Recall that the value we write down for the enthalpy of a reaction depends on how we have written the balanced chemical equation for that reaction. Hence, we either need to include the complete balanced chemical reaction with our values of ΔrH° or we need to define a way of writing the equations so we don’t have to actually specify them. The latter is the way we do it in many cases. Two common defined reaction types and their associated enthalpy changes are given here.
- Standard Formation reactions are written by default such that exactly one mole of the compound of interest is formed from its constituent elements in their thermodynamic standard states. It is important to remember what the standard state of an element is when we try to write a particular standard formation reaction. For example, water if formed from hydrogen and oxygen. So the standard formation reaction for water is:
H2(g) + 1/2 O2(g) → H2O(l) ΔfH° = -286. kJ/mol.
See how the standard form of hydrogen and oxygen is diatomic gas while that of water is molecular liquid. This is because the normal temperature of T=25℃ (and standard P=1 bar) is used unless otherwise specified. For example, had we been seeking the enthalpy change for the same reaction at a temperature T=120℃ (and P=1bar), we would still have calculated a standard enthalpy of formation but the standard states would be different.
H2(g) + 1/2 O2(g) → H2O(g) ΔfH° = -327. kJ/mol
Note that value for the standard enthalpy change at this temperature is not the same as at room temperature because we had to incorporate the enthalpy of vaporization of ~41 kJ/mol for water. This is still a ‘standard enthalpy change’ because all the chemicals are at standard state for T = 120℃ and P = standard pressure of 1 bar.
Note also that we assumed the heat capacity of the products and reactants don’t contribute significantly to the overall enthalpy change for this reaction under the temperature change from 25℃ to 120℃. This is not necessarily good assumption as we’ll see later.
- Standard Combustion Reactions involve the complete oxidation of one mole of the compound of interest to its fully oxidized products. The word standard means everything is in its standard state for the temperature specified.
C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(l) ΔcombH° = -1560.4 kJ/mol
Since both of these types of reactions are now well defined, we can re-create the chemical equation exactly knowing only the compound of interest and the type of reaction. Hence, the values for the standard enthalpy changes for these defined types of reactions can be tabulated with need to only specify the chemical of interest. Thus, the following two specifications are complete because the reaction type and the chemical for each is specified so we can duplicate the appropriate balanced chemical equation that goes with each enthalpy change value.
ΔfH°(H2O)= –286. kJ/mol}
ΔcombH°(C2H6) = –1560.4 kJ/mol}
(since T was not specified for these two values, you must assume T=25℃)