CHEMISTRY 202 Practice Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (40 pts.) 22 (20 pts.)

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1 CHEMISTRY 202 Practice Hour Exam II Fall 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 7 numbered pages. Check now to make sure you have a complete exam. You have two hours to complete the exam. Determine the best answer to the first 20 questions and enter these on the special answer sheet. Also, circle your responses in this exam booklet. Show all of your work and/or provide complete answers to questions 21 and (60 pts.) 21 (40 pts.) 22 (20 pts.) Total (120 pts.) Useful Information: Always assume ideal behavior for gases (unless explicitly told otherwise). STP = standard temperature and pressure = 0 C and 1.00 atm 760 torr = 1.00 atm R = Latm/molK = J/Kmol K = C N A = x E = q + w S = q rev /T H = E + PV G = H TS Here are some of the formulas we used/derived in studying thermodynamics. An individual formula may or may not apply to a specific problem. This is for you to decide! S = nrln(v 2 /V 1 ) S = H/T C v = (3/2)R C p = (5/2)R S = ncln(t 2 /T 1 ) G = G + RTln(Q) S surr = q/t w = P V q rev = nrtln(v 2 /V 1 ) q = nc T ln(k) = H R 1 + T S R K ln K 2 1 H = R 1 T2 1 T 1

2 Practice Hour Exam II Page No Given that ΔH fusion for Fe(s) = 13.6 kj/mol and the following data: ΔH 0 f Fe 2 O 3 (s) Al 2 O 3 (s) (kj/mol) Determine ΔH rxn for the thermite reaction (at constant temperature) we saw in lecture as represented by the equation: Fe 2 O 3 (s) + 2Al(s) Al 2 O 3 (s) + 2Fe(l) a) kj b) kj c) kj d) kj e) kj 2. You pour ml of water at 90. C into a one liter Thermos initially containing air at 1 atm 25 C and seal the Thermos. You are to estimate to the nearest degree the final temperature of the water in the Thermos. Make the following assumptions: I. The heat capacity of water does not change with temperature and is 4.18 J/g C. II. The density of water is exactly 1g/mL and does not change with temperature. III. Air has the following heat capacities (assume they do not change with temperature): C v = J/Kmol, C p = J/Kmol. IV. The Thermos is a perfect insulator. a) 50. C b) 60. C c) 70. C d) 80. C e) 90. C 3. At 25 C, the following enthalpies of reaction are known: 2C 2 H 2 (g) + 5O 2 (g) 4CO 2 (g) + 2H 2 O(l) C(s) + O 2 (g) CO 2 (g) 2H 2 (g) + O 2 (g) 2H 2 O(l) H = kj H = 394 kj H = 572 kj Calculate ΔH for the following reaction: 2C(s) + H 2 (g) C 2 H 2 (g) a) 452 kj b) 452 kj c) 1634 kj d) 226 kj e) 226 kj 4. Using the information below, calculate H f for 1.00 mole of PbO(s). PbO(s) + CO(g) Pb(s) + CO 2 (g) H f for CO 2 (g) = kj/mol H f for CO(g) = kj/mol H = kj a) kj b) kj c) kj d) kj e) kj 5. For how many of the following processes is the entropy change of the system equal to q/t? I. An ideal gas expands into a vacuum. II. Hydrogen gas reacts with oxygen gas to produce water. III. Water evaporates isothermally at room temperature and pressure. IV. Water freezes at 273K and 1 atm. a) 0 b) 1 c) 2 d) 3 e) 4

