CHAPTER 0 THERMODYNAMICS: ENTROPY, FREE ENERGY, AND THE DIRECTION OF CHEMICAL REACTIONS FOLLOW UP PROBLEMS 0.1A Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, is positive. a) PCl 5 (g). For substances with the same type of atoms and in the same physical state, entropy increases with increasing number of atoms per molecule because more types of molecular motion are available. b) BaCl (s). Entropy increases with increasing atomic size. The Ba + ion and Cl ion are larger than the Ca + ion and F ion, respectively. c) Br (g). Entropy increases from solid liquid gas. 0.1B Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, is positive. a) LiBr(aq). For substances with the same number of atoms and in the same physical state, entropy increases with increasing atomic size. LiBr has the lower molar mass of the two substances and, therefore, the lower entropy. b) Quartz. Quartz has a crystalline structure and the particles have less freedom (and lower entropy) in that structure than in glass, an amorphous solid. c) Cyclohexane. Ethylcyclobutane has a side chain, which has more freedom of motion than the atoms in the ring. In cyclohexane, there are no side chains. Freedom of motion is restricted by the ring structure. 0.A Plan: Predict the sign of rxn by comparing the randomness of the products with the randomness of the reactants. Calculate rxn using Appendix B values and the relationship rxn = m S products n S reactants. a) 4NO (g) N O(g) + N O 3 (g) The rxn is predicted to decrease ( rxn < 0) because four moles of random, gaseous product are transformed into two moles of random, gaseous product (the change in gas moles is ). rxn = m S products n S reactants rxn = [(1 mol N O 3 )(Sº of N O 3 ) + (1 mol N O)(Sº of N O)] [(4 mol NO)(Sº of NO)] rxn = [(1 mol N O 3 )(314.7 J/mol K) + (1 mol N O)(19.7 J/mol K)] [(4 mol NO)(10.65 J/mol K)] = 308. J/K rxn < 0 as predicted. 0-1
b) CH 3 OH(g) CO(g) + H (g) The change in gaseous moles is +, so the sign of rxn = [(1 mol CO)(Sº of CO) + ( mol H )(Sº of H )] [(1 mol CH 3 OH)(Sº of CH 3 OH)] rxn = [(1 mol CO)(197.5 J/mol K) + ( mol H )(130.6 J/mol K)] [(1 mol CH 3 OH)(38 J/mol K)] = 0.7 = 1 J/K rxn > 0 as predicted. rxn is predicted to be greater than zero. 0.B Plan: Predict the sign of rxn by comparing the randomness of the products with the randomness of the reactants. Calculate rxn using Appendix B values and the relationship rxn = m S products n S reactants. a) NaOH(s) + CO (g) Na CO 3 (s) + H O(l) The rxn is predicted to decrease ( rxn < 0) because the more random, gaseous reactant is transformed into a more ordered, liquid product. rxn = m S products n S reactants rxn = [(1 mol Na CO 3 )(Sº of Na CO 3 ) + (1 mol H O)(Sº of H O)] [( mol NaOH)(Sº of NaOH) + (1 mol CO )(Sº of CO )] rxn = [(1 mol Na CO 3 )(139 J/mol K) + (1 mol H O)(69.940 J/mol K)] [( mol NaOH)(64.454 J/mol K) + (1 mol CO )(13.7 J/mol K)] = 133.668 = 134 J/K rxn < 0 as predicted. b) Fe(s) + 3H O(g) Fe O 3 (s) + 3H (g) The change in gaseous moles is zero, so the sign of rxn is difficult to predict. Iron(III) oxide has greater entropy than Fe because it is more complex, but this is offset by the greater molecular complexity of H O versus H. rxn = [(1 mol Fe O 3 )(Sº of Fe O 3 ) + (3 mol H )(Sº of H )] [( mol Fe)(Sº of Fe) + (3 mol H O)(Sº of H O)] rxn = [(1 mol Fe O 3 )(87.400 J/mol K) + (3 mol H )(130.6 J/mol K)] [( mol Fe)(7.3 J/mol K) + (3 mol H O)(188.7 J/mol K)] = 141.56 = 141.6 J/K The negative rxn shows that the greater entropy of H O versus H does outweigh the greater entropy of Fe O 3 versus Fe. 0.3A Plan: Write the balanced equation for the reaction and calculate the rxn using Appendix B. Determine the surr by first finding rxn. Add surr to rxn to verify that univ is positive. P 4 (s) + 6Cl (g) 4PCl 3 (g) rxn = [(4 mol PCl 3 )(Sº of PCl 3 )] [(1 mol P 4 )(Sº of P 4 ) + (6 mol Cl )(Sº of Cl )] rxn = [(4 mol PCl 3 )(31 J/mol K)] [(1 mol P 4 )(41.1 J/mol K) + (6 mol Cl ) (3.0J/mol K)] rxn = 131J/K (The entropy change is expected to be negative because the change in gas moles is negative.) 0-
rxn = m f (products) n f rxn = [(4 mol PCl 3 )( f of PCl 3 )] [(1 mol P 4 )( f of P 4 ) + (6 mol Cl )( f of Cl )] rxn = [(4 mol PCl 3 )( 87 kj/mol)] [(1 mol P 4 )(0 kj/mol) + (6 mol Cl )(0 kj/mol)] rxn = 1148 kj 1148 kj surr = = = 3.853 kj/k(10 3 J/1 kj) = 3850 J/K H rxn T 98 K S univ = rxn + surr = ( 131 J/K) + (3850 J/K) = 3719 J/K Because univ is positive, the reaction is spontaneous at 98 K. 0.3B Plan: Write the balanced equation for the reaction and calculate the rxn using Appendix B. Determine the surr by first finding rxn. Add surr to rxn to verify that univ is positive. FeO(s) + 1/O (g) Fe O 3 (s) rxn = [(1 mol Fe O 3 )(Sº of Fe O 3 )] [( mol FeO)(Sº of FeO) + (1/ mol O )(Sº of O )] rxn = [(1 mol Fe O 3 )(87.400 J/mol K)] [( mol FeO)(60.75 J/mol K) + (1/ mol O ) (05.0J/mol K)] rxn = 136.6 J/K (entropy change is expected to be negative because gaseous reactant is converted to solid product). rxn = m f (products) n f rxn = [(1 mol Fe O 3 )( f of Fe O 3 )] [( mol FeO)( f of FeO) + (1/ mol O )( f of O )] rxn = [(1 mol Fe O 3 )( 85.5 kj/mol)] [( mol FeO)( 7.0 kj/mol) + (1/ mol O )(0 kj/mol)] rxn = 81.5 kj surr = rxn 81.5 kj = = 0.94463 kj/k(10 3 J/1 kj) = 944.63 J/K T 98 K S univ = rxn + surr = ( 136.6 J/K) + (944.63 J/K) = 808.03 = 808 J/K Because univ is positive, the reaction is spontaneous at 98 K. This process is also known as rusting. Common sense tells us that rusting occurs spontaneously. Although the entropy change of the system is negative, the increase in entropy of the surroundings is large enough to offset rxn. 0.4A Plan: Calculate the rxn using f values from Appendix B. Calculate rxn use the relationship rxn = rxn Trxn. rxn = m f (products) n f from tabulated Sº values and then rxn = [( mol NOCl)( f of NOCl)] [( mol NO)( f of NO) + (1 mol Cl )( f of Cl )] 0-3
