Using the bond enthalpy data given below, estimate the enthalpy of formation of gaseous isoprene
`CH_(2)=underset(CH_(3))underset(|)C-CH=CH_(2)`
`{:("Data","Bond enthalpy of",C-H"bond"=413.38KJmol^(-1),,),(,"Bond enthalpy of",C-C"bond"=347.69KJmol^(-1),,),(,"Bond enthalpy of",C=C"bond"=615.05 KJ mol^(-1),,):}`
Enthalpy of sublimation of carbon (graphite) `=718.39 KJ mol^(-1)`
Enthalpy of dissociation of `H_(2)(g)=435.97 KJ mol^(-1)`

Explanation

For isoprene, we have to form
`2C-C "bonds" , 2C=C "bonds"` and `8C-H "bonds"`
Method-1 For which energy released is
`[(2(+347.69)+2(+615.05)+8(+413.38)]KJ mol^(-1)=5232.52 KJ mol^(-1)`
that is, `Delta H` (from gaseous atoms)`=+5232.52 KJ mol^(-1)`
The reaction corresponding to this is
`5C(g)+8H(g)rarrC_(5)H_(8)(g)" "Delta_(f)H_(1)= -5232.52 KJ mol^(-1)`
But we want `Delta_(f)H` corresponding to the following equation
`5C("graphite")+4H_(2)(g)rarrC_(6)H_(8)(g)" "Delta_(f)H=?`
This can be obtained by the following manipulations:
`{:(5C(g)+8H(g)rarrC_(5)H_(8)(g),,,,Delta_(r )H_(2)= -5232.52KJmol^(-1)),(5C("graphite")rarr5C(g),,,,Delta_(r )H_(3)=5xx718.39KJmol^(1-)),(4H_(2)(g)rarr8H(g),,,,Delta_(r )H_(4)=4xx435.97KJmol^(-1)):}`
Adding, we get
`5C ("graphite")+4H_(2)(g)rarr C_(5)H_(8)(g)" "Delta_(f)H=103.31KJmol^(-1)`
Method-2 Diagrammatically, the above calculations may be represented as follows.
Applying Hess's law, we get
`Delta_(f)H=5DeltaH_(3)+4DeltaH_(4)-2DeltaH_(C=C)-2DeltaH_(c-c)-8Delta_(C-H)`
`=(5xx718.39+4xx435.97-2xx615.05-2xx347.69-8xx413.38)KJ mol^(-1)`
`Delta_(f)H=103.31 KJmol^(-1)`