Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume cnstant spacing between levels t determine the energies f higher energy levels. 8 6 A B C D Use the letter U fr the energy unit. 1. What are the energies f systems A and B? Which system is at the higher temperature? E A = +++ = 8U E B = 3+++1 = 6U System A has the greater energy, s it is at the higher temperature. 8 6 3. In hw many ways can the energies f systems A and B be distributed? Which system has the greater entrpy? 8 6 System A has 8U f energy and A mlecules have energy levels at,, 6, and 8U. The pssible cmbinatins that sum t 8U and the number f each are: ne mlecule with 8U {}, One with 6U and ne with U {1}, tw mlecules with U each {6}; ne mlecule with U and tw with U each {1}, all fur mlecules with U each {1}. The ttal number f ways = + 1 + 6 + 1 + 1 = 35. System B has 6U f energy and B mlecules have energy levels at 1,, 3,, 5, and 6U. The pssible cmbinatins that sum t 6U and the number f each are: ne mlecule with 6U {}, One with 5U and ne with 1U {1}, ne mlecule with U and ne with U {1}; ne mlecule with U and tw with 1U each {1}, tw mlecules with 3U each {6}; ne mlecule each with 3U, U, and 1U {}; three mlecules with U each {}. The ttal number f ways = + 1 + 1 + 1 + 6 + + = 7. There are mre ways t distribute the energy in System B, s it has the higher entrpy. Nte System A has mre energy, but System B has the higher entrpy because the energy levels are clser in System B. 5. List the systems in rder f increasing entrpy at a temperature where they all had 1 U f energy. The entrpies f a grup f systems at the same temperature increase as the spacing between their energy levels decreases. The energy levels are clsest in System B, s it wuld have the highest energy at any given temperature. 7. State the secnd law f thermdynamics. The entrpy f the universe must increase during a spntaneus prcess. 9. What effect des the enthalpy change f a prcess have n the entrpy f the universe in prcesses carried ut at cnstant T and P? The heat absrbed (r given ff) by a system cmes frm (r enters) the surrundings. The heat flw causes an entrpy in the surrundings: S sur = - H/T, which impacts the entrpy f the universe: S univ = S + S sur. 11. Cnsider the reactin H + I HI G =.6 kj at 98 K. Criticize and crrect the fllwing statement: G >, s the reactin is nt spntaneus and HI cannt be made frm this reactin at 98 K. ` G indicates the extent f the reactin, nt the spntaneity. G > means that K < 1, s the equilibrium cncentratin f HI will be less than at least ne f the reactants, but it will nt be zer. 13. Explain why disslving sugar in water always results in a hmgeneus slutin. Disslving sugar in water will result in a hmgenus mixture because the water mlecules have the highest entrpy when they are farthest frm each ther. 15. Indicate which member f each pair has the higher entrpy and indicate the reasn a) PF 3 (g) r PF 5 (g) at 75 C PF 5 (g) has higher entrpy because it has mre atms and mre degrees f freedm. b) I (s) r I (g) at 3 K I (g) has higher entrpy as entrpy increases frm slid t liquid t gas. c) He(g) at K r He(g) at 6 K He(g) at 6K has higher entrpy as entrpy increases with increase in temperature. d) A piece f tin r a piece f 6: slder (a slid slutin that is 6% Pb and % Sn. 6: slder has higher entrpy than a piece f tin as entrpy decreases as perfectin f crystal increases. 8 6
