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Chem301 - HW13

# Chem301 - HW13 - segal(mas6582 H13 Thermo 2 mccord(50970...

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segal (mas6582) – H13: Thermo 2 – mccord – (50970) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the reaction 4 FeO(s) + O 2 (g) -→ 2 Fe 2 O 3 (s) and heat-of-formation data Fe + 1 2 O 2 (g) -→ FeO Δ H = - 274 kJ / mol 2 Fe + 3 2 O 2 -→ Fe 2 O 3 Δ H = - 826 kJ / mol Find the change in enthalpy. Your answer must be within ± 0.1% Correct answer: - 556. Explanation: 4 FeO(s) + O 2 (g) -→ 2 Fe 2 O 3 (s) Reaction Δ H 0 f (kJ/mol) 4 [FeO Fe + 1 2 O 2 ] 4 (+274) = +1096 2 [2 Fe + 3 2 O 2 Fe 2 O 3 ] 4 ( - 826) = - 1652 4 FeO + O 2 2 Fe 2 O 3 - 556 002 10.0 points Calculate the standard enthalpy of formation of bicyclo[1.1.0]butane H H C C CH 2 H 2 C given the standard enthalpies of formation of 717 kJ · mol 1 for C(g) and 218 kJ · mol 1 for H(g) and the average bond enthalpies of 412 kJ · mol 1 for C H and 348 kJ · mol 1 for C C. 1. - 472 kJ · mol 1 2. +312 kJ · mol 1 3. - 36 kJ · mol 1 correct 4. +175 kJ · mol 1 5. - 124 kJ · mol 1 Explanation: We can write an equation in which we com- pletely decompose bicyclo [1,1,0] butane: C 4 H 6 (bicyclobutane , g) 4 C(g) + 6 H(g) Δ H rxn = 5 (BE C C ) + 6 (BE C H ) (The comment on the right comes from dissecting the structure given in the question and noting how many of each kind of bond is present). To find Δ H for this reaction we can use Hess’ Law with formation enthalpies: Δ H rxn = bracketleftBig 4 Δ H f C(g) + 6 Δ H f H(g) bracketrightBig - bracketleftBig 1 Δ H f C 4 H 6 (g) bracketrightBig = [4 (717 kJ / mol) + 6 (218 kJ / mol)] - Δ H f C 4 H 6 (g) = 4176 kJ / mol - Δ H f C 4 H 6 (g) Now we use the comment on which bonds were broken: Δ H rxn = 6 BE C C + 6 BE C H = 5 (348 kJ / mol) + 6 (412 kJ / mol) = 4212 kJ / mol We can set the two sides of the equation equal since they represent the same reaction: 4212 kJ / mol = 4176 kJ / mol - Δ H f C 4 H 6 (g) Δ H f C 4 H 6 (g) = - 36 kJ / mol . 003 10.0 points Calculate the enthalpy change for the reaction 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) Δ H f for SO 2 (g) = - 16 . 9 kJ/mol; Δ H f for SO 3 (g) = - 21 . 9 kJ/mol.

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segal (mas6582) – H13: Thermo 2 – mccord – (50970) 2 1. - 77 . 6 kJ/mol rxn 2. - 10 . 0 kJ/mol rxn correct 3. +10 . 0 kJ/mol rxn 4. - 5 . 0 kJ/mol rxn 5. +5 . 0 kJ/mol rxn Explanation: Reactants: Δ H f SO 2 (g) = - 16 . 9 kJ/mol Δ H f O 2 (g) = 0 kJ/mol Products: Δ H f SO 3 (g) = - 21 . 9 kJ/mol Δ H rxn = summationdisplay n Δ H f products - summationdisplay n Δ H f reactants = (2 mol)( - 21 . 9 kJ / mol) - (2 mol)( - 16 . 9 kJ / mol) - (1 mol)(0 kJ / mol) = - 10 . 0 kJ / mol rxn 004 10.0 points Consider the following substances: HCl(g) F 2 (g) HCl(aq) Na(s) Which response includes ALL of the sub- stances listed that have Δ H 0 f = 0? 1. Na(s) 2. HCl(g) and Na(s) 3. HCl(g), Na(s) and F 2 (g) 4. HCl(g), Na(s), HCl(aq) and F 2 (g) 5. Na(s) and F 2 (g) correct Explanation: Δ H 0 f = 0 for elements in their standard states. This would be true for both Na(s) and F 2 (g).
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