# Standard Enthalpies of Formation

```Standard Enthalpies of Formation
Chapter 8.5
Standard Enthalpies of Formation
“Heat of Formation”

The standard enthalpy of formation of a
substance, denoted DHfo, is the enthalpy
change for the formation of one mole of a
substance in its standard state from its
component elements in their standard
state.
– Note that the standard enthalpy of
formation for a pure element in its standard
state is zero.
NH3 = -46.1
You have a
similar table
on page 209
Using DHfo to Calculate DHo for a Rxn

The law of summation of heats of
formation states that the enthalpy of a
reaction is equal to the total formation energy
of the products minus that of the reactants.
DH o   nDH of (products )   mDH of (reactants )
S is the mathematical symbol meaning “the
sum of”, and m and n are the coefficients of
the substances in the chemical equation.
Why use DHfo to Calculate DHo when
we can use Hess’s Law?
Simply put it is easier
 Finding combinations of reactions to
tedious
 We have very accurate data tables
that give us the DHfo values
 Pure elements are 0!

Let’s do a Practice Problem!!!
4NH 3 (g )  5O 2 (g )  4NO(g )  6H 2O(g )
4( 45.9)
5(0)
4(90.3)
6( 241.8)
-183.6 + 0
361.2 + (-1450.8)
-183.6 for reactants
-1089.6 for products
Prod. – React. = -1089.6 – (-183.6) = -906 kJ
*Hint: First record the values of DHfo under the formulas in the
equation then multiplying them by the coefficients in the
equation. You can then determine DHo by subtracting the
values for the reactants from the values for the products.
Homework for 8.5
Problems #38 (don’t forget DH in
equations!), 39, 43, and 45
Chem Dollar problem #52 (\$2)
```