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Chem301 - Exam 4

# Chem301 - Exam 4 - Version 346 Exam 4 mccord(50970 This...

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Version 346 – Exam 4 – mccord – (50970) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the following data at 25.0 C and 1 atm. Species Δ H 0 f S 0 Δ G 0 f kJ/mol J/mol · K kJ/mol N 2 (g) 0 . 0 191 . 5 0 . 0 O 2 (g) 0 . 0 205 . 0 0 . 0 N 2 O 5 (g) 11 . 0 ? 115 . 0 Calculate the absolute entropy of N 2 O 5 at 25.0 and 1 atm. 1. +396 . 5 J · mol 1 · K 1 2. +355 . 0 J · mol 1 · K 1 correct 3. - 704 J · mol 1 · K 1 4. +1053 . 0 J · mol 1 · K 1 5. +47 . 5 J · mol 1 · K 1 6. +252 . 5 J · mol 1 · K 1 7. - 349 . 0 J · mol 1 · K 1 Explanation: Δ G rxn = Δ H rxn - T Δ S rxn so we must find the Δ H and Δ S of the re- action. To do this, we need to write out the balanced equation: 2 N 2 + 5 O 2 2 N 2 O 5 Δ H rxn = Δ H products - Δ H reactants = (2 mol)(11 kJ / mol) - 0 = 22 kJ Δ S rxn = Δ S products - Δ S reactants = 2 x - (5 mol)(205 J / mol · K) - (2 mol)(191 . 5 J / mol · K) = 2 x - 1408 J / K = 2 x - 1408 1000 kJ / K Now substitute into Δ G rxn = Δ H rxn - T Δ S rxn : (2 mol)(115 kJ / mol) = 22 - (298 K)(2 x - 1408) kJ / K 1000 Solving for x gives x = 355 . 007J/mol · K. 002 10.0 points Which of the following has the smallest molar entropy at 298 K? 1. He(g) correct 2. N 2 (g) 3. Ne(g) 4. Cl 2 (g) 5. F 2 (g) Explanation: To find the entropy we can use Boltzmann’s formula S = k ln W where W is the number of microstates. At zero K, there is no mo- tion and W is given by the number of possible discernible orientations of the molecule in a crystal raised to the power of the number of molecules (1 mole for each in this case). If the substance forms a perfect crystal there should be zero residual entropy. We would expect this for these materials since they are either homonuclear diatomics or monoatomic species; in the solid, there is only one dis- cernible arrangement. We now consider the effect of being above 0 K. The greater the mass of the species, the closer together are the vibrational energy levels, so at a given temperature, more vibrational energy levels are occupied, thereby increasing the number of microstates. Thus the substance with the

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Version 346 – Exam 4 – mccord – (50970) 2 lowest standard molar entropy would be He, the lightest monoatomic species. 003 10.0 points Which of the following reactions is an en- thalpy of formation reaction? 1. C diamond (s) + 2H 2 (g) ←→ CH 4 (g) 2. 2Fe(s) + 3 / 2O 2 (g) ←→ Fe 2 O 3 (s) cor- rect 3. CH 4 (g) + 2O 2 (g) ←→ CO 2 (g) + 2H 2 O 4. Hg(s) + 1 / 2O 2 (g) ←→ HgO(s) Explanation: Formation reactions describe production of exactly one mole of one product from stoichio- metric quantities of elements in their standard states. 004 10.0 points Reactions with positive values of Δ S r always become spontaneous at low temperatures. 1. True 2. False correct Explanation: Δ G = Δ H - T Δ S is used to predict spon- taneity. G is negative for a spontaneous reaction.) T is always positive. When Δ S is positive and T is small, then in order for Δ G to be negative, Δ H must be negative, zero or have a very small positive value. But this is not always the case, so not ALL reactions with a positive Δ S
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Chem301 - Exam 4 - Version 346 Exam 4 mccord(50970 This...

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