rxn = [( mol NOCl)(51.71 kj/mol)] [( mol NO)(90.9 kj/mol) + (1 mol Cl )(0 kj/mol)] rxn = 77.16 kj rxn = m S products n S reactants rxn = [( mol NOCl)(Sº of NOCl)] [( mol NO)(Sº of NO) + (1 mol Cl )(Sº of Cl )] rxn = [( mol NOCl)(61.6 J/mol K)] [( mol NO)(10.65 J/mol K) + (1 mol Cl )(3.0 J/mol K)] rxn = 11.1 J/K rxn = rxn Trxn = 77.16 kj [(98 K)( 11.1 J/K)(1 kj/10 3 J)] = 41.07 = 41.1 kj 0.4B Plan: Calculate the rxn using f values from Appendix B. Calculate rxn use the relationship rxn = rxn Trxn. rxn = m f (products) n f rxn = [( mol NO )( f of NO )] [( mol NO)( f of NO) + (1 mol O )( f of O )] rxn = [( mol NO )(33. kj/mol)] [( mol NO)(90.9 kj/mol) + (1 mol O )(0 kj/mol)] rxn = 114.18 = 114. kj rxn = m S products n S reactants from tabulated Sº values and then rxn = [( mol NO )(Sº of NO )] [( mol NO)(Sº of NO) + (1 mol O )(Sº of O )] rxn = [( mol NO )(39.9 J/mol K)] [( mol NO)(10.65 J/mol K) + (1 mol O )(05.0 J/mol K)] rxn = 146.5 J/K rxn = rxn Trxn = 114. kj [(98 K)( 146.5 J/K)(1 kj/10 3 J)] = 70.543 = 70.5 kj 0.5A Plan: Use f values from Appendix B to calculate rxn using the relationship rxn = m f (products) n f. a) rxn = [( mol NOCl)( f of NOCl)] [( mol NO)( f of NO) + (1 mol Cl )( f of Cl )] rxn = [( mol NOCl)(66.07 kj/mol)] [( mol NO)(86.60 kj/mol) + (1 mol Cl )(0 kj/mol)] rxn = 41.06 kj b) rxn = [( mol Fe)( f of Fe) + (3 mol H O)( f of H O)] [(3 mol H )( f of H ) + (1 mol Fe O 3 )( f of Fe O 3 )] rxn = [( mol Fe)(0 kj/mol) + (3 mol H O)( 8.60 kj/mol)] [(3 mol H )(0 kj/mol) + (1 mol Fe O 3 )( 743.6 kj/mol)] rxn = 57.8 kj 0.5B Plan: Use f values from Appendix B to calculate rxn using the relationship rxn = m f (products) n f. a) rxn = [( mol NO )( f of NO )] [( mol NO)( f of NO) + (1 mol O )( f of O )] 0-4
rxn = [( mol NO )(51 kj/mol)] [( mol NO)(86.60 kj/mol) + (1 mol O )(0 kj/mol)] rxn = 71. = 71 kj b) rxn = [( mol CO)( f of CO)] [( mol C)( f of C) + (1 mol O )( f of O )] rxn = [( mol CO)( 137. kj/mol)] [( mol C)(0 kj/mol) + (1 mol O )(0 kj/mol)] rxn = 74.4 kj 0.6A Plan: Predict the sign of rxn by comparing the randomness of the products with the randomness of the reactants. Use the relationship rxn = rxn Trxn to answer b). a) The reaction is X Y (g) X (g) + Y (g). Since there are more moles of gaseous product than there are of gaseous reactant, entropy increases and > 0. b) The reaction is only spontaneous above 35ºC or in other words, at high temperatures. In the relationship rxn = rxn Trxn, when > 0 so that T is < 0, will only be negative at high T if > 0. 0.6B Plan: Predict the sign of rxn by comparing the randomness of the products with the randomness of the reactants. Use the relationship rxn = rxn Trxn to answer b). a) A solid forms a gas and a liquid, so > 0. A crystalline array breaks down, so > 0. b) For the reaction to occur spontaneously ( < 0), T must be greater than, which would occur only at higher T. 0.7A Plan: Use the equation = T to determine if the reaction is spontaneous ( < 0). Then examine the same equation to determine the effect of raising the temperature on the spontaneity of the reaction. a) rxn = rxn Trxn = 19.7 kj [(98 K)( 308. J/K)(1 kj/10 3 J)] = 100.8564 = 100.9 kj Because < 0, the reaction is spontaneous at 98 K. b) As temperature increases, T becomes more positive, so the reaction becomes less spontaneous at higher temperatures. c) rxn = rxn Trxn = 19.7 kj [(773 K)( 308. J/K)(1 kj/10 3 J)] = 45.5386 = 45.5 kj 0.7B Plan: Examine the equation = T and determine which combination of enthalpy and entropy will describe the given reaction. Two choices can already be eliminated: 1) When > 0 (endothermic reaction) and < 0 (entropy decreases), the reaction is always nonspontaneous, regardless of temperature, so this combination does not describe the reaction. ) When < 0 (exothermic reaction) and > 0 (entropy increases), the reaction is always spontaneous, regardless of temperature, so this combination does not describe the reaction. Two combinations remain: 3) > 0 and > 0, or 4) < 0 and < 0. If the reaction becomes spontaneous at 40 C, this means that becomes negative at lower temperatures. Case 3) becomes spontaneous at higher temperatures, when the T term is larger than the positive enthalpy term. By process of elimination, Case 4) describes the reaction. At a lower temperature, the negative becomes larger than the positive ( T ) value, so becomes negative. 0.8A Plan: To find the temperature at which the reaction becomes spontaneous, use rxn = 0 = rxn Trxn and solve for temperature. 0-5
The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0. rxn = rxn Trxn 0 = 19.7x10 3 J (T)( 308. J/K) T = 65. K 73.15 = 35.0 C 0.8B Plan: To find the temperature at which the reaction becomes spontaneous, use rxn = 0 = rxn Trxn and solve for temperature. rxn can be calculated from the individual f values of the reactants and products by using the relationship rxn = m f (products) n f. rxn can be calculated from the individual S values of the reactants and products by using the relationship CaO(s) + CO (g) CaCO 3 (s) rxn = m f (products) n f rxn = m S products n S reactants. rxn = [(1 mol CaCO 3 )( f of CaCO 3 )] [(1 mol CaO)( f of CaO) + (1 mol CO )( f of CO )] rxn = [(1 mol CaCO 3 )( 106.9 kj/mol)] [(1 mol CaO)( 635.1 kj/mol) + (1 mol CO )( 393.5 kj/mol)] rxn = 178.3 kj rxn = m S products n S reactants rxn = [(1 mol CaCO 3 )(Sº of CaCO 3 )] [(1 mol CaO)(Sº of CaO) + (1 mol CO )(Sº of CO )] rxn = [(1mol CaCO 3 )(9.9 J/mol K)] [(1 mol CaO)(38. J/mol K) + (1 mol CO )(13.7 J/mol K)] rxn = 159.0 J/K = 0.159 kj/k rxn = 0 = rxn Trxn rxn = Trxn 178.3 kj T = = = 111.384 = 111 K 0.1590 kj/k The reaction becomes spontaneous at temperatures < 111 K. 0.9A Plan: First find, then calculate K from = RT ln K. Calculate using f values in the relationship rxn = m f (products) n f. C(graphite) + O (g) CO(g) rxn = m f (products) n f rxn = [( mol CO)( 137. kj/mol)] [( mol C)(0 kj/mol) + (1 mol O )( 0 kj/mol)] = 74.4 kj 74.4 kj/mol 1000 J ln K = = RT ( 8.314 J/mol K)( 98 K ) 1 kj = 110.7536 = 111 K = e 111 = 1.6095x10 48 = 1.6x10 48 0-6
