Spring 09 Recitation Quiz 5

Spring 09 Recitation Quiz 5 - CHEM 0120 Golde SPRING 2009...

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CHEM 0120 SPRING 2009 Monday 1:00 PM Golde QUIZ 5 WEEK 7 February 16, 2009 Name_______________________________ G ° = H ° - T S ° , G ° = -RTln(K), R = 8.31 J/mol K, G = G o + RTln(Q) 1. Given the following data for the process in which solid NH 4 Cl dissolves in water: NH 4 Cl(s) NH 4 + (aq) + Cl - (aq) H o f (kJ/mol) -314.4 -133.3 -167.1 S o (J/mol K) +94.6 +111.2 +56.6 a) Calculate H o and S o for this reaction. b) Calculate G o of reaction at 25 o C. Do you expect NH 4 Cl to be very soluble at 25 o C? c) In Expt. #1, your lab studied solutions of NH 4 Cl at lower temperatures. Using your answers to (a) and/or (b), do you expect dissolving NH 4 Cl at -10 o C to be more or less favorable than at 25 o C? Give your reasoning. ANSWERS 1. a) H o = [-167.1 -133.3] - [-314.4] kJ = +14.0 kJ S o = [111.2 + 56.6] - [94.6] J/K = +73.2 J/K b) G o = 14000 J - 298K (73.2 J/K) = -7.81 kJ The negative sign indicates that this process is spontaneous and that NH 4 Cl is very soluble. c) At -10 o C, ΔG o = -5.25 kJ. This is less negative than at 25 o C, so we expect that NH 4 Cl is less soluble at -10 o C than at 25 o C. [One can also argue from the signs of ΔH o and ΔS o : at lower temperature, the ΔH o term becomes more important. As ΔH o > 0, dissolution becomes less favorable under these conditions]. CHEM 0120 SPRING 2009 Monday 6:00 PM Golde QUIZ 5 WEEK 7 February 16, 2009 For Make-up Only Regular recitation Day: Time: instructor: Quiz Score: Group score: Scores recorded by:
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Name_______________________________ G ° = H ° - T S ° , G ° = -RTln(K), R = 8.31 J/mol K, G = G o + RTln(Q) 1. In each of the following pairs, state which species has the higher absolute entropy. Give your reasoning. a) 1 mole of solid ice at 0 o C or 1 mole of liquid water at 0 o C b) 0.1 mole of CO 2 gas at 2 atm pressure and 25 o C or 0.1 mole of CO 2 gas at 1 atm pressure and 25 o C. 2. Given the following information for the vaporization of bromine: Br 2 (l) Br 2 (g) G f o (kJ/mol) 0 +3.11 at 25 o C a) Calculate the equilibrium constant, K p , at 25 o C for this process. b) From your answer to (a), deduce whether the boiling point of Br 2 lies below or above 25 o C. ANSWERS 1. a) Liquid water. The molecules have more randomness (more freedom of motion) in the liquid than in the solid. b) 0.1 mole of CO 2 gas at 1 atm pressure. At the lower pressure, the gas volume is larger, so the randomness of the gas molecules is larger than at the higher pressure. 2. a) G o of vaporization = +3.11 kJ = 3110 J ln (K p ) = K 298 J/K x 8.31 J 3110 - = -1.256 K p = inv ln (-1.256) = 0.285 b) K p = 1 at the boiling point, and K p increases with T for this endothermic process. Hence the boiling point is above 25 o C. OR
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This note was uploaded on 04/06/2009 for the course CHEM 0120 taught by Professor Golde during the Spring '07 term at Pittsburgh.

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Spring 09 Recitation Quiz 5 - CHEM 0120 Golde SPRING 2009...

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