17. What is an extensive prcess? What thermdynamic prperty indicates the extent f a reactin? An Extensive Prcess is ne which ges essentially t cmpletin, i.e., a prcess in which K >> 1. The standard free energy change ( G ) indicates the extent f the reactin. 19. Indicate whether each f the fllwing statements must be, can be, r cannt be true fr a spntaneus endthermic reactin at cnstant pressure: a) H > This statement must be true because the reactin is endthermic. b) S univ = This statement cannt be true because S univ > fr a spntaneus prcess. c) G < This statement can be true, but it cannt be determined frm the given infrmatin. d) G < Must be true because the reactin is spntaneus at cnstant T and P. e) S > Must be true. G = H - T S, s S must be psitive if G < and H >. f) E univ = Must be true; it is the first law f thermdynamics. 1. Indicate the sign f the entrpy change fr each f the fllwing prcesses: a) Increasing the temperature f a pt f water. Mlecular mtin (disrder) increases with thermal energy, S >. b) Cndensing a gas The material becmes mre rdered as it is cndensed frm a gas t a liquid; S <. c) Clearing a field and planting rws f crn This prcess results in mre rder; S <. d) NH NO 3 (s) N O(g) + H O(g) 3 mles f gas are prduced per mle f slid that reacts; S >. 3. Use data in Appendix G t calculate the standard entrpy change fr a) the rusting f irn: Fe(s) + 3 O (g) Fe O 3 (s) S RXN = x S (Fe O 3 ) [ x S (Fe) + 3 x S (O )] = (87.) [(7.3) + 3(5.3)] = -59.5 J/K b) the decmpsitin reactin: CaCO 3 (s) CaO(s) + CO (g) S RXN = S (CO ) + S (CaO) - S (CaCO 3 )] = 38. + 13.6 9.9 = 158.9 J/K 5. Gashl is a mixture f ethanl (C H 5 OH) and gasline. Write the chemical equatin fr the cmbustin f ethanl and determine the maximum amunt f wrk that can be btained frm the cmbustin f 1 gal f ethanl at 98 K and standard cnditins. The density f ethanl is.789 g. ml, and 1 gal is 3.79 L. Accrding t Equatin.3, the maximum wrk dne by a system = - G The chemical equatin fr the cmbustin and the standard free energies are CH 3 CH OH (l) + 3 O (g) 3 H O (l) + CO (g) G f 7.76-37. -39.36 G = [(-39.36) + 3(-37.)] - [(7.76] = 36 kj T calculate the free energy change, G, fr 1 galln f ethanl, cnvert t mles and multiply. The mlecular weight f ethanl (C H 5 OH) is 6.69 g ml. 3.79 Liters 1 ml.789 g 1 ml 136 kj 1 gal. CH5OH = 8.61 1 kj 1galln 1 L 1 ml 6.69 g ml 7. Determine the standard entrpy f frmatin fr each f the fllwing substances at 98 K: a) H (g) H (g) H (g); S = ; n change in prducts frm reactants. b) H O(g) H (g) + 1 / O (g) H O (g) S 13.57 5.3 188.7 S = 188.7 1(13.57) 1 / (5.3) = -. J/K fr the abve reactin, s S = -. J ml K c) NH 3 (g) 1 / N (g) + 3 / H (g) NH 3 (g) S 191.5 13.57 19.3 J. ml. K f S = 19.3 1 / (191.5) 3 / (13.57) = -99.3 J/K fr the abve reactin, S = -99.3 J ml K f 9. Which f the fllwing is the mst stable under standard cnditins: Al O 3 (s), NO(g) r FeO(s)? The mre psitive the value f G, the greater the tendency t decmpse, and the less stable the cmpund is. Al O 3 (s) 58 kj/ml NO(g) 86.6 kj/ml FeO(s) -55. kj/ml Al O 3 (s) is the mst stable. -
31. Draw diagrams that indicate the relative psitins f the free energy minima fr reactins with the fllwing G values. Refer t Figure.3. a) small and negative b) large and negative a) % 1 % Extent f reactin is slightly greater than 5% because DG is negative but small b) % 1 % Extent f reactin is great because DG is large and negative 33. When gaseus zne, O 3, is frmed frm O by the reactin 3 / O (g) O 3 (g). What is the equilibrium cnstant f this reactin at 98 K? We must first find G RXN, using Appendix G: G RXN = 163-3/() = 163 kj/ml Frm equatin.11, K= exp(- G /RT). Therefre, K = exp[-(163)/.78)] =.68 x 1-9. 