The standard enthalpies of combustion of `C_(6)H_(6)(l), C("graphite") and H_(2)(g)` are respectively `-3270 kJ "mol"^(-1), -394 kJ "mol"^(-1)` and `-286 kJ "mol"^(-1)`. What is the standard enthalpy of formation of `C_(6)H_(6)(l)` in kJ ` "mol"^(-1)`?

Options

A. `-48`
B. `+48`
C. `-480`
D. `+480`

Answer

B

Explanation

(i) `C_(6)H_(6)(l) + 15/2O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l) DeltaH = -3270 kJ "mol"^(-1)`
(ii) `C(gr) + O_(2)(g) to CO_(2)(g) DeltaH = -394 kJ "mol"^(-1)`
(iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH = -286 kJ "mol"^(-1)`
The required equation is
(iv) `6C(gr) + 3H_(2)(g) to C_(6)H_(6)(l),DeltaH` = ?
Multiply eq.(ii) by 6 and eq(ii) by 3 and adding we get :
(v) `6C(gr) + 15/2O_(2)(g) + 3H_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l), DeltaH = -3222 kJ`
Subtract eq.(i) from eq.(v)
`6C(gr) + 3H_(2)(g) to C_(6)H_(6)(l) DeltaH = + 48 kJ "mol"^(-1)`.