3 Practice Hour Exam II Page No , 7. For substance X ΔH vaporization = 42.8 kj/mol at its normal boiling point (375 C and 1 atm). For the process X(l) X(g) at 1 atm and 375 C, calculate the value of: 6. ΔS surr a) 0 b) 114 J/Kmol c) 114 J/Kmol d) 66 J/Kmol e) 66 J/Kmol 7. ΔS univ a) 0 b) 114 J/Kmol c) 114 J/Kmol d) 66 J/Kmol e) 66 J/Kmol For the isothermal expansion of an ideal monatomic gas against a constant pressure of 2 atm, which of the following statements (a-d) is true? a) Because the expansion is isothermal, ΔH = 0, and because ΔS = ΔH/T, ΔS = 0. b) ΔS = nc p ln(t 2 /T 1 ) and because T 2 = T 1, ΔS = 0. c) ΔS = nc p ln(v 2 /V 1 ) and because P 1 V 1 = P 2 V 2, ΔS = nc p ln(p 1 /P 2 ). Since pressure is constant, P 2 = P 1, so ΔS = 0. d) ΔS = q rev /T and because the expansion is isothermal and ΔS is a state function, q rev = 0, so ΔS = 0. e) None of the above statements (a-d) is true. 9. Determine the signs of ΔH, ΔS and ΔG for the freezing of liquid water at 2 C and 1 atm. ΔH ΔS ΔG a) b) + c) d) + 0 e) At 298K, ΔG = 30.3 kj for the reaction Fe 2 O 3 (s) + 3CO(g) 3CO 2 (g) + 2Fe(s). If we carry out the reaction at 298K but with pressures of 10.0 atm instead of 1.00 atm, ΔG would equal: a) 36.0 kj b) 30.3 kj c) 24.6 kj d) 18.9 kj e) 13.2 kj 11. The molar enthalpy of fusion of water is 6.02 kj at 0 C. If S of ice is 41.3 J/molK at 0 C determine S of liquid water at 0 C. a) 19.2 J/molK b) 21.1 J/molK c) 22.1 J/molK d) 61.5 J/molK e) 63.3 J/molK 12. SO 2 (g) and O 2 (g) react to form SO 3 (g). The partial pressures at equilibrium are as follows: P(SO 2 ) = atm, P(O 2 ) = atm, and P(SO 3 ) = atm. Calculate ΔG at K for the reaction 2SO 2 (g) + O 2 (g) 2SO 3 (g). a) 0 b) 14.7 kj c) 14.7 kj d) 10.3 kj e) 10.3 kj

4 Practice Hour Exam II Page No Which of the following statements are correct if the volume of 1.0 mole of an ideal gas is decreased at constant temperature? I. G gas increases IV. G gas decreases II. S gas increases V. S gas decreases III. H gas increases VI. H gas decreases a) IV, V b) II, IV c) I, V d) II, III, IV e) IV, V, VI 14. Calculate the magnitude of the maximum work produced when 1.00 mol of an ideal monatomic gas expands isothermally from 2.00 L to 6.50 L at a final pressure of 1.50 atm. a) 473 J b) 684 J c) 915 J d) 1.16 kj e) 2.22 kj 15. Sodium bromide dissolves in water according to the equation: NaBr(s) Na + (aq) + Br (aq). Determine the value of the equilibrium constant for this process at 25.0 C given the following data. ΔG 0 f (kj/mol) Na + (aq) 262 Br (aq) 121 NaBr(s) 360 a) 9.35 x 10-5 b) 1.01 c) 2.56 d) 80.4 e) 1.08 x Each of the following chemical reactions represented below is spontaneous at some temperature: I. 2KClO 3 (s) 2KCl(s) + 3O 2 (g) II. 2H 2 O 2 (l) 2H 2 O(l) + O 2 (g) III. Pb(NO 3 ) 2 (aq) + 2NaCl(aq) PbCl 2 (s) + 2NaNO 3 (aq) IV. 2SO 3 (g) 2SO 2 (g) + O 2 (g) From just this fact (that each is spontaneous) and the balanced chemical equations, how many of the reactions can you conclude must be exothermic? Note: do not use any background knowledge about the reactions (that is, the question is not asking which are actually exothermic but which must be exothermic given the information provided). a) 0 b) 1 c) 2 d) 3 e) 4

