Applications to Chemical Reactions

The following information helps us considerably in understanding the applications of thermochemistry to chemical reactions:

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

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

    NO(g) --> N2(g) + O2(g), deltaH = -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, deltaH 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 deltaH and deltaE 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 deltaq nor can we write deltaW -- 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 deltaH. As long as we end up with the desired reactants on the left and the desired products on the right then the deltaH will be correct.

So let's apply all of this to an example. Calculate deltaH 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:

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

Note that I have multiplied each deltaHf 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):

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

or deltaH = -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

deltaH = -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 deltaH 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) deltaH = 49.0 kJ

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

3. H2 + 1/2 O2(g) --> H2O(l) deltaH = -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 deltaH's by the same number). Then we have

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

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

3H2 + 3/2 O2(g) --> 3H2O(l), deltaH = -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 deltaH's gives

deltaHcombustion = -3267.4 kJ

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

help Now you can practice Hess's law problems.

help Here you can practice some other types of thermochemistry problems.

help Here is a good tutoring site for thermochemistry.

Now take a practice quiz to help you understand if you understand the basic concepts.
You must use your real name when it asks for a name.
The test will only submit when you have answers all of the questions correctly.
If you are not taking this course for credit please do not answer all the questions correctly for I don't want to be flooded with email answers to the tests.




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