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 DE ~ DH.

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 DH:

• Exothermic Reaction ® DH < 0 (negative) ® reaction gives off heat
• Endothermic Reaction ® DH > 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) ........................ DH = -483.6 kJ

You can see that DH < 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, DH, of a chemical reaction by subtracting the enthalpy of the reactants from the enthalpy of the products:

DHrxn = SHproducts - SHreactants

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

2H2(g) + O2(g) ® 2H2O(g) ........................ DH = -483.6 kJ

4H2(g) + 2O2(g) ® 4H2O(g) ......................DH = -967.2 kJ

Example

What is DH associated with the production of 6.14 g of KCl according to the following reaction?

2KClO3(s) ® 2KCl(s) + 3O2(g) ........................ DH = -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 DH remains the same, but the sign changes.

2H2(g) + O2(g) ® 2H2O(g) ........................ DH = -483.6 kJ

2H2O(g) ® 2H2(g) + O2(g) ........................ DH = 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) ........................ DH = -483.6 kJ

2H2(g) + O2(g) ® 2H2O(l) ......................... DH = -571.6 kJ

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) ........................ DH = -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) ........................ DH = 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) ........................ DH = 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) .................. DH = -112 kJ

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) .................................................... DH = 180 kJ

2NO(g) + O2(g) ® 2NO2(g) .............................................. DH = -112 kJ

N2(g) + 2 O2(g) + 2NO(g) ® 2NO(g) + 2NO2(g) ........... DH = 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) .............................................. DH = 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) ........................ DH = -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) ........................ DH = -802 kJ

2H2O(g) ® 2H2O(l) ...................................................... DH = -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) ........ DH = -802-88 = -890 kJ

Another possible reaction pathway could involve forming carbon monoxide first

2CH4(g) + 3O2(g) ® 2CO(g) + 4H2O(g) ............ DH = -1039 kJ

2CO(g) + O2(g) ® 2CO2(g) .................................. DH = -566 kJ

4H2O(g) ® 4H2O(l) ............................................. DH = -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) ................ DH = -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) ..................... DH = -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 DH 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).

Standard Enthalpy of Formation (DHf° ) ® 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) ....................................... DH = -393.5 kJ/mol

H2(g) + ½O2(g) ® H2O(l) .............................................. DH = -286 kJ/mol

Note that for elements already in their standard state (i.e. N2(g), Au(s), Cl2(g), etc.) DHf° = 0.

DHf° 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:

DHrxn = S n Hf° (products) - S m Hf° (reactants)

where n and m are the number of moles of each product and reactant.

Example

What is DH° 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 DHf° (kJ/mol) C3H8(g) -103.85 O2(g) 0 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:

DHrxn = 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) ........................ DH = -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) ................................ DH = ?

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) ........................... DH = 1939.1 kJ

3[Cgraphite(s) + O2(g) ® CO2(g)] ................................... DH = 3(-393.5 kJ)

2[H2(g) + ½O2(g) ® H2O(l)] ............................................. DH = 2(-285.8 kJ)

3Cgraphite(s) + 2H2(g) ® C3H4(g) .......................................... DH = 187.0 kJ

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