CHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 31 (20 pts.) 32 (40 pts.)

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1 CHEMISTRY 202 Hour Exam II October 27, 2015 Dr. D. DeCoste Name Signature T.A. This exam contains 32 questions on 11 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 30 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 31 and (60 pts.) 31 (20 pts.) 32 (40 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 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. An ideal gas is compressed isothermally against a constant pressure until internal and external pressures reach equilibrium. Which of the following statements (a-d) is true? a) ΔG = 0 since the system reaches equilibrium. b) ΔG = 0 since the process is isothermal. c) ΔG < 0 since the process is spontaneous. d) ΔG > 0 since a gas will not spontaneously be compressed. e) None of the statements (a-d) is true.

3 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. Consider the following system at equilibrium at 25 C: PCl 3 (g) + Cl 2 (g) PCl 5 (g). For the reaction as written, ΔG = kj. When the temperature is raised the ratio of the partial pressure of PCl 3 (g) to the partial pressure of PCl 5 (g) will a) decrease. b) increase. c) not change. d) This is impossible to answer without additional information. 10. Which of the following statements is true for the vaporization of a liquid at a given pressure? a) ΔG is positive at all temperatures. b) ΔG is negative at all temperatures. c) ΔG is negative at high temperatures and positive at low temperatures. d) ΔG is positive at high temperatures and negative at low temperatures. e) The answer depends on the nature of the liquid. 11. 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) + 0

4 Hour Exam II Page No 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) 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. 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 15. 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 , 17. Answer the questions based on the following data: H 2 O(l) H 2 O(g) H f (kj/mol) S (J/mol K) Determine ΔG for the vaporization of 1.00 mole of H 2 O(l) at 25 C a) 0 b) 35.4 kj c) 35.4 kj d) 8.57 kj e) 8.57 kj 17. As we discussed when considering equilibrium, when water is placed in a sealed container it will evaporate until the vapor reaches the equilibrium vapor pressure. Determine the equilibrium vapor pressure of H 2 O(l) at 25 C in torr (1 atm = 760 torr). a) 3.14 torr b) 14.3 torr c) 23.9 torr d) 31.8 torr e) 128 torr

5 Hour Exam II Page No 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 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) 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

6 Hour Exam II Page No Two samples of 1.00 mole of helium gas (considered to be an ideal, monatomic gas) are in separate containers at the same conditions of pressure, volume, and temperature (V i = 2.00 L and P i = 9.00 atm). Both samples undergo changes in conditions and finish with V f = 3.00 L and P f = 6.00 atm. They do so in different pathways as described below: Pathway I Step 1a: Pressure remains constant, final conditions are 9.00 atm, 3.00 L. Step 1b: Volume remains constant; final conditions are 6.00 atm, 3.00 L. Pathway II Step 2a: Volume remains constant; final conditions are 6.00 atm, 2.00 L. Step 2b: Pressure remains constant, final conditions are 6.00 atm, 3.00 L. 21. Which of the following is true and best explains the value of ΔH for Pathway I overall? a) ΔH = 0 for Pathway I because ΔH is a state function. b) ΔH = 0 for Pathway I because the process is isothermal. c) ΔH < 0 because heat is released into the surroundings. d) ΔH > 0 because heat is taken in from the surrounding. e) ΔH cannot be determined because both pressure and volume change. 22. For which step is the magnitude of work (that is, w ) the greatest? a) Pathway I step 1a b) Pathway I step 1b c) Pathway II step 2a d) Pathway II step 2b e) The magnitude of work is equally large in at least two of the steps. 23. True or false? In step 1a of Pathway I, w = 0. a) True. This is because the process is isothermal. b) True. This is because it is working against a constant pressure so ΔP = 0. c) True. This is because w = PΔV and ΔV=0. d) False. In step 1a of Pathway I w > 0. e) False. In step 1a of Pathway I w < For which method is the magnitude of heat (that is, q ) the greatest? a) Pathway I b) Pathway II c) The magnitude of q is equal in each pathway (though opposite in sign) because q is a state function. d) The magnitude of q is equal in each pathway (though opposite in sign) and q 0. This is true even though q is not a state function, because the process is isothermal. e) For both pathways q = 0, because the process is isothermal. 25. Determine the value of ΔE for Pathway I step 1b. a) 1.37 kj b) 1.37 kj c) 2.29 kj d) 2.29 kj e)

7 Hour Exam II Page No 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) 26. ΔG for a reaction vs. number of moles of product formed a 27. w vs. the number of steps in the isothermal compression of an ideal gas d 28. w vs. q for the isothermal expansion of an ideal gas a 29. H vs. V at constant temperature for 1.0 mole of an ideal gas c 30. G vs. S for 1.0 mole of an ideal gas at constant T e

