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Chapter20_1_chm2046 - ll CHAPTER 20 THERMODYNAMICS ENTROPY...

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Unformatted text preview: ll! CHAPTER 20 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND THE 201 20.2 20.3 20.4 20.5 20.6 20.7 20.8 ‘DIRECTION OF CHEMICAL REACTIONS Spontaneous processes proceed without outside intervention. The fact that a process is spontaneous does not mean that it will occur instantaneously or even at an observable rate. The rusting of iron is an example of a process that is spontaneous but very slow. The ignition of gasoline is an example of a process that is not spontaneous but very fast. A spontaneous process occurs by itself (possibly requiring an initial input of energy) whereas a nonspontaneous process requires a continuous supply of energy to make it happen. It is possible to cause a nonspontaneous process to occur, but the process stops once the energy source is removed. A reaction that is found to be nonspontaneous under one set of conditions may be spontaneous under a different set of conditions (different temperature, different concentrations). ' a) The energy of the universe is constant. b) Energy cannot be created or destroyed. C) AEsystem = “ surroundings The first law is concerned with balancing energy for a process but says nothing about whether the process can, in fact, occur. Entropy is related to the freedom of movement of the particles. A system with greater freedom of movement has higher entropy. a) and b) Probability is so remote as to be virtually impossible. Both would require the simultaneous, coordinated movement of a large number of independent particles, so are very unlikely. Vaporization is the change of a liquid substance to a gas so ASvapofizafiDn = Sgas —— 5‘“un Fusion is the change of a solid substance into a liquid so ASfusion = Suquid - Ssolid- Vaporization involves a greater change in volume than fusion. Thus, the transition from liquid to gas involves a greater entropy change than the transition from solid to liquid. In an exothermic process, the system releases heat to its surroundings. The entropy of the surroundings increases because the temperature of the surroundings increases (ASsm > 0). In an endothermic process, the system absorbs heat from the surroundings and the surroundings become cooler. Thus, the entropy of the surroundings decreases (ASsm< 0). A chemical cold pack for injuries is an example of a spontaneous, endothermic chemical reaction as is the melting of ice cream at room temperature. a) According to the Third Law the entropy is zero. b) Entropy will increase with temperature. c) The third law states that the entropy of a pure, perfectly crystalline element or compound may be taken as zero at zero Kelvin. Since the standard state temperature is 25°C and entropy increases with temperature, 5" must be greater than zero for an element in its standard state. d) Since entropy values have a reference point (0 entropy at 0 K),'actual entropy values can be determined, not just entropy changes. a) Spontaneous, evaporation occurs because a few of the liquid molecules have enough energy to break away from the intermolecular forces of the other liquid molecules and move spontaneously into the gas phase. b) Spontaneous, a lion spontaneously chases an antelope Without added force. This assumes that the lion has not just eaten. ~ - . c) Spontaneous, an unstable substance decays spontaneously to a more stable substance. 