Calculate the standard enthalpy of formation of `CH_(3)OH(l)` from the following data:
`CH_(3)OH(l)+3/2O_(2)(g) rarr CO_(2)(g)+2H_(2)O(l), …(i), Delta_(r)H_(1)^(Θ)=-726 kJ mol^(-1)`
`C(g)+O_(2)(g) rarr CO_(2)(g), …(ii), Delta_(c )H_(2)^(Θ)=-393 kJ mol^(-1)`
`H_(2)(g)+1/2O_(2)(g) rarr H_(2)O(l), ...(iii), Delta_(f)H_(3)^(Θ)=-286 kJ mol^(-1)`

Answer

`-239 kJ mol^(-1)`

Explanation

The reaction that takes place during the formation of `CH_(3)OH_((l))` can be written as :
`C_((s)) +2H_(2O_((g))+(1)/(2)O_(2(g))rarrCH_(3)OH_((l))(1)`
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as :
Equation (ii) +2xequations (iii) -equation (i)
`Delta_(f)H^(theta)[CH_(3)OH_((l))]Delta_(c)H^(theta)+2Delta_(f)H^(theta)[H_(2)O_((l))]-Delta_(r)H^(theta)`
`=(-393 kJ "mol"^(-1))+2(-286 kJ "mol"^(-1))-(-726 kJ "mol"^(-1))`
`=(-393-572+726)kJ "mol"^(-1)`
`:. Delta(f)H^(theta)[CH_(3)OH_((l))]=-239 kJ "mol"^(-1))`