0.9B Plan: The equilibrium constant, K, is related to through the equation = RT ln K. = RT ln K = (8.314 J/mol K)(98 K) ln (.x10 15 ) = 8.35964x10 4 J = 83.6 kj/mol 0.10A Plan: Write the equilibrium expression for the reaction and calculate Q c for each scene. Remember that each particle represents 0.10 mol and that the volume is 1.0 L. A reaction that is proceeding to the right will have < 0 and a reaction that is proceeding to the left will have > 0. A reaction at equilibrium has = 0. a) A(g) + 3B(g) AB 3 (g) b) Q c = [AB 3 ] [A][B] 3 Mixture 1: Q c = [AB 3 ] [A][B] 3 = [0.0] [0.40][0.80] 3 = 0.98 Mixture : Q c = [AB 3 ] [A][B] 3 = [0.30] [0.30][0.50] 3 = 8 Mixture 3: Q c = [AB 3 ] [A][B] 3 = [0.40] [0.0][0.0] 3 = 50 Mixture is at equilibrium since Q c = K c c) Q c < K c for Mixture 1 and the reaction is proceeding right to reach equilibrium; thus < 0. Mixture is at equilibrium and = 0. Mixture 3 proceeds to the left to reach equilibrium since Q c > K c and > 0. The ranking for most positive to most negative is 3 > > 1. 0.10B Plan: Write the equilibrium expression for the reaction and calculate Q c for each scene. A reaction that is proceeding to the right will have < 0 and a reaction that is proceeding to the left will have > 0. A reaction at equilibrium has = 0. a) X (g) + Y (g) XY (g) [XY ] Q c = [X ][Y ] Mixture 1: Q c = Mixture : Q c = Mixture 3: Q c = [XY ] [X ][Y ] [XY ] [X ][Y ] [XY ] [X ][Y ] = = = [5] [][1] [4] [][] [] [4][] = 1.5 = = 0.5 Mixture is at equilibrium since Q c = K c b) Q c > K c for Mixture 1 and the reaction is proceeding left to reach equilibrium; thus > 0. Mixture is at equilibrium and = 0. Mixture 3 proceeds to the right to reach equilibrium since Q c < K c and < 0. The ranking for most negative to most positive is 3 < < 1. c) Any reaction mixture moves spontaneously towards equilibrium so both changes have a negative. 0.11A Plan: The equilibrium constant, K, is related to through the equation = RT ln K. The free energy of the reaction under non-standard state conditions is calculated using = + RT ln Q. 33.5 kj/mol 1000 J a) ln K = = RT ( 8.314 J/mol K)( 98 K ) 1 kj = 13.513 K = e 13.513 = 7.451x10 5 = 7.45x10 5 0-7
[C H 5 Cl] b) Q = [C H 4 ][HCl] = (1.5) (0.50)(1.0) = 3.0 = + RT ln Q = ( 33.5 kj/mol) + (8.314 J/mol K) (1 kj/1000 J) (98 K) ln (3.0) = 30.7781 = 30.8 kj/mol 0.11B Plan: Write a balanced equation for the dissociation of hypobromous acid in water. The free energy of the reaction at standard state is calculated using = RT ln K. The free energy of the reaction under non-standard state conditions is calculated using = + RT ln Q. HBrO(aq) + H O(l) BrO (aq) + H 3 O + (aq) a) = RT ln K = (8.314 J/mol K)(98)ln (.3x10 9 ) = 4.97979x10 4 = 4.9x10 4 J/mol = 49 kj/mol + H3O BrO b) Q = 6.0x10 4 [ 0.10 ] = HBrO 0.0 [ ] [ ] = + RT ln Q = (4.97979x10 4 J/mol) + (8.314 J/mol K) (98 K) ln 4 [ ] [ 0.0] 6.0x10 0.10 =.918399x10 4 =.9x10 4 J/mol = 9 kj/mol The value of K a is very small, so it makes sense that is a positive number. The natural log of a negative exponent gives a negative number (ln 3.0 x 10 4 ), so the value of decreases with concentrations lower than the standard state 1 M values. CHEMICAL CONNECTIONS BOXED READING PROBLEMS B0.1 Plan: Convert mass of glucose (1 g) to moles and use the ratio between moles of glucose and moles of ATP to find the moles and then molecules of ATP formed. Do the same calculation with tristearin. a) Molecules of ATP/g glucose = 3 1 mol glucose 36 mol ATP 6.0x10 molecules ATP ( 1 g glucose) 180.16 g glucos e 1 mol glucose 1 mol ATP = 1.0333x10 3 = 1.03x10 3 molecules ATP/g glucose b) Molecules of ATP/g tristearin = 3 1 mol tristearin 458 mol ATP 6.0x10 molecules ATP ( 1 g tristearin) 897.50 g tristearin 1 mol tristearin 1 mol ATP = 3.073065x10 3 = 3.073x10 3 molecules ATP/g tristearin B0. Plan: Add the two reactions to obtain the overall process; the values of the two reactions are then added to obtain for the overall reaction. creatine phosphate creatine + phosphate = 43.1 kj/mol ADP + phosphate ATP = +30.5 kj/mol creatine phosphate + ADP creatine + ATP = 43.1 kj/mol + 30.5 kj/mol = 1.6 kj/mol END OF CHAPTER PROBLEMS 0.1 Spontaneous processes proceed without outside intervention. The fact that a process is spontaneous does not mean that it will occur instantaneously (in an instant) 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. 0-8
0. 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). 0.3 a) The energy of the universe is constant. b) Energy cannot be created or destroyed. c) E system = E surroundings The first law is concerned with balancing energy for a process but says nothing about whether the process can, in fact, occur. 0.4 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. 0.5 Vaporization is the change of a liquid substance to a gas so vaporization = S gas S liquid. Fusion is the change of a solid substance into a liquid so fusion = S liquid S solid. 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. 0.6 In an exothermic process, the system releases heat to its surroundings. The entropy of the surroundings increases because the temperature of the surroundings increases ( surr > 0). In an endothermic process, the system absorbs heat from the surroundings and the surroundings become cooler. Thus, the entropy of the surroundings decreases ( surr < 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. 0.7 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 5 C and entropy increases with temperature, S 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. 0.8 Plan: A spontaneous process is one that occurs by itself without a continuous input of energy. 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. 0.9 a) The movement of Earth about the Sun is spontaneous. b) The movement of a boulder against gravity is nonspontaneous. c) The reaction of an active metal (sodium) with an active nonmetal (chlorine) is spontaneous. 0.10 Plan: A spontaneous process is one that occurs by itself without a continuous input of energy. 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. c) Not spontaneous, a cooked egg will not become raw again, no matter how long it sits or how many times it is mixed. 0-9