20- l 20.9 20.10 20.11 20.12 20.13 20.14 20.15 20.16 20.17 20.18 20.19 a) Spontaneous, with a small amount of energy input, methane will continue to burn without additional energy (the reaction itself provides the necessary energy) until it is used up. b) Spontaneous, the dissolved sugar molecules have more states they can occupy than the crystalline sugar, so the reaction proceeds in the direction of dissolution. 0) Not spontaneous, a cooked egg will not become raw again, no matter how long it sits or how many times it is mixed. a) AS,ys positive, melting is the change in state from solid to liquid, The solid state of a particular substance always has lower entropy than the same substance in the liquid state. Entropy increases during melting. b) AS”, negative, the entropy of most salt solutions is greater than the entropy of the solvent and solute separately, so entropy decreases as a salt precipitates. c) AS,” negative, dew forms by the condensation of water vapor to liquid. Entropy of a substance in the gaseous state is greater than its entropy in the liquid state. Entropy decreases during condensation. a) AS,,, positive, the process described is liquid alcohol becoming gaseous alcohol. The gas molecules have greater entropy than the liquid molecules. b) ASSyS positive, the process described is a change from solid to gas, an increase in possible energy states for the system. 0) AS”, positive, the perfume molecules have more possible locations in the larger volume of the room than inside the bottle. A system that has more possible arrangements has greater entropy. a) AS,ys negative, reaction involves a gaseous reactant and no gaseous products, so entropy decreases. The number of particles also decreases, indicating a decrease in entropy. b) AS,,, negative, gaseous reactants form solid product and number of particles decreases, so entropy decreases. c) AS,ys positive, when salts dissolve in water, entropy generally increases. a) ASSy, negative b) AS,“ negative 0) AS,ys negative a) 1353,, positive, the reaction produces gaseous C02 molecules that have greater entropy than the physical states of the reactants. b) ASsys negative, the reaction produces a net decrease in the number of gaseous molecules, so the system’s entropy decreases. c) AS,ys positive, the reaction produces a gas from a solid. a) AS,ys negative b) AS,in positive c) AS595 negative a) ASSys positive, decreasing the pressure increases the volume available to the gas molecules so entropy of the system increases. b) ASSys negative, gaseous nitrogen molecules have greater entropy (more possible states) than dissolved nitrogen molecules. 0) AS,” positive, dissolved oxygen molecules have lower entropy than gaseous oxygen molecules. a) AS,” negative b) ASSy, positive c) AS,yS negative a) Butane has the greater molar entropy because it has two additional C—H bonds that can vibrate and has greater rotational freedom around its bond. The presence of the double bond in 2~butene restricts rotation. b) Xe(g) has the greater molar entropy because entropy increases with atomic size. c) CH4(g) has the greater molar entropy because gases in general have greater entropy than liquids. a) Ethanol, C2H50H(l), is a more complex molecule than methanol, CH30H, and has the greater molar entropy. b) A salt dissolves, giving an increase in the number of possible states for the ions increases. Thus, KClO‘3(aq) has the greater molar entropy. . ' c) K(s) has greater molar entropy because K(s) has greater mass than Na(s). 