Thermochemistry Unit Section 16.2

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

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 = KJ 6C (s) + 3H 2(g)  C 6 H 6(l) ΔH o f = +49 KJ

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

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) KJ C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 0 (l) 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) KJ c. Pentane, C 5 H 12(l) C 5 H 12(l) + 8O 2(g)  5CO 2(g) + 6H 2 0 (l) KJ

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) KJ C 7 H 6(s) + 11O 2(g) ΔH comb = KJ reactants products Enthalpy, H 7CO 2(g) + 8H 2 0 (l)

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 g of methane forms from the elements? q = ? m = g ΔH o f = KJ/mol q = nΔH o f n methane = m/M = ( g) / (16.05 g/mol) = mol  q = nΔH o f = (3.115 mol)(-74.6 KJ/mol) = KJ

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

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

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

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

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

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

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 = X 10 4 kJ ΔH o f = KJ/mol m methanol = (n)(M) = (98.07 mol)(32.05 g/mol) = g  n = q / ΔH o f =(-2.34 X 10 4 kJ)( KJ/mol) = mol M methanol = 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?

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) = mol  q = nΔH o fus = (0.455 mol)(6.02 KJ/mol) = 2.74 KJ M ice cube = g/mol q = nΔH o fus

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

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

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) kJ  NaCl (aq)

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) = 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 = g/mol q = nΔH o sol

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.

Practice Problem #28: What mass of diethyl ether, C 4 H 10 O, can be vaporized by adding 80.7 kJ of heat? q = 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) = g M diethyl ether = g/mol q = nΔH o vap

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 = 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) = J/mol  n = m/M= (100 g) / (78.12 g/mol) = 1.28 mol M benzene = g/mol q = nΔH o vap