35. Write the expressin fr the reactin qutient and determine the value f the equilibrium cnstant at 98 K fr each f the fllwing reactins: (P HBr ) a) H (g) + Br (l) HBr(g) Q = (P )(P ) H Br The chemical equatin is fr the frmatin f HBr, s G = G f (HBr) = ( ml)(-53.5 kj/ml) = 7. kj Frm Eqn.11: K= exp(- G /RT); RT = (.831 kj/ml. K)(98 K) =.8 kj/ml K = exp[-(7. kj/ml)/(.8/ml)] = e 3.15 = 5.66 x 1 18 b) I (s) I (g) Q = P I ; G = 19.36 = 19.36 kj/ml; K = exp[-(19.36)/(.8)] =. x 1 - c) CH 3 OH(l) + 3O (g) CO (g) + H O(l) (P CO ) Q = (P 3 O ) G RXN = (-39.36) + (-37.) - [ + (66.3)] = 5.1 kj/ml [ ( 15.1)] 567. 6 K = exp = e = 1.78. If yur calculatr des nt give this result, try slving fr the lg K lg K = 567/.33 = 6, which means that K = 1 6 d) NH 3 (g) + HCl(g) NH Cl(s) Q = (P 1 )(P ) G RXN = -3. [6.5 + -95.31] = -91. kj/ml, s K = exp [-(-91.)/.8] = e 36.8 = 9.6 x 1 15 37. Indicate the activity f each f the fllwing: The activity f a species is the rati f the cncentratin f the species t its cncentratin in its standard state. Fr pure slids and liquids in their standard state, their activity is unity (1). The activity f a gas in its standard state is 1 atm and fr a slute it is 1 M. Thus, the activity f slute A is [A]/1M, which is unitless but numerically equal t the mlarity f A. Similarly, the activity f gas B is P B /1atm, which is als unitless but numerically equal its partial pressure in atmspheres. a) NH 3 gas at. atm a = (. atm)/(1 atm) =.. b) Cl 1- in in a.11-m slutin f NaCl a =.11 M / 1 M =.11. c) A crystal f AgCl sitting in liquid water Pure slids in their standard states have an activity f 1 (unity). d) H gas at 31 trr Change trr t atm: (31 trr)(76 trr/1 atm) =. atm. a =. atm/1 atm =.. 39. K and K p have the same value fr the fllwing. Hwever, K has n units, while K p des. What are the units f K c fr each f the fllwing the reactins? NH3 HCl a) ClF(g) + O (g) Cl O(g) + OF (g) K atm atm = = atm atm P atm -3
b) HI(g) H (g) + I (g) K atm atm = = atm P n units c) NH NO 3 (s) N O(g) + H O(g) K = P atm 1. Cnsider the fllwing gas phase equilibrium at 98 K: N (g) + 3 H (g) Í NH 3 (g) a) Frm standard free energies f frmatin, calculate the equilibrium cnstant K at 98 K. This is the frmatin reactin fr ammnia. G = Gf ( NH 3) = ( ml)(6.5 kj/ml) = -33. kj G - -33. kj/ml RT 5 Use Eq..11 t slve fr K: K = e = exp -3 = 6.9 x 1 (8.31 1 kj/ml K)( 98 K) b) Calculate G when a reactin mixture cnsists f 1 atm N, 1 atm H, and 1 atm NH 3. PNH 1 3 Determine the reactin qutient with the given pressures: Q = = = 1. 1 3 3 PH P N 1 1 - Q 1. 1 Use Eq..1 t calculate G: G = RT ln = (.831 kj/ml K)(98 K)(ln ) = -55.8 kj/ml 5 K 6.9 1 3. Cnsider the fllwing acid-base reactin: NH 1+ (aq) + CN 1- (aq) NH 3 (aq) + HCN(aq) K = 1.7 a) What is the value f G? G = -RTlnK = -(8.31 x 1-3 kj.ml.k)(98 K)(ln 1.7) =.3 kj/ml What are the values f Q and G at 98 K and in which directinis the reactin prceeding? [NH 1+ ] [CN 1- ] [ NH 3 ] [HCN] b).1.1.1.1 Q [ NH3 ][ HCN ] (.1)(.1) G = RTln Q = = = 1+ 1- K NH CN (.1)(.1) 1 1 G = (8.31 J ml K )(98 K)(ln ) =.3 kj 1.7 ( G <, s Rxn ) c).1.1.17.1 (.17)(.1) 1.7 Q = = 1.7; G = (8.31 J ml K )(98 K)( ln ) = ( G =, s Rxn ) (.1)(.1) 1.7 d).18.18.1.1 (.1)(.1).31 Q = =.31; G = (8.31 J ml K )(98 K)( ln ) = -.kj ( G <, s Rxn ) (.18)(.18) 1.7 5. The prcess f disslving a gas in a liquid is exthermic because f slvent-slute interactins. Cnsider the prcess f making carbnated water, CO (g) CO (aq) H = 9. kj a) Write the equilibrium cnstant expressin fr the carbnatin prcess and slve it fr the equilibrium cncentratin f CO in slutin in terms f the partial pressure f the gas in equilibrium with it. [CO (aq)] K = [CO (aq)] = KPCO P CO b) Use Equatin.19 and the expressin fr the equilibrium cnstant fr the carbnatin prcess t explain why a carbnated drink gets flat when it is allwed t warm. Accrding t Eq.13, the equilibrium cnstant fr an exthermic reactin decreases as the temperature increases. Thus, warming the drink decreases the equilibrium cnstant, which means that the pressure f the gas increases while the cncentratin f the gas in the slutin decreases. Lss f gas frm the slutin int the vapr causes the drink t g flat. c) Explain why a carbnated drink gets flat when allwed t sit in an pen cntainer. Fr an pen cntainer, P CO is lwer than in a clsed cntainer (P CO is quite lw in the atmsphere). As a result, the equilibrium will shift t the left reducing the cncentratin f CO in slutin. This can als be determined by rearranging the expressin frm Part b: [ CO (aq) ] = K P CO. Frm this equatin, ne can see that [CO (aq)] decreases with P CO. -
7. The pressure f water vapr in a clsed cntainer is.3 atm. If the vapr pressure f water at this temperature is 1.5 atm, what are the signs f G and G fr H O(l) H O(g)? P HO= 1.5 atm, s K = 1.5 P HO =.3 atm, s Q =.3 K > 1, s G <. Q < K s G < 9. Use the data in Appendix G t estimate the vapr pressure f water at 5. C. Express yur answer in trr. Recall that the vapr pressure is equal t the equilibrium cnstant (K) in atmspheres. Therefre, we need t calculate G vap at 5 C. Apply equatin. t the vaprizatin prcess: G vap = H vap - T S vap First, we need t calculate the H vap and S vap by taking the difference between the values fr H O(l) and H O(g) using the data fund in Appendix G (See Example. fr a similar prblem) H vap = -1.8 kj/ml - (-85.83 kj/ml) =.1 kj/ml S vap = 188.7 J/ml K - 69.1 J/ml K = 118.8 J/ml K Substituting int Eq.., we get: G vap =.1 kj/ml - (33 K)(.1188 kj/ml K) = 5.6 kj/ml Then, use eqn..6 t calculate the equilibrium cnstant (See Example.3 fr a similar prblem): G 5,6 J/ml -.1 K = exp - = exp - = e =.1 RT (8.31 J/ml K)(33 K) K has n units as derived abve, but is equal t the pressure in atmspheres, which can be cnverted t trr by multiplying by 76 trr/atm: (.1 atm)(76 trr/atm) = 93. trr 51. The heat f vaprizatin f SiCl at 3. K is 9.7 kj/ml and its vapr pressure is 3. trr. a) What are G and S f vaprizatin f SiCl at 3 K? (3/76) G = -RT ln K K = 1 G = ( 8.31J/ml K)(3K)(ln.5) K =.5 G = 7.75 kj = H T S G + H (7.75 kj) + 9.7 kj 1J S = = = 73. J/K T 3 K 1 kj b) Assume H and S are temperature independent and estimate the biling pint f SiCl. H vap 9.7 kj/ml G vap = H vap T S vap = T bp = = =6K r 133 C S.73 kj/ml K vap 53. The enthalpy f vaprizatin f H S at 1.8 K is 18.67 kj/ml. Calculate the mlar entrpy f vaprizatin f.5 mles f H S at this temperature. J H 1867 ml T slve this prblem, rearrange Eq..11: S= = =87.73 J ml K T 1.8 K Taking int accunt the number f mles: S = (.5 ml)(87.73 J.ml.K ) = 19 J.K. 55. The equilibrium cnstant f a reactin is 3.x1 3 at 358 K and 1.7x1 at 56K. What are H and S fr the reactin? Slve tw equatins in tw unknwns as fllws. Nte extra sig figs are used in intermediate steps. G = -RT ln K = H - T S G 1 = -(8.31 J ml K )(358 K)(ln 3) = -. 1 J = H - 358 S G = -(8.31 J ml K )(56 K)(ln 17) =.97 1 J = H - 56 S Subtract the tw standard free energies t eliminate the standard enthalpy then slve fr the entrpy. -55 J G 1 - G = (-. + 1.97) 1 = (-358 + 56) S S = = -6 J K 98 K Substitute this value int the expressin fr G t btain.97 1 J = H - (56 K)(-6. J K ) s H =.97 1 J -.116 1 J = -.1 1 J = -1 kj -5
