Enthalpies of Formation

by Andrea Badua and Raul Cabrera

What is Enthalpy Formation?

• Enthalpy Formation (Heat of Formation): is the enthalpy change (i.e. heat absorbed) when some amount of a compound is formed from its elements in their standard states under standard conditions.

different forms of enthalpy are:

-enthalpies of vaporization (ΔH for converting liquids to gases)-enthalpies of fusion (ΔH for melting solids)-enthalpies of combustion (ΔH for combusting a substance in oxygen)

• Enthalpy is a state function, meaning that its value is fixed when temperature, pressure, composition, and physical form are specified.

• Second, at a constant pressure, DH = q, meaning that at constant pressure, the enthalpy of a system is equal to the heat, in joules, of a system.

● The magnitude of any enthalpy change depends on the temperature, pressure, and state (gas, liquid, or solid crystalline form) of the reactants and products.

● To compare enthalpies of different reactions, we use a set of conditions called a standard state, at which most enthalpies are tabulated (condensed)

● The standard state is its pure form at (1 atm) while the temperature is set at

(298 K or 25 degrees Celsius)

● Standard Enthalpy Change: the enthalpy change when all reactants and products are in their standard states. ΔH°, where the superscript ° indicates standard- state conditions

● Standard Enthalpy of Formation (a compound: ΔHf°): is the change in

enthalpy for the reaction that forms one mole of the compound from its elements with all substances in their standard states

Standard Enthalpies of Formation

● We usually report ΔHf° values at 298 K. If an element exists in more

than one form under standard conditions, the most stable form of the element is usually used for the formation reaction.

● For example, the standard enthalpy of formation for ethanol, C2 H5 OH is the enthalpy change for the reaction:

2C (graphite) + 3 H2(g) + ½ O2(g) ---> C2 H5 OH(l) ΔHf° = -277.7 kJ

● The elemental source of oxygen is O2, not O or O3, because O2 is the stable form of oxygen at 298 K and atmospheric pressure.

● The elemental source of carbon is graphite and not diamond because graphite is the more stable (lower- energy) form at 298 K and atm pressure.

● The most stable form of hydrogen under standard condition is H2(g), that is why it is used as the source of hydrogen in the equation above.

2C (graphite) + 3 H2(g) + ½ O2(g) ---> C2 H5 OH(l) ΔHf° = -277.7 kJ

● The stoichiometry of formation reactions always shows that one mole of the desired substance is produced, just like the equation above.

● As a result, standard enthalpies of formation are reported in kJ/mol of the substance being formed.

● Standard enthalpy formation of the most stable form of any element is zero because there is no formation reaction needed when the element is already in its standard state.

● Thus the values of ΔHf° for C(graphite), H2(g),O2(g), and the

standard states of other elements are zero by definition.

Plan:

We need to examine each equation to determine1) whether the reaction is one in which one mole of substance is formed from

the elements2) whether the reactant elements are in their standard states

In a) 1 mol NA2 O is formed from the elements sodium and oxygen in their proper states, solid Na and O2 gas, respectively. Therefore, the enthalpy change for reaction a) corresponds to a standard enthalpy of formation. This is fine as is.

In b) potassium is given as a liquid. It must be changed to the solid form, its standard state at room temperature. Furthermore, two moles of product are formed, so the enthalpy change for the reaction as written is twice the standard enthalpy of formation of KCl(s). The equation for the formation reaction of 1 mol of KCl(s) is:

K(s) + ½ Cl(g) --->KCl(s)

Reaction c) does not form a substance from its elements. Instead, a substance decomposes to its elements, so this reaction must be reversed. Next, the element carbon is given as diamond, whereas graphite is the standard state of carbon at room temperature and 1 atm pressure. The equation that correctly represents the enthalpy of formation of glucose from its element is:

6C (graphite) + 6 H2(g) + 3 O2(g) ---> C6H12O6(s)

We can use Hess’s law and tabulations of ΔHf° values to calculate the standard enthalpy change

for any reaction for which we know the ΔHf° values for all reactants and products. For example,

consider the combustion of propane gas, C3H8(g), to CO2(g) and H2O(l) under standard conditions:

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

We can use Hess’s law in this way where n and m are the stoichiometric coefficients.

The symbol Σ (sigma) means “the sum of.”

The first term on the right in the equation above represents the formation reactions of the products, which are written in the “forward” direction in the chemical equation, that is, elements reacting to form products. The second term on the right in the equation represents the reverse of the formation reactions of the reactants which is why this term is preceded by a minus sign.

Decomposition: the reverse of the formation reaction for C3H8(g), so the

enthalpy change for this decomposition reaction is the negative of the ΔHf°

value for the propane formation reaction: -ΔHf° [C3H8(g)]

Formation of CO2: This equation 3C(s) + 3 O2(g) -> 3CO2(g) is the formation reaction for 3 mol of CO2(g). Because enthalpy is an extensive property, the enthalpy change for this step is

3 ΔHf° [CO2(g)]

Formation of H2O: The enthalpy change for formation 4 H2(g) + 2 O2(g) ---

> 4 H2O(l), formation of 4 mol of H2O, is 4ΔHf° [H2O(l)]. The reaction

specifies that H2O(l) is produced.

Summary• Standard enthalpy change of formation is the enthalpy change (i.e. heat absorbed) when

some amount of a compound is formed from its elements in their standard states under standard

conditions.

• If heat is released in the process of formation (for example burning carbon in oxygen to form

carbon dioxide), then the sign will be negative.

• It is typically given the symbol ΔHf0 or ΔfH0, and measured in kJ/mol (kilojoules per mole).

• Due to conservation of energy, enthalpies of formation can be used to calculate the heat

absorbed or released in any chemical reaction.

• We can break down any reaction into formation reactions. When we do, we obtain the general

result that the standard enthalpy change of a reaction is the sum of the standard enthalpies of

formation of the products minus the standard enthalpies of formation of the reactants.

Questions?