1. Thermochemistry Unit Section 16.2

2. Practice Problem #15: a. H 2 O (g) H 2(s) + 1/2O 2(g)  H 2 O (g) + 241.8 KJ b. CaCl 2(s) Ca (s) + Cl 2(g)  CaCl 2(s) + 795.4 KJ c. CH 4(g) C (s) + 2H 2(g)  CH 4(g) + 74.6 KJ

3. d. C 6 H 6(l) 6C (s) + 3H 2(g) + 49 KJ  C 6 H 6(l) e. Show the standard molar enthalpy of parts c) and d) using another method C (s) + 2H 2(g)  CH 4(g) ΔH o f = -74.6 KJ 6C (s) + 3H 2(g)  C 6 H 6(l) ΔH o f = +49 KJ

4. Practice Problem #16: Draw an enthalpy diagram to represent the standard molar enthalpy of formation of sodium chloride. NaCl (s) ΔH o f = -411.1 KJ/mol Exothermic Na (s) + 1/2Cl 2(g)  NaCl (s) + 411.1 KJ Na (s) + 1/2Cl 2(g) NaCl (s) ΔH = -411.1 KJ reactants products Enthalpy, H

5. Practice Problem #17: a. Ethane, C 2 H 6(g) C 2 H 6(g) + 7/2O 2(g)  2CO 2(g) + 3H 2 0 (l) + 1250.9 KJ C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 0 (l) + 2323.7 KJ b. Propane, C 3 H 8(g) c. Butane, C 4 H 10(g) C 4 H 10(g) + 13/2O 2(g)  4CO 2(g) + 5H 2 0 (l) + 3003.0 KJ c. Pentane, C 5 H 12(l) C 5 H 12(l) + 8O 2(g)  5CO 2(g) + 6H 2 0 (l) + 3682.3 KJ

6. Practice Problem #18: Draw an enthalpy diagram to represent the standard molar enthalpy of combustion of heptane, C 7 H 16(l) (use Table 16.3).  Exothermic reaction, so the enthalpy of reactants is higher than the enthalpy of the products. C 7 H 16(l) + 11O 2(g)  7CO 2(g) + 8H 2 0 (l) + 5040.9 KJ C 7 H 6(s) + 11O 2(g) ΔH comb = - 5040.9 KJ reactants products Enthalpy, H 7CO 2(g) + 8H 2 0 (l)

7. Sample Problem (page 644): Methane is the main component of natural gas. Natural gas undergoes combustion to provide energy for heating homes and cooking food. a) How much heat is released when 500.00 g of methane forms from the elements? q = ? m = 500.0 g ΔH o f = -74.6 KJ/mol q = nΔH o f n methane = m/M = (500.00 g) / (16.05 g/mol) = 3.115 mol  q = nΔH o f = (3.115 mol)(-74.6 KJ/mol) = -232.4 KJ

8. b) How much heat is released when 50.00 g of methane undergoes complete combustion? q = ? m = 50.0 g ΔH o comb = -965.1 KJ/mol q = nΔH o comb n methane = m/M = (500.00 g) / (16.05 g/mol) = 3.115 mol  q = nΔH o comb = (3.115 mol)(-965.1 KJ/mol) = -3006.29 KJ

9. Practice Problem #19: a) Hydrogen gas and oxygen gas react to form 0.534 g of liquid water. How much heat is released to the surroundings? q = ? m = 0.534 g ΔH o f = -285.8 KJ/mol q = nΔH o f n water = m/M = (0.534 g) / (18.02 g/mol) = 0.0296 mol  q = nΔH o f = (0.0296 mol)(-285.8 KJ/mol) = -8.47 KJ

10. b) Hydrogen gas and oxygen gas react to form 0.534 g of gaseous water. How much heat is released to the surroundings? q = ? m = 0.534 g ΔH o f = -241.8 KJ/mol q = nΔH o f n water = m/M = (0.534 g) / (18.02 g/mol) = 0.0296 mol  q = nΔH o f = (0.0296 mol)(-241.8 KJ/mol) = -7.16 KJ

11. Practice Problem #21: a)Determine the heat released by the combustion of 56.78 g of pentane, C 5 H 12(l) q = ? m = 56.78 g ΔH o comb = -3682.3 KJ/mol n water = m/M = (56.78 g) / (72.17 g/mol) = 0.787 mol  q = nΔH o comb = (0.787 mol)(-3682.3 KJ/mol) = -2897.06 KJ M pentane = 72.17 g/mol q = nΔH o comb

12. b) Determine the heat released by the combustion of 56.78 g of pentane, C 5 H 12(l) q = ? m = 1.36 Kg = 1360 g ΔH o comb = -5720.2 KJ/mol n water = m/M = (1360 g) / (114.26 g/mol) = 11.90 mol  q = nΔH o comb = (11.90 mol)(-5720.2 KJ/mol) = -68070.38 KJ = -6.81 X 10 4 KJ M octane = 114.26 g/mol q = nΔH o comb

