Standard enthalpy of formation Knowledge-Based Study

Study notes

1. The standard enthalpy of formation, denoted as $\Delta H_f^\circ$, is the enthalpy change that accompanies the formation of exactly one mole of a compound from its constituent elements in their standard states.

   Standard State Conditions

   By international convention, the standard state refers to:

   • A pressure of exactly 1 bar (or traditionally 1 atm).
   • A specified temperature, typically 298.15 K (25 °C).
   • The most stable physical form and allotrope of the element at that temperature and pressure.

   Zero Enthalpy of Formation by Convention

   An element in its standard state is assigned a standard enthalpy of formation of zero:

   $$\Delta H_f^\circ = 0\text{ kJ/mol}$$

   Examples of substances with $\Delta H_f^\circ = 0$:

   • $\text{O}_2(g)$, but not $\text{O}_3(g)$ (ozone is less stable).
   • $\text{C}(\text{graphite})$, but not $\text{C}(\text{diamond})$.
   • $\text{Br}_2(l)$, but not $\text{Br}_2(g)$.
   • $\text{Fe}(s)$, $\text{Na}(s)$, and $\text{N}_2(g)$.

   A chemical equation represents a true formation reaction only when it satisfies two strict criteria:

   1. The reactants must be exclusively pure elements in their standard states.
   2. The stoichiometric coefficient of the target product must be exactly 1.

   Examples of Formation Equations

   Formation of liquid water:

   $$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = \Delta H_f^\circ[\text{H}_2\text{O}(l)]$$

   Formation of gaseous carbon dioxide:

   $$\text{C}(\text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = \Delta H_f^\circ[\text{CO}_2(g)]$$

   Formation of liquid ethanol ($\text{C}_2\text{H}_5\text{OH}$):

   $$2\text{C}(\text{graphite}) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)$$

   Note: Fractional coefficients for elements are fully acceptable in formation equations because the product quantity must remain fixed at one mole.

   Using Hess's Law, the standard enthalpy change of any chemical reaction ($\Delta H_{rxn}^\circ$) can be determined from the standard enthalpies of formation of all participating substances:

   $$\Delta H_{rxn}^\circ = \sum n\Delta H_f^\circ(\text{products}) - \sum m\Delta H_f^\circ(\text{reactants})$$

   where $n$ and $m$ are the stoichiometric coefficients of the products and reactants in the balanced chemical equation.

   Worked Example

   Consider the complete combustion of methane:

   $$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$

   The calculation is structured as:

   $$\Delta H_{rxn}^\circ = \left[ \Delta H_f^\circ[\text{CO}_2(g)] + 2\Delta H_f^\circ[\text{H}_2\text{O}(l)] \right] - \left[ \Delta H_f^\circ[\text{CH}_4(g)] + 2\Delta H_f^\circ[\text{O}_2(g)] \right]$$

   Because $\text{O}_2(g)$ is an element in its standard state, $\Delta H_f^\circ[\text{O}_2(g)] = 0\text{ kJ/mol}$.

   To avoid common errors when working with standard enthalpies of formation, keep the following principles in mind:

   • States of matter matter: $\Delta H_f^\circ[\text{H}_2\text{O}(l)] \neq \Delta H_f^\circ[\text{H}_2\text{O}(g)]$. The gas phase has a higher enthalpy than the liquid phase by an amount equal to the enthalpy of vaporization.
   • Do not confuse synthesis with formation: The reaction $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$ produces two moles of water; therefore, its $\Delta H^\circ$ equals $2\Delta H_f^\circ[\text{H}_2\text{O}(l)]$, not $\Delta H_f^\circ$.
   • Allotropes require precision: Only the most stable allotropic form has $\Delta H_f^\circ = 0$. For carbon, graphite is assigned zero, while diamond has $\Delta H_f^\circ = +1.9\text{ kJ/mol}$.
   • Sign convention: Negative values of $\Delta H_f^\circ$ indicate that the compound is thermodynamically more stable than its constituent elements under standard conditions.