Exam 4 301 TTh-solutions

Exam 4 301 TTh-solutions - Version 346 Exam 4 301 TTh...

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Unformatted text preview: Version 346 Exam 4 301 TTh sutcliffe (50985) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Choose the formula for the compound peri- odic acid. (HIO 3 is iodic acid.) 1. H 2 IO 2 2. HIO 3 3. HIO 4 correct 4. HI 5. H 2 IO 3 6. HIO 2 7. H 3 IO 8. HIO Explanation: 002 10.0 points Given the absolute entropies (in J/mol K) NO 2 = 240 H 2 O = 69.91 HNO 3 = 146 J/mol K NO = 210.7 at 298 K and 1 atm, evaluate S for the reaction 3 NO 2 (g) + H 2 O( ) 2 HNO 3 (aq) + NO(g) 1. S = +257 J / mol K at 25 C 2. S =- 770 J / mol K at 25 C 3. S =- 287 J / mol K at 25 C correct 4. S = +144 J / mol K at 25 C 5. S = +1009 J / mol K at 25 C Explanation: S rxn = summationdisplay nS f prod- summationdisplay nS f rct = bracketleftBig 2 (146 J / mol K) +210 . 7 J / mol K bracketrightBig- bracketleftBig 3 (240 J / mol K) +69 . 91 J / mol K bracketrightBig =- 287 . 21 J / mol K 003 10.0 points Is the reaction Cl 2 (g) + 2 HBr(g) Br 2 ( ) + 2 HCl(g) spontaneous at 25 C and one atmosphere pressure? The standard molar Gibbs free en- ergy of formation of HBr(g) at 25 C is- 53 . 43 kJ/mol and that of HCl(g) is- 95 . 30 kJ/mol. 1. Yes, because G for the reaction is neg- ative. correct 2. No, because G for the reaction is posi- tive. 3. Yes, because G for the reaction is pos- itive. 4. Cannot tell unless we are given the stan- dard molar Gibbs free energies of formation at 25 C of Br 2 ( ) and Cl 2 (g). 5. No, because G for the reaction is neg- ative. Explanation: G f HBr(g) =- 53 . 43 kJ/mol G f HCl(g) =- 95 . 30 kJ/mol G f Br 2 = 0 kJ/mol G f Cl 2 = 0 kJ/mol G rxn = summationdisplay n G f prod- summationdisplay n G f rct = 2(- 95 . 30 kJ / mol)- 2(- 53 . 43 kJ / mol) =- 83 . 74 kJ / mol For spontaneous reactions, G is negative. Version 346 Exam 4 301 TTh sutcliffe (50985) 2 004 10.0 points Calculate the standard enthalpy change for the reaction 2 C(s) + 2 H 2 O(g) CH 4 (g) + CO 2 (g) given C(s) + H 2 O(g) CO(g) + H 2 (g) H = +131 . 3 kJ/mol rxn CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) H =- 41 . 2 kJ/mol rxn CO(g) + 3H 2 (g) CH 4 (g) + H 2 O(g) H =- 206 . 1 kJ/mol rxn 1. H = +15 . 3 kJ/mol rxn correct 2. H =- 378 . 6 kJ/mol rxn 3. H =- 116 . 0 kJ/mol rxn 4. H =- 157 . 2 kJ/mol rxn 5. H = +378 . 6 kJ/mol rxn Explanation: The first equation needs to be multiplied by two in order to get the equation were interested in. Thus its H is multiplied by two as well. 2 C(s) + 2 H 2 O(g) 2 CO(g) + 2 H 2 (g) H = +262 . 6 kJ/mol rxn Once thats done, the three equations are just summed together to get the desired equa- tion. Showing all reactants leading to all products, 2 C(s) + 3 H 2 O(g) + 2 CO(g) + 3 H 2 (g) 2 CO(g) + 3 H 2 (g) + CO 2 (g) + CH 4 (g) + H 2 O(g)...
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This note was uploaded on 12/11/2010 for the course CH 301 taught by Professor Fakhreddine/lyon during the Fall '07 term at University of Texas.

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Exam 4 301 TTh-solutions - Version 346 Exam 4 301 TTh...

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