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Thermochemistry Unit Section 16.2

Thermochemistry Unit Section 16.2. 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. d . C 6 H 6(l). 6C (s) + 3H 2(g) + 49 KJ  C 6 H 6(l).

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Thermochemistry Unit Section 16.2

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  1. Thermochemistry Unit Section 16.2

  2. Practice Problem #15: a. H2O(g) H2(s) + 1/2O2(g) H2O(g) + 241.8 KJ b. CaCl2(s) Ca(s) + Cl2(g) CaCl2(s) + 795.4 KJ c. CH4(g) C(s) + 2H2(g) CH4(g) + 74.6 KJ

  3. d. C6H6(l) 6C(s) + 3H2(g) + 49 KJ  C6H6(l) e. Show the standard molar enthalpy of parts c) and d) using another method C(s) + 2H2(g) CH4(g)ΔHof= -74.6 KJ 6C(s) + 3H2(g) C6H6(l) ΔHof= +49 KJ

  4. Na(s) + 1/2Cl2(g) reactants Enthalpy, H ΔH = -411.1 KJ NaCl(s) products Practice Problem #16: Draw an enthalpy diagram to represent the standard molar enthalpy of formation of sodium chloride. NaCl(s)ΔHof = -411.1 KJ/mol Exothermic Na(s) + 1/2Cl2(g) NaCl(s) + 411.1 KJ

  5. Practice Problem #17: a. Ethane, C2H6(g) C2H6(g) + 7/2O2(g) 2CO2(g) + 3H20(l) + 1250.9 KJ b. Propane, C3H8(g) C3H8(g) + 5O2(g) 3CO2(g) + 4H20(l) + 2323.7 KJ c. Butane, C4H10(g) C4H10(g) + 13/2O2(g) 4CO2(g) + 5H20(l) + 3003.0 KJ c. Pentane, C5H12(l) C5H12(l) + 8O2(g) 5CO2(g) + 6H20(l) + 3682.3 KJ

  6. C7H6(s) + 11O2(g) reactants Enthalpy, H ΔHcomb= - 5040.9 KJ 7CO2(g) + 8H20(l) products Practice Problem #18: Draw an enthalpy diagram to represent the standard molar enthalpy of combustion of heptane, C7H16(l) (use Table 16.3). C7H16(l) + 11O2(g) 7CO2(g) + 8H20(l) + 5040.9 KJ Exothermic reaction, so the enthalpy of reactants is higher than the enthalpy of the products.

  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 =nΔHof q= ? nmethane=m/M m = 500.0 g = (500.00 g) / (16.05 g/mol) ΔHof= -74.6 KJ/mol = 31.15 mol q =nΔHof = (31.15 mol)(-74.6 KJ/mol) = -2323.99 KJ

  8. b) How much heat is released when 50.00 g of methane undergoes complete combustion? q =nΔHocomb q= ? nmethane=m/M m = 50.0 g = (500.00 g) / (16.05 g/mol) ΔHocomb= -965.1 KJ/mol = 3.115 mol q =nΔHocomb = (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 =nΔHof q= ? nwater =m/M m = 0.534 g = (0.534 g) / (18.02 g/mol) ΔHof= -285.8 KJ/mol = 0.0296 mol q =nΔHof = (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 =nΔHof q= ? nwater =m/M m = 0.534 g = (0.534 g) / (18.02 g/mol) ΔHof= -241.8 KJ/mol = 0.0296 mol q =nΔHof = (0.0296 mol)(-241.8 KJ/mol) = -7.16 KJ

  11. nwater =m/M = (56.78 g) / (72.17 g/mol) = 0.787 mol • Practice Problem #21: • Determine the heat released by the combustion of 56.78 g of • pentane, C5H12(l) q= ? q =nΔHocomb m = 56.78 g ΔHocomb= -3682.3 KJ/mol Mpentane= 72.17 g/mol q =nΔHocomb = (0.787 mol)(-3682.3 KJ/mol) = -2897.06 KJ