8 Hour Exam II Page No We discussed in lecture how, for small temperature changes, we can assume the values of ΔH and ΔS are independent of temperature. Also, when determining boiling or melting points, we can often assume this because the changes in H and S somewhat cancel out. Let s test these assumptions. a. Given the following data (at standard conditions of 25 C and 1 atm), and assuming H and S are independent of temperature, calculate the normal (at 1 atm) boiling point for bromine. Show all work and briefly explain what you are doing and why. [6 pts.] Br 2 (l) Br 2 (g) H f (kj/mol) S (J/mol K) At the boiling point the system between liquid and vapor is at equilibrium, so G = 0. Since G = H T S, H T S = 0, thus T = H/ S. Since we are assuming H and S are independent of temperature we can use T = H / S. T = [31,000 J - 0 J] [245 J/K -152 J/K] = 333 K b. Now that you have determined the boiling point of bromine, calculate H for the vaporization of Br 2 (l) at this temperature and 1 atm. The heat capacity for Br 2 (l) = J/molK and the heat capacity for Br 2 (g) = J/Kmol. Assume these heat capacities are independent of temperature. Show all work. In addition, briefly justify the relative values of ΔH vaporization at the different temperatures (that is, which is larger and why?). [6 pts.] ΔH at 298K = 31.0 kj (from above) which is for Br 2 (l) Br 2 (g) ΔH for Br 2 (l) at 333K to Br 2 (l) at 298K = (1 mole)(75.69 J/molK)(298K 333K) = 2.65 kj ΔH for Br 2 (g) at 298K to Br 2 (l) at 333K = (1 mole)(36.02 J/molK)(333K 298K) = 1.26 kj So, ΔH at 333K for Br 2 (l) Br 2 (g) = 31.0 kj + ( 2.65 kj) kj = 29.6 kj Thus, ΔH at 333K is less than ΔH at 298K which makes sense because at the higher temperature, less energy is required to vaporize the liquid since the liquid is already at a higher temperature.

9 Hour Exam II Page No (con t) c. Now that you have determined the boiling point of bromine, calculate S for the vaporization of Br 2 (l) at this temperature and 1 atm. The heat capacity for Br 2 (l) = J/molK and the heat capacity for Br 2 (g) = J/Kmol. Assume these heat capacities are independent of temperature. Show all work. [4 pts.] ΔS at 298K = 93.0 J/K (from part a) which is for Br 2 (l) Br 2 (g) ΔS for Br 2 (l) at 333K to Br 2 (l) at 298K = (1 mole)(75.69 J/molK)(ln{298K/333K}) = 8.4 J/K ΔS for Br 2 (g) at 298K to Br 2 (l) at 333K = (1 mole)(36.02 J/molK)(ln{333K/298K}) = 4.0 J/K So, ΔS at 333K for Br 2 (l) Br 2 (g) = 93.0 J/K + ( 8.4 J/K) J/K = 88.6 J/K d. Determine the normal boiling point for bromine from the H and S values calculated in parts b and c. Show all work. [2 pts.] T = [ H] [ S] = [29,600 J ] [88.6 J/K] = 334 K e. Briefly comment on our assumptions. Are they reasonable for the vaporization of bromine? Explain. [2 pts.] The values for ΔH and ΔS did change slightly but not dramatically. The calculated boiling temperature was within a rounding error. Yes, the assumptions are reasonable. [4-5% error for ΔH, ~5% error for ΔS, ~0.3% error for T]

10 Hour Exam II Page No On the second day of lecture (back in August) I mentioned that we will be studying thermodynamics this semester and that the knowledge and understanding we gain will provide information on why processes happen and will allow us to make predictions about whether a process will occur. Now is the time for you to show what you know about this for two scenarios that are described on the following pages. In the first portion you are to discuss the scenario in words (without equations). Make sure to address the following: Discuss what happens during the process. o For example, if one of the scenarios was You add a drop of red food coloring to water, you should respond with The food coloring will spread in the water until the solution is equally red throughout. As long as the food coloring and water began at the same temperature, the temperature will not change. List which factor(s) make the process more thermodynamically favorable and discuss why each particular factor is thermodynamically favorable. List which factor(s) (if any) make the process less thermodynamically favorable and discuss why each particular factor is thermodynamically unfavorable. If there is at least one thermodynamically unfavorable factor, explain how the process you have described can be spontaneous. Make sure to use the terms ΔS, ΔS surr, and ΔS univ in your discussion. Clearly specify the system and the surroundings. In the second portion you are to use the appropriate data and equations to calculate ΔS, ΔS surr, ΔS univ and explain how these support the fact that the process you have described in the first portion is spontaneous. NOTE: You will be provided with all of the data you need, but you may not need all of the data you are provided. LIMIT EACH ANSWER TO THE SPACE PROVIDED. THINK ABOUT WHAT YOU WILL WRITE BEFORE YOU START WRITING.