20-2 20.20 20.21 20.22 20.23 20.24 20.25 20.26 20.27 20.28 a) Diamond < graphite < charcoal. Diamond has an ordered, 3~dimensional crystalline shape, followed by graphite with an ordered 2—dimensional structure, followed by the amorphous (disordered) structure of charcoal. b) Ice < liquid water < water vapor. Entropy increases as a substance changes from solid to liquid to gas. c) O atoms < 02 < 03. Entropy increases with molecular complexity because there are more modes of movement (e. g., bond vibration) available to the complex molecules. 3) Ribose < glucose < sucrose; entropy increases with molecular complexity. b) Ca,CO3(s) < (Ca0(s) + C02(g)) < (Ca(s) + C(s) + 3/2 02(g)); entropy increases with moles of gas particles. c) SF4(g) < Sng) < SZF10(g); entropy increases with molecular complexity. a) ClO4'(aq) > ClO3'(aq) > ClOflaq). The decreasing order of molar entropy follows the order of decreasing molecular complexity. b) N02(g) > N0(g) > N2(g). N2 has lower molar entropy than NO because N2 consists of two of the same atoms while NO consists of two different atoms. N02 has greater molar entropy than NO because N02 consists of three atoms while NO consists of only two. ‘ c) Fe304(s) > Fe203(s) > A1203(s). Fe304 has greater molar entropy than Fe203 because Fe304 is more complex and more massive. Fe203 and A1203 contain the same number of atoms but Fe203 has greater molar entropy because iron atoms are more massive than aluminum atoms. a) Ba(s) > Ca(s) > Mg(s); entropy decreases with lower mass. b) C6131” > C6111; > C6116; entropy decreases with lower molecular complexity and lower molecular flexibility. c) PF2C13(g) > PF5(g) > PF3(g); entropy decreases with lower molecular complexity. 3) X2(g) + 3 Y2(g) —+ 2 XY3(g) b) AS < 0 since there are fewer moles of gas in the products than in the reactants. c) XY3 is the most complex molecule and thus will have the highest molar entropy. Entropy, like enthalpy, is a state function, which is a property of a system determined by the state of the system and not in the process by which it achieved that state. A system at equilibrium does not spontaneously produce more products or more reactants. For either reaction direction, the entropy change of the system is exactly offset by the entropy change of the surroundings. Therefore, for system at equilibrium, ASuniv = ASSyS + ASsulT = 0. However, for a system moving to equilibrium, ASlmiv > 0, because the Second Law states that for any spontaneous process, the entropy of the 'universe increases. Since entropy is a State function, the entropy changes can be found by summing the entropies of the products and subtracting the sum of the entropies of the reactants. For the given reaction 11ng“ = 2 S°(HCIO-(g)) — (S°H20(g)) + S °(C120(g)). Rearranging this expression to solve for S°(C120(g)) gives S°(C120(g)) = 2 S°(HCIO(g)) —- S°(H20(g)) —AS;xn a) Prediction: AS" negative because number of moles of (An) gas decreases. AS" = [(1 m01N20(g)) (S°(N20)) + (1 m01N02(g)) (S°(N02))l . — [(3 H101 N0(g)) (S°(NO))] AS° = [(1 mol N20(g)) (219.7 J/mol-K) + (1 mol N02(g)) (239.9 J/mol-K)] — [(3 mol NO(g)) (210.65 J/mol-K)] AS0 = —172.35 = ~172.4 J/K b) Prediction: Sign difficult to predict because An = 0, but possibly AS" positive because water vapor has greater complexity than H2 gas. AS" = [(2 11101 Fe(S)) (S°(F6)) + (3 mol 1120(8)) (3°(H20))l ~ [(3 H101 H2(g)) (S°(H2)) + (1 111011362030» (S°(F6203))] AS" = [(2 mol Fe(s)) (27.3 J/mol-K) + (3 mol H20(g)) (188.72 J/mol-K)] — [(3 mol H2(g)) (130.6 J/mol-K) + (1 mol Fe203(s)) (87.400‘J/mol-K)] AS° = 141.56 = 141.6 J/K ‘ 20-3 20.29 20.30 20.31 20.32 20.33 0) Prediction: AS535 negative because a gaseous reactant forms a solid product and also because the number of moles of gas (An) decreases. 135° = [(1 111011340106» (5°(P40101)1 — [(1 H101 P4(S)) (3°(P411 + (5 H101 02(3)) (5°(02))] A30 = [(1 11101 P4010(S)) (229 J/mol-K)] -— [(1 mol P4(S)) (41.1 J/mOI'K) + (5 mol 02(g)) (205.0 J/mol'K)] AS° = ~837.1 = ~837 J/K a) 3 N02(g) + H200) —9 2 1111030) + NO(g) AS negative AS;xn = [2 mol I-lNO3(l) (155.6 J/K-mol) + 1 mol NO(g) (210.65 J/K-mol)] — [3 mol N02(g) (239.9 J/K-rnol) + 1 mol H200) (69.940 J/K-rnoD] = ~267.79 = ~267.8 J/K b) N2(g) + 3 F2(g) a 2 NF3(g) AS negative 11ng = [2 mol NF3(g) (260.6 J/K'mol)] — [1 mol N2(g) (191.5 J/K-mol) + 3 mol F2(g) (202.7 J/K-mol)] = —278.4 J/K c) (251112043 + 6 02(g) —> 6 C02(g) + 6 H20(g) AS positive ASE.n = [6 mol C02(g) (213.7 J/K-mol) + 6 mol H20(g) (188.72 J/K'mol)] —- [1 mol C6H1206(s) (212.1 J/K-mol) + 6 mol 02(g) (205.0 J/K-mol)] = 972.42 = 972.4 J/K The balanced combustion reaction is - 2 C2H6(g) + 7 02(8) —> 4 C0209) + 5 H2003) AS;xn = [4 mol (S°(C02)) + 6 mol (S°(H20))] — [2 mol (S°(C2H6) + 7 mol (S°(C02)] = [4 mol (213.7 J/mol-K) + 6 mol (188.72 J/mol-K)] — [2 11101 (229.5 J/rnol-K) + 7 mol (205.0 J/mol-K)] = 93.12 = 93.1 J/K The entropy value is not per mole of C2H5 but per 2 moles. Divide the calculated value by 2 to obtain entropy per mole of C2115. Yes, the positive sign of AS is expected because there is a net increase in the number of gas molecules from 9 moles as reactants to 10 moles as products. The balanced chemical equation for the described reaction is: 2 N0(g) + 5 H2(g) -> 2 NH3(g) + 2 1120(8) Because the number of moles of gas decreases, i.e., An = 4 - 7 = —3, the entropy is expected to decrease. AS" = {(2 mol NHg) (193 J/rnol-K) + (2 mol H20) (188.72 J/mol-K)} —- {(2 mol NO) (210.65 J/mol-K) + (5 mol Hg) (130.6 J/mol-K)} AS° = ~310.86 = ~311 J/K Yes, the calculated entropy matches the predicted decrease. The reaction for forming CuZO from copper metal and oxygen gas is 2 Cu(s) + 1/2 02(g) —9 Cu20(s) 11ng = [1 mol (S°(Cu20))] -— [2 mol (S°(Cu)) + 1/2 mol (S°(Oz))] = [1 mol (93.1 J/mol-K)] —— [2 mol (33.1 J/mol-K) + 1/2 mol (205.0 J/mol-K)] = —75.6 J/K One mole of methanol is formed from its elements in their standard states according to the following equation: C(g) + 2 H2(g) + 1/2 02(g) -> CH30H(Z) AS" = [(S°(CH30H))1 - [S°(C(grflphit€)) + 2 (S°(H2)) + 1/2 (5°(02)l AS" = [(1 mol CH30H) (127 J/mol-K)] — [(1 mol C) (5.686 J/mol-K) + (2 mol H2) (130.6 J/mol-K) + (1/2 mol 02) (205.0 J/mol-K)] AS" = ~242.386 = —242 J/K 20~4 [H 20.34 20.35 20.36 20.37 20.38 20.39 20.40 502(g) + Ca(OH)2(s) —> Ca803(s) + H200) AS;xn = [1 mol CaS03(s) (101.4 J/K-Inol) + 1 mol H200) (69.940 