#Chem 162-2008 homework 11th week

#Chem 162-2008 homework 11th week - Chem 162-2008 Hill...

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Free Energy Change and Equilibrium 51. Write K eq expressions for the following reactions. Which, if any, of these expressions correspond to equilibrium constants that we have previously denoted as K c , K p , K a , and so on? (a) 2 NaHSO 3 (s) Na 2 SO 3 (s) + H 2 O(g) + SO 2 (g) K eq = P H2O P SO2 corresponds to K p (b) Mg(OH) 2 Mg 2+ (aq) + 2OH - (aq) K eq = [Mg 2+ ][OH - ] 2 corresponds to K sp (c) CH 3 COO - (aq) + H 2 O(l) CH 3 COOH(aq) + OH - (aq) K eq = [CH 3 COOH][OH - ]/[CH 3 COO - ] corresponds to K cb 55. The following data are given for the sublimation of napthalene at 298K: ∆S o = 168.7 J mol -1 K -1 , ∆H o = 73.6 kJ mol -1 . Calculate the pressure of napthalene vapor in equilibrium with solid napththalene at 298 K. C 10 H 8 (s) C 10 H 8 (g) ∆G o = ∆H o – T∆S o ∆G o = 73600 – (298 x 168.7) = 2.333 x 10 4 J ∆G o = -RTln(K) 2.333 x 10 4 = -8.314 x 298 x ln(K) (((2.333 x 10 4 ) = ((-8.314) x 298 x (ln(K)))),X) K eq = 8.14 x 10 -5 K p = P C10H8 K eq = K p K p = 8.14 x 10 -5 atm G o and K eq as Functions of Temperature 61. The following thermodynamic data are for 298K. Use these data and data from Appendix C to determine K eq at 45 o C for the reaction CO 2 (g) + SF 4 (g) CF 4 (g) + SO 2 (g) ∆H o f , J/mol S o , Jmol -1 K -1 SF 4 (g) -763000 299.6 CF 4 (g) -925000 261.6 From Appendix C: CO 2 (g) -393500 213.6 SO 2 (g) -296800 248.1 ∆G o = ∆H o – T∆S o ∆H o = (-925000 + -296800) – (-393500 + -763000) = -65300 J ∆S o = (261.6 + 248.1) – (213.6 + 299.6) = -3.5 J 1
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o = -65300 – (318 x -3.5) = -64187 J ∆G o = -RTln(K) -64187 = -8.314 x 318 x ln(K) (((-64187) = (-8.314 x 318 x (ln(K)))),X) K = 3.50 x 10 10 63. The normal boiling point of 1-butanol is 117.8 o C. CH 3 (CH 2 ) 2 CH 2 OH(l, 1atm) CH 3 (CH 2 ) 2 CH 2 OH(g, 1 atm) ∆H o = 43.82 kJ Calculate the boiling point of 1-butanol when the barometric pressure is 747 mm Hg Clausius-Clapeyron equation: ln(P 2 /P 1 ) = -∆H o /R x (1/T 2 – 1/T 1 ) “normal” means at 1 atm P 1 = 760 mm T 1 = 117.8 o C = 390.95K P 2 = 747 mm T 2 = X o C ln(747/760) = -43820/8.314 x (1/T 2 – 1/390.95) (((ln(747/760)) = ((-43820/8.314) x ((1/T 2 ) – (1/390.95)))),X) T 2 = 390.45K = 117.3 o C 65. Use data from Appendix C to calculate K p at 155 o C for the following reaction. 2NO(g) + O 2 (g) 2NO 2 (g) ∆G o f NO(g) 86570 Jmol -1 O 2 (g) 0 NO 2 (g) 51300 ΔG o = Σn p ΔG o f(products) - Σn r ΔG o f(reactants) ΔG o = (2 x 51300) – ((2 x 86570) + 0) = -7.054 x 10 4 J ∆G o = -RT ln(K) -7.054 x 10 4 = -8.314 x (273 + 155) x ln(K) (((-7.05 x 10 4 ) = ((-8.314) x (273 + 155) x (ln(K)))),X) K = 4.07 x 10 8 The Solutions Manual did the problem a different way, and got a different answer. I don’t see anything wrong with their approach to solving the problem, or to my approach, so I can’t explain the difference in answers. 2
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#Chem 162-2008 homework 11th week - Chem 162-2008 Hill...

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