Good Morning! 8/25/2014

1. Good Morning! 8/25/2014 • 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. Complete the equation for finding heat energy. • H = m C T

4. 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/g•ºC 150 C  H = 47,925 J = 48 kJ

5. H m x T = C 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 m x T 770 J = 0.14 J/g· C 50.0 g 110C

6. 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/g•ºC 103 C  H = 6,939.62 J = 6.94 kJ

7. H m x C = ∆T 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 m x C 372 J = 17.8 C 4.184 J/g·C 5.00 g ∆Tf = 40.8 C

8. 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/g•ºC 30 C  H = 5,268 J = 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/g•ºC 3 C  H = 1,300 J

10. H m x T = C 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 m x T 1300 J = 0.21 J/g· C 85.0 g 74C

11. In the following reaction: H2 (g) + 1/2 O2 (g) → H2O(l)H = -286 kJ/mol H2How much heat is produced when 5.00g of H2 (at STP) is reacted with excess O2?

12. Find the standard heat of formation for the following chemical reaction. Use the table provided.4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)H = Hf (products) - Hf (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. How many joules are there in 165 calories? (1 cal = 4.184 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 • Cice 150g 2.1 J/g•ºC 20 C  H = 6,300 J = 6.3 kJ

16. Melting: (Temperature stays at 0 C)

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

18. Boiling: (Temperature stays at 100 C)

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

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

21. (Hess’s Law) Show All Work! 2 pointsCalculate the enthalpy change (H) in kJ for the following reaction.2 Al(s) +Fe2O3(s)  2 Fe(s) +Al2O3(s) Flip the Combustion of Fe and add them together. • 2Al(s) +1.5 O2(g)Al2O3(s)H = -1669.8 kJ • Fe2O3(s) 2 Fe(s) +1.5 O2(g)H = +824.2 kJ • 2 Al(s) +Fe2O3(s)  2 Fe(s) +Al2O3(s) • H = -1669.8 kJ + 824.2 kJ = - 845.6 kJ