exam 4 spring 2007 - Kolar, Leah Exam 4 Due: May 2 2007,...

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Kolar, Leah – Exam 4 – Due: May 2 2007, noon – Inst: McCord 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Benzoic Acid, C 6 H 5 COOH = 122.12 g/mol 001 (part 1 oF 1) 10 points What would be the signs oF Δ H and Δ S For the reaction CH 4 (g) CH 3 (g) + H(g) ? 1. Δ H = 0, Δ S = - 2. Both zero. 3. Both negative. 4. Δ H = - , Δ S = 0 5. Δ H = - , Δ S = + 6. Δ H = 0, Δ S = + 7. Δ H = +, Δ S = 0 8. Δ H = +, Δ S = - 9. Impossible to tell From the inFormation given. 10. Both positive. correct Explanation: C H H H H ----→ C H H H + H Δ H 0 rxn = X BE rct - X BE prod = £ 4 (C H) / - £ 3 (C H) / = (C H) = 414 kJ / mol Δ H is positive (endothermic, net bond break- ing). Δ S is also positive because the random- ness oF the system has increased. (1 mole oF gaseous reactants Forms 2 moles oF gaseous products.) 002 (part 1 oF 1) 10 points Evaluate Δ G 0 For the reaction 2 N 2 (g) + 3 O 2 (g) 2 N 2 O 3 (g) at 25 C. Substance Δ H 0 f S 0 kJ/mol J/mol · K N 2 0 . 0 191 . 5 O 2 0 . 0 205 . 0 N 2 O 3 83 . 72 312 . 2 1. +56 . 1 kJ/mol rxn 2. - 540 . 0 kJ/mol rxn 3. +278 . 7 kJ/mol rxn correct 4. +540 . 0 kJ/mol rxn 5. - 561 . 0 kJ/mol rxn Explanation: In general, it is possible to obtain Δ G 0 in several ways. The inFormation available to us suggests we should use the standard state Gibbs equation: Δ G 0 = Δ H 0 - T Δ S 0 , However, we have to calculate Δ H 0 and Δ S 0 frst. Δ H 0 can be determined using Hess’s Law: Δ H 0 = X n Δ H 0 f products - X n Δ H 0 f reactants = 2(Δ H 0 f For N 2 O 3 ) - 2 ( Δ H 0 f For N 2 ) - 3 ( Δ H 0 f For O 2 ) = 2(83 . 72 kJ / mol) - 0 - 0 = 167 . 44 kJ / mol
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Kolar, Leah – Exam 4 – Due: May 2 2007, noon – Inst: McCord 2 Then Δ S 0 can be determined from the equation similar to Hess’s Law: Δ S 0 = X n Δ S 0 products - X n Δ S 0 reactants = 2 ( S 0 for N 2 O 3 ) - £ 2 ( S 0 for N 2 ) + 3 ( S 0 for O 2 )/ = 2(312 . 2 J / mol · K) - 2(191 . 5 J / mol · K) - 3(205 . 0 J / mol · K) = - 373 . 6 J / mol · K With Δ H 0 and Δ S 0 in hand, we can now use the Gibbs equation. Remember that the absolute (Kelvin) temperature must be used in this equation: T = 25 C + 273 = 298 K Thus Gibbs equation yields Δ G 0 = (167 . 4 kJ / mol) - (298 K) µ - 373 . 6 J mol · K · 1 kJ 1000 J = 278 . 773 kJ / mol 003 (part 1 of 1) 10 points For the reaction H 2 (g) + Cl 2 (g) 2 HCl(g) at constant temperature and pressure, we would expect 1. Δ H Δ E . 2. Δ H > Δ E . 3. Δ H Δ E . 4. Δ H < Δ E . 5. Δ H = Δ E . correct Explanation: The equation that relates Δ H to Δ E is Δ E = Δ H - n ) R T . As 2 moles of gaseous reactants (1 mol of H 2 and 1 mol of Cl 2 ) become 2 moles of gaseous product, Δ n must be 0. This means that the term (Δ
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This test prep was uploaded on 03/19/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.

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exam 4 spring 2007 - Kolar, Leah Exam 4 Due: May 2 2007,...

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