Compute the heat of formation of liquid methyl alcohol in 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

Given `CH_(3) OH(g) to CH_(3)OH(l) , Delta H = -38 kJ`
`C(g) + 4H(g) rarr CH_(3)OH(g) , DeltaH = -(3 xx 415 + 356 + 463)`
`:. H = H_(1) + H_(2) = - 2064 kJ`
`C(s) to C(g) , Delta H = 715 kJ`
`2H_(2)(g) rarr 4H(g) , Delta H = 2 xx 2 xx 218 = 872 kJ`
Adding : `(1/2O_(2)(g) to O(g) , Delta H = 249 kJ)/(C(gr) + 2H_(2)(g) + 1/2 O_(2) (g) to CH_(3)OH(l))`
`Delta H = -265 kJ//mol`.