Calculate the enthalpy of formation of methanol from the following data,
`CH_(3)OH(l) + (3)/(2)O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) ….(i) " "Delta H^(@) = -726.4 kJ mol^(-1)`
`C"(graphite)" + O_(2)(g) rarr CO_(2)(g) ….(iii) " "Delta H^(@) = -393.5 kJ mol^(-1)`
`H_(2)(g) + (1)/(2)O_(2)(g) rarr H_(2)O(l) ....(iii) " "Delta H^(@) = -285.8 kJ mol^(-1)`.

Explanation

Required equation,
`C "(graphite)" + 2H_(2)(g) + (1)/(2)O_(2)(g) rarr CH_(3)OH(l)`
Reverse Eq. (1) multiply Eq. (3) by 2 and add to Eq. (2)
`{:(CO_(2)(g)+2H_(2)O(l) rarr CH_(3)OH(l)+ (3)/(2)O_(2)(g) " " Delta H^(@) = +726.4 kJ mol^(-1)),(2H_(2)(g) + O_(2)(g) rarr 2H_(2)O(l) " "Delta H^(@) = -571.6 kJ mol^(-1)),("C(graphite)" + O_(2)(g) rarr CO_(2)(g) " "Delta H^(@) = -393.5 kJ mol^(-1)):}/("C(graphite)" + 2H_(2)(g) + (1)/(2)O_(2)(g) rarr CH_(3)OH(l)Delta H^(@) = -238.5 kJ mol^(-1))`
`Delta_(f)H^(@)` of `CH_(3)OH = -238.5 kJ mol^(1)`.