H12 Thermo 2

# H12 Thermo 2 - saldana (avs387) H12: Thermo 2 Mccord...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: saldana (avs387) H12: Thermo 2 Mccord (52385) 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 H =- 269 kJ / mol 2 Fe + 3 2 O 2- Fe 2 O 3 H =- 824 kJ / mol Find the change in enthalpy. Your answer must be within 0.1% Correct answer:- 572. Explanation: 4 FeO(s) + O 2 (g)- 2 Fe 2 O 3 (s) Reaction H f (kJ/mol) 4 [FeO Fe + 1 2 O 2 ] 4 (+269) = +1076 2 [2 Fe + 3 2 O 2 Fe 2 O 3 ] 4 (- 824) =- 1648 4 FeO + O 2 2 Fe 2 O 3- 572 002 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.- 36 kJ mol 1 correct 3.- 124 kJ mol 1 4. +312 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) 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 H for this reaction we can use Hess Law with formation enthalpies: H rxn = bracketleftBig 4 H f C(g) + 6 H f H(g) bracketrightBig- bracketleftBig 1 H f C 4 H 6 (g) bracketrightBig = [4 (717 kJ / mol) + 6 (218 kJ / mol)]- H f C 4 H 6 (g) = 4176 kJ / mol- H f C 4 H 6 (g) Now we use the comment on which bonds were broken: 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- H f C 4 H 6 (g) 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) H f for SO 2 (g) =- 16 . 9 kJ/mol; H f for SO 3 (g) =- 21 . 9 kJ/mol. saldana (avs387) H12: Thermo 2 Mccord (52385) 2 1. +10 . 0 kJ/mol rxn 2.- 5 . 0 kJ/mol rxn 3. +5 . 0 kJ/mol rxn 4.- 10 . 0 kJ/mol rxn correct 5.- 77 . 6 kJ/mol rxn Explanation: Reactants: H f SO 2 (g) =- 16 . 9 kJ/mol H f O 2 (g) = 0 kJ/mol Products: H f SO 3 (g) =- 21 . 9 kJ/mol H rxn = summationdisplay n H f products- summationdisplay n 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 H f = 0?...
View Full Document

## This note was uploaded on 10/25/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

### Page1 / 7

H12 Thermo 2 - saldana (avs387) H12: Thermo 2 Mccord...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online