When 2 moles of `C_(2)H_(6)(g)` are completely burnt `3120` kJ of heat is liberated. Calculate the enthyalpy of formation, of `C_(2)H_(6)(g)`. Given `Delta_(f)H` for `CO_(2)(g)` & `H_(2)O(l)` are `-395` & `-286` kJ respectively.

Answer

`-88` kJ/mol

Explanation

`{:(C_(2)H_(6)+7/2 O_(2) rarr 2CO_(2)+3H_(2)O,DeltaH=-1560,.........(1)),(C+O_(2) rarr CO_(2),DeltaH=-395,.........(2)),(H_(2)+1/2 O_(2) rarr H_(2)O,DeltaH=-286,.........(3)):}`
`rArr" "` Target reaction `2C+3H_(2) rarr C_(2)H_(6) " "DeltaH=?`
`DeltaH` can be obtained by `(2) xx2+(3)xx3-(1)`
`DeltaH=-88" kJ mol"^(-1)`