J/K-mol)] . — [1 mol 802(g) (248.1 J/K'mol) + 1 mol Ca(OH)2(s) (83.39 J/K-mol)] = -160.15 = ~160.2 J/K Complete combustion of a hydrocarbon includes oxygen as a reactant and carbon dioxide and water as the products. C2H2(8) + 5/ 2 (Mg) —> 2 (302(8) + H20(g) AS;xn = [2 mol (S°(C02)) + 1 mol (S°(H20))] —— [1 mol (S°(C2H2)) + 5/2 mol (S°(02))] = [2 mol (213.7 J/mol-K) + 1 mol (188.72 J/mol'K)] - [1 mol (200.85 J/mol-K) + 5/2 mol (205.0 J/mol-K)] = ~97.23 = -—97.2 J/K Reaction spontaneity may now be predicted from the value of only one variable (AGsys) rather than two (ASSys and ASSUR)‘ A spontaneous process has ASH“;V > 0. Since the Kelvin temperature is always positive, AGsys must be negative (AGsys < 0) for a spontaneous process. a) AG: AH— TAS. Since TAS > AH for an endothermic reaction to be spontaneous the reaction is more likely to be spontaneous at higher temperatures. b) The change depicted is the phase change of a solid converting to a gas (sublimation). 1. Energy must be absorbed to overcome intermolecular forces to convert a substance in the solid phase to the gas phase. This is an endothermic process and AH is positive. 2. Since gases have higher entropy values than solids, the process results in an increase in entropy and AS is positive. ‘ 3. This is an endothermic process so the surroundings loses energy to the system. ASsurr is negative. . 4. AG = AH — TAS. Both AH and AS are positive. At low temperature, the AH term will predominate and AG will be positive; at high temperatures, the TAS term will predominate and AG will be negative. ' 0 AH m positive and Aszys positive. The reaction is endothermic (AH an > 0) and requires a lot of heat from its surroundings to be spontaneous. The removal of heat from the surroundings results in ASS"urr < 0. The only way an endothermic reaction can proceed spontaneously is if ASgys >> O, effectively offsetting the decrease in the entropy of the surroundings. In summary, the values of AH ;,m and AS:yS are both positive for this reaction. Melting is an example. The AG?“ can be calculated from the individual AG; ’3 of the reactants and products found in Appendix B. AG;m = Z[m AG; (products)] -— 23[n AG; (reactants)] a) AG:Kn = [(2 mol MgO) (AG; Mg0)] — [(2 mol Mg) (AG; Mg) + (1 mol 02) (AG; 02)] Both Mg(s) and 02(g) are the standard state forms of their respective elements, so their AG; ’8 are zero. AG;m = [(2 mol MgO) (—569.0 kJ/mol)] — [(2 mol Mg) (0) + (1 mol 02) (0)] = ~-1138.0 k} b) AG;xn = [(2 mol C02) (AG; C02) + (4 mol H20)( AG; H20)] —— [(2 mol CH30H) (AG; CH3OH) + (3 mol 02) (AG; 02)] AG:7m = [(2 mol C02) (-394.4 kJ/mol) + (4 mol H20) (—228.60 kJ/mol)] ~ [(2 mol CH30H) (—161.9 kJ/mol) + (3 mo] 02) (0)] Aegm = 4379.4 kJ 20~5 20.41 20.42 c) AGfxn = [(1 mol BaC03) (AG; BaC03)] —~ [(1 mol BaO) (AG; BaO) + (1 mol C02) (AG; cog] A016“, == [(1 mol BaCO3) (—1139 kJ/mol)] — [(1 mol BaO) (~520.4 kJ/mol) + (1 mol cog) (494.4 kJ/mol)] Aegim = .2242 = —224 k] a) H2(g) + I2(s) -> 2 HI(g) aagm = [2 mol HI(g) (1.3 kJ/mol)] — [1 m01H2(g) (0 kJ/mol) + 1 mol I2(s) (0 kJ/mol)] = 2.6 kJ b) Mn02(s) + 2 CO(g) ~—-> Mn(s) + 2 C02(g) AGE“, = [1 mol Mn(s) (0 kJ/mol) + 2 mol C02(g) (~394.4 kJ/mol)] -— [1 mol Mn02(s) (~466.1 kJ/mol) + 2 mol CO(g) (~137.2 kJ/mol)] = —-48.3 k] c) NH4C1(S) —~> NH3(g) + HC1(g) AG;m = [1 mol NH3(g) (—16 kJ/mol) + 1 mol HC1(g) (—95.30 