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Sum: 2C 6 H 14 (l) + 19O 2  12CO 2 + 14H 2 O(g) D H o = - 2 D H f o C6H14 + 1 2 D H f o CO2 + 14 D H f o H2O

2. Energy Level Diagram. Elements: 12C(s), 14H 2 (g), 19O 2 (g). 0. D H f o kJ. Reactants : 2C 6 H 14 (l). 2 D H f o C 6 H 14 -334. S n D H f o (react) = -334. Products : 14H 2 O(g). Net D -7782 or -3891 kJ/mol. 14 D H f o H2O -3388. 1. Using Formula.

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Sum: 2C 6 H 14 (l) + 19O 2  12CO 2 + 14H 2 O(g) D H o = - 2 D H f o C6H14 + 1 2 D H f o CO2 + 14 D H f o H2O

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  1. 2. Energy Level Diagram Elements: 12C(s), 14H2(g), 19O2(g) 0 DHfo kJ Reactants: 2C6H14(l) 2DHfo C6H14 -334 SnDHfo(react) = -334 Products: 14H2O(g) Net D -7782 or -3891 kJ/mol 14DHfoH2O -3388 1. Using Formula Calculating DHo :An Example DHo = SnDHfo(prod) - SnDHfo(react) = 12DHfoCO2 + 14DHfoH2O- 2DHfoC6H14 = 12(-394) + 14(-242) – 2(-167) = -7782 kJ or –3891 kJ/mol DHo = -8116 – (-334) = -7782 kJ Products: 12CO2(g) . 12DHfoCO2 -4728 2C6H14(l) + 19O2 12CO2 + 14H2O(g) SnDHfo(prod) = -8116 3. Adding appropriate equations: Hess’s Law 2C6H14(l)  12C(s) + 14H2(g) DHo = -2DHfoC6H14 12C(s) + 12O2(g)12CO2(g)DHo = 12DHfoCO2 14H2(g) + 7O2 14H2O(g)DHo = 14DHfo H2O(g) 5. Use average Bond energies: 4. Hess’s Law Diagram 12-C’s, 28-H’s, 38-O’s 2 C6H14 2DHfo C6H14 19O2(g) 24,440 kJ d 12C(s),14H2(g) 19O2(g) DHo. e 28-H’s 38-O’s 12DHfo CO2 + 14DHfo H2O 12CO2 + 14H2O(g) 28 O-H d +e 30.796 kJ a+b+c b c 12-C’s 2DHfo C6H14 + DHcomb = a 24 O=C 12DHfoCO2 + 14DHfoH2O Or DH0 = 12DHfoCO2 + 14DHfoH2O - 2DHfo C6H14 = -7782 kJ or –3891 kJ/mol Net D -6356 or -3178 kJ/mol 10 C-C 28C-H 19O=0 DHo= S (nD)broken bonds- S (nD) bonds made (a + b + c) - (d + e) = 10(348) + 28(412) + 19(496) – 28(463) – 24(743 = 24,440 - 30,796 = -6356 kJ = -3178 kJ/mol C6H14 Sum: 2C6H14(l) + 19O2 12CO2 + 14H2O(g) DHo = -2DHfoC6H14 + 12DHfoCO2 +14DHfo H2O(g) = -2(-167) + 12(-394) + 14(-242) = -7782 kJ or –3891 kJ/mol

  2. CH123: Gen. Chem 1 12/3/01 Dr. J DH group problems The oxidation of long-chain carbon molecules powers our society and our bodies. Consider the following combustion reactions of two different molecules each containing a 6C chain: hexane (C6H14) and glucose (C6H1206). Their structures are shown to the right. • Calculate the standard enthalpy (of combustion) of hexane, as depicted in the reaction below, given the following information:DHfoC6H14 = -167 kJ/mol; DHfoCO2 = -394 kJ/mol; DHfoH2O = -242 kJ/mol; • 2 C6H14(l) + 19O2(g)  12CO2(g) + 14H2O(g) • Calculate the standard enthalpy of combustion of glucose, as depicted in the reaction below, given the following information:DHfoglc = -1260 kJ/mol; DHfoCO2 = -394 kJ/mol; DHfoH2O = -242 kJ/mol; • C6H12O6 (s) + 6O2(g)  6CO2(g) + 6H2O(g) • Compare the DHocombustion of each (kJ/mol). Which is higher? Would you predict this from their chemical structures? Explain.

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