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Unformatted text preview: Version 047 – Exam 4 – McCord – (91750) 1 This printout should have 26 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Water data is on the back of your bub blesheet. 001 10.0 points Consider the reaction PCl 3 ( ℓ ) → PCl 3 (g) at 298 K. If Δ H ◦ is 32.5 kJ/mol, Δ S ◦ is 93.3 J/K mol, and Δ G ◦ is 4.7 kJ/mol, what would be the boiling point of PCl 3 at one atmosphere? 1. 298 K 2. 200 K 3. 369 K 4. 348 K correct 5. 438 K 6. 187 K 7. 596 K 8. 50.4 K 9. 153 K 10. 19.9 K Explanation: Δ H = 32 . 5 kJ/mol Δ S = 93 . 3 J/K mol Δ G = 4 . 7 kJ/mol T = 298 K The boiling point is a point of equilibrium between the liquid and gaseous states. At any equilibrium, Δ G is 0. So the challenge in this problem is to discover the temperature at which Δ G is 0. Luckily for us, Δ G is dependent on tem perature, but Δ H and Δ S do not vary signif icantly with temperature. So we should use the Gibbs equation: Δ G = Δ H T Δ S with the Δ H and Δ S values given, but we should set Δ G equal to 0 and solve for T . (Note that once we move away from 25 ◦ C, we are no longer under standard conditions and thus we lose the superscript “0” in our terms.) T = Δ H Δ G Δ S = 32 . 5 kJ / mol . 0933 kJ / mol · K = 348 . 339 K 002 10.0 points For the reaction 2 C(s) + 2 H 2 (g) → C 2 H 4 (g) Δ H ◦ r = +52 . 3 kJ · mol − 1 and Δ S ◦ r = 53 . 07 J · K − 1 · mol − 1 at 298 K. The reverse reaction will be spontaneous at 1. temperatures above 985 K. 2. temperatures below 985 K. 3. all temperatures. correct 4. no temperatures. 5. temperatures below 1015 K. Explanation: Δ G = Δ H T Δ S is used to predict spon taneity. (Δ G is negative for a spontaneous reaction.) T is always positive; for the reverse reaction, we reverse the sign of Δ H and Δ S . We thus have Δ G = ( ) T (+) for the re verse reaction, so Δ G will be negative for any physically possible value of T . 003 10.0 points 3.5 g of a hydrocarbon fuel is burned in a ves sel that contains 250. grams of water initially at 25.00 C. After the combustion, the temper ature of the water is 26.55 C. How much heat is evolved per gram of fuel burned? 1. 511 J/g 2. 463 J/g correct Version 047 – Exam 4 – McCord – (91750) 2 3. 1764 J/g 4. 26555 J/g 5. 143 J/g 6. 1263 J/g 7. 73.5 J/g 8. 1105 J/g Explanation: q = m Cp Δ T q = 250 (4.184) 1.55 = 1621.3 J per gram will be 1621.3 J / 3.5 g = 463 J/g 004 10.0 points Calculate the final temperature when 2.50 kJ of energy is transferred as heat to 1.50 mol N 2 at 298 K and 1.0 atm at constant pressure. 1. 159 ◦ C 2. 82 ◦ C correct 3. 57 ◦ C 4. 355 ◦ C 5. 75 ◦ C Explanation: 005 10.0 points When 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston at an external pressure of 2.00 atm, the nitro gen gas expands from 2.00 to 5.00 L against this constant pressure. What is Δ U for the process?...
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 Fall '07
 Fakhreddine/Lyon
 Thermodynamics, Enthalpy, Entropy, heavy metal, kJ/mol, H H C H C H H

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