11 Hour Exam II Page No Address both of these scenarios. Please see page 9 for what is required. a. You take liquid water from your freezer that is set to 0. C and place the water outside on a winter day in which the temperature is 15 C. i. What happens and why? Please address all bullet points listed on page 9. [10 pts.] The system is the water and the surroundings are the rest of outside. The water will freeze (turn to ice) at 0. C and the ice will eventually reach a temperature of 15 C. This is thermodynamically favorable because: 1) the process of freezing is exothermic, which will increase ΔS surr, and 2) thermal equilibrium increases ΔS univ (heat is released as the temperature of the ice decreases). This is thermodynamically unfavorable because 1) when a liquid turns into a solid, ΔS decreases since a solid is more ordered than a liquid and 2) when the ice cools its entropy (ΔS) decreases as well. Overall ΔS univ must be positive for the process to occur so the ΔS surr must be greater in magnitude overall than that of ΔS. This makes sense because the process is exothermic and at low temperatures exothermicity is thus a more important driving force (the temperature is below the freezing point of water). ii. The magnitude for the standard (25 C and 1 atm) change in enthalpy of fusion water is 6.03 kj/mol and the freezing point of water is 0. C, 1 atm. The heat capacity of ice is 37.5 J/Kmol, and the heat capacity of liquid water is 75.3 J/Kmol. Assume ΔH, ΔS, and the heat capacities are independent of temperature. Determine ΔS, ΔS surr, and ΔS univ for the process that occurs when you place 1.00 mole of 0. C water outside when the temperature is 15 C and explain how they support what you explained above. [10 pts.] Freezing of water at 0. C: ΔS = ( 6030J/273K) = 22.1 J/K ΔS surr = ( 6030J/258K) = 23.4 J/K Cooling of ice: heat released = (1 mole)(37.5 J/Kmol)( 15K) = J = q ΔS = (1 mole)(37.5 J/Kmol)(ln{258/273}) = 2.12 J/K ΔS surr = ( 562.5J/258K) = 2.18 J/K Overall: ΔS = J/K ΔS surr = J/K ΔS univ = 1.36 J/K; since ΔS univ is greater than zero, the process is spontaneous. [Note: both processes, freezing and cooling, are spontaneous]

12 Hour Exam II Page No (con t). c. Consider the two-bulb system as shown below. You have a sample of N 2 gas at 75.0 C in the left bulb and you have a sample of He gas at 25.0 C in the right bulb. You open the stopcock between the two bulbs. Assume this system is perfectly insulated. i. What happens and why? Please address all bullet points listed on page 9. [10 pts.] The system is the gases in the two-bulb container. The surroundings are outside of the container. The gases will mix and spread throughout the container so that they are evenly distributed (in terms of pressure) and will eventually reach the same temperature. This is thermodynamically favorable because: 1) the entropy of each gas will increase by taking up a larger volume (more microstates), 2) the temperature of the helium will increase, thus increasing its entropy. [could also mention thermal equilibrium] This is thermodynamically unfavorable because 1) the temperature of the nitrogen will decrease, thus decreasing its entropy. Overall ΔS univ must be positive for the process to occur and since ΔS surr = 0 (perfectly insulated) ΔS must be positive (the mixing and increase in temperature of the helium outweighs the decrease in temperature of the nitrogen). ii. In the left bulb (3.00 L) you have 1.00 mole N 2 gas at 75.0 C and in the right bulb (1.00 L) you have 3.00 moles He gas at 25 C. For N 2, C v = J/Kmol and C p = J/Kmol; for He: C v = J/Kmol and C p = J/Kmol. Assume the heat capacities are independent of temperature. Determine ΔS, ΔS surr, and ΔS univ for the process that occurs when you open the stopcock and explain how they support what you explained above. [10 pts.] Solve for T f : (1.00 mol)(20.71 J/Kmol)(75 T f ) = (3.00 mol)(12.47 J/Kmol)(T f 25) T f = 42.8 C = 315.8K Nitrogen gas: Decreasing temperature: ΔS = (1 mole)(20.71 J/Kmol)(ln{315.8K/348K})= 2.01 J/K Spreading out in the container: ΔS = (1 mole)(8.314 J/Kmol)(ln{4.00L/3.00L})= 2.39 J/K Helium gas: Increasing temperature: ΔS = (3 mole)(12.47 J/Kmol)(ln{315.8K/298K})= 2.17 J/K Spreading out in the container: ΔS = (3 mole)(8.314 J/Kmol)(ln{4.00L/1.00L})= 34.6 J/K Overall: ΔS = J/K ΔS surr = 0 ΔS univ = J/K ; since ΔS univ is greater than zero, the process is spontaneous.