GENERAL CHEMISTRY TOPICAL: Thermodynamics and Thermochemistry Test Time: 22 Minutes* Number of Questions: 17 * The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Rf (261) 105 Ha (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Thermodynamics and Thermochemistry Test Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a negligible effect on the temperature The specific heat of pewter is 0.17 J/g•K and that of water is 4.184 J/g•K The entire procedure is done under a pressure of 1.0 atm temperature C time B D time After the tea is added to the thermos, the temperature of the liquid quickly falls from 80°C to 76.9°C as it reaches thermal equilibrium with the thermos flask What is the heat capacity of the thermos? A 9.9 J/K B 14 J/K C 58 J/K D 878 J/K An alternative method for keeping the tea hot would be to place the teapot on a 10 pound block that has been heated in an oven to 300°C A block of which of the following substances would best be able to keep the tea hot? A B C D copper (specific heat = 0.39 J/g•K) granite (specific heat = 0.79 J/g•K) iron (specific heat = 0.45 J/g•K) pewter (specific heat = 0.17 J/g•K) time temperature A 500 g pewter teapot and an insulated thermos are in a 21°C room The teapot is filled with 1000 g of the boiling water 12 tea bags are then placed into the teapot The brewed tea is allowed to cool to 80°C, and then 250 g of the tea is poured from the teapot into the thermos The teapot is then kept on an insulated warmer that transfers 500 cal/min to the tea A temperature A 1008 g sample of water is heated to boiling During the heating, g of water is lost from vaporization Most of the water is lost near the boiling point (The heat of fusion of water is 80 cal/g, and the heat of vaporization of water is 540 cal/g There are 4.184 J in one calorie.) Which of the following plots best represents the cooling curve for the contents of the teapot if it is placed outside at –23°C? temperature Passage I (Questions 1–8) time If, after some of the tea has been transferred to the thermos (as described in the passage), the teapot with its contents (at a temperature of 80°C) was placed on the insulated warmer for minutes, what would be the temperature at the end of this 5-minute period? (Assume that no significant heat transfer occurs with the surroundings.) A B C D 80.7°C 82.5°C 83.2°C 95.2°C What is the relationship between the temperature of the tea and its heat content? A Heat and temperature are the same; they just use different numerical scales B Heat is the energy transferred between the two species as a result of differences in their temperatures C Heat and temperature are not related D Heat is a measure of the stability of the molecules of the species, and temperature is a measure of the average potential energy of the molecules GO ON TO THE NEXT PAGE KAPLAN MCAT The heat that is transferred from the boiling water to the pewter is referred to as: A B C D entropy specific heat enthalpy heat capacity Questions through 12 are NOT based on a descriptive passage What is the amount of heat given off by 100 g of oxygen gas when it is used to burn an excess of sulfur according to the following reaction? S(s) + O2(g)→ SO2(g) Which of the following systems has the lowest entropy? A B C D Water at 80°C Water at 100°C Steam at 100°C Pewter at 80°C How much energy was required to evaporate the g of water during boiling? A 4320 kcal B 4320 kJ C 640 cal D 18.1 kJ ∆H = –296 kJ/mole A 925,000 J B 29,600 J C 1850 J D 925 J An insulated tube with a movable piston at one end had 500 J of heat added to it If, during the experiment, the piston moves and does 75 J of work on the atmosphere, what is the change in the energy of the tube system? A B C D 575 500 425 –75 J J J J 1 Which of the following is NOT a state function? A B C D Temperature Density Work Entropy Which of the following reactions shows a decrease in entropy? A B C D C(s) + 2H2(g) → CH4(g) N2O(g) → N2(g) + 1/2 O2(g) 2NI3(s) → N2(g) + 3I2(g) 2O3(g) → 3O2(g) GO ON TO THE NEXT PAGE as developed by Thermodynamics and Thermochemistry Test Passage II (Questions 13–17) A chemist was interested in producing acetic acid by combining inexpensive starting materials She decided that the least expensive compounds available were carbon dioxide and water These compounds normally react to produce carbonic acid according to the following reaction: H2O(l) + CO2(g) → H2CO3(aq) ∆G° = 8.5 kJ/mol Reaction She reasoned that if the correct catalyst could be