0.11 a) If a satellite slows sufficiently, it will fall to Earth s surface through a spontaneous process. b) Water is a very stable compound; its decomposition at 98 K and 1 atm is not spontaneous. c) The increase in prices tends to be spontaneous. 0.1 Plan: Particles with more freedom of motion have higher entropy. Therefore, S gas > S liquid > S solid. If the products of the process have more entropy than the reactants, sys is positive. If the products of the process have less entropy than the reactants, sys is negative. a) sys 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) sys 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) sys 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. 0.13 a) sys positive b) sys positive c) sys negative 0.14 Plan: Particles with more freedom of motion have higher entropy. Therefore, S gas > S liquid > S solid. If the products of the process have more entropy than the reactants, sys is positive. If the products of the process have less entropy than the reactants, sys is negative. a) sys positive, the process described is liquid alcohol becoming gaseous alcohol. The gas molecules have greater entropy than the liquid molecules. b) sys positive, the process described is a change from solid to gas, an increase in possible energy states for the system. c) sys 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. 0.15 a) sys negative b) sys negative c) sys negative 0.16 Plan: sys is the entropy of the products the entropy of the reactants. Use the fact that S gas > S liquid > S solid ; also, the greater the number of particles of a particular phase of matter, the higher the entropy. a) sys 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) sys negative, gaseous reactants form solid product and number of particles decreases, so entropy decreases. c) sys positive, when a solid salt dissolves in water, entropy generally increases since the entropy of the aqueous mixture has higher entropy than the solid. 0.17 a) sys negative b) sys negative c) sys negative 0.18 Plan: sys is the entropy of the products the entropy of the reactants. Use the fact that S gas > S liquid > S solid ; also, the greater the number of particles of a particular phase of matter, the higher the entropy. a) sys positive, the reaction produces gaseous CO molecules that have greater entropy than the physical states of the reactants. b) sys negative, the reaction produces a net decrease in the number of gaseous molecules, so the system s entropy decreases. c) sys positive, the reaction produces a gas from a solid. 0.19 a) sys negative b) sys positive c) sys negative 0-10
0.0 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. If the entropy of the products is greater than that of the reactants, is positive. a) sys positive, decreasing the pressure increases the volume available to the gas molecules so entropy of the system increases. b) sys negative, gaseous nitrogen molecules have greater entropy (more possible states) than dissolved nitrogen molecules. c) sys positive, dissolved oxygen molecules have lower entropy than gaseous oxygen molecules. 0.1 a) sys negative b) sys positive c) sys negative 0. Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. 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 -butene restricts rotation. b) Xe(g) has the greater molar entropy because entropy increases with atomic size. c) CH 4 (g) has the greater molar entropy because gases in general have greater entropy than liquids. 0.3 a) N O 4 (g); it has greater molecular complexity. b) CH 3 OCH 3 (l); hydrogen bonding in CH 3 CH OH would increase order. c) HBr(g); it has greater mass. 0.4 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. a) Ethanol, C H 5 OH(l), is a more complex molecule than methanol, CH 3 OH, and has the greater molar entropy. b) When a salt dissolves, there is an increase in the number of possible states for the ions. 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). 0.5 a) P 4 (g); it has greater molecular complexity. b) HNO 3 (aq); because S(solution) > S(pure). c) CuSO 4 5H O; it has greater molecular complexity. 0.6 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. a) Diamond < graphite < charcoal. Diamond has an ordered, three-dimensional crystalline shape, followed by graphite with an ordered two-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 < O < O 3. Entropy increases with molecular complexity because there are more modes of movement (e.g., bond vibration) available to the complex molecules. 0.7 a) Ribose < glucose < sucrose; entropy increases with molecular complexity. b) CaCO 3 (s) < (CaO(s) + CO (g)) < (Ca(s) + C(s) + 3/O (g)); entropy increases with moles of gas particles. c) SF 4 (g) < SF 6 (g) < S F 10 (g); entropy increases with molecular complexity. 0-11
