H13- Thermo 2-solutions

# H13- Thermo 2-solutions - phan (dtp377) – H13: Thermo 2...

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Unformatted text preview: phan (dtp377) – H13: Thermo 2 – mccord – (51600) 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 =- 273 kJ / mol 2 Fe + 3 2 O 2-→ Fe 2 O 3 Δ H =- 822 kJ / mol Find the change in enthalpy. Your answer must be within ± 0.1% Correct answer:- 552. Explanation: 4 FeO(s) + O 2 (g)-→ 2 Fe 2 O 3 (s) Reaction Δ H f (kJ/mol) 4 [FeO → Fe + 1 2 O 2 ] 4 (+273) = +1092 2 [2 Fe + 3 2 O 2 → Fe 2 O 3 ] 4 (- 822) =- 1644 4 FeO + O 2 → 2 Fe 2 O 3- 552 002 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.- 36 kJ · mol − 1 correct 2.- 472 kJ · mol − 1 3. +312 kJ · mol − 1 4.- 124 kJ · mol − 1 5. +175 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. phan (dtp377) – H13: Thermo 2 – mccord – (51600) 2 1.- 77 . 6 kJ/mol rxn 2. +10 . 0 kJ/mol rxn 3. +5 . 0 kJ/mol rxn 4.- 10 . 0 kJ/mol rxn correct 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 f = 0?...
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## This note was uploaded on 03/26/2012 for the course CHEM 301 taught by Professor Wandelt during the Spring '08 term at University of Texas at Austin.

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H13- Thermo 2-solutions - phan (dtp377) – H13: Thermo 2...

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