Chapter 11 (Practice Test)

1. Chapter 11 (Practice Test) Thermochemistry

2. = m • C • ∆T ∆H 1. The specific heat capacity of graphite is 0.71 J/g•°C. Calculate the energy required to raise the temperature of 450 g of graphite by 150 °C. ∆H = m = C = ∆T = 450 g 0.71 J/g•°C 150 °C ∆H = (150 °C) (450g) (0.71 J ) g•°C Calculator 450 x 0.71 x 150 = 47,925 J ∆H= 48,000 J or 48 kJ Answer w/ Sig. Figs.

3. = m • C • ∆T ∆H ∆H m x ∆T ∆H C = m x ∆T 2. 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? 2. 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 = C = ∆T = 770 J 50 g m x ∆T m x ∆T _____ J/g°C 110 °C C = Calculator 770 ÷ 50 ÷ 110 = 0.14 or 770 ÷ (50 x 110) = 0.14 770 J = 50 g x 110 °C C = 0.14 J/g•°C

4. = m • C • ∆T ∆H 3. Calculate the heat absorbed by the water in a calorimeter when 175.0 grams of copper cools from 125.0 °C to 22.0 °C The specific heat capacity of copper is 0.385 J/g•°C. 3. Calculate the heat absorbed by the water in a calorimeter when 175.0 grams of copper cools from 125.0 °C to 22.0 °C The specific heat capacity of copper is 0.385 J/g•°C. ∆H = m = C = ∆T = 175.0 g 0.385 J/g•°C 22.0°C – 125.0°C ∆T = – 103.0°C ∆H = (-103 °C) (175g) (0.385 J ) g•°C Calculator 175 x 0.385 x -103 = -6939.625 J ∆H= + 6939.625 J But water Absorbed the heat so…

5. = m • C • ∆T ∆H ∆H mx C ∆H ∆T= mx C 4. 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? (Note: Find ∆T first then find the final temperature) The specific heat capacity of water = 4.184 J/g•°C? 4. 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? (Note: Find ∆T first then find the final temperature) The specific heat capacity of water = 4.184 J/g•°C? ∆H = m = C = ∆T = 372 J 5.00 g mx C m x C 4.184 J/g°C _____ °C ∆T= Ti = Tf = _____°C 23.0 °C 372 = = 17.8 °C (5x 4.184) Calculator 372 ÷ 5 ÷ 4.184 = 17.78 Heat was added so temp. increases by 17.8. Tf = 23.0 + 17.8 = 40.8 °C

6. = m • C • ∆T ∆H 5. How much heat is required to raise the temperature of 2.0 x 102 g of aluminum by 30 °C. (specific heat of Al = 0.878 J/g•°C) ∆H = m = C = ∆T = 2.0 x 102 g = 200 g 0.878 J/g•°C 30°C ∆H = (30 °C) (200g) (0.878 J ) g•°C Calculator (2.0 x 102) x 0.878 x 30 = 5268 J ∆H= 5 kJ Answer to correct sig.figs.

7. = = Tf- Ti m• C •∆T ∆T ∆H ∆H C = mx∆T 6. 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. Water so…C is known Water ∆H = m = C = ∆T = _________ J 99.5 g ∆H = (3.0 °C) (99.5g) (4.184 J ) g•°C 4.184 J/g•°C ∆H = 1,248.924 J 1,248.924 25-22=3.0°C Energy water gained from the hot metal so… Lead ∆H = m = C = ∆T = - _________ J - 1248.924 J = 85.0 g (85.0)(-74.0 °C) ______ J/g•°C C = 0.199 = 0.20 J/g•°C 25-99=-74.0°C

8. 7. Find the standard heat of formation for the following reaction. ∆H° = ∆Hf°(products) - ∆Hf°(reactants) 4 NH3(g) + 5 O2 (g) 4 NO(g) + 6 H2O (l) -46.19 x 4 -184.76 kJ 0.0 x 5 0 kJ 90.37 x 4 361.48 kJ -285.8 x 6 -1714.8 kJ + + -184.76 kJ -1,353.32 kJ (reactants) (products) ∆H°=∆Hf°(products) - ∆Hf°(reactants) ∆H° = -1,353.32 kJ – (-184.76 kJ) = - 1168.56 kJ ∆H° = - 1168.56 kJ

9. 8. Heat of Reaction: 2 H2(g) + O2 (g) 2 H2O (l)∆H =-572 kJ 2 -572 kJ How much heat is produced when 5.00 g of H2 (at STP) is reacted with excess O2? 5.00 g H2 1 mol H2 x x = -708 kJ mol H2 2.02 g H2 1.01 x 2 2.02 1 H Hydrogen 1.01 Calculator: 5.00 ÷ 2.02 x -572 ÷ 2 = - 707.9

10. 9. Heat of Solution: ∆Hsoln =-445.1 kJ/mol -445.1 kJ Determine the heat of solution when 40.00 g of NaOH is dissolved in water. 40.00 g NaOH 1 mol NaOH x x = -445.1 kJ mol NaOH 40.00 1 g NaOH 11 Na sodium 22.99 8 O Oxygen 16.00 1 H Hydrogen 1.01 Calculator: 40.00 ÷ 40.00 x -445.1 = - 445.1

11. 10. Heat of Solution: ∆Hsoln =25.7 kJ/mol 25.7 kJ Determine the heat of solution when 25.58 g of NH4NO3 is dissolved in water. 25.58 g NH4NO3 1 mol NH4NO3 x x = 8.211 kJ 1 mol NH4NO3 80.06 g NH4NO3 7 N Nitrogen 14.01 8 O Oxygen 16.00 1 H Hydrogen 1.01 Calculator: 25.58 ÷ 80.06 x 25.7 = 8.211416438