0.8 Plan: Particles with more freedom of motion have higher entropy. In general the entropy of gases is greater than that of liquids, and the entropy of liquids is greater than that of solids. Entropy increases with temperature. For substances in the same phase, entropy increases with atomic size and molecular complexity. a) ClO 4 (aq) > ClO 3 (aq) > ClO (aq). The decreasing order of molar entropy follows the order of decreasing molecular complexity. b) NO (g) > NO(g) > N (g). N has lower molar entropy than NO because N consists of two of the same atoms while NO consists of two different atoms. NO has greater molar entropy than NO because NO consists of three atoms while NO consists of only two. c) Fe 3 O 4 (s) > Fe O 3 (s) > Al O 3 (s). Fe 3 O 4 has greater molar entropy than Fe O 3 because Fe 3 O 4 is more complex and more massive. Fe O 3 and Al O 3 contain the same number of atoms but Fe O 3 has greater molar entropy because iron atoms are more massive than aluminum atoms. 0.9 a) Ba(s) > Ca(s) > Mg(s); entropy decreases with lower mass. b) C 6 H 14 > C 6 H 1 > C 6 H 6 ; entropy decreases with lower molecular complexity and lower molecular flexibility. c) PF Cl 3 (g) > PF 5 (g) > PF 3 (g); entropy decreases with lower molecular complexity. 0.30 a) X (g) + 3Y (g) XY 3 (g) b) ΔS < 0 since there are fewer moles of gas in the products than in the reactants. c) XY 3 is the most complex molecule and thus will have the highest molar entropy. 0.31 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, univ = sys + surr = 0. However, for a system moving to equilibrium, univ > 0, because the second law states that for any spontaneous process, the entropy of the universe increases. 0.3 Plan: 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. rxn = [( mol HClO)(Sº of HClO)] [(1 mol H O)(Sº of H O) + (1 mol Cl O)(Sº of Cl O)] Rearranging this expression to solve for Sº of Cl O gives: Sº of Cl O = (Sº of HClO) Sº of H O rxn 0.33 Plan: To calculate the standard entropy change, use the relationship rxn = m S products n S reactants. To predict the sign of entropy recall that in general S gas > S liquid > S solid, and entropy increases as the number of particles of a particular phase of matter increases, and with increasing atomic size and molecular complexity. a) Prediction: negative because number of moles of ( n) gas decreases. = [(1 mol N O)(S of N O) + (1 mol NO )(S of NO )] [(3 mol NO)(S of NO)] = [(1 mol)(19.7 J/mol K) + (1 mol)(39.9 J/mol K)] [(3 mol)(10.65 J/mol K)] = 17.35 = 17.4 J/K b) Prediction: Sign difficult to predict because n = 0, but possibly positive because water vapor has greater complexity than H gas. = [( mol Fe)(S of Fe) + (3 mol H O)(S of H O)] [(3 mol H )(S of H ) + (1 mol Fe O 3 )(S of Fe O 3 )] = [( mol)(7.3 J/mol K) + (3 mol)(188.7 J/mol K)] [(3 mol)(130.6 J/mol K) + (1 mol)(87.400 J/mol K)] = 141.56 = 141.6 J/K c) Prediction: negative because a gaseous reactant forms a solid product and also because the number of moles of gas ( n) decreases. = [(1 mol P 4 O 10 )(S of P 4 O 10 )] [(1 mol P 4 )(S of P 4 ) + (5 mol O )(S of O)] = [(1 mol)(9 J/mol K)] [(1 mol)(41.1 J/mol K) + (5 mol)(05.0 J/mol K)] = 837.1 = 837 J/K 0-1
0.34 a) 3NO (g) + H O(l) HNO 3 (l) + NO(g) negative = [( mol HNO 3 )(S of HNO 3 ) + (1 mol NO)(S of NO)] [(3 mol NO )(S of NO ) +( 1 mol H O)(S of H O)] = [( mol)(155.6 J/K mol) + (1 mol)(10.65 J/K mol)] [(3 mol)(39.9 J/K mol) + (1 mol)(69.940 J/K mol)] = 67.79 = 67.8 J/K b) N (g) + 3F (g) NF 3 (g) negative = [( mol NF 3 )(S of NF 3 ] [(1 mol N )(S of N ) + (3 mol F )(S of F )] = [( mol)(60.6 J/K mol)] [(1 mol)(191.5 J/K mol) + (3 mol)(0.7 J/K mol)] = 78.4 J/K c) C 6 H 1 O 6 (s) + 6O (g) 6CO (g) + 6H O(g) positive = [(6 mol CO )(S of CO ) + (6 mol H O)(S of H O)] [(1 mol C 6 H 1 O 6 )(S of C 6 H 1 O 6 ) + (6 mol O )(S of O )] = [(6 mol)(13.7 J/K mol) + (6 mol H O)(188.7 J/K mol)] [(1 mol)(1.1 J/K mol) + (6 mol)(05.0 J/K mol)] = 97.4 = 97.4 J/K 0.35 Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship rxn = m S products n S reactants. To predict the sign of entropy recall that in general S gas > S liquid > S solid, entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with increasing atomic size and molecular complexity. The balanced combustion reaction is: C H 6 (g) + 7O (g) 4CO (g) + 6H O(g) = [(4 mol CO )(S of CO ) + (6 mol H O)(S of H O)] [( mol C H 6 )(S of C H 6 ) + (7 mol O )(S of O )] = [(4 mol)(13.7 J/mol K) + (6 mol)(188.7 J/mol K)] [( mol)(9.5 J/mol K) + (7 mol)(05.0 J/mol K)] = 93.1 = 93.1 J/K The entropy value is not per mole of C H 6 but per two moles. Divide the calculated value by two to obtain entropy per mole of C H 6. Yes, the positive sign of is expected because there is a net increase in the number of gas molecules from nine moles as reactants to ten moles as products. 0.36 CH 4 (g) + O (g) CO (g) + H O(l) = [(1 mol CO )(S of CO ) + ( mol H O)(S of H O)] [(1 mol CH 4 )(S of CH 4 ) + ( mol O )(S of O )] = [(1 mol)(13.7 J/K mol) + ( mol)(69.940 J/K mol)] [(1 mol)(186.1 J/K mol) + ( mol)(05.0 J/K mol)] = 4.5 = 4.5 J/K Yes, a decrease in the number of moles of gas should result in a negative value. 0.37 Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship rxn = m S products n S reactants. To predict the sign of entropy recall that in general S gas > S liquid > S solid, entropy increases as the number of particles of a particular phase of matter increases, and entropy increases with increasing atomic size and molecular complexity. The balanced chemical equation for the described reaction is: NO(g) + 5H (g) NH 3 (g) + H O(g) Because the number of moles of gas decreases, i.e., n = 4 7 = 3, the entropy is expected to decrease. = [( mol NH 3 )(S of NH 3 ) + ( mol H O)(S of H O)] [( mol NO)(S of NO) + (5 mol H )(S of H )] = [( mol)(193 J/mol K) + ( mol)(188.7 J/mol K)] [( mol)(10.65 J/mol K) + (5 mol)(130.6 J/mol K)] = 310.86 = 311 J/K Yes, the calculated entropy matches the predicted decrease. 0-13
