Lesson 6.7 Standard Enthalpies of Formation
Suggested Reading

Zumdahl Chapter 6 Section 6.4
Essential Question

How can enthalpy changes for formation reactions be used to
calculate the enthalpy of reaction?
Learning Objectives

Calculate the heats of reaction using standard enthalpies of
formation.
Introduction
Because Hess's law relates the enthalpy changes of some reactions to the
enthalpy changes of others, we only need to tabulate the enthalpy changes
of certain types of reactions. We also generally list enthalpy changes only
for certain standard thermodynamic conditions. These are called standard
enthalpies of formation.
Standard Enthalpies of Formation
The term standard state refers to the standard thermodynamic conditions,
such as temperature and pressure, chosen for substances when listing or
comparing thermodynamics data (Watch out, because the term standard
state has different meanings throughout chemistry). In thermochemistry,
these conditions are usually 1 atm of pressure and 25°C. These standard
conditions are indicated by a superscript degree sign (°). The enthalpy
changed for a reaction in which the reactants in their standard states yield
products in their standard states is denoted by the symbol ∆H° (read as
"delta H degree" or "delta H zero"). The quantity ∆H° is called the standard
enthalpy of reaction.
The standard enthalpy for formation (∆H°f) is defined as the change in
enthalpy that accompanies the formation of one mole of a compound from
its elements with all elements in their reference (standard ) forms under
standard state conditions. The reference (standard) form of an element
for the purpose of specifying the formation reaction is usually the stablest
form (physical state and allotrope) of the element under standard
thermodynamic conditions. For example, the reference form of oxygen at
25°C is O2(g). Similarly, the reference form for carbon is graphite. As you
will see, standard enthalpies of formation can be used to calculate ∆H for
chemical reactions.
An example will help you to understand the concept of standard enthalpy of
formation. Lets look at the standard enthalpy of formation for liquid water
from its elements hydrogen and oxygen. The stables forms of hydrogen
and oxygen at 1 atm and 25°C are H2(g) and O2(g), respectively. These are
therefore the reference forms of the elements. You write the formation
reaction for m1 mol of liquid water as follows:
H2(g) + ½O2(g) → H2O(l)
This reaction is called a formation reaction because we are forming 1 mol
of water from its elements. The standard enthalpy change for this reaction
is -285.8 kJ per mole of H2O. Therefore, the thermochemical equation is
H2(g) + ½O2(g) → H2O(l); ∆H°f = -285.8 kJ
Values of standard enthalpies of formations like the one above are listed in
tables containing thermodynamic table. Refer to Appendix 4 on page A19
of your textbook for an example. You must know how to use and table such
as the one given there to obtain standard enthalpies of formation as well as
other thermodynamic quantities, which we will learn about later. These
values are either obtained by direct measurement or by applying Hess's
law.
Please note that the value of ∆H°f = 0 for elements in their reference
forms.
Now let us see how to use standard enthalpies of formation to find the
standard enthalpy change for a reaction. In general, you can calculate
the ∆H° for a reaction by the equation
∆H° = ∑ n ∆H°f(products) - ∑ n ∆H°f(reactants)
Here ∑ is the mathematical symbol meaning "the sum of" and n is the
coefficients of the substances from the balanced chemical equation.
The following video will show you how to use this equation.
YouTube Video
Students often make the following mistakes when using standard
enthalpies of formation to calculate standard enthalpies of reaction, so
check yourself before you wreck yourself!
1. They reverse the equation and subtract products from reactions. This
equation is given of the AP equation sheet, so use it to verify your
work.
2. They forget that the standard enthalpy of formation for an element in
its reference state under standard conditions is O.
Lets look at an textbook example.
Calculating the Heat of Phase Transition form Standard Enthalpies of
Formation
Use the values of ∆H°f to calculate the heat of vaporization, ∆H°vap of
carbon disulfide at 25°C. The vaporization process is
CS2(l) → CS2(g)
Solution
The vaporization process can be treated like a chemical reaction, with
CS2(l) the reactant and CS2(g) the product. It is convenient to record the
values of ∆H°f under the formulas in the equation, multiplying them by the
coefficients in the equation. You can then calculate ∆H°f by subtracting the
values for the "reactant" from values for the "product" (Be careful not to
reverse the equation!). Lets first write the values, which were obtained from
a table like the one in appendix 4, under the equation.
CS2(l) → CS2(g)
1(88 kJ) 1(117 kJ)
Then,
∆H°vap = ∆H° = ∑ n ∆H°f(products) - ∑ n ∆H°f(reactants)
= ∆H° = ∆H°f [CS2(g)] - ∆H°f [CS2(l)]
= (117 - 88) kJ = 29 kJ