found, the following reaction might occur, and acetic acid could be produced at very low cost What is the value of ∆G°f for H2O(l)? A –237.5 kJ B –286.5 kJ C –229 kJ D kJ Would the decomposition of H2O(g) into its elements proceed spontaneously under standard conditions? A B C D Yes, because the entropy increases Yes, because H2O does not exist as a gas at 25°C No, because the reaction is endothermic No, because ∆G° for the reaction is positive 2CO2(g) + 2H2O(l) → CH3COOH(l) + 2O2(g) ∆H° = 873 kJ/mol ∆S° = J/mol•K Reaction Table shows the enthalpies and free energies of formation for several compounds Substance CO2(g) H2O(g) CH3COOH(l) H2CO3(aq) Table ∆H°f (kJ/mol) –394 –242 –484 –700 What condition must be satisfied for Reaction to occur spontaneously? (Note: Assume that ∆H and ∆ S are independent of temperature.) A B C D The temperature is sufficiently high The correct catalyst is found The temperature is sufficiently low It will never occur spontaneously because ∆G° is positive ∆G°f (kJ/mol) –394 –229 –389 –623 Under standard conditions, what is the entropy change for the formation of acetic acid from its elements? A –389 kJ/mol•K B –0.319 kJ/mol•K C 319 kJ/mol•K D 484 kJ/mol•K What is the value of the thermodynamic equilibrium constant for Reaction at 298 K? (Note: ln x = 2.3 log 10 x The universal gas constant, R, is 8.3145 J/mol•K.) A B C D × 10–17 × 10–2 25 KAPLAN END OF TEST MCAT ANSWER KEY: C C B D D D C A B 10 C 11 12 13 14 15 C A B B A 16 D 17 A as developed by Thermodynamics and Thermochemistry Test EXPLANATIONS Passage I C Given the mass of the tea poured into the thermos, and its temperature change, one can determine the heat lost by the tea from q = mc∆T: qtea = (250 g)(4.184 J/g•K)(76.9°C – 80°C) ≈ (250 × × –3) J = –3000 J Note that since the “size” of one degree Celsius and one Kelvin are the same, the two can be used interchangeably in cases like the above, where we deal only with changes in temperature Don’t waste time converting here! This heat that is lost by the tea is gained by the thermos: qthermos = –qtea ≈ +3000 J The thermos had an initial temperature of 21°C, and must have a final temperature equal to that of the tea inside, since it is stated in the question stem that the two reach thermal equilibrium The heat capacity of the thermos can therefore be determined: 3000 J = (mc)(76.9°C – 21°C) ≈ (mc)56°C = (mc)(56 K) 3000 J heat capacity = mc ≈ = 60 J/K 50 K Choice C is the only value that is close enough to this result that we obtained with quite a few approximations Wrong answer choices on this problem are likely to result from using wrong mass values for the tea (1000 g in the teapot or the 1008 g present initially), from inconsistency with energy units (confusing calories with joules), or from an inability to determine the correct initial and/or final temperature of the thermos from the information given in the passage B To keep the tea hot, it needs to be placed on an object that can transfer large amounts of heat to it as they reach thermal equilibrium The substance with the highest specific heat (granite) is the one that can transfer the most heat to the tea and teapot per degree drop in its own temperature D The tea inside the teapot is going to cool from 80°C to –23°C if placed outside under these conditions Initially, the temperature of the tea decreases Then , when the tea reaches its freezing point, its temperature remains constant as it changes from the liquid into the solid phase Once the entire sample is frozen, the temperature of the solid tea will decrease until it reaches thermal equilibrium with its surroundings The cooling curve will therefore show a segment with negative slope first, then a horizontal region, and finally another segment with negative slope (If we follow the process for a long enough period of time, we will see the curve approach –23°C as the final equilibrium value.) Since the question asks for the “cooling” curve, choice C, which shows an increase in temperature, can be eliminated immediately Choice A is incorrect since it indicates that the temperature will initially remain constant, even though exposed to a surrounding with a very different temperature Choice B fails to show the constant temperature region over which freezing takes place, and so is also incorrect C The insulated warmer transfers heat to the teapot and its contents for minutes at a rate of 500 cal/min The total heat transferred is: 500 cal/min × = 2500 cal The heat absorbed