0.38 4NH 3 (g) + 7O (g) 4NO (g) + 6H O(g) = [(4 mol NO )(S of NO ) + (6 mol H O)(S of H O)] [(4 mol NH 3 )(S of NH 3 ) + (7 mol O )(S of O )] = [(4 mol)(39.9 J/K mol) + (6 mol)(188.7 J/K mol)] [(4 mol)(193 J/K mol) + (7 mol)(05.0 J/K mol)] = 115.08 = 115 J/K Yes, a loss of one mole of a gas should result in a small negative value. 0.39 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxn = m S products n S reactants. To calculate ΔS o of the universe in part b, first calculate ΔH o of the reaction using the relationship rxn = m f (products) n f Use ΔH o of the reaction to calculate ΔS of the surroundings. surr = rxn T Add ΔS of the surroundings and ΔS o of the reaction to calculate ΔS of the universe. If ΔS of the universe is greater than zero, the reaction is spontaneous at the given temperature. S univ = rxn + surr a) The reaction for forming Cu O from copper metal and oxygen gas is: Cu(s) + 1/O (g) Cu O(s) = [(1 mol Cu O)(S of Cu O)] [( mol Cu)(S of Cu) + (1/ mol O )(S of O )] = [(1 mol)(93.1 J/mol K)] [( mol)(33.1 J/mol K) + (1/ mol)(05.0 J/mol K)] = 75.6 J/K b) rxn = m f (products) n f rxn = [(1 mol Cu O)( f of Cu O)] [( mol Cu)( f of Cu) + (1/ mol O )( f of O )] rxn = [(1 mol Cu O)( 168.6 kj/mol)] [( mol Cu)(0 kj/mol) + (1/ mol O )(0 kj/mol)] rxn = 168.6 kj surr = rxn 168.6 kj = T 98 K = 0.56577 kj/k(103 J/1 kj) = 565.77 J/K S univ = rxn + surr = ( 75.6 J/K) + (565.77 J/K) = 490.17 = 490. J/K Because univ is positive, the reaction is spontaneous at 98 K. 0.40 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxn = m S products n S reactants. To calculate ΔS o of the universe in part b, first calculate ΔH o of the reaction using the relationship rxn = m f (products) n f Use ΔH o of the reaction to calculate ΔS of the surroundings. surr = rxn T Add ΔS of the surroundings and ΔS o of the reaction to calculate ΔS of the universe. If ΔS of the universe is greater than zero, the reaction is spontaneous at the given temperature. S univ = rxn + surr 0-14
a) One mole of hydrogen iodide is formed from its elements in their standard states according to the following equation: 1/H (g) + 1/I (s) HI(g) = [(1 mol HI)(S of HI)] [(1/ mol H )(S of H ) + (1/ mol I )(S of I )] = [(1 mol)(06.33 J/K mol)] [(1/ mol)(130.6 J/K mol) + (1/ mol)(116.14 J/K mol)] = 8.96 = 83.0 J/K b) rxn = m f (products) n f rxn = [(1 mol HI)( f of HI)] [(1/ mol H )( f of H ) + (1/ mol I )( f of I )] rxn = [(1 mol HI)(5.9 kj/mol)] [(1/ mol H )(0 kj/mol) + (1/ mol I )(0 kj/mol)] rxn = 5.9 kj surr = rxn 5.9 kj = T 98 K = 0.086918 kj/k(103 J/1 kj) = 86.918 J/K S univ = rxn + surr = (83.0 J/K) + ( 86.918 J/K) = 3.918 = 3.9 J/K Because univ is negative, the reaction is not spontaneous at 98 K. 0.41 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxn = m S products n S reactants. To calculate ΔS o of the universe in part b, first calculate ΔH o of the reaction using the relationship rxn = m f (products) n f Use ΔH o of the reaction to calculate ΔS of the surroundings. surr = rxn T Add ΔS of the surroundings and ΔS o of the reaction to calculate ΔS of the universe. If ΔS of the universe is greater than zero, the reaction is spontaneous at the given temperature. S univ = rxn + surr a) One mole of methanol is formed from its elements in their standard states according to the following equation: C(g) + H (g) + 1/O (g) CH 3 OH(l) = [(1 mol CH 3 OH)(S of CH 3 OH)] [(1 mol C)(S of C) + ( mol H )(S of H ) + (1/ mol O )(S of O )] = [(1 mol)(17 J/mol K)] [(1 mol)(5.686 J/mol K) + ( mol)(130.6 J/mol K) + (1/ mol)(05.0 J/mol K)] = 4.386 = 4 J/K b) rxn = m f (products) n f rxn = [(1 mol CH 3 OH)( f of CH 3 OH)] [(1 mol C)( f of C) + ( mol H )( f of H ) + (1/ mol O )( f of O )] rxn = [(1 mol CH 3 OH)( 38.6 kj/mol)] [(1 mol C)(0 kj/mol) + ( mol H )( 0 kj/mol) + (1/ mol O )( 0 kj/mol)] rxn = 38.6 kj surr = rxn 38.6 kj = T 98 K = 0.80067 kj/k(103 J/1 kj) = 800.67 J/K 0-15
S univ = rxn + surr = ( 4 J/K) + (800.67 J/K) = 558.67 = 559 J/K Because univ is positive, the reaction is spontaneous at 98 K. 0.4 Plan: Write the balanced equation. To calculate the standard entropy change (part a), use the relationship rxn = m S products n S reactants. To calculate ΔS o of the universe in part b, first calculate ΔH o of the reaction using the relationship rxn = m f (products) n f Use ΔH o of the reaction to calculate ΔS of the surroundings. surr = rxn T Add ΔS of the surroundings and ΔS o of the reaction to calculate ΔS of the universe. If ΔS of the universe is greater than zero, the reaction is spontaneous at the given temperature. S univ = rxn + surr a) One mole of dinitrogen oxide is formed from its elements in their standard states according to the following equation: N (g) + 1/O (g) N O(g) = [(1 mol N O)(S of N O] [(1 mol N )(S of N ) + (1/ mol O )(S of O )] = [(1 mol)(19.7 J/K mol)] [(1 mol)(191.5 J/K mol) + (1/ mol)(05.0 J/K mol)] = 14.775 = 74.3 J/K b) rxn = m f (products) n f rxn = [(1 mol N O)( f of N O)] [(1 mol N )( f of N ) + (1/ mol O )( f of O )] rxn = [(1 mol N O)(8.05 kj/mol)] [(1 mol N )(0 kj/mol) + (1/ mol O )(0 kj/mol)] rxn = 8.05 kj surr = rxn 8.05 kj = T 98 K = 0.7534 kj/k(103 J/1 kj) = 75.34 J/K S univ = rxn + surr = ( 74.3 J/K) + ( 75.34J/K) = 349.64 = 340.6 J/K Because univ is negative, the reaction is not spontaneous at 98 K. 0.43 SO (g) + Ca(OH) (s) CaSO 3 (s) + H O(l) = [(1 mol CaSO 3 )(S of CaSO 3 ) + (1 mol H O)(S of H O)] [(1 mol SO )(S of SO ) + (1 mol Ca(OH) )(S of Ca(OH) )] = [1 mol)(101.4 J/K mol) + (1 mol)(69.940 J/K mol)] [(1 mol)(48.1 J/K mol) + (1 mol)(83.39 J/K mol)] = 160.15 = 160. J/K 0.44 Plan: Write the balanced equation. To calculate the standard entropy change, use the relationship rxn = m S products n S reactants. Complete combustion of a hydrocarbon includes oxygen as a reactant and carbon dioxide and water as the products. C H (g) + 5/O (g) CO (g) + H O(g) 0-16
