Applications to Chemical Reactions

1. H for chemical reactions is just H_products - H_reactants where the H of each is a sum of the H_formation of the individual reactants and products.
2. H is an extensive property so if we know H per mole then we can calculate H for any quantity of the compound.
3. Reversing a chemical reaction just changes the sign of H:

   1/2N₂(g) + 1/2O₂(g) --> NO(g), H = 90.37 kJ/mol

   NO(g) --> N₂(g) + O₂(g), H = -90.37 kJ/mol

4. A formation reaction consists of the elements in their standard state on the left and only one mole of a single product on the right.
5. Enthalpy has no unique zero point
6. Enthalpy is a state function (Hess's law)

The last three items require a little discussion.

Formation reaction

An example of a formation reaction would be the above reaction for both nitrogen and oxygen are in their standard states (gases) and there is only one mole of product on the right (now you see why I put 1/2's on the reactant side.

1/2N₂(g) + 1/2O₂(g) --> NO(g)

Enthalpy has no unique zero point

Remember that H = E + PV and remember that E is the sum of kinetic and potential energies. Also remember that potential energy has no unique zero point and thus E has no unique zero point. So since E has no unique zero point then H has no unique zero point. Therefore we can define the zero point of enthalpy to be whatever we wish. Of course we will want this zero point to be the most useful for our purposes. It turns out that the most useful definition of zero enthalpy is for elements in their standard state (gas, solid, or liquid and perhaps diatomics) at STP where STP stands for standard conditions of temperature and pressure and this T is zero C and this P is 1 atmosphere.

Then the standard enthalpy for H₂(g) would be zero. The standard enthalpy of Na(s) would be zero. The standard enthalpy of Cl₂(g) would be zero. And so forth.

State Functions

A state function is a function whose change during a change of state is not dependent upon the path involved in the change. So going from state A to state B, H is independent of the path of going from A to B. Energy is also a state function, but q and W are not state functions. Thus we can write H and E for remember the delta just means final state minus initial state and since the changes are path independent it is only the initial and final values that are important. However we cannot write q nor can we write W -- remember this.

Applying Hess's law to a chemical reaction is just to realize that enthalpy is a state function so that it does not matter how we get to the reaction for which we want to determine H. As long as we end up with the desired reactants on the left and the desired products on the right then the H will be correct.

So let's apply all of this to an example. Calculate H for the following reaction (not yet using Hess's law):

How much heat is released upon the burning of one pound (454g) of propane?

C₃H₈(g) + 5 O₂(g) --> 3 CO₂(g) + 4 H₂O(l)

Apply 1 from above:

H = 3H_f(CO₂(g)) + 4 H_f(H₂O(l)) - [5H_f(O₂(g)) + H_f(C₃H₈(g))]

Note that I have multiplied each H_f by the appropriate stoichiometric coefficient because enthalpy is an extensive property and we must use the proper amounts of each. Now we just look up all of these values in the back of your book (table beginning on A21 -- be careful to use the proper states of the substances):

H = 3moles(-393.5 kJ/mol) + 4moles(-286 kJ/mol) - [5moles(0kJ/mol) + 1mole(-104 kj/mol)] = -2324.5 kJ - (-104 kJ)

or H = -2220.5 kJ

But remember that enthalpy is an extensive variable and this value is for burning one mole of propane (the stoichiometric number of propane in the reaction). So this result is for one mole of propane, but the question asked for the burning of one pound of propane (molar mass of 44.0 g/mol). So the answer will be

H = -2220.5 kJ/mol * (454g/44.0g/mol) = -22,900 kJ

This is a rather large amount of energy given off in the form of heat which is why we use propane as a fuel.

Now let's look at a problem involving Hess's law.

Calculate H of combustion for benzene:

C₆H₆(l) + 15/2 O₂(g) --> 6 CO₂(g) + 3 H₂O(l)

Given that

1. 6C(graphite) + 3H₂(g) --> C₆H₆(l) H = 49.0 kJ

2. C(graphite) + O₂(g) --> CO₂ H = -393.5 kJ

3. H₂ + 1/2 O₂(g) --> H₂O(l) H = -285.8 kJ

You can try just adding these reactions, but you can easily see that just adding them will not give us the needed reaction. So what do we need to do to the reactions? Just look at them and look at your needed reaction and determine what you need to do. You see that reaction 1 is going the wrong way for what we need so we will have to reverse it. You see that if we multiply reaction 2 by 6 we will have what we need and if we multiply reaction 3 by 3 we will have what we need (don't forget to multiply the H's by the same number). Then we have

C₆H₆(l) --> 6C(graphite) + 3H₂(g), H = -49.0kJ

6C(graphite) + 6O₂(g) --> 6CO₂, H = -2361.0 kJ

3H₂ + 3/2 O₂(g) --> 3H₂O(l), H = -857.4 kJ

Now if we add these three equations we do get the needed equation:

C₆H₆(l) + 15/2 O₂(g) --> 6 CO₂(g) + 3 H₂O(l)

And adding all of the H's gives

H_combustion = -3267.4 kJ

So you can now work all of the homework problems for this chapter. Practice, practice, practice.

Now you can practice Hess's law problems.

Here you can practice some other types of thermochemistry problems.

Here is a good tutoring site for thermochemistry.

Web Author: Dr. Leon L. Combs
Copyright ©1999 by Dr. Leon L. Combs - ALL RIGHTS RESERVED