# How can you find heat of formation of water?

Feb 21, 2017

-280KJ

#### Explanation:

triangle"H^of = $\triangle \text{H of products}$ - $\triangle \text{H"^of "reactants}$

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 $\triangle \text{H of reaction}$ 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 $\triangle \text{H of reaction}$

For example

$\text{Bond energies of reactant - (Bond energies of product)}$

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

$2 {H}_{2} + {O}_{2} = 2 {H}_{2} O$

[Bond energy of product that is${H}_{2} O$]

Bond energy of ${H}_{2} O$ Bond energy of O-H = 463KJ
No. of O-H bonds in water = 2
Bond energy of ${H}_{2} O$ = 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${H}_{2} O$]

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

$\text{Bond energy of oxygen + Bond energy of hydrogen - (Bond energy of water)}$

$\text{triangleH} = \left(872 K J + 499 K J\right) - 1852 K J = - 481 K J$

For the reaction

${H}_{2} + \frac{1}{2} {O}_{2} = {H}_{2} O$

the $\triangle H \text{of formation}$ would be of $\frac{1}{2}$ value = -240.1KJ

because all the products have become halve