Compute the heat of formation of liquid methyl alcohol in kJ mol^(-1) , using the following : Heat of vaporisation of liquid methyl alcohol=38 kJ mol^(-1). Heat of formation of gaseous atoms from the elements in their standard states : H, 218 kJ//mol , C,715 kJ // mol, O, 249 kJ //mol. Averagebond energies : C-H, 415 kJ //mol ,C-O kJ //mol, O-H, 463 kJ //mol

Compute the heat of formation of liquid methyl alcohol in kJ mol^(-1)

1.	Compute the heat of formation of liquid methyl alcohol in kJ mol^(-1) , using the following : Heat of vaporisation of liquid methyl alcohol=38 kJ mol^(-1). Heat of formation of gaseous atoms from the elements in their standard states : H, 218 kJ//mol , C,715 kJ // mol, O, 249 kJ //mol. Averagebond energies : C-H, 415 kJ //mol ,C-O kJ //mol, O-H, 463 kJ //mol
Answer» Solution :We aim at `: C(s) +2H_(2)(g) +(1)/(2)O_(2) RARR CH_(3)OH(l), DeltaH =?` We are given`:(i) CH_(3)OH(l) rarr CH_(3) OH (g), DeltaH = 38 kJ mol^(-1)`(III) `C(s)rarr C(g), DeltaH =715 kJ mol^(-1)` (IV) `(1)/(2)O_(2)(g) rarrO(g) , DeltaaH -249 kJ mol^(-1)` Also fromthe givenbond energies, we have (V)`CH_(3)OH(g)(H- underset(H) underset(\|) overset(H)overset(\|) (C) -O-H) rarr C(g) + 4H(g)+O(g), DeltaH= 3 xx 415 +356 +463 = 2064kJ mol^(-1)` Eqn. (iii) `+4 ` Eqn. (ii) `+` Eqn. (iv)- Eqn.(i) - Eqn. (v) gives the requried RESULT i.e., `DeltaH = 715 + 4( 218) + 249-38 -2064 = -266 kJ mol^(-1)`