJ released by burning propane +1 kJ water = –1 kJ propane

20. Enthalpy Calculations 1.00 grams of solid KClO 3 (122.55 g/mol) is dropped into 25.00 grams of water at 25.0 o C. If the heat of solution for KClO 3 is +41.38 kJ/mol, what is the final temperature of the water? What is the system?Dissolving of KClO 3 What is the surrounding?Water!! What would we measure?Temperature of WATER!! What would we observe?Temperature of water decreasing! Did the water gain or lose energy?Water LOST energy!! Where did it go?Into the process of dissolving the KClO 3(s)

21. Enthalpy Calculations 1.00 grams of solid KClO 3 (122.55 g/mol) is dropped into 25.00 grams of water at 25.0 o C. If the heat of solution for KClO 3 is +41.38 kJ/mol, what is the final temperature of the water? What is the system?Dissolving of KClO 3 What is the surrounding?Water!! What would we measure?Temperature of WATER!! What would we observe?Temperature of water decreasing! Did the water gain or lose energy?Water LOST energy!! Where did it go?Into the process of dissolving the KClO 3(s)

22. Enthalpy (  H) When 10.0 g KOH is dissolved in 100.0 g of water in a coffee ‐ cup calorimeter, the temperature rises from 25.18 °C to 47.53 °C. What is the enthalpy change in kJ per mole of KOH dissolved in the water? Assume that the solution has a specific heat capacity of 4.18 J/g ⋅ K.

23. Enthalpy Calculations When 2.10 grams of solid KOH (56.11 g/mol) is dropped into 55.00 grams of water at 22.5 o C, the water raises to 31.9 o C. What is the heat of solution for KOH in kJ/mol? What is the system?Dissolving of KOH What is the surrounding?Water!! What would we measure?Temperature of WATER!! What would we observe?Temperature of water increasing! Did the water gain or lose energy?Water GAINED energy!! Where did it go?From the process of dissolving the KOH (s)