12. 11. Heat of Combustion: 2 C2H2 + 5 O2  4 CO2 + 2 H2O ∆H =-2600 kJ 2 -2600 kJ How much heat is produced when 35.00 g of C2H2 is reacted with excess O2? 35.00 g C2H2 1 mol C2H2 x x = -1747 kJ mol C2H2 26.04 g C2H2 6 C Carbon 12.01 1 H Hydrogen 1.01 Calculator: 35.00 ÷ 26.04 x -2600 ÷ 2 = - 1747.311828

13. 12. (Hess’s Law) Calculate the enthalpy change (∆H) in kJ for the following reaction. 2 Al(s)+ Fe2O3 (s)2 Fe(s)+ Al2O3(s) ∆H =______ kJ Use the enthalpy changes for the combustion of aluminum and iron. 1) 2 Al(s)+1.5 O2 (g) Al2O3(s) ∆H1 =-1,669.8 kJ 2) 2 Fe(s)+1.5 O2 (g) Fe2O3(s) 2 Fe(s)+1.5 O2 (g) Fe2O3(s) ∆H2 =+824.2 kJ ∆H2 =-824.2 kJ Iron is supposed to be a product so reverse 2nd reaction and change sign for ∆H. It is called Hess’s Law of Heat SUMMATION Aluminum is supposed to be a reactant so leave 1st reaction alone. 1) 2 Al(s)+1.5 O2 (g) Al2O3(s) ∆H1 =-1,669.8 kJ +  2) -845.6 2 Al(s)+ Fe2O3 (s)2 Fe(s)+ Al2O3(s) ∆H =-845.6 kJ

14. = = Tf- Ti m • C • ∆T ∆T ∆H 13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C? ∆H(vap) = 40.7 kJ/mol ∆H(fus) = 6.01 kJ/mol Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C Solid phase: Temp. from -20.0 °C to 0 °C. ∆T = 0°C – (-20.0°C) = 20.0°C m = C(ice)= ∆H = 150.0 g 2.1 J/g•°C ______ J ∆H = (20 °C) (150.0g) (2.1 J ) g•°C ∆H= 6.3 kJ ∆H = 150 x 2.1 x 20 = 6,300 J

15. H2O 1.01 x 2 2.02 16.00 x 1 +16.00 = 18.02 g 13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C? ∆H(vap) = 40.7 kJ/mol ∆H(fus) = 6.01 kJ/mol Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C Melting: Temperature stays at 0 °C. ∆T = 0°C – 0°C = 0°C Can’t use ∆H = m • C • ∆T It takes 6.01 kJ to melt 1 mole of water. 1 H Hydrogen 1.01 8 O Oxygen 16.00 Use ∆H(fus.) = 6.01 kJ/mol 6.01 kJ 1 mol H2O Set up unit conversions to solve: mol H2O 1 150.0 g H2O = 50.0 kJ x x g H2O 18.02 Calculator: 150 ÷ 18.02 x 6.01 = 50.02774695 kJ

16. = = Tf- Ti m • C(liquid) • ∆T ∆T ∆H 13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C? ∆H(vap) = 40.7 kJ/mol ∆H(fus) = 6.01 kJ/mol Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C Liquid phase: Temp. from 0 °C to 100 °C. ∆T = 100°C – 0°C = 100°C m = C(liquid)= ∆H = 150.0 g 4.184 J/g•°C ______ J ∆H = (100 °C) (150.0g) (4.184 J ) g•°C ∆H = 150 x 4.184 x 100 = 62,760 J ∆H= 62.8 kJ

17. H2O 1.01 x 2 2.02 16.00 x 1 +16.00 = 18.02 g 13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C? ∆H(vap) = 40.7 kJ/mol ∆H(fus) = 6.01 kJ/mol Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C Boiling: Temperature stays at 100 °C. ∆T = 100°C – 100°C = 0°C Can’t use ∆H = m • C • ∆T It takes 40.7 kJ to boil 1 mole of water. 1 H Hydrogen 1.01 8 O Oxygen 16.00 Use ∆H(vap.) = 40.7 kJ/mol 40.7 kJ 1 mol H2O Set up unit conversions to solve: mol H2O 1 150.0 g H2O = 338.8 kJ x x g H2O 18.02 Calculator: 150 ÷ 18.02 x 40.7 = 338.790231 kJ

18. = = Tf- Ti m • C(steam) • ∆T ∆T ∆H 13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C? ∆H(vap) = 40.7 kJ/mol ∆H(fus) = 6.01 kJ/mol Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C Gas phase: Temp. from 100 °C to 120.0 °C. ∆T = 120.0°C – 100 °C = 20.0°C m = C(steam)= ∆H = 150.0 g 1.7 J/g•°C ______ J ∆H = (20 °C) (150.0g) (1.7 J ) g•°C ∆H= 5.1 kJ ∆H = 150 x 1.7 x 20 = 5,100 J

19. 13. How much heat is absorbed by 150.0 g of ice at -20.0 °C to steam at 120 °C? ∆H(vap) = 40.7 kJ/mol ∆H(fus) = 6.01 kJ/mol Cice= 2.1 J/g°C Cliquid= 4.184 J/g°C Csteam= 1.7 J/g°C 6.3 kJ Solid phase 50.0 kJ Melting 62.8 kJ Liquid phase 338.8 kJ Boiling + 5.1 kJ Gas phase 463.0 kJ Total energy =