Enthalpy is an important concept in thermochemistry that represents the heat transferred during a chemical reaction under constant pressure. It can be used to determine if a reaction is exothermic or endothermic based on the sign of ΔH. Hess's law states that the enthalpy change of a reaction is independent of the pathway and can be calculated from the enthalpies of the steps. Properties of enthalpy include it being a state function, extensive property, and dependent on the states of the reactants and products.

1. Enthalpy
In the previous section we defined enthalpy as being the heat
transferred during a constant pressure process. Enthalpy is a very
important concept in thermochemistry for the following reasons:
 Most reactions occur under constant pressure conditions (i.e. rxn’s
in open containers such as test tubes and beakers, rxn’s that take
place in biological systems, etc.).
 Many reactions do not involve gaseous reactants or products, as a
consequence there is very little work associated with the reaction
and E ~ H.
Note that since the enthalpy is equal to the heat transfer (provided we
have constant pressure) so we can determine the nature of a reaction
from the sign of H:
 Exothermic Reaction  H < 0 (negative)  reaction gives off heat
 Endothermic Reaction  H > O (positive)  reaction absorbs heat
Reaction Enthalpy
Some reactions give off so much energy (primarily as heat) that they are
explosive, other reactions give off only a little bit of heat, still other
reactions don’t take place unless we add heat from the surroundings. It’s
very useful information to know how much heat a reaction will give
off/absorb (if nothing else so you won’t blow yourself up in the
laboratory). This quantity is called reaction enthalpy.
Example
Consider the combination reaction between two moles of hydrogen and
one mole of oxygen to create two moles of water:
2H2(g) + O2(g)  2H2O(g)H = -483.6 kJ

2. You can see that H < 0 so this is an exothermic reaction (the reaction
gives off heat). Of course if you were in class on the day that we ignited
balloons filled with H2(g)/O2(g) mixtures you know that this reaction
releases heat to the surroundings.
We can view this reaction on a molecular level to understand why this
reaction releases so much heat:
 Step 1 – Break 2 H-H bonds and 1 O-O bond (energy in)
 Step 2 – Form 4 O-H bond (energy out)
Energy must be supplied to the system to break chemical bonds, whereas
energy is released when chemical bonds are formed. In this example the
energy released by bond formation is much greater than the energy
required to break the bonds present in the products. Because enthalpy is
a state function (doesn’t depend upon the path of the reaction only the
initial and final states) it doesn’t matter whether the reaction occurs in
two steps as shown above or all at once.
Properties of Enthalpy
When calculating enthalpy changes associated with chemical reactions we
will make use of several properties of enthalpy.
1. Enthalpy is a state function. Therefore, we can determine the enthalpy
change, H, of a chemical reaction by subtracting the enthalpy of the
reactants from the enthalpy of the products:
Hrxn = Hproducts - Hreactants
The importance of this property will become evident when we discuss
Hess’ Law and standard enthalpy of formation.
2. Enthalpy is an extensive property.
Example

3. 2H2(g) + O2(g)  2H2O(g)H = -483.6 kJ
4H2(g) + 2O2(g)  4H2O(g)H = -967.2 kJ
Example
What is H associated with the production of 6.14 g of KCl according to
the following reaction?
2KClO3(s)  2KCl(s) + 3O2(g)H = -84.9 kJ
The equation tells us that the decomposition of 2 moles of potassium
chlorate to form 2 moles of potassium chloride will release 84.9 kJ of
heat. However, we are producing 6.14 g rather than 2 moles, so we first
convert to moles and then use the conversion –84.9 kJ per 2 moles KCl(s).
6.14 g 2KCl(s)  [1 mol KClO3/74.55 g KClO3]  [-84.9 kJ/2 mol KClO3]
= -3.49 kJ
3. If we reverse the reaction the magnitude of H remains the same, but
the sign changes.
2H2(g) + O2(g)  2H2O(g)H = -483.6 kJ
2H2O(g)  2H2(g) + O2(g)H = 483.6 kJ
Recall in class when we used electricity to decompose water into hydrogen
and oxygen. The input of electrical energy to drive the reaction is
consistent with the fact that the reaction is highly endothermic.
4. The enthalpy change of a reaction depends upon the state of the
reactants and products.
2H2(g) + O2(g)  2H2O(g)H = -483.6 kJ
2H2(g) + O2(g)  2H2O(l)H = -571.6 kJ

4. This implies that the enthalpy of 2 moles of water in the liquid state is 88
kJ lower than 2 moles of water in the gaseous state.
2H2O(g)  2H2O(l)H = -88 kJ
This shouldn’t be surprising, after all we all know that you have to add
heat to liquid water in order to make it evaporate (boil) into the gas
phase.
Hess’ Law
Because enthalpy is a state function the enthalpy change associated with
a reaction depends only upon the identity of the products and the
reactants. It is completely independent of whether the reaction takes
place in a series of steps or all at once. This principle is known as Hess’
Law.
We can use this principle to calculate the enthalpy of unknown or even
hypothetical reactions from the enthalpies of known reactions.
Example
The reaction between nitrogen and oxygen to form nitrogen dioxide has
the following enthalpy change:
N2(g) + 2O2(g)  2NO2(g)H = 68 kJ
However, if we were to carry the reaction out in two steps by adding first
half of the oxygen:
(a) N2(g) + O2(g)  2NO(g) H = 180 kJ
Then reacting the nitrogen monoxide that would form in (a) with the
remaining oxygen to form nitrogen dioxide:
(b) 2NO(g) + O2(g)  2NO2(g) H = -112 kJ

