1. Good Morning! 9/20/2015  Today we will be… Preparing for tomorrow’s test by going through the answers to the Practice Test  Before we get into the practice test turn in your lab 19.  Test tomorrow.

2. Practice Test: Ch. 11 (Thermochemistry)

3. 1.Complete the equation for finding heat energy.   H = m C  T

4. 2.The specific heat capacity of graphite is 0.71 J/(g x °C). Calculate the energy required to raise the temperature of 450 g of graphite by 150°C.  H = m  T C 450g 0.71 J / gºC 150  C  H = 47,925 J = 48 kJ = 48 kJ

5. 3.It takes 770 joules of energy to raise the temperature of 50 g of mercury by 110°C. What is the specific heat capacity of mercury?  H = m x C x  T m x  T  H m x  T = C 770 J = 0.14 J/g·  C 50.0 g 110  C

6. 4.Calculate the heat absorbed by the water in a calorimeter when 175 grams of copper cools from 125.0°C to 22.0°C. The specific heat capacity of copper is 0.385 J/(g x °C).  H = m  T C 175g 0.385 J / gºC 103  C  H = 6,939.62 J = 6.94 kJ = 6.94 kJ

7. 5.Assume 372 joules of heat are added to 5.00 g of water originally at 23.0°C. What would be the final temperature of the water? The specific heat capacity of water = 4.184 J/(g x °C).  H = m x C x  T m x C  H m x C = ∆T 372 J = 17.8  C 5.00 g 4.184 J / g·  C ∆T f = 40.8  C

8. 6.How much heat is required to raise the temperature of 2.0 x 102 g of aluminum by 30°C? (specific heat of aluminum = 0.878 J/(g x °C))  H = m  T C 200g 0.878 J / gºC 30  C  H = 5,268 J = 5.3 kJ = 5.3 kJ

9. 7. Find the specific heat capacity of Lead if an 85.0 g sample of lead with an initial temperature of 99.0°C is placed into 99.5 g of water with an initial temperature of 22.0 °C. The final temperature of the water and the lead is 25.0 °C. First find the  H for water  H = m  T C 99.5g 4.184 J / gºC 3  C 3  C  H = 1,300 J

10. On the test I need you to show that you know how to do the algebra below, but you only have to do it once  H = m x C x  T m x  T  H m x  T = C 1300 J = 0.21 J/g·  C 85.0 g 74  C

11. 8.In the following reaction: H 2 (g) + 1/2 O 2 (g) → H 2 O (l)  H = -286 kJ/mol H 2 How much heat is produced when 5.00g of H2 (at STP) is reacted with excess O 2 ? 5.00g H 2 x 1 mol H 2 x 2.02 g H 2 -286 kJ 1 mol H 2 = 708 kJ

12. 9.Find the standard heat of formation for the following chemical reaction. Use the table provided. 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (g)  H =  H f (products) -  H f (reactants) Hf Prod.=(4mols ● 90.37 kJ/mol) + (6mols ● - 285.8 kJ/mol) Hf React. = (4mols ● - 46.19 kJ/mol)+(5mols ● 0 kJ/mol) H = (361.48kJ – 1714.8kJ) – (-184.76kJ) H = -1168.56 kJ

13. 10.How many joules are there in 165 calories? (1 cal = 4.184 J) 165 cal x 4.184 J = 1 cal 690 J 690 J

14. 11. How much heat is absorbed by 150.0 g of ice at - 20.0 °C to steam at 120 °C?  Hfus= 6.01 kJ/mol  Hvap = 40.7 kJ/mol Cice= 2.1 J/(g °C) Cliquid= 4.184 J/(g °C) Csteam= 1.7 J/(g°C) Hint: Calculate the energy for each phase then find the total energy.

15. Solid phase: (Temperature increase: - 20.0  C to 0  C)  H = m  T C ice 150g 2.1 J / gºC 20  C  H = 6,300 J = 6.3 kJ = 6.3 kJ

16. Melting: (Temperature stays at 0  C) 8.32 mols H 2 O x 6.01 kJ 1 mol H 2 O = 50 kJ 150g H 2 O x 1 mol H 2 O = 18.02 g H 2 O 8.32 mols H 2 O

17. Liquid phase: (Temperature increase 0  C to 100  C)  H = m  T C liquid 150g 4.184 J / gºC 100  C  H = 62,800 J = 62.8 kJ = 62.8 kJ

18. Boiling: (Temperature stays at 100  C) 8.32 mols H 2 O x 40.7 kJ 1 mol H 2 O = 333 kJ

19. Gas phase: (Temperature increase 100  C to 120.0  C)  H = m  T C steam 150g 1.70 J / gºC 20  C  H = 5,100 J = 5.1 kJ = 5.1 kJ

20. So how much energy was used to heat the water from -20  C to 120  C ?  Warming Ice =  Melting Ice =  Warming Water =  Boiling Water =  Warming Steam =  Total = 6.3 kJ 6.3 kJ 50.0 kJ 50.0 kJ 62.8 kJ 62.8 kJ 333.0 kJ 333.0 kJ 5.1 kJ 5.1 kJ + 457.2 kJ 457.2 kJ

21. 12.(Hess’s Law) Show All Work! 2 points Calculate the enthalpy change (  H) in kJ for the following reaction. 2 Al (s) +Fe 2 O 3(s)  2 Fe (s) +Al 2 O 3(s) Flip the Combustion of Fe and add them together.  2Al (s) +1.5 O 2(g)  Al 2 O 3(s)  H = -1669.8 kJ  Fe 2 O 3(s)  2 Fe (s) +1.5 O 2(g)  H = +824.2 kJ  2 Al (s) +Fe 2 O 3(s)  2 Fe (s) +Al 2 O 3(s)   H = -1669.8 kJ + 824.2 kJ = - 845.6 kJ