1.饱和溶解度法:SatmB△,G%(B,Soln, T) =△,Gm(B,T)- RT In(ym,BhoAG >甘(aq,m=1 m°)甘(S)A,G%(甘)A,G%(aq)△G△G,甘(sat, m=3.33m2)△G, = 0△G = △G + △G, = △G,= μ° -(μ° + RTInasa) = RTInasat△G= △,G(甘)-△,G(aq)
1. 饱和溶解度法: , ( ,Soln, ) ( , ) ln( ) sat B f m f m m B m G B T G B T RT m ΔG (S) (aq, =1 ) m m 甘 甘 ( ) ( ) f m f m G G 甘 aq m m 甘 (sat, =3.33 ) G1 G2 G 0 1 1 2 2 sat sat ( ln ) ln G G G G RT a RT a ( ) ( ) G G G f m f m 甘 aq
:. △,G(甘)-△,G%(aq)=RT In asat(-370.7+372.9)x103In asat=.0.888298×8.314asat=2.43qsat2.43=0.7298Y-m/m°3.332.在298K时,从大量的等物质的量的C,H,Br,和C,H,Brz溶1717 J液中分离出1mol纯C,H,Br,所需做的最小功为W'= △G= μ-(μ +RTInxB)=-RT InXg= 8.314 x 298ln0.5 = 1717J只有1mol C,H,Br2始终态发变化了
2. 在298 K 时,从大量的等物质的量的C2H4Br2和C3H6Br2溶 液中分离出 1 mol 纯C2H4Br2所需做的最小功为_。 0 888 sat ( ) ( ) ln × ln . × sat f m f m G G RT a a 3 甘 aq = (-370.7+372.9)10 = = 298 8.314 2.43 sat a = 2.43 0.7298 3.33 sat a m m 1717 J 8 314 298 0 5 1717 ' ( ln ) ln . ln . J W G RT x RT x B B B B 只有1 mol C2H4Br2 始终态发变化了