5. summing the reactants, products and enthalpy changes from reactions (a)
and (b) gives the same results as the one step reaction:
N2(g) + O2(g)  2NO(g) H = 180 kJ
2NO(g) + O2(g)  2NO2(g) H = -112 kJ
N2(g) + 2 O2(g) + 2NO(g)  2NO(g) + 2NO2(g) H = 180-112 kJ
Since there are 2 moles of NO(g) on either side of the equation we can
cancel these out and add the enthalpies to get:
N2(g) + 2O2(g)  2NO2(g) H = 68 kJ
Which is identical to the one step reaction, as promised by Hess’ Law.
It is important to remember that Hess’ Law still works even if we
choose the wrong steps.
Example
Consider the reaction enthalpy associated with the combustion of
methane:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ
One possible reaction pathway would be to first form carbon dioxide and
water vapor, then let the water vapor condense to form liquid water:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -802 kJ
2H2O(g)  2H2O(l) H = -88 kJ
CH4(g) + 2O2(g) + 2H2O(g)  2CO2(g) + 2H2O(l) + 2H2O(g)
Which gives a net equation of
CH4(g) + 2O2(g)  2CO2(g) + 2H2O(l) ........ H = -802-88 = -890 kJ

6. Another possible reaction pathway could involve forming carbon monoxide
first
2CH4(g) + 3O2(g)  2CO(g) + 4H2O(g) H = -1039 kJ
2CO(g) + O2(g)  2CO2(g) H = -566 kJ
4H2O(g)  4H2O(l) H = -176 kJ
2CH4(g) + 4O2(g) + 2CO(g) + 4H2O(g)  2CO(g) + 4H2O(g) + 2CO2(g) +
4H2O(l)
Which gives a net equation of
2CH4(g) + 4O2(g)  2CO2(g) + 2H2O(l) ................ H = -1039-566-176 = -
1781 kJ
Dividing everything by 2, which we can do because enthalpy is an
extensive property.
CH4(g) + 2O2(g)  CO2(g) + H2O(l) H = -890 kJ
So we see that when we use Hess’ law it doesn’t matter what series of
steps we choose only that we have the right starting (reactants) and
ending (products) points.
Standard Enthalpy of Formation
Hess’ Law is very powerful because it allows us to calculate the H of a
reaction even if we can’t measure it directly. To extend the power of the
Hess’ Law approach further, consider the following definitions.
Standard State  The standard state of an element or compound is its
stable form at 25 C (~ room temperature) under a pressure of 1
atmosphere (ambient pressure).

7. Standard Enthalpy of Formation (Hf

)  The change in enthalpy which
accompanies the formation of 1 mole of a compound from its elements
(with all substances in their standard state).
Examples
Cgraphite(s) + O2(g)  CO2(g) H = -393.5 kJ/mol
H2(g) + �O2(g)  H2O(l) H = -286 kJ/mol
Note that for elements already in their standard state (i.e. N2(g), Au(s),
Cl2(g), etc.) Hf

= 0.
Hf

values have been experimentally determined for a vast number of
compounds (see table 5.3 in your text for a very limited sample of such
values, a more extensive list is given in Appendix C). We can use these
values to calculate the enthalpies of many reactions, using the following
formula:
Hrxn =  n Hf

(products) -  m Hf

(reactants)
where n and m are the number of moles of each product and reactant.
Example
What is H rxn for the combustion of propane?
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
I go to table 5.3 to get standard enthalpies of formation for all of the
products and reactants.
Compound Hf (kJ/mol)
C3H8(g) -103.85
O2(g) 0

8. CO2(g) -393.5
H2O(l) -285.8
Note the standard enthalpy of oxygen is zero by definition, since it is in
its standard elemental form.
Now we use the formula above:
Hrxn = 3(-393.5) + 4(-285.8) – [-103.85 + 5(0)] = -2220 kJ
Finally lets use all of the properties of enthalpy we have thus far in a
single example.
Example
Using the standard enthalpies of formation of CO2(g) and H2O(l), which
were given above, and the enthalpy of the following combustion reaction:
C3H4(g) + 4O2(g)  3CO2(g) + 2H2O(l) ........................ H = -1939.1 kJ
Calculate the enthalpy change that occurs when 25.5 g of propyne (C3H4)
is formed from graphite and hydrogen under standard conditions?
First we need to write an equation for the formation of propyne
3Cgraphite(s) + 2H2(g)  C3H4(g) H = ?
Now we can use Hess’ Law and standard enthalpies of formation to
determine the enthalpy change associated with the above reaction:
3CO2(g) + 2H2O(l)  C3H4(g) + 4O2(g) ........................... H = 1939.1 kJ
3[Cgraphite(s) + O2(g)  CO2(g)] H = 3(-393.5 kJ)
2[H2(g) + �O2(g)  H2O(l)] H = 2(-285.8 kJ)
3Cgraphite(s) + 2H2(g)  C3H4(g) .......................................... H = 187.0 kJ

9. Finally, we calculate the amount of heat required (this is an endothermic
reaction) to form 25.5 g of C3H4(g).
25.5 g C3H4(g)  [1 mol C3H4/40.07 g]  [187.0 kJ/1 mol C3H4(g)] = 119 kJ