O )] = [( mol CO )(S of CO ) + (1 mol H O)(S of H O)] [(1 mol C H )(S of C H ) + (5/ mol O )(S of = [( mol)(13.7 J/mol K) + (1 mol)(188.7 J/mol K)] [(1 mol)(00.85 J/mol K) + (5/ mol)(05.0 J/mol K)] = 97.3 = 97. J/K 0.45 Reaction spontaneity may now be predicted from the value of only one variable ( sys ) rather than two ( sys and surr ). 0.46 A spontaneous process has univ > 0. Since the Kelvin temperature is always positive, sys must be negative ( sys < 0) for a spontaneous process. 0.47 a) = T. Since T > 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 is positive.. Since gases have higher entropy values than solids, the process results in an increase in entropy and is positive. 3. This is an endothermic process so the surroundings lose energy to the system. surr is negative. 4. = T. Both and are positive. At low temperature, the term will predominate and will be positive; at high temperatures, the T term will predominate and will be negative. 0.48 Plan: Examine the provided diagrams. Determine whether bonds are being formed or broken in order to determine the sign of ΔH. Determine the relative number of particles before and after the reaction in order to determine the sign of ΔS. a) The sign of ΔH is negative. Bonds are being formed, so energy is released and ΔH is negative. b) The sign of ΔS is negative. There are fewer particles in the system after the reaction, so entropy decreases. c) The sign of ΔS surr is positive. surr = rxn Since ΔH rxn is negative, ΔS surr will be positive. T d) Because both ΔH and ΔS are negative, this reaction will become more spontaneous (ΔG will become more negative) as temperature decreases. 0.49 is positive and is positive. The reaction is endothermic ( > 0) and requires a lot of heat from its surroundings to be spontaneous. The removal of heat from the surroundings results in < 0. The only way an endothermic reaction can proceed spontaneously is if > 0, effectively offsetting the decrease in surroundings entropy. In summary, the values of and are both positive for this reaction. Melting is an example. 0.50 For a given substance, the entropy changes greatly from one phase to another, e.g., from liquid to gas. However, the entropy changes little within a phase. As long as the substance does not change phase, the value of is relatively unaffected by temperature. 0.51 Plan: can be calculated with the relationship m f (products) n f. a) = [( mol MgO)( f of MgO)] [( mol Mg)( f of Mg) + (1 mol O )( f of O )] Both Mg(s) and O (g) are the standard-state forms of their respective elements, so their f values are zero. = [( mol)( 569.0 kj/mol)] [( mol)(0) + (1 mol)(0)] = 1138.0 kj b) = [( mol CO )( f of CO ) + (4 mol H O)( f of H O)] [( mol CH 3 OH)( f of CH 3 OH) + (3 mol O )( f of O )] 0-17
= [( mol)( 394.4 kj/mol) + (4 mol)( 8.60 kj/mol)] [( mol)( 161.9 kj/mol) + (3 mol)(0)] = 1379.4 kj c) = [(1 mol BaCO 3 )( f of BaCO 3 )] [(1 mol BaO)( f of BaO) + (1 mol CO )( f of CO )] = [(1 mol)( 1139 kj/mol)] [(1 mol)( 50.4 kj/mol) + (1 mol)( 394.4 kj/mol)] = 4. = 4 Kj 0.5 a) H (g) + I (s) HI(g) = [( mol HI)( f of HI] [(1 mol H )( f of H ) + (1 mol I )( f of I )] = [( mol)(1.3 kj/mol)] [(1 mol)(0 kj/mol) + (1 mol)(0 kj/mol)] =.6 kj b) MnO (s) + CO(g) Mn(s) + CO (g) = [(1 mol Mn)( f of Mn) + ( mol CO )( f of CO )] [(1 mol MnO )( f of MnO ) + ( mol CO)( f of CO)] = [(1 mol)(0 kj/mol) + ( mol)( 394.4 kj/mol)] [(1 mol)( 466.1 kj/mol) + ( mol)( 137. kj/mol)] = 48.3 kj c) NH 4 Cl(s) NH 3 (g) + HCl(g) = [(1 mol NH 3 )( f of NH 3 ) + (1 mol HCl)( f of HCl)] [(1 mol NH 4 Cl)( f of NH 4 Cl)] = [(1 mol)( 16 kj/mol) + (1 mol)( 95.30 kj/mol)] [1 mol)( 03.0 kj/mol)] = 91.7 = 9 kj 0.53 Plan: rxn can be calculated from the individual f values of the reactants and products by using the relationship rxn = m f (products) n f. rxn can be calculated from the individual S values of the reactants and products by using the relationship rxn = m S products n S reactants. Once rxn and rxn are known, can be calculated with the relationship rxn = rxn Trxn. rxn values in J/K must be converted to units of kj/k to match the units of rxn. a) rxn = [( mol MgO)( f of MgO)] [( mol Mg)( f of Mg) + (1 mol O )( f of O )] rxn = [( mol)( 601. kj/mol)] [( mol)(0 kj/mol) + (1 mol)(0 kj/mol)] rxn = 10.4 kj rxn = [( mol MgO)( S of MgO)] [( mol Mg)( S of Mg) + (1 mol O )( S of O )] rxn = [( mol)(6.9 J/mol K)] [( mol)(3.69 J/mol K) + (1 mol)(05.0 J/mol K)] rxn = 16.58 J/K rxn = rxn Trxn = 10.4 kj [(98 K)( 16.58 J/K)(1 kj/10 3 J)] = 1137.859 = 1138 kj b) rxn = [( mol CO )( f of CO ) + (4 mol H O)( f of H O)] [( mol CH 3 OH)( f of CH 3 OH) + (3 mol O )( f of O )] rxn = [( mol)( 393.5 kj/mol) + (4 mol)( 41.86 kj/mol)] [( mol)( 01. kj/mol) + (3 mol)(0 kj/mol)] rxn = 1351.904 kj rxn = [( mol CO )( S of CO ) + (4 mol H O)( S of H O)] [( mol CH 3 OH)( S of CH 3 OH) + (3 mol O )( S of O )] rxn = [( mol)(13.7 J/mol K) + (4 mol)(188.7 J/mol K)] 0-18
[( mol)(38 J/mol K) + (3 mol)(05.0 J/mol K)] = 91.8 J/K rxn = rxn Trxn = 1351.904 kj [(98 K)(91.8 J/K)(1 kj/10 3 J)] = 1379.105 = 1379 kj c) rxn = [(1 mol BaCO 3 )( f of BaCO 3 )] [(1 mol BaO)( f of BaO) + (1 mol CO )( f of CO )] rxn = [(1 mol)( 119 kj/mol)] [(1 mol)( 548.1 kj/mol) + (1 mol)( 393.5 kj/mol)] rxn = 77.4 kj rxn = [(1 mol BaCO 3 )( S of BaCO 3 )] [(1 mol BaO)( S of BaO) + (1 mol CO )( S of CO )] rxn = [(1 mol)(11 J/mol K)] [(1 mol)(7.07 J/mol K) + (1 mol)(13.7 J/mol K)] rxn = 173.77 J/K rxn = rxn Trxn = 77.4 kj [(98 K)( 173.77 J/K)(1 kj/10 3 J)] = 5.665 = 6 kj 0.54 a) rxn = [( mol HI)(5.9 kj/mol)] [(1 mol H )(0 kj/mol) + (1 mol I )(0 kj/mol)] rxn = 51.8 kj rxn = [( mol HI)(06.33 J/K mol)] [(1 mol H )(130.6 J/K mol) + (1 mol I )(116.14 J/K mol)] rxn = 165.9 J/K rxn = rxn Trxn = 51.8 kj [(98 K)(165.9 J/K)(1 kj/10 3 J)] =.3558 =.4 kj b) rxn = [(1 mol Mn)(0 kj/mol) + ( mol CO )( 393.5 kj/mol)] [(1 mol MnO )( 50.9 kj/mol) + ( mol CO)( 110.5 kj/mol)] rxn = 45.1 kj rxn = [(1 mol Mn)(31.8 J/K mol) + ( mol CO )(13.7 J/K mol)] [(1 mol MnO )(53.1 J/K mol) + ( mol CO)(197.5 J/K