  12. q= ? q =nΔHocomb nwater =m/M m = 1.36 Kg = 1360 g = (1360 g) / (114.26 g/mol) ΔHocomb= -5720.2 KJ/mol = 11.90 mol Mpentane= 114.26 g/mol q =nΔHocomb = (11.90 mol)(-5720.2 KJ/mol) = -68070.38 KJ = -6.81 X 104 KJ b) Determine the heat released by the combustion of 56.78 g of pentane, C5H12(l)

  13. q= ? q =nΔHocomb nwater =m/M m = 2.344 X 104 g = (2.344 X 104g) / (86.20 g/mol) ΔHocomb= -4361.6 KJ/mol = 271.93 mol Mhexane= 86.20 g/mol q =nΔHocomb = (271.93 mol)(-4361.6 KJ/mol) = -1 186 031.37 KJ = -1.186 X 106 KJ c) Determine the heat released by the combustion of 2.344 X 104 g of hexane, C6H14(l)

  14. mmethanol=(n)(M) = (98.07 mol)(32.05 g/mol) = 3143.21 g Practice Problem #23: What mass of methanol, CH3OH(l), is formed from its elements if 2.34 X 104 kJ of energy is released during the process? Practice Problem #23: What mass of methanol, CH3OH(l), is formed from its elements if 2.34 X 104 kJ of energy is released during the process? m= ? q =nΔHof q = -2.34 X 104 kJ n =q / ΔHof =(-2.34 X 104 kJ)/(-238.6 KJ/mol) = 98.07 mol ΔHof= -238.6 KJ/mol Mmethanol= 32.05 g/mol

  15. nwater =m/M = (8.2g) / (18.02 g/mol) = 0.455 mol 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= ? q =nΔHofus mice cube= 8.2 g ΔHomelt= ΔHofus=6.02 KJ/mol Mice cube= 18.02 g/mol q =nΔHofus = (0.455 mol)(6.02 KJ/mol) = 2.74 KJ

  16. nwater =m/M = (100 g) / (18.02 g/mol) = 5.55 mol 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= ? q =nΔHofus mwater= 0.100 kg = 100 g ΔHofreezing= ΔHofus=-6.02 KJ/mol Mwater= 18.02 g/mol q =nΔHofus= (5.55 mol)(-6.02 KJ/mol) = -33.41 KJ b)qmelting=33.41 KJ qfreezing=-33.41 KJ

  17. nwater =m/M = (0.325 g) / (200.59 g/mol) = 0.00162 mol 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= ? q =nΔHovap mmercury= 0.325 g ΔHovap=59 KJ/mol Mmercury= 200.59 g/mol q =nΔHovap = (0.00162 mol)(59 KJ/mol) = 0.0956 KJ This is then an endothermic reaction since heat energy is absorbed.

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

  19. nNaCl=m/M = (25.3 g) / (58.44 g/mol) = 0.433 mol 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= ? q =nΔHosol mNaCl= 25.3 g ΔHosol=3.9 KJ/mol MNaCl= 58.44 g/mol q =nΔHosol= (0.433 mol)(3.9 KJ/mol) = 1.69 KJ This is then an endothermic reaction since heat energy is absorbed.

  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. nNaCl=q / ΔHovap = (+80.7 kJ) / (29 kJ/mol) = 2.78 mol Practice Problem #28: What mass of diethyl ether, C4H10O, can be vaporized by adding 80.7 kJ of heat? q= +80.7 kJ q =nΔHovap mdiethyl ether= ? ΔHovap=29 KJ/mol Mdiethyl ether= 74.14 g/mol m=nM= (2.78 mol)(74.14 g/mol) = 206.08 g

  22. ΔHovap=q/n = (+3.97 X 104 kJ)/(1.28 mol) = 31 015.63 J/mol Practice Problem #29: 3.97 X 104 J of heat is required to vaporize 100 g of benzene, C6H6. What is the molar enthalpy of vaporisation of benzene? q= +3.97 X 104 J q =nΔHovap mbenzene= 100 g ΔHovap=? Mbenzene= 78.12 g/mol n=m/M= (100 g) / (78.12 g/mol) = 1.28 mol

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