Compute the heat of formation of liquie methyl alcohol is kilojoule per mol using the following data. Heat of vaporisation of liquid methyl alcohol `=38kJ//mol`. Heat of formation of gaseous atoms from the elements in their standard states `:H=218kJ // mol, C=715kJ //mol, O=249kJ //mol.`
Average bond energies`:`
`C-H415kJ//mol, C-O 356kJ//mol, O-H 463kJ //mol.`

Explanation

`CH_(3)OH(1) rarr CH_(3)OH(g), DeltaH = 38 kJ mol^(-1)`
`(1)/(2)H_(2)(g) rarr H(g), DeltaH = 218 kJ mol^(-1)`
`2H_(2)(g) rarr 4H(g), DeltaH = (218xx4) kJ mol^(-1)`
`C("graphite") rarr C(g), DeltaH = 715 kJ mol^(-1)`
`(1)/(2) O_(2)(g) rarr O(g), DeltaH = 249 kJ mol^(-1)`
`C -H(g) rarr C(g) +H(g), DeltaH = 415 kJ mol^(-1)`
`C -O (g) rarr C(g) +O(g), 356 kJ mol^(-1)`
`O-H(g) rarr O(g) +H(g), DeltaH = 463 kJ mol^(-1)`
`C(s) +2H_(2)+(1)/(2)O_(2)(g) rarr CH_(3)OH(g)`
The following energies are required:
`2H_(2)(g) rarr 4H(g)`
`C("graphite") rarr C(g)`
`(1)/(2)O_(2)(g) rarr O(g)`
Total energy required is
`(218 xx4) +715 +249 = 1836 kJ`
`H-overset(H)overset(|)underset(H)underset(|)(C)-O-H`
The following bonds are formed.
Three `C -H` bonds, one `C -O` bond, are `O -H` bond Total energy released is
`(3 xx 415) +356 +464 = 2064 kJ`
Net energy released for `CH_(3)OH(g) = 2064 - 1836 = 228 kJ`
For `CH_(3)OH(l)` another `38kJ` is released.
Net energy released for `CH_(3)OH (1) = 228 +38 = 266 kJ`
Therefore,
`DeltaH_(f)CH_(3)OH(l) =- 266 kJ mol^(-1)`