Question

How do you calculate standard molar enthalpy of formation?

Answer 1

Hint: The given thermochemical equation has to be checked, and the equation has to be divided with the coefficient of the final product as to make it 1 mole. Once that is done, the value of enthalpy also needs to be divided and this resultant enthalpy is the enthalpy of formation.

Complete answer:
In order to answer our question, we need to learn about what enthalpy actually is. Let us consider that in a system, pressure is kept constant. So the heat content of this particular system is called the enthalpy. It shows how much heat is gained or lost in a particular process. The standard enthalpy of formation is denoted by ${{\Delta }_{f}}H$ where 'f' is the subscript that one mole of the compound has been formed in its standard state from its elements in their most stable states of aggregation. Standard enthalpy of formation of a substance is defined as the enthalpy change, i.e., accompanied in the formation of one mole of a compound from its elements in their most stable states of aggregation (also known as reference states). For example, the reference state of hydrogen is ${{H}_{2}}$ gas and that of dioxygen, carbon and sulphur are ${{O}_{2}}$ gas,${{C}_{graphite}}$ and rhombic sulphur respectively. Now, let us consider the following reaction:
\[{{H}_{2}}(g)+B{{r}_{2}}(g)\to 2HBr(g);{{\Delta }_{r}}H=-72.8kJ\,mo{{l}^{-1}}\]
The enthalpy change does not represent the enthalpy of formation of the gaseous HBr. Here, 2 moles of product are formed instead of one mole. So, we can write that ${{\Delta }_{r}}H=2{{\Delta }_{f}}H$. As 1 mole of the product is desired, so we divide the whole equation by 2. The coefficients, along with the value of enthalpy also gets divided, so we obtain the final equation as:
\[\dfrac{1}{2}{{H}_{2}}(g)+\dfrac{1}{2}B{{r}_{2}}(g)\to HBr(g),{{\Delta }_{f}}H=\dfrac{-72.8}{2}=-36.4kJ\,mo{{l}^{-1}}\]
So, this is how we calculate the standard molar enthalpy of combustion, for a given reaction.

Note:
It becomes important to understand that a standard molar enthalpy of formation,${{\Delta }_{f}}H$ is a special case of ${{\Delta }_{r}}H$, where one mole of a compound is formed from its constituent elements in their most stable states.