Question

How can you find heat of formation of water?

Answer 1

-280KJ

Explanation:

`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