How can you find heat of formation of water?

#triangle"H^of# = #triangle"H of products"# - #triangle"H"^of "reactants"#

#triangle"H^of# = #triangle"H^of H_2O# - #triangle"H^of H_2# - #triangle"H^of O_2#

#triangle"H^of# = -280KJ - 0 + 0
= -280KJ

The difference between #triangle"H^of# and #triangle"H of reaction"# in this reaction is that #triangle"H^of# doesn't rely on the real reaction like here actually gaseous water is formed.

We have to calculate #triangle"H of reaction"#

For example

#"Bond energies of reactant - (Bond energies of product)"#

To calculate heat of formation we need a reaction that only forms water

#2H_2 + O_2 = 2H_2O#

[Bond energy of product that is# H_2O#]

Bond energy of #H_2O #

Bond energy of O-H = 463KJ
No. of O-H bonds in water = 2
Bond energy of #H_2O# = 463KJ * 2 = 926KJ

There are 2 moles of water formed so bond energy will be twice

926KJ * 2 = 1852KJ

[Bond energy of products that is #O_2# and# H_2O#]

Lewis dot structure of oxygen

Bonds found in oxygen is 1 oxygen double bond

O = O = 499KJ

Lewis structure of Hydrogen

Bonds found in hydrogen is 1 H-H bond

H-H = 436KJ

There are 2 moles of hydrogen so the bond energy would be twice

436KJ * 2 = 872KJ

#"Bond energy of oxygen + Bond energy of hydrogen - (Bond energy of water)"#

#"triangleH" = (872KJ + 499KJ) - 1852KJ = -481KJ #

For the reaction

#H_2+ 1/2O_2 = H_2O #

the #triangle H "of formation"# would be of #1/2# value = -240.1KJ

because all the products have become halve