by both the teapot and the tea is equal to the heat given off by the warmer The tea and the teapot both have an initial temperature of 80°C, and will have a common final temperature They will thus have the same value for ∆T Since the specific heats for the tea and the teapot are expressed in J/g•K, the calorie units must be converted to joules: mpot cpot∆T + mteactea∆T = 2500 cal × 4.184 J/cal ≈ 2500 cal × 4.2 J/cal = 10300 J [(500 g)(0.17 J/g•K) + (750 g)(4.184 J/g•K)] ∆T = 10300 J (85 + 3150) ∆T = 10300 10300 ∆T = 3235 KAPLAN MCAT This is between three and four and so added to the initial temperature of 80°C, we arrive at choice C as the only possible answer Notice how since the values of the answer choices are closer apart than in question 1, we have been more careful in making approximations, using a value of 4.2 rather than for 4.184 B Heat is defined as a transfer of thermal energy as a result of differences in temperature Choice D may be incorrectly chosen if not read carefully: Temperature is a measure of the average kinetic energy of the molecules, not potential energy as stated in D C For any system at constant pressure, the heat lost or gained is equal to the enthalpy, so choice C is the correct answer Even if you had been unaware of this fact, you should know that out of the answer choices, only this one has the same unit as heat or energy, and so it has to be choice C Entropy, choice A, is a measure of the disorder of a system Specific heat is defined as the amount of heat needed to raise the temperature of one unit mass of a substance one degree centigrade Heat capacity is defined as the amount of heat needed to raise the temperature of a substance one degree centigrade D Entropy is a measure of disorder, or randomness, of a system The more disordered a system, the greater its entropy Gases have the highest entropy, then liquids, then solids—solids being the most ordered of the three phases Pewter at 80°C is a solid, so it has the lowest entropy Of the answer choices, choice C, steam, has the highest entropy since it is a gas D The passage states that the heat of vaporization of water is 540 calories per gram of water So, 540 cal/g × g = 4,320 cal (NOT kilocalories) Since the answer choice is not here in calories, it must be in joules 4,320 cal × 4.184 J/cal ≈ 18,000 J or 18 kJ Choice D is therefore the correct answer Make sure you know what units you are working with If you didn’t keep track of this, you could have chosen choice A or B Independent Questions A This is a simple stoichiometry problem The first thing that you need to is convert 100 grams of oxygen into moles of oxygen Using your periodic table, one mole of oxygen gas weighs 32 grams, so we have just over three moles of oxygen gas From the stoichiometry of the reaction, one mole of oxygen gives off 296 kilojoules, or 296,000 joules So, three moles will give off three times this amount: just about 900,000 joules Since we underestimated the number of moles of oxygen, choice A is the correct answer 10 C In order to answer this question correctly, you need to know the first law of thermodynamics: the change in internal energy of a system is equal to the heat added to the system plus the work done on the system (or MINUS the work done by the system) Mathematically, you may see it either as ∆E = q + w, or as ∆E = q – w The difference between the two is again how we define w in each case: as work done on the system in the former, and as work done by the system as the latter There should be no confusion as long as we always keep in mind that a system that does work on its surroundings will expend some energy, leading to a decrease in energy, while doing work on a system will add energy to it In this case, we add 500 J of heat to a system The internal energy is therefore expected to increase by 500 J Subsequently, the system does 75 J of work on its surroundings, and so we expect the energy to decrease by 75 J The net effect on the energy is thus an increase of (500 – 75) = 425 J 11 C A state function is ones whose value depends only on the initial and final states of the system; it is independent of the path taken in going from the initial to the final states Examples of state functions are pressure, volume, temperature, density, enthalpy, internal energy, entropy, and free energy Two important function that are not state functions are work and q, the amount of heat transferred 12 A An increase or decrease of entropy is easy to predict