InterviewSolution
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InterviewSolution
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When `1`pentyne `(A)` is treated with `4N` alcoholic `KOH` at `175^(@)C`, it is slowly converted into an equilibrium mixture of `1.3%` of `1`pentyne `(A), 95.2% 2`-pentyne `(B)` and `3.5%` of `1,2`-pentandiene `(C )`. The equilibrium was maintained at `175^(@)C`. calculate `DeltaG^(Theta)` for the following equilibria: `B hArr A, DeltaG^(Theta)underset(1) = ?` `B hArr C, DeltaG^(Theta)underset(2) =?` From the calculated value of `DeltaG^(Theta)underset(1)`and `DeltaG^(Theta)underset(2)`, indicate the order of stability of `A,B` and `C`. |
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Answer» `{:(,"Pentyne"-1,hArr,"Pentyne"-2,+,1","2-"pentadiene"),(,(A),,(B),,(C)),(t_(eq),1.3,,95.2,,3.5):}` `K_(eq)=([B][C])/([A])=(95.2xx3.5)/(1.3)=256.31` for `BhArrA` `K_(1)=([A])/([B])=([C])/(K_(eq))=(3.5)/(256.31)=0.013` `DeltaG_(1)^(@)=-2.303RT" log "K_(1)` `=-2.303xx8.314xx448" log "0.013` `=16178J=16.178kJ` for `BhArrC` `K_(2)=([C])/([B])=(K_(eq)[A])/([B]^(2))=(256.31xx1.3)/((95.2)^(2))=0.037` `DeltaG_(2)^(@)=-2.303RT" log "K_(2)` `=-2.303xx8.314xx448" log "0.037` `=12282J=12.282kJ` Stability will lie in the order `B gt C gt A` |
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