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The freezing pointof a solution of acetic acid (mole fraction0.02) in benzene is 277.K. Acetic acid exists party as a dimmer, 2A hArr A_(2). Calculate equilibrium constant for dimerization. Freezing point of benzene is 278.4 K and K_(t) for benzen is 5. |
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Answer» Solution :LET acetic acid =A Benzene =B ASSUME `alpha` part of A forms dimer `{:(2A ,hArr, A_(2)),(1,0,"Intial moles"),(1-alpha, alpha//2 ,"moles after at eqm"):}` `:.I=((1-alpha)+alpha//2)/(1)=1-alpha//2` Mol. Fraction of `A=X_(A)=0.02` Mol. fraction of `B=X_(B)=0.98` Molality of A in B `=(X_(A))/(m_(1))XX(1000)/(X_(B))=(0.02)/(78)xx(1000)/(0.98)= 0.262 "mol" kg^(-1) (m_(1)="mol"` wt of solvent) Since `DeltaT_(t)=K_(1)xxlxx"molality"` `278.4-277.4=5xxixx0.262` or `1=5xxixx0.262` `i=(1)/(5)xx0.262=0.763` `1-alpha//2=0.763 rArr alpha=0.48` Hencethe molality of A after dimer is formed `(1-alpha)xx` initial molality `=0.52xx0.262` Molality of `A_(2)` after dimer formed `=(alpha)/(@)xx"molality" =(0.48)/(2)xx 0.262` `=0.24xx0.26=0.06288` The equilibrium constant `K_(eq)=([A_(2)])/([A]^(2))=(0.06288)/((0.13624)^(2))=3.39` |
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