The standard enthalpies of combustion of `C_(6)H_(6(l)),C_(("graphite"))and H_(2(g))` are respectively `-3270 kJmol^(-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

`C_(6)H_(6(l))+(15)/(2)O_(2(g))rarr6CO_(2(g))+3H_(2)O_((l)), DeltaH=-3270 kJ mol^(-1)" ...(i)"`
`C_((gr))+O_(2(g))rarrCO_(2(g)), DeltaH=-394 kJ mol^(-1)" ....(ii)"`
`H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)), DeltaH=-286 kJ mol^(-1)" ...(iii)"`
Formation of `C_(6)H_(6)`
`6C_((gr))+3H_(2(g))rarrC_(6)H_(6(l)),DeltaH=? " ...(iv)"`
By multiplying eq. (ii) with 6 and eq. (iii) with 3 and adding we get,
`6C_((gr))+6O_(2(g))+3H_(2(g))+(3)/(2)O_(2(g))rarr6CO_(2(g))+6H_(2)O_((l)) " ...(v)"`
`DeltaH=6(-394)+3(-284)=(-2364)+(-858)=-3222 kJ//mol`
Now, by substracting eq. (i), from (v) we get
`6C_((gr))+3H_(2(g))rarrC_(6)H_(6(l))`
`DeltaH=-3222-(-3270)=+48kJ//mol`