when the chemical reactions involve gases If there are more moles of gas on the product side than on the reactant side, there is an increase in entropy If there are fewer moles of gas on the product side than on the reactant side, there is a decrease in entropy For choice A, there are two moles of gas on the reactant side and one mole of gas on the product side—there is, therefore, a decrease in entropy for this reaction Choice A is correct For choice B, there is one mole of gas on the reactant side and one and a half moles on the product side—there is a corresponding increase in entropy for this reaction For choice C, there is no gas on the reactant side and moles of gas on the product side—there is as developed by Thermodynamics and Thermochemistry Test an increase in entropy for this reaction as well For choice D, there are two moles of gas on the reactant side and three moles of gas on the product side—there is an increase in entropy of this system as well Passage II 13 B For this question, you need to employ the relationship ∆G = ∆H – T∆S In this case, we are applying it to a formation reaction under standard conditions, hence ∆G°f = ∆H°f – T∆S°f From Table 1, you can see that the third reaction down is the reaction that we’re looking for: the formation of acetic acid from its elements –389 kJ/mol = –484 kJ/mol – (298 K)∆S°f –389 + 484 95 ∆S° f = =– kJ/mol•K –298 298 which is about 1/3 kJ/mol•K Choice B is therefore the correct answer 14 B The value of Keq can be determined using the equation ∆G° = –RTlnKeq ∆G° for reaction is given as 8.5 kJ/mol, or 8500 J/mol Therefore, 8500 8500 ≈ – ≈ –3.3 8.3145 × 298 8.5 × 300 2.3 log10 K eq ≈ –3.3 log10 K eq ≈ –1.5 K eq ≈ 10–1.5 ln K eq = – The value of the thermodynamic equilibrium constant, then, has to be between 10–2 (= 0.01), and 10–1 (= 0.1) Choice B is the only possible response This problem can also be approached from a qualitative perspective The value of ∆G is positive, which tells us that the reaction is not favored thermodynamically, so Keq should be less than one At the very least, then, we should have been able to eliminate choices C and D 15 A The value cannot simply be read off from table 1, since it only contains the entry for water vapor, which is different from the value for liquid water asked for Instead, we must apply to reaction the following relationship: ∆G° = free energy of formation of products – free energy of formation of reactants 8.5 kJ = free energy of formation of carbonic acid – (free energy of formation of liquid water + free energy of formation of carbon dioxide) = –623 kJ – (free energy of formation of liquid water – 394 kJ) 8.5 + 623 – 394 = ∆G°f (H2O, l) ∆G°f (H2O, l) = –237.5 kJ Choice C is the incorrect answer one would have gotten if attention were not paid to the phase of H2O Choice D may have been tempting if one assumes that since water is a liquid in its most stable form under standard conditions, it has a free energy of formation of zero This line of reasoning is correct only for elements, not compounds 16 D The reverse of the second reaction listed in Table is the decomposition of water vapor into its elements If the equation is reversed, the signs of ∆G and ∆H change Since for the reverse reaction, ∆G° is positive, decomposition is nonspontaneous 17 A Since the ∆H° and ∆S° values for Reaction are both positive, the reaction will be spontaneous only at high temperatures This is because only at high temperatures would the free energy change, ∆H° – T∆S°, be negative Choice B is incorrect because a catalyst does not effect the spontaneity of a reaction Choice C is incorrect because a reaction that is spontaneous only at low temperatures will have negative ∆H° and ∆S° Choice D is incorrect because even though the free energy change for the reaction is positive at 298 K, its value may (and does) change as the temperature changes KAPLAN ... Thermodynamics and Thermochemistry Test Assume that the specific heat of brewed tea is the same as that of pure water, and that the tea bags have a very small mass compared to that of the water, and a... 10–17 × 10–2 25 KAPLAN END OF TEST MCAT ANSWER KEY: C C B D D D C A B 10 C 11 12 13 14 15 C A B B A 16 D 17 A as developed by Thermodynamics and Thermochemistry Test EXPLANATIONS Passage I C... choice C, there is no gas on the reactant side and moles of gas on the product side—there is as developed by Thermodynamics and Thermochemistry Test an increase in entropy for this reaction as