Calculate the standard enthaply of formation of `CH_(3) OH(l)` from the following data
`CH_(3) OH (l) + (3)/(2) O_(2) (g) to CO_(2) (g) + 2H_(2)O (l) , Delta_(r) H^(@) = - 726` kj/mol
`C(g) + O_(2) (g) to CO_(2) (g), Delta_(c) H^(@) = - 393` kj/mol
`H_(2)(g) + (1)/(2) O_(2) (g) to H_(2) O(l) : Delta_(f) H^(@) = - 286` kj /mol

Explanation

We are given that
`CO_(3)OH(l)+(3)/(2)O_(2)(g)toCO_(2)(g)+2H_(2)O(l),Delta H^(@) = - 726 kj /mol`
`C(g) + O_(2) (g) to CO_(2) (g), Delta_(f)H^(@) = - 393 kj/mol`
`H_(2)(g)+(1)/(2) O_(2)(g) to H_(2) O(l) , Delta_(c)H6(@) = - 286 kj/mol`
We are to calculate the standard enthalpy of formation of `CH_(3)OH`. Adding eq. (ii) and (iii), we get
C(g) +`H_(2) (g) + (3)/(2) O_(2) (g) to CO_(2) (g) + H_(2) O (l) Delta H^(@) = - 679`
Subtracting equation (i) from equation (iv), we get
`C(g) + H_(2) (g) to CH_(3)OH _ H_(2) O Deta H^(@) = 47 kj /mol`
`C(g) + H_(2) + H_(2) O to CH_(3) OH Delta H^(@) = 47` kj /mol
The standard enthalpy of formation of methanol is 47 kj/mol