Chem 338 Hmewrk Set #4 slutins September 25, 21 Frm Atkins: 4.2, 4.4, 4.6, 4.12, 4.18, 4.19, 4.22 4.2) Suppse yu put a cube f ice f mass 1 g int glass f water at just abve ºC. When the ice melts, abut 33 kj f energy is absrbed frm the surrundings as heat. What is the change in entrpy f (a) the sample (the ice), (b) the surrundings (the glass f water)? a) S sys b) S surr = + 33 kj =+ 12. kj K = 33 kj = 12. kj K 4.4) Calculate the change in entrpy f 1 g f ice at ºC as it is melted, heated t 1ºC, and then vaprized at that temperature. Suppse that the changes are brught abut by a heater that supplies heat at a cnstant rate, and sketch a graph shwing (a) the change in temperature f the system, (b) then enthalpy f the system, (c) the entrpy f the system as a functin f time. S S 1 2 S3 1 g H2O(s) C H2O(l) C H2O(l) 1 C H2O(g) 1 C 1 ml amunt f ice: n = 1 g = 5. 559 ml 18.15 g q n H S rev fus 1 = 5 559 6 1 T = T = (. )(. ) 15 =.. 12213 kj/k Tf 373. 15 S2 = ncpm, ln = ( 5. 559)(. 7529) ln =. 1338 kj/k Ti. 15 S3 = n Hvap 5 559 4 7 T = (. )(.) 373 15 =.. 6544 kj/k Sttal = S1 + S2 + S3 =. 858 kj / K = 86 J / K
Ttal heat input int the system: q = H = n H fus + nc p, m T + n Hvap = 33. 361 + 41. 793 + 225. 92 = 31 kj Fr a cnstant rate f heat input int the system: Temperature vs. time (as in the last hmewrk set) 1ºC liquid t gas phase change Temp warming f liquid frm ºC t 1ºC ºC ice t liquid phase change Time
Enthalpy vs. time (nte: q = H and q is being delivered at a cnstant rate) 31 H (J) Time Entrpy vs. time 86 liquid t gas phase change S (J) warming f liquid frm ºC t 1ºC (varies as ln T) ice t liquid phase change Time
4.6) A sample f carbn dixide that initially ccupies 15. L at 25 K and 1. atm is cmpressed isthermally. Int what vlume must the gas be cmpresed t reduce its entrpy by 1. J K? First calculate the number f mles f gas: pv n = RT = (. 1)( 15.) (. )( ) =. 7312 ml 82578 25 Fr the isthermal expansin/cmpressin f a perfect gas: S nr V f = ln Vi Slving fr V f : Vf = S nr Ve i Fr an entrpy change f 1. J/K : Vf = ( 15. ) e = 29. L 1. (. 7312 )( 8. 31451 ) 4.12) Calculate the change in entrpy when 1 g f water at 8ºC is pured int 1 g f water at 1ºC in an insulated vessel given that C p,m = 75.5 J K ml. We first need t find the final temperature. Since the amunts f ht and cld water are the same, the final temp is just the average f the initial temperatures: (8 + 1)/2 = 45ºC. Mre rigrusly, this is btained by nting that the heat lst by the ht water is exactly equal t the heat gained by the cld water since the cntainer is insulated: qlst by ht water = qgained by cld water nhcpm, ( Tf 353) = nccpm, ( Tf 283) and thus, Tf = 318 K (45 C) since nh = nc
Since entrpy is a state functin we can break dwn this prcess int tw steps, the cling f the ht water and the heating f the cld water: S = Sht + Scld 318 318 = nhcpm, ln + nccpm, ln 353 283 1 ml In each case, the amunt f water is: n = 1 g = 5. 559 ml 18.15 g S = ( 5. 559)( 75. 5) [. 144 +. 1166]= 5. 11 J / K 4.18) Calculate the standard reactin entrpy at 298 K f (a) 2CH 3 CHO(g) + O 2 (g) 2CH 3 COOH(l) In each case, S r is given by ν S ν S prd p m reac r m S r = 2( 159. 8) [ 2( 25. 3) + ( 25. 138) ]= 386. 1 J K ml (b) 2AgCl(s) + Br 2 (l) 2AgBr(s) + Cl 2 (g) S r = 2( 17. 1) + ( 223. 7) [ 2( 96. 2) + ( 152. 23) ]= 92. 6 J K ml (c) Hg(l) + Cl 2 (g) HgCl 2 (s) S r = ( 146. ) [( 76. 2) + ( 223. 7) ]= 153. 1 J K ml (d) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) S r = ( 112. 1) + ( 33. 15) [( 41. 63) + ( 99. 6) ]= 21. J K ml (e) C 12 H 22 O 11 (s) + 12O 2 (g) 12CO 2 (g) + 11H 2 O(l) S r = 12( 213. 74) + 11( 69. 91) [( 36. 2) + 12( 25. 138) ]= 512. J K ml
4.19) The cnstant-pressure mlar heat capacities f linear gaseus mlecules are apprximately 7/2 R and thse f nnlinear gaseus mlecules are apprximately 4R. Estimate the change in standard reactin entrpy f the fllwing tw reactins when the temperature is increased by 1 K frm K at cnstant pressure. (a) 2H 2 (g) + O 2 (g) 2H 2 O(g) This prblem is analgus t the derivatin f Kirchhff s law fr the determinatin f the temperature dependence f the enthalpy. Using this first reactin as an example, here is the path that we shuld cnsider: S 283 2H 2(g) + O 2(g) 2H2O(g) S 1 S 2 S 3 S 2H 2(g) + O 2(g) 2H2O(g) Where, lin S nc 7 1 = pm, ln = 2( R) ln 283 2 283 lin S nc 7 2 = pm, ln = ( R) ln 283 2 283 nn lin 283 283 S3 = ncpm, ln = 24 ( R) ln S283 = S1 + S2 + S + S3 r, S283 S = S1 + S2 + S3 But we can write the sum f the three entrpy terms as: 283 S 1 + S 2 + S 3 = r C p ln prd where, r C p = ν p C pm, νrcpm reac, (as in Kirchhff s law) prd reac
r C p = 2 4R 2 7 R 7 ( ) ( ) ( R) = 2. 5R= 2. 786 J K ml 2 2 283 S283 S = ( 2. 786) ln = 75. J K ml (b) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Fr this reactin, lin nn lin nn lin r C p = C p + 2Cp Cp + 2Cp lin = R 7 ( + 8 4 7 1 )= R 2 2 S S 1 283 283 = ( R) ln = +15. J K ml 2 [ ] 4.22) The change in Gibbs energy that accmpanies the xidatin f C 6 H 12 O 6 (s) t carbn dixide and water vapur at 25ºC is 2828 kj/ml. Hw much glucse des a persn f mass 65 kg need t cnsume t climb thrugh 1 m? C 6 H 12 O 6 (s) + 9O 2 (g) 6CO 2 (g) + 6H 2 O(g) G r yields the maximum nn-expansin wrk (like climbing), wrk = mgh, fr 65 kg thrugh 1 m : wrk = (65)(9.81)(1) = 6376.5 J # mles f glucse = 6376 5 1 ml glucse 3. = 225. 1 3 2828 1 J 3 18. 16 g 225. 1 ml = 41. g f glucse ml ml