Chem 162-2010 homework 11th week

Chem 162-2010 homework 11th week - Chem 162-2010 Hill &...

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th Week Standard Free Energy Change 39. Use data from Appendix C to determine ∆G o values for the following reactions at 25 o C. (a) FeO(s) + H 2 (g) → Fe(s) + H 2 O(g) (b) CdO(s) + 2HCl(g) → CdCl 2 (s) + H 2 O(l) ∆G o f FeO(s) -251.5 kJ/mol H 2 (g) 0 Fe(s) 0 H 2 O(g) -228.6 CdO(s) -228 HCl(g) -95.30 CdCl 2 (s) -344.0 H 2 O(l) -237.2 (a) (0 + -228.6) – (-251.5 + 0) = 22.9 kJ (b) (-344.0 + (-237.2)) – (-228 + (2 x (-95.30))) = -162.6 kJ 41. Use data from Appendix C to detemine ∆H o and ∆S o , at 298K, for the following reaction. Then determine ∆G o in two ways and compare the results. CS 2 (l) + 3O 2 (g) → CO 2 (g) + 2SO 2 (g) ∆H o f ∆G o f S o CS 2 (l) 89700J 65270J 151.3J O 2 (g) 0 0 205.0 CO 2 (g) -393500 -394400 213.6 SO 2 (g) -296800 -300200 248.1 ∆H o : (-393500 + (2 x -296800)) – (89700 + (3 x 0)) = -1.077 x 10 6 J ∆S o : (213.6 + (2 x 248.1)) – (151.3 + (3 x 205.0)) = -56.5 J ∆G o = ∆H o – T∆S o = (-1.077 x 10 6 J) – (298 x (-56.5J)) = -1.060 x 10 6 J ∆G o = (-394400 + (2 x (-300200))) – (65270 + (3 x 0)) = -1.060 x 10 6 J 43. Why is the standard free energy change, ∆G o , so important in dealing with the question of spontaneous change, even though the conditions in a chemical reaction are usually nonstandard ? ∆G o provides a reference point. ∆G (i.e., ∆G o under nonstandard conditions), can be calculated by adding a correction factor to the reference calculation. Free Energy Change and Equilibrium 45. Estimate the normal boiling point of heptane, C 7 H 16 , given that at this temperature ∆H o vapn = 31.69 kJ/mol. Trouton’s law: ∆S o vapn = ∆H o vapn /T bp ≈ 87 J/mol 31690/T = 87 T = 364K 1
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47. The normal boiling point of sulfuryl chloride, SO 2 Cl 2 (l), is 69.3 o C. Estimate ∆H o vapn of sulfuryl chloride. Compare your result with a value based on data from Appendix C. Appendix C ∆H o f SO 2 Cl 2 (g) -364.0 kJ SO 2 Cl 2 (l) -394.1 kJ Trouton’s law: ∆S o vapn = ∆H o vapn /T bp ≈ 87 J/mol ∆H o vapn /T bp ≈ 87 J/mol ∆H o vapn /342.3 ≈ 87 J/mol ∆H o vapn = 2.98 x 10 4 J From Appendix C: ∆H o = ∆H o p - ∆H o r = ∆H o SO2Cl2(g) - ∆H o SO2Cl2(l) = -364000 – (-394100) = 3.01 x 10 4 J Additional Problems 75. For the decomposition of NaHCO 3 (s), estimate the temperature at which the total pressure of the
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Chem 162-2010 homework 11th week - Chem 162-2010 Hill &...

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