Energy Relations in Chemistry: Thermochemistry

Enthalpies of Formation

Using Hess's Law we can calculate reaction enthalpies for a variety of reactions using tables of known enthalpies

Many experimentally determined enthalpies are listed by the type of process

• DH for converting various liquids to the gas phase are listed in tables of enthalpies of vaporization
• DH for melting solids to liquids are listed in tables of enthalpies of fusion
• DH for for combusting a substance in oxygen are listed in tables of enthalpies of combustion

The enthalpy change associated with the formation of a compound from its constituent elements is called the enthalpy of formation (DH_f )

Conditions which influence enthalpy changes include:

• temperature
• pressure
• state of reactants and products (s, g, l, aq)

The standard state of a substance is the form most stable at 298 °K (25 °C, or standard "room temperature") and 1 atmosphere (1 atm) of pressure

When a reaction occurs with all reactants and products in their standard states, the enthalpy change is the standard enthalpy of reaction (DH°)

Thus, the standard enthalpy of formation (DH°_f)of a compound is the change in enthalpy that accompanies the formation of 1 mole of that substance from its elements, with all substances in their standard states

The standard enthalpy of formation for ethanol (C₂H₅OH) is the enthalpy change for the following reaction

Notes:

• Elemental source of oxygen is O₂ and not O because O₂ is the stable form of oxygen at 25 °C and 1 atm, likewise with H₂
• Elemental source of carbon is specified as graphite (and not, for example, diamond) because graphite is the lowest energy form of carbon at room temp and 1 atm
• Why is the O₂ stoichiometry left at "1/2"? The stoichiometry of formation reactions always indicates the formation of 1 mol of product. Thus, DH°_f values are reported as kJ / mole of the substance produced

If C(graphite) is the lowest energy form of carbon under standard conditions, then what is the DH°_f for C(graphite)?

• By definition, the standard enthalpy of 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
• DH°_f for C(graphite), H₂(g) and O₂(g) = 0

Consider the following combustion reaction of propane:

C₃H₈(g) + 5O₂(g) -> 3CO₂(g) + 4H₂O(l)

The reactants:

• The standard heat of formation (DH°_f) of propane gas from its elemental constituents in the standard state is ­103.85 kJ/mole

3C(graphite) + 4H₂(g) -> C₃H₈(g) DH°_f = -103.85 kJ

• The standard heat of formation (DH°_f) for O₂(g) is zero

O₂(g) -> O₂(g) DH°_f = 0 kJ

and so...

5O₂(g) -> 5O₂(g) DH°_f = 0 kJ

Overall, therefore, the standard heat of formation (DH°_f) for the reactants is:

3C(graphite)+4H₂(g)+5O₂(g) -> C₃H₈(g)+5O₂(g)

DH°_f = -103.85 kJ

• The standard heat of formation (DH°_f) of CO₂(g) from its elemental constituents in the standard state is ­393.5 kJ/mole

C(graphite) + O₂(g) -> CO₂(g) DH°_f = ­393.5 kJ

so, for 3 moles of CO₂ molecules the standard heat of formation would be:

3C(graphite) + 3O₂(g) -> 3CO₂(g) DH°_f = ­1180.5 kJ

• The standard heat of formation (DH°_f) of H₂O(l) from its elemental constituents in the standard state is ­285.8 kJ/mole

and so the DH°_f for 4 waters would be:

combining the DH°_f for both products yields:

3C(graphite)+4H₂(g)+5O₂(g) -> 3CO₂(g)+4H₂O(l)

with DH°_f = (-1180.5) + (-1143.2) = -2323.7 kJ

Let's summarize what we have determined so far:

Overall the standard heat of formation (DH°_f) for the reactants is:

3C(graphite)+4H₂(g)+5O₂(g) -> C₃H₈(g)+5O₂(g)

DH°_f = -103.9 kJ

Overall the standard heat of formation (DH°_f) for the products is:

3C(graphite)+4H₂(g)+5O₂(g) -> 3CO₂(g)+4H₂O(l)

DH°_f = - 2323.7 kJ

Lets plot these on a relative scale of enthalpy:

This information can be used to determine the relative enthalpy difference under standard conditions (DH°) between the reactants and products:

This enthalpy difference (-2219.8 kJ) is the enthalpy of the reaction for the combustion of propane under standard conditions (DH°_rxn)

Calculate the enthalpy change (DH°_rxn) for the combustion of 1 mol of ethanol

C₂H₅OH(l) + 3O₂(g) -> 2CO₂(g) + 3H₂O(l)

heat of formation for reactants

2C(graphite)+3H₂(g)+(1/2)O₂(g) -> C₂H₅OH(l)

DH°_f = -277.7 kJ

plus

3O₂(g) -> 3O₂(g) DH°_f = 0 kJ

gives:

2C(graphite)+3H₂(g)+(7/2)O₂(g) -> C₂H₅OH(l)+ 3O₂(g)

DH°_f = -277.7 kJ

heat of formation for products

C(graphite)+O₂(g) -> CO₂(g) DH°_f = -393.5 kJ

therefore

2C(graphite)+2O₂(g) -> 2CO₂(g) DH°_f = -787 kJ

H₂(g) + (1/2)O₂(g) -> H₂O(l) DH°_f = -285.8 kJ

therefore

3H₂(g) + (3/2)O₂(g) -> 3H₂O(l) DH°_f = -857.4 kJ

combining gives:

2C(graphite)+3H₂(g)+(7/2)O₂(g) -> 2CO₂(g)+ 3H₂O(l)

DH°_f = (-857.4)+(-787) = -1644.4kJ

DH°_rxn

DH°_rxn = DH°_f (products) - DH°_f (reactants)

(-1644.4) - (-277.7) = -1366.7 kJ

1996 Michael Blaber