kJ/m01)] — 1 mol NH4C1(s) (—203.0 kJ/mol) = 91.7 = 92 k] The AH:x11 can be calculated from the individual AH; ’s of the reactants and products found in Appendix B. AH gm = Z[m AH f (products)] —- Z[n AH 12’ (reactantsfl The Ang can be calculated from the individual S°’s of the reactants and products found in Appendix B. AS;xn = 2[m S°(products)] — Z[n S°(reactants)] AG:xn can be calculated using Angn = AH gm — T AS;xn a) AHI‘TXH = [2 mol ( AH; (MgO))] — [2 mol (AI-1 E (Mg)) + 1 mol (AH }’ (02))] = [2 mol (~601.2 kJ/mol)] — [2 mol (0) + 1 mol (0)] = —1202.4 1d AS;m = [2 mol (S°(MgO))] -— [2 mol (S°(Mg)) + 1 mol S°(Og))] = [2 mol (26.9 J/mol-K)] —- [2 mol (32.69 J/mol-K) + 1 mol (205.0 J/mol-K)] = ~216.58 J/K (unrounded) AGE,“1 =AH;xn — TASS“, = —1202.4 k] — [(298 K) (-216.58 J/K) (1 kJ/IO3 J)] = ~1137.859 = ~1138 M b) AHA“ = [2 mol (AH; (COZD + 4 mol (AH; (H20(g))] — [2 mol (AH;z (CH3OH)) + 3 mol (AH; (02))] = [2 mol (—393.5 kJ/mol) + 4 mol (~241.826 kJ/mol)] ~ [2 mol (~201._2 kJ/mol) + 3 mol (0)] = 4351.904 kl (unrounded) ' » , ‘ - ' AS;xn = [2 mol (S°(COZ)) + 4 mol (S°(H20(g))] —— [2 mol (3°(CH3OH)) + 3 mol (S°(02))] = [2 mol (213.7 J/mol-K) + 4 mol (188.72 J/mol-K)] — [2 mol (238 J/mol-K) + 3 mol (205.0 J/mol-K)] = 91.28 J/K (unrounded) Aagm = 2111an — TASan = 4351904 kJ — [(293 K) (91.23 J/K) (1 11/103 1)] = 4379105 = —1379 RJ c) 21113,, = [1 mol (AH; (BaC03(s))] — [1 mol (AH; (BaO)) + 1 mol (AH; (C0;))] = [1 mol (—1219 kJ/mol)] — [1 mol (~548.1 kJ/mol) + 1 mol (~3935 kJ/mol)] = —277.4 k] (unrounded) Asgm = [1 mol (S°(BaCO3(s))] - [1 mol (S°(Ba0)) + 1 m01(S°(C02))] = [1 11101 (112 J/mol-K)] —— [1 mol (72.07 J/mol-K) + 1 mol (213.7 J/mol-K)] = —173.77 J/K (unrounded) AGE“, = AH gm — TASTE = ~277.4 kJ - [(298 K) (~173.77 J/K) (1 HI 103 1)] = ~225.6265 = 3-226 kJ 20-6 111 20.43 20.44 20.45 a) 1111;“ = [2 mol HI(g) (25.9 ch/mol)] — [1 mol H2(g) (o kJ/rnol) + 1 m0112(s)(0 kJ/mol)] = 51.8 M ASfixn = [2 mol HI(g) (206.33 J/K-mol)] —— [1 mol H2(g) (130.6 J/K-mol) + 1 111011;“) (116.14 J/K-rnom = 165.92 J/K (unrounded) Acgm = 1111;“ — 1115;” = 51.8 M — [(298 K) (165.92 J/K) (1 kJ/ 103 1)] = 2.3558 = 2.4 M b) AH]:m = [1 mol Mn(s) (0 kJ/mol) + 2 mol C01(g) (—393.5 kJ/mol)] — [1 mol Mn02(s) (~520.9 kJ/mol) + 2 mol CO(g) (—110.5 kJ/mol)] = —45 .1 k] AS:xn = [1 mol Mn(s) (31.8 J/K-mol) + 2 mol C02(g) (213.7 J/K-mol)] - [1 mol Mn02(s) (53.1 J/K-mol) + 2 mol CO(g) (197.5 J/K-mol)] = 11.1 J/K A013,“, = 7111;“ ~ T ASS“, = 415.110 — [(298 K) (11.1 J/K) (1 k1 / 103 1)] = 418.4078 = —48.4 k] c) Afon = [1 mol NH3(g) (—45.9 kJ/mol) + 1 mol HC1(g) (92.3 kJ/mol)] — [1 mol NILCKS) (414.4 kJ/mol)] = 176.2 k1 AS;xn = [1 mol NH3(g) (193 J/K-rnol) + 1 mol HCl(g) (186.79 J/K-mol)] -'[1 mol NH4C1(S) (94.6 J/K-mol)] = 285.19 J/K (unrounded) Aagm = AHgm — Tnsgm = 176.2 kJ ~ [(298 K) (285.19 J/K) (1 k1 / 103 1)] = 91213 = 91.2 U a) Entropy decreases (AS° negative) because the number of moles of gas decreases fiom reactants ( 1% mol) to products (1 mole). The oxidation (combustion) of CO requires initial energy input to start the reaction, but then releases energy (exothermic, AH ° negative) which is typical of all combustion reactions. b) Method 1: Calculate AG;xn from AG? ’5 of products and reactants. AG:xn = 2m AG; (products)] —— Z[n AG; (reactants)] AG;m = [(1 mo...
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