mol)] rxn = 11.1 J/K rxn = rxn Trxn = 45.1 kj [(98 K (11.1 J/K)(1 kj/10 3 J)] = 48.4078 = 48.4 kj c) rxn = [(1 mol NH 3 )( 45.9 kj/mol) + (1 mol HCl)( 9.3 kj/mol)] [(1 mol NH 4 Cl)( 314.4 kj/mol)] rxn = 176. kj rxn = [(1 mol NH 3 )(193 J/K mol) + (1 mol HCl)(186.79 J/K mol)] [(1 mol NH 4 Cl)(94.6 J/K mol)] rxn = 85.19 J/K rxn = rxn Trxn = 176. kj [(98 K)(85.19 J/K)(1 kj/10 3 J)] = 91.13 = 91. kj 0.55 Plan: rxn can be calculated with the relationship m f (products) n f. Alternatively, rxn can be calculated with the relationship rxn = rxn Trxn. Entropy decreases (is negative) when there are fewer moles of gaseous products than there are of gaseous reactants. o a) Entropy decreases ( ΔS negative) because the number of moles of gas decreases from reactants (1 1/ mol) to products (1 mole). The oxidation (combustion) of CO requires initial energy input to start the reaction, but then o releases energy (exothermic, ΔH negative) which is typical of all combustion reactions. b) Method 1: Calculate rxn from f values of products and reactants. rxn = m f (products) n f rxn = [(1 mol CO )( f of CO )] [(1 mol CO)( f of CO) + (1/ mol)( f of O )] 0-19
rxn = [(1 mol)( 394.4 kj/mol)] [(1 mol)( 137. kj/mol) + (1/ mol)(0 kj/mol)] = 57. kj Method : Calculate rxn from rxn and rxn at 98 K (the degree superscript indicates a reaction at standard state, given in the Appendix at 5 C). rxn = m f (products) n f rxn = [(1 mol CO )( f of CO )] [(1 mol CO)( f of CO) + (1/ mol)( f of O )] rxn = [(1 mol)( 393.5 kj/mol)] [(1 mol)( 110.5 kj/mol) + (1/ mol)(0 kj/mol)] = 83.0 kj rxn = m S products n S reactants rxn = [(1 mol CO )( S of CO )] [(1 mol CO)( S of CO) + (1/ mol)( S of O )] rxn = [(1mol)(13.7 J/mol K)] [(1mol)(197.5 J/mol K) + (1/ mol)(05.0 J/mol K)] rxn = 86.3 J/K rxn = rxn Trxn = ( 83.0 kj) [(98 K)( 86.3 J/K)(1 kj/10 3 J)] = 57.86 = 57.3 kj 0.56 C 4 H 10 (g) + 13/O (g) 4CO (g) + 5H O(g) a) An increase in the number of moles of gas should result in a positive value. The combustion of C 4 H 10 (g) will result in a release of energy or a negative value. b) rxn = m f (products) n f rxn = [(4 mol CO )( 393.5 kj/mol) + (5 mol H O)( 41.86 kj/mol)] [(1 mol C 4 H 10 )( 16 kj/mol) + (13/ mol O )(0 kj/mol)] rxn = 657.13 kj rxn = m S products n S reactants rxn = [(4 mol CO )(13.7 J/K mol) + (5 mol H O)(188.7 J/K mol)] [(1 mol C 4 H 10 )(310 J/K mol) + (13/ mol O )(05.0 J/K mol)] rxn = 155.9 J/K rxn = rxn Trxn = 657.13 kj [(98 K)(155.9 J/K)(1 kj/10 3 J)] = 703.588 = 704 kj rxn = m f (products) n f rxn = [(4 mol CO )( 394.4 kj/mol) + (5 mol H O)( 8.60 kj/mol)] [(1 mol C 4 H 10 )( 16.7 kj/mol) + (13/ mol O )(0 kj/mol)] rxn = 703.9 kj 0.57 Plan: Use the relationship rxn = rxn Trxn to find rxn, knowing rxn and rxn. This relationship is also used to find rxn at a different temperature. Reaction is Xe(g) + 3F (g) XeF 6 (g) a) rxn = rxn Trxn H G 40 kj/mol ( 80. kj/mol) = = = 0.40939597 = 0.409 kj/mol K T 98 K b) rxn = rxn Trxn = ( 40 kj/mol) [(500. K)( 0.40939597 kj/mol K)] = 197.30 = 197 kj/mol 0.58 a) = H G ( 0. kj/mol) ( 06 kj/mol) = = 0.046979865 = 0.047 kj/mol K T 98 K 0-0
b) rxn = rxn Trxn = 0. kj/mol (450. K)( 0.046979865 kj/k mol) = 198.859 = 199 kj/mol 0.59 Plan: rxn can be calculated from the individual f values of the reactants and products by using the relationship rxn = m f (products) n f. rxn can be calculated from the individual S values of the reactants and products by using the relationship n S reactants. Once rxn and rxn are known, can be calculated with the relationship rxn = rxn Trxn. rxn values in J/K must be converted to units of kj/k to match the units of rxn. The temperature at which a reaction becomes spontaneous can be calculated by setting ΔG to zero in the free energy equation (assuming the reaction is at equilibrium) and solving for T. rxn = m S products a) rxn = [(1 mol CO)( f of CO) + ( mol H )( f of H )] [(1 mol CH 3 OH)( f of CH 3 OH)] rxn = [(1 mol)( 110.5 kj/mol) + ( mol)(0 kj/mol)] [(1 mol)( 01. kj/mol)] rxn = 90.7 kj rxn = [(1 mol CO)( S of CO) + ( mol H )( S of H )] [(1 mol CH 3 OH)( S of CH 3 OH)] rxn = [(1 mol)(197.5 J/mol K) + ( mol)(130.6 J/mol K)] [(1 mol)(38 J/mol K)] rxn = 0.7 = 1 J/K b) rxn = rxn Trxn T 1 = 8 + 73 = 301 K = 90.7 kj [(301 K)(0.7 J/K)(1 kj/10 3 J)] = 4.693 = 4.3 kj T = 18 + 73 = 401 K = 90.7 kj [(401 K)(0.7 J/K)(1 kj/10 3 J)] =.1993 =. kj T 3 = 8 + 73 = 501 K = 90.7 kj [(501 K)(0.7 J/K)(1 kj/10 3 J)] = 19.8707 = 19.9 kj c) For the substances in their standard states, the reaction is nonspontaneous at 8 C, near equilibrium at 18 C, and spontaneous at 8 C. Reactions with positive values of rxn and rxn become spontaneous at high temperatures. d) The reaction will become spontaneous when ΔG changes from being positive to being negative. This point occurs when ΔG is 0. rxn = rxn Trxn 0 = 90.7 x 10 3 J (T)(1 J/K) T = 410. K At temperatures above 410. K, this reaction is spontaneous. (Because both ΔH and ΔS are positive, the reaction becomes spontaneous above this temperature.) 0.60 Plan: rxn can be calculated from the individual f values of the reactants and products by using the relationship rxn = m f (products) n f. rxn can be calculated from the individual S values of the reactants and products by using the relationship rxn = m S products n S reactants. Once rxn and rxn are known, can be calculated with the relationship rxn = rxn Trxn. rxn values in J/K must be converted to units of kj/k to match the units of rxn. The temperature at which a reaction becomes spontaneous can be calculated by setting ΔG to zero in the free energy equation (assuming the reaction is at equilibrium) and solving for T. a) N (g) + O (g) NO(g) rxn = [( mol NO)(90.9 kj/mol)] [(1 mol N )(0 kj/mol) + (1 mol O )(0 kj/mol)] rxn = 180.58 kj rxn = [( mol NO)(10.65 J/K mol)] [(1 mol N )(191.5 J/K mol) + (1 mol O )(05.0 J/K mol)] 0-1