57. The vapr pressure f slid CO (dry ice) is 8. trr at -9. C and 15 trr at. C. a) What is the value f G fr the reactin CO (s) CO (g) at each temperature? The first step is t cnvert mm Hg t atm. 1 atm 1 atm 8 mm Hg =.368 atm 15 mm Hg =.138 atm 76 mm Hg 76 mm Hg Thus, K =.368 at 9 C and K =.138 at 1 C. Use Eq..8 with these values f K and T. At -9 C: G =-RTlnK= -(.831 kj ml K )(183 K)(ln.368)=1.5 kj ml At C: G =-RTlnK=-(.831 J ml K )( 173 K)(ln.138)=.85 kj ml b) What are H and S fr the reactin CO (s) CO (g) at these temperatures? Assume that H and S are cnstant ver this temperature range. Use Eq.. fr each temperature t prduce tw equatins with tw unknwns ( H and S ). At -9 C: 1.5J ml = H 83 K S at C:.85 J ml = H 73 K S Subtract the equatin n the left frm the ne n the right and slve fr S. 1.33 kj ml 1.33 kj ml = (1K) S s S = =.13 kj ml K 1 K Substitute S int either f the riginal equatins. H =1.5 kj ml + (183 K)(.13 kj ml K ) = 6 kj ml c) A dry ice bath is used rutinely in the labratry t keep things cld. It is made by making a pwder f the dry ice and then mixing the pwder with a slvent t make a slurry. Estimate the temperature f a dry ice bath by determining the temperature at which the vapr pressure f CO is 1 atm. kj H 5.9 ml Apply Equatin.11 t the sublimatin f dry ice. T= = =195 K=-78 C kj S.133 59. Write the reactin that wuld result when the xidatin f S t SO is cupled t each f the fllwing reactins. Calculate G at 98 fr each cupled reactin and indicate whether it is extensive. The xidatin f S t SO is shwn as fllws: S(g) + O (g) SO (g) G = -3 kj/ml Each reactin is the reverse f the frmatin reactin f the substance, s use Appendix G t find G. a) Cr O 3 (s) Cr(s) + 3 / O (g) Cr O 3 (s) Cr(s) + 3 / O (g) ml K G = -(59 kj) = +159 kj 3 / S(s) + 3 / O (g) 3 / SO (g) G = 3 / (-3) = -5 kj Cr O 3 (s) + 3 / S Cr(s) + 3 / SO (g) b) SiO (s) Si(s) + O (g) SiO (s) Si(s) + O (g) S(s) + O (g) SO (g) Cr O 3 (s) + 3 / S Cr(s) + 3 / SO (g) G = 159 5 = +69 kj G >, s reactin is nt extensive. G = -(-857 kj) = +857 kj G = -3 kj G = 857 3 = +557 kj G >, s reactin is nt extensive. 61. Calculate H, S, and G at 98 K fr the fllwing reactin: Al O 3 (s) + 3C(s,graphite) + 3Cl (g) AlCl 3 (s) + 3CO(g) H = (-7.) + 3(11.5) [676 + 3() + 3()] = -63.96 kj/ml S = (11.7) + 3(197.56) [5.9 + 3(5.7) + 3(3.)] = 76.9 J/ml. K G = (-68.8) + 3(37.15) [58 + 3() + 3()] = -87.5 kj/ml a) Is the reactin extensive at standard cnditins and 98 K? The reactin is extensive at standard cnditins and 98 K because G <. b) Write the expressin fr K and calculate its value at 98 K. 3 (P CO) G 87.5 K = = exp = exp = e = 1.8 1 3 (P Cl ) RT.78 35.13 15-6
c) Estimate the value f K at 6 K. Thermdynamics and Equilibrium G = H -T S = -63.96 kj/ml (6K)(.769kJ/ml. K) = 1.1 kj/ml. K Then, K = exp-( G /RT) = exp-(1.1/.988) = 3.9 x 1 9 63. Cnsider the fllwing equilibrium, which defines the slubility f lead chlride in water at 5 C: PbCl (s) Pb + (aq) + Cl 1- (aq) K = 1.7x1-5 a) What is the value f G fr the reactin at 5 C? use Eq..8 t calculate G : G = -RTln(K) = -(8.31 J/ml K)(98K)ln(1.7x1-5 ) = 7. kj/ml Determine Q and G fr the slubility f lead chlride under each f the fllwing circumstances and indicate whether mre slid culd disslve r mre slid wuld precipitate: b) [Pb + ] = 1. mm and [Cl 1- ] = 15. mm Q = [Pb + ][Cl 1- ] =[.1 M][.15 M] =.8x1-7 Use Eqn..1 t calculate G fr this prcess: G = RTln(Q/K) = (8.31 J/ml K)(98)ln(.8x1-7 /1.7x1-5 ) =.1 kj/ml Because G <, mre slid will disslve. c) [Pb + ] =.75 M and [Cl 1- ] =. M Q = [Pb + ][Cl 1- ] =[.75 M][. M] = 1.x1 - G =.8 kj/ml G >, s n mre slid will disslve. 