Enthalpy of Formation
By Jamie Leopold
Keep an eye out for the
yummy food!
In this PPT, you’ll learn all about…
1. 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
what is enthalpy?
– Enthalpy is the measure of the total energy of
a thermo chemistry system or reaction.
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.
STANDARD ENTHALPY OF
FORMATION
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:
http://chemistry.about.com/library/weekly
/blheatform.htm
- 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.
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
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)
From website #5
H
= H(products) - H(reactants)
“The standard state of any given
substance is: 298 K (or 25 degrees C)
and 1 atmosphere (1 atm) of pressure.”
From website #5
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 B5H9 73.2 kj + 12 mol O2 0kj
---------------mol
mol
Website
#2/6
Starting with the reactants:
5 mol B2O3 -1272.77kJ
+ 9 mol H2O
----------------mol
-241.82 KJ
-------------mol
Part 2:
5 mol B2O3 -1272.77kJ + 9 mol H2O -241.82 KJ
-----------------mol
mol
- 2 mol B H
73.2 kj + 12 mol O2 0kj
--------------mol
mol
- 2 mol B H
73.2 kj + 12 mol O2 0kj
--------------mol
mol
5
9
Now, its time to cross out!
5 mol B2O3 -1272.77kJ + 9 mol H2O -241.82 KJ
-----------------mol
mol
5
9
You’re now left with only the numbers:
You multiply the # of moles by the kJ/mol
Reactants:
5 x -1272.77 = -6363.85
9 x -241.82 = -2176.38
Products:
2 x 73.2 = 146.4
12 x 0 = 0
Then you add and subtract them, like this:
(-6363.85 + -2176.38) – 146.4 = -8686.6 kJ  final answer!!
BIBLIOGRAPHY:
1. http://chemed.chem.wisc.edu/chempaths/GenChemTextbook/Standard-Enthalpies-of-Formation-551.html
2. http://chemistry.about.com/library/weekly/blheatform.htm
3.
http://www.mikeblaber.org/oldwine/chm1045/notes/Energy/HeatF
orm/Energy05.htm
4. http://en.wikipedia.org/wiki/Standard_enthalpy_of_formation
5. http://www.ausetute.com.au/enthchan.html
6.
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/enthalp
y.htm
7. http://en.wikipedia.org/wiki/Enthalpy
8. If you want the website for the great food: www.foodporn.com