1. Enthalpy of Formation By Jamie Leopold
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2. In this PPT, you’ll learn all about…
What the enthalpy of formation (Heat of Formation) is.
2. What the standard enthalpy of formation is.
3. How to use the enthalpy of formation to calculate ∆H of a reaction

3. what is enthalpy?
Enthalpy is the measure of the total energy of a thermo chemistry system or reaction.

4. Enthalpy of Formation – Heat of Formation
The enthalpy of formation is the heat released or absorbed, depending on what type of reaction it is, in a chemical reaction. The heat that is being either released or absorbed is doing so at a constant rate.
If heat is released, then the sign of the enthalpy (or change in heat) will be negative. If heat is absorbed, then the sign will be positive.

5. STANDARD ENTHALPY OF FORMATION
- For example: how we can burn carbon in oxygen to form carbon dioxide.
- The unit of measure for standard enthalpy of formation is either: ΔHf0 or ΔfH0 (measured in kJ/mol)
Thanks to the conservation of energy, standard enthalpy of formation AND enthalpy of formation can be used to calculate the heat absorbed/released in any chemical reaction.
The standard enthalpy of formation is the change in enthalpy (total energy, ie. heat absorbed or heat given off) to form a more complicated compound from its elements. For this to work, the elements must be in their most natural state (liquid, solid, gas, etc).
A chart of where to find ALL of the standard enthalpy of formation numbers:

6. How to use the enthalpy of formation to calculate ∆H of a reaction:
The basics - word equation for Hess’ Law: Enthalpy change = standard enthalpy of formation of product – standard enthalpy of formation of reactant
H = H(products) - H(reactants)
General background Info:
In an exothermic reaction: Energy of the reactants is greater than the energy of the products: H(reactants) > H(products)
In an endothermic reaction: Energy of the reactants is less than the energy of the products: H(reactants) < H(products)
“The standard state of any given substance is: 298 K (or 25 degrees C) and 1 atmosphere (1 atm) of pressure.”
From website #5
From website #5

7. an example problem and how to do it step by step!
Use:
an example problem and how to do it step by step!
Example by website 6: 1.
If we break this equation up into its many different elements and parts we would get:
2 moles of B5H9 reacts with 12 moles of O2 to yield (or make) five moles of B2O3 and 9 moles of H2O.
Now, as our formula says, we must add up the sums of our products and then subtract that from the sum of our reactants.
Now the products: 2 mol B5H kj mol O kj mol mol
Website #2/6
Starting with the reactants:
5 mol B2O kJ mol H2O KJ mol mol

8. Part 2:
2 mol B5H kj mol O kj mol mol -
5 mol B2O kJ mol H2O KJ mol mol
Now, its time to cross out!
2 mol B5H kj mol O kj mol mol -
5 mol B2O kJ mol H2O KJ mol mol
You’re now left with only the numbers:
You multiply the # of moles by the kJ/mol
Reactants: 5 x = x = Products: 2 x 73.2 = x 0 = 0
Then you add and subtract them, like this:
( ) – = kJ  final answer!!

9. BIBLIOGRAPHY:
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