5 Practice Hour Exam II Page No We all love the pop bottles that begin most of the lectures. A mixture of hydrogen gas and oxygen gas remains unreacted until it is exposed to burning methane from a Bunsen burner. Then the reaction occurs very rapidly (at constant pressure). Given the data provided, how many of the following statements help to explain this behavior? 2H 2 (g) + O 2 (g) 2H 2 O(l) G = 474 kj H = 572 kj S = 327 kj I. The mixture remains unreacted because the reactants are thermodynamically more stable than the products. II. The mixture remains unreacted because the reaction has a small equilibrium constant. III. The mixture remains unreacted because the negative value for S slows down the reaction. IV. The mixture remains unreacted until the heat from the Bunsen burner raises the temperature of the system and this makes the reaction more thermodynamically favorable. a) 0 b) 1 c) 2 d) 3 e) Choose the most appropriate plot for each of the following. A plot may be used once, more than once, or not at all. a) b) c) d) e) 18. ΔG for a reaction vs. number of moles of product formed a 19. w vs. the number of steps in the isothermal compression of an ideal gas d 20. H vs. V at constant temperature for 1.0 mole of an ideal gas c

6 Practice Hour Exam II Page No When we first started studying thermodynamics, I stated that we will be able to explain observations such as the fact that a hot cup of coffee will eventually reach room temperature. You all knew that we do not observe a cup of hot coffee spontaneously heat up (that is, increase in temperature), but by studying thermodynamics we are able to support this both qualitatively (conceptually) and quantitatively. The time to do so is now! a. For this portion of the explanation, do not use sample calculations (the math will come in part b). i. If the temperature of a hot cup of coffee increased in a cooler room, would this violate the first law of thermodynamics? Why or why not? Make sure to state the first law of thermodynamics in your discussion. [4 points] NO ii. If the temperature of a hot cup of coffee increased in a cooler room, would this violate the second law of thermodynamics? Why or why not? Make sure to state the second law of thermodynamics in your discussion. [4 points] YES iii. There is a factor in this process (hot coffee increasing in temperature) that would favor the heating up of the coffee. What is it? Then why doesn t the process happen? [4 points] SEE CHAPTER 10

7 Practice Hour Exam II Page No (con t) b. Suppose you have hot coffee at 80.0 C in a room at 25.0 C. Provide quantitative support in both of these answers and provide brief explanations of the significance of your answers. Assume you have about a cup of coffee (250.0 g) and that the heat capacity of the coffee is the same as that of water (4.18 J/g C). i. Show that it is not thermodynamically favorable for the coffee to spontaneously increase in temperature even by 1.0 C and that it is less favorable to increase by 10.0 C. Make sure to explain the significance of your calculations and answers. [14 points] If the coffee heats up from 80.0 C to 81.0 C: ΔS univ = = J/K: not-spontaneous since ΔS univ is negative If the coffee heats up from 80.0 C to 90.0 C: ΔS univ = -5.9 J/K: not-spontaneous since ΔS univ is negative ΔS univ is more negative (less favorable) for coffee heating up 10.0 C than 1.0 C ii. Show that it is most thermodynamically favorable for the coffee to spontaneously cool to 25.0 C as opposed to slightly above or below this temperature. Make sure to explain the significance of your calculations and answers. [14 points] If the coffee cools from 80.0 C to 25.0 C: ΔS univ = J/K: spontaneous since ΔS univ is positive If the coffee cools from 80.0 C to 24.0 C (for example): ΔS univ = J/K: spontaneous since ΔS univ is positive If the coffee cools from 80.0 C to 26.0 C (for example): ΔS univ = J/K: spontaneous since ΔS univ is positive ΔS univ is most positive (most favorable) for coffee cooling to 25.0 C than 24.0 C or 26.0 C.

8 Practice Hour Exam II Page No You have 1.00 mol of a monatomic ideal gas in a container fitted with a massless, frictionless piston at 1.00 atm and heat it to 400.0K. The gas, as expected, eventually cools to room temperature (in this case, 300.0K). [Note: at 400.0K, S = J/Kmol for this gas] a. Calculate ΔG (in kj) for this process. Show all work. [9 points] ΔG = J = kj b. Why isn t ΔG an indication about the spontaneity of the cooling of the gas? [4 points] SEE CHAPTER 10 c. Provide mathematical support for the fact that the cooling of the gas is a spontaneous process. [7 points] ΔS univ = J/K

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