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TA practice quiz 6 key

# TA practice quiz 6 key - CH301 Fall 2010 Practice Quiz...

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Unformatted text preview: CH301 Fall 2010 Practice Quiz 6—Answer Key 1. Rank the following molecules in terms of increasing real {experimentally‐observed) positional entropy at 0 K. H2O, H2S, Ar 1. H2O < H2S < Ar 2. Ar < H2S < H2O 3. Ar < H2O < H2S correct 4. H2S < CO2 < Ar Explanation. Helium only has one possible orientation, so its entropy at 0 K will be very close to 0. Although it initially appears that hydrogen sulfide and water have the same number of possible orientations, water actually has a lower value for W because it has a larger dipole moment. Some of the orientations are preferred over others, reducing the entropy. 2. If no vibrational modes are active (internal energy only comes from translation and rotation), which of the following molecules will have the largest molar heat capacity. 1. He 2. N2 3. C2H2 4. CO 5. C2H6 correct E C T , where C is the heat capacity. Thus, the molecule that has the largest coefficient in its internal energy expression will be the molecule with the highest heat capacity. If no vibrational modes are active (which is actually quite common), then the internal energy only comes from rotational and translational degrees of freedom. The internal energy of each of these molecules without any vibrational consideration is 3 He = kT 2 5 N2 = kT 2 5 C2H2 = kT 2 5 CO = kT 2 C2H6 = 3kT Thus, the molecule with the largest heat capacity is ethane, C2H6. 3. Imagine that you have a system consisting of just 3 molecules: HCl, HF, and HBr. Imagine that you have 3 spots on a solid for these molecules to stick to when the system is cooled. What is the maximum theoretical positional entropy of the system at 0 K? Assume each of the molecules can land with two possible orientations. (I am sure you know that ln(2) is about .7 and ln(3) is about 1. 1. 0 J K 1 2. 1.3 10 23 J K 1 3. 2.0 10 23 J K 1 4. 5.2 10 23 J K 1 correct Explanation. The best way to do this problem is probably brute force. The goal is to write down all possible arrangements. First, you want to write down all of the orders that HCl, HF, and HBr can bind to these three sites, independent of orientation. These are HCl HF HBr HCl HBr HF HF HCl HBr HF HBr HCl HBr HF HCl HBr HCl HF If you have taken probability, you also might recognize this as 3! = 6. However, we are not done because each of molecules has 2 possible orientations. For each individual molecule, there are two orientations. Which means there are 8 ways to have each arrangement above. So we have HCl HF HBr 8 ways HCl HBr HF 8 ways HF HCl HBr 8 ways HF HBr HCl 8 ways HBr HF HCl 8 ways HBr HCl HF 8 ways Add them all up, which means that W = 48. Now plug into the formula S k ln W S k ln 48 S k ln 4 4 ln 3 S k 4 ln 2 ln 3 S k 4 .7 1 S 3.8 k S 5.2 10 23 J K 1 4. Two different ideal gases, A and B, are initially placed into two pistons under identical conditions (P,V,T,n,etc). Which of the following will be the same for any isothermal (T = 0) compression of the gases? I. w II. q E III. a. b. c. d. e. f. g. I only II only III only correct I and II only II and III only I, II, and III None of these Explanation‐For an ideal gas, E is only a function of the change intemperature. Thus, for any isothermal compression, the change in energy will be 0. So, E is the same for any isothermal compression. However, w is not a state function, it depends on the pressure and the change in volume. So w is not the same for all of these processes. For an isothermal process, q = ‐w. Thus, q can also be different. 5. A system composed of one mole of gas is in thermal equilibrium. The gas is then compressed from 4.0 L to 2.0 L under constant external pressure of 2 atm. The change in internal energy of the system is 10 L atm. What is the change in heat for the system in kJ? Assume 1 L atm = .1 kJ I. 6 kJ II. ‐6 kJ III. 0.6 kJ correct IV. ‐0.6 kJ Explanation‐ wsys = ‐PextV = ‐2(2‐4) = 4 L atm and U = q + w qsys = U – w = 10 – 4 = 6 L atm Converting Q = 6 L/atm = 0.6 kJ 6. Rank the entropies of the following systems in terms of decreasing entropy: I. Pure ethanol (liquid) II. Table salt dissolved in ethanol III. Pure ethanol (solid) IV. A 50/50 mixture of ethanol and methanol 1. II > IV > I > III correct 2. IV > II > I > III 3. II > IV > III > I 4. III > I > IV > II 5. III > I > II > IV 6. I > III > IV > II 7. II > IV > I > III Explanation‐solutions composed of just one molecule are always less entropic than a solution with some other molecule dissolved in it. For example, pure ethanol is less entropic than table salt dissolved in ethanol. This explains why pure ethanol is less entropic than the solutions of pure ethanol. Also, the solid phase is less entropic than the liquid phase, which is less entropic than the gas phase. This is simply because molecules in more condensed phases have less energy and cannot move as freely. The less movement a molecule has, the less options the molecule has in terms of where it can be. Therefore, molecules in condensed phases cannot occupy different spaces and is less entropic. For the solutions of ethanol, this can be simply solved by looking at how many different types of molecules, ions, or atoms are dissolved within the solvent. For the mixture of ethanol and methanol, the solution is composed of two molecules. For the solution of ethanol and table salt, the table salt will dissolve into its component ions of sodium cation and chloride anion. Therefore, there are three molecules and ions in solution. The more types of molecules, the more entropic a solution is. Therefore, system II is more entropic than system IV. 7. A process has a Ssurr of 8 J K 1 at 27 C . Estimate the value of Ssurr when the temperature is lowered to 173 C . (Hint: find Hsys first.) Would you have predicted this new Ssurr to be larger or smaller at this colder temperature? 1. 45 J K 1 2. 16 J K 1 3. 24 J K 1 Correct 4. 42 J K 1 Explanation. Since Ssurr H , Tsurr H Tsurr Ssurr H 300 K 8 J K 1 H 2400 J Thus, at this new temperature, H Ssurr Tsurr 2400 J Ssurr 100 K Ssurr 24 J K 1 This makes sense given that as the temperature goes down, the amount of heat dumped into the surroundings will have a bigger impact (create more disorder.) 8. Assume that G for the process in question 7 at 25 oC is ‐165 J. What is Suniv for this process? (Hint: find Ssys first.) 1. 7.5 J/K 2. .5 J/K correct 3. ‐.5 J/K 4. ‐7.5 J/K Explanation: Hsys = ‐2400 J as was calculated in question 7. So: G = H ‐ TS ‐165J = ‐2400J ‐ 298Ssys Rearranging Ssys = ‐7.5J at room temperature. Now calculate Suniv Ssurr = 8 J/K as was given in question 7 So Suniv = Ssys + Ssurr 0.5 J = ‐7.5 J + 8.0 J This makes sense—the free energy change was negative meaning the reaction happened. So Suniv should be positive. ...
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