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Find the bond energy of S-S bond from the following data: {:(C_(2)H_(5)-S-C_(2)H_(5)(g),,,,Delta_(f)H^(@)=-147KJmol^(-1)),(C_(2)H_(5)-S-S-C_(2)H_(5)(g),,,,Delta_(f)H^(@)=-201KJmol^(-1)),(S(g),,,,Delta_(f)H^(@)=222 KJ mol^(-1)):} |
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Answer» Solution :Given that (i) `4C(s)+5H_(2)(g)+S(s)rarrC_(2)H_(5)-S-C_(2)H_(5)(g)""Delta_(F)H^(@)= -147KJ mol^(-1)` (ii) `4C(s)+5H_(2)(g)+2S(s)rarrC_(2)H_(5)-S-S-C_(2)H_(5)(g)""Delta_(r )H^(@)-201 KJ mol^(-1)` Substracting Eq.(i) from Eq.(ii), we get `C_(2)H_(5)-S-C_(2)H_(5)(g)+S(s)rarrC_(2)H_(5)-S-S-C_(2)H_(5)(g)"" Delta_(r )H^(@)=-54KJmol^(-1)` ADDING to this, the following equation `S(g)rarrS(s)""Delta_(r )H^(@)=-222 KJ mol^(-1)` we get `C_(2)H_(5)-S-C_(2)H_(5)(g)+S(g)rarrC_(2)H_(5)-S-S-C_(2)H_(5)(g)""Delta_(r )H^(@)= -276 KJ mol^(-1)` In the least equation `276 KJ` of heat evolved because of the `S-S` bond formation. Hence, the bond enthalpy of `S-S` is `276 KJ mol^(-1)` Method-2 Diagrammatically, we may represent the above calculation as follows: According to Hess's LAW `Delta_(r )H^(@)=` "Enthalpy INVOLVED in bond breaking `Delta_(vap)H^(@)(s)`-Enthalpy involved in bond making `Delta_(r )H^(@)=(2DeltaH_(C-c)+10DeltaH_(C-H)+2DeltaH_(c-s)+Delta_(vap)H(S)]+(-2DeltaH_(c-c)-10DeltaH_(C-H)-2DeltaH_(c-s)-DeltaH_(s-s)]` `Delta_(vap)H^(@)(S)-DeltaH_(s-s)` or `DeltaH_(s-s)=Delta_(vap)H^(@)(S)-DeltaH^(@)` `=Delta_(vap)H^(@)(S)-[Delta_(r )H^(@)(C_(2)H_(5)-S-S-C_(2)H_(5)]=Delta_(f)H^(@)(C_(2)H_(5)-S-C_(2)H_(5))]` `=[222-{-201-(-174)}]KJmol^(-1)=276KJ mol^(-1)` |
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