13. c) Determine the heat released by the combustion of 2.344 X 10 4 g of hexane, C 6 H 14(l) q = ? m = 2.344 X 10 4 g ΔH o comb = -4361.6 KJ/mol n water = m/M = (2.344 X 10 4 g) / (86.20 g/mol) = 271.93 mol  q = nΔH o comb = (271.93 mol)(-4361.6 KJ/mol) = -1 186 031.37 KJ = -1.186 X 10 6 KJ M octane = 86.20 g/mol q = nΔH o comb

14. Practice Problem #23: What mass of methanol, CH 3 OH (l), is formed from its elements if 2.34 X 10 4 kJ of energy is released during the process? m = ? q = -2.34 X 10 4 kJ ΔH o f = -238.6 KJ/mol m methanol = (n)(M) = (98.07 mol)(32.05 g/mol) = 3143.21 g  n = q / ΔH o f =(-2.34 X 10 4 kJ)(-238.6 KJ/mol) = 98.07 mol M methanol = 32.05 g/mol q = nΔH o f Practice Problem #23: What mass of methanol, CH 3 OH (l), is formed from its elements if 2.34 X 10 4 kJ of energy is released during the process?

15. Practice Problem #24: An ice cube with a mass of 8.2 g is placed in some lemonade. The ice cube melts completely. How much heat does the ice cube absorb from the lemonade as it melts? q = ? m ice cube = 8.2 g ΔH o melt = ΔH o fus = 6.02 KJ/mol n water = m/M = (8.2g) / (18.02 g/mol) = 0.455 mol  q = nΔH o fus = (0.455 mol)(6.02 KJ/mol) = 2.74 KJ M ice cube = 18.02 g/mol q = nΔH o fus

16. Practice Problem #25: A teacup contains 0.100 kg of water at its freezing point. The water freezes solid. a) How much heat is released to its surroundings? q = ? m water = 0.100 kg = 100 g ΔH o freezing = ΔH o fus = -6.02 KJ/mol n water = m/M = (100 g) / (18.02 g/mol) = 5.55 mol  q = nΔH o fus = (5.55 mol)(-6.02 KJ/mol) = -33.41 KJ M water = 18.02 g/mol q = nΔH o fus  q freezing = -33.41 KJ b)  q melting = 33.41 KJ

17. Practice Problem #26: A sample of mercury vaporizes. The mercury is at its boiling point and has a mass of 0.325 g. How much heat is absorbed or released to the surroundings? q = ? m mercury = 0.325 g ΔH o vap = 59 KJ/mol n water = m/M = (0.325 g) / (200.59 g/mol) = 0.00162 mol  q = nΔH o vap = (0.00162 mol)(59 KJ/mol) = 0.0956 KJ This is then an endothermic reaction since heat energy is absorbed. M mercury = 200.59 g/mol q = nΔH o vap

18. Practice Problem #27: The molar enthalpy of solution for sodium chloride, NaCl, is 3.9 kJ/mol. a)Write a thermochemical reaction to represent the dissolution of sodium chloride? Dissolution: Solid state  Liquid state NaCl (s) + 3.9 kJ  NaCl (aq)

19. b) Suppose you dissolve 25.3 g of sodium chloride in a glass of water at room temperature. How much heat is absorbed or released by the process? q = ? m NaCl = 25.3 g ΔH o sol = 3.9 KJ/mol n NaCl = m/M = (25.3 g) / (58.44 g/mol) = 0.433 mol  q = nΔH o sol = (0.433 mol)(3.9 KJ/mol) = 1.69 KJ This is then an endothermic reaction since heat energy is absorbed. M NaCl = 58.44 g/mol q = nΔH o sol

20. c) Do you expect the glass containing the salt solution to feel warm or cool? Explain your answer. Answer: Since heat is absorbed to dissolve the salt, heat is removed from the glass and it will feel cold.

21. Practice Problem #28: What mass of diethyl ether, C 4 H 10 O, can be vaporized by adding 80.7 kJ of heat? q = +80.7 kJ m diethyl ether = ? ΔH o vap = 29 KJ/mol n NaCl = q / ΔH o vap = (+80.7 kJ) / (29 kJ/mol) = 2.78 mol  m = nM= (2.78 mol)(74.14 g/mol) = 206.08 g M diethyl ether = 74.14 g/mol q = nΔH o vap

22. Practice Problem #29: 3.97 X 10 4 J of heat is required to vaporize 100 g of benzene, C 6 H 6. What is the molar enthalpy of vaporisation of benzene? q = +3.97 X 10 4 J m benzene = 100 g ΔH o vap = ? ΔH o vap = q/n = (+3.97 X 10 4 kJ)/(1.28 mol) = 31 015.63 J/mol  n = m/M= (100 g) / (78.12 g/mol) = 1.28 mol M benzene = 78.12 g/mol q = nΔH o vap