65. Indicate whether G increases, decreases, r remains the same as the partial pressure f CO is increased in each f the fllwing. If CO is a prduct then increasing its partial pressure increases Q, but if it is a reactant then the increase in pressure decreases Q. Using that lgic and G = RTln(Q/K), we cnclude that G becmes mre negative (decreases) if the CO is a reactant, but it gets mre psitve (increases) if CO is a prduct. a) CaCO 3 (s) CaO(s) + CO (g) decreases b) CO (g) CO (s) increases c) C 3 H 8 (g) + 5O (g) 3CO (g) + H O(l) increases 67. When bth driving frces in a reactn are favrable ( H < and S > ) the reactin is extensive (K > 1) at all temperatures, and when bth are unfavrable, the reactin is never extensive (K < 1). Hwever, when ne driving frce is favrable and the ther is nt, the reactin can be either extensive r nt depending upn the temperature. Determine the temperature at which K ~ 1 fr each f the fllwing reactins and discuss the cnditins at which the reactin is likely t be extensive. K = 1 when G = H T S =, s we determine H and S at 98. If they have the same sign, ne is favrable ane the ther is nt, s there is a temperature at which G =. We btain the temperature by setting H = T S and slve fr T ~ H / S. We use the apprximately equals sign because we are assuming that H and S are temperature independent. The effect f temperature is determined by the sign f S. When the entrpy term is favrable ( S > ), the extent f reactin is increased by increasing the temperature, s the reactin is extensive at temperatures abve the ne at which K ~ 1. When the entrpy term is unfavrable ( S < ), reducing the temperature increases the extent f reactin. a) NO (g) N O (g) H = 9.66 (33.8) = -58. kj; S = 3.3 (.5) = 76.6 J.K T ~ H / S = -58./-.1766 = 39K. The entrpy term is unfavrable ( S < ), s the reactin is extensive at temperatures belw ~3 K. b) SO 3 (g) 1 / O (g) + SO (g) H = ( 1 / )() 96.83 + 395.7 = +98.89 kj; S = ( 1 / )(5.3) + 8.1 56.6 = 9. J.K T ~ H / S = 98.89/.9 = 1.5x1 3 K. The entrpy term is favrable ( S > ), s the reactin is extensive at temperatures abve ~11 K. -7
c) NH 3 (g) + HCl(g) NH Cl(s) H = -31. + 6.11 + 9.3 = 75.99 kj; S = 9.6 19.3 186.8 = -8.5 J.K T ~ H / S = 75.99/-.85 = 618 K. S <, s the reactin is extensive at temperatures belw ~6 K d) N (g) + O (g) NO (g) H = (33.8) - = 67.7 kj; S = (.5) 191.5 5.3 = 8. J.K T ~ H / S = 67.7/.8= 83 K. S >, s the reactin is extensive at temperatures abve ~8 K 69. Determine S and H frm the fllwing equilibrium cnstant/temperature data: T( C) 5 3 5 6 K 19.6 1.9 11.6 7.8 5.6 T use all f the data, we must plt RlnK versus 1/T and determine the slpe ( S ) and intercept (- H ) f the best straight line thrugh the data. First, cnvert then plt the data. Recall the T must always be n the Kelvin scale. t ( C) T (K) 1/T (x1 3 ) R ln K 73.11.7 5 98.8 19.86 3 53 1.988.38 5 53 1.91 17.8 6 533 1.876 1.3 R lnk 18 16 1.18.19..1. 1/T The slpe and intercept f the best line (shwn in plt) are 1513 J. ml and -6.9 J. ml. K. Therefre, H = -1.5 kj. ml and S = -6.9 J. ml. K 71. Use the fllwing equilibrium cnstant-temperature data t determine H and S fr the reactin. T (K) 3 35 375 6 K. 87.7 33. 7. 1.8 Cnvert the given data t that required in Equatin.18. The resulting table f 1/T and R ln K values is 1/T x 1 3 R ln K 3.33 63.19.86 37..67 9.1.5 16.18.17.89 7 6 5 3 1..5.3.35 1/ T Linear regressin yields an intercept f 9 and a slpe f 5.13x1, s S = 9 J. K and H = -51.3 kj. If nly the last tw data pints are used, the slpe = 3.x1 J.K, which which differs frm the best fit slpe by 33%. Such large deviatins can ccur when ne r bth pints are nt n the best fit line. -8