Enthalpy of Formation

Standard Enthalpy of Formation

So, what is the enthalpy of formation? Let's take a look!

The standard enthalpy of formation, ΔH_f^° - for a given molecule, or compound, is:

First, let's talk about where the Standard Enthalpy of Formation, ΔH_f°, fits into the bigger picture. The overall enthalpy of a chemical reaction (also called the standard enthalpy of reaction, ΔH °) is given by the following equation:

$$\Delta{H^\circ}=\Sigma_{i=1}^n\,[q\Delta{H_f^\circ}(Products)]_i-\Sigma_{i=1}^n\,[r\Delta{H_f^\circ}(Reactants)]_i$$

Where,

• Σ, is the summation symbol

• q and r are the stoichiometric coefficients of the balanced equation for products and reactants, respectively.

• The standard enthalpy of formation of products is ΔH_f° (Products)

• The standard enthalpy of formation of reactants is, ΔH_f° (Reactants)

Standard Enthalpy of Formation Table

Now we may ask, "How to find the standard enthalpy of formation?" To do this, we must refer to a standard enthalpy of formation table. Here we present the standard enthalpies of formation for just a few commonly used compounds.

Table 1: Standard Enthalpies of Formation ¹ (at 25°C)

Notice that in the above list of compounds, those in their elemental state (reference form) have a standard enthalpy of formation equal to zero.

Enthalpy of Formation Equation

As noted earlier, the standard enthalpy of formation terms for products, ΔH_f °(Products), and the standard enthalpy of formation terms for reactants, ΔHf °(Reactants), are used to calculate the standard enthalpy of reaction, ΔH °:

$$\Delta{H^\circ}=\Sigma_{i=1}^n\,[q\Delta{H_f^\circ}(Products)]_i-\Sigma_{i=1}^n\,[r\Delta{H_f^\circ}(Reactants)]_i$$

Where, Σ, is the summation symbol and, q, and, r, are the stoichiometric coefficients of the balanced equation for products and reactants, respectively.

For example, consider the following reaction between methane gas, CH₄ (g), and chlorine gas, Cl₂ (g), to yield carbon tetrachloride liquid, CCl₄ (l), and hydrochloric acid gas, HCl (g):

Notice that in the table above, those compounds that are in their elemental state have a standard enthalpy of formation that is equal to zero.

Calculate the Enthalpy of Formation

Now you may ask, "How to calculate the enthalpy of formation?"

1. Let's consider the enthalpy of formation of the product hydrogen chloride, HCl (g), from the reactants, hydrogen gas, H₂ (g), and chlorine gas, Cl₂ (s), under standard conditions:

$$H_2\,(g)+Cl_2\,(g) \rightarrow HCl\,(g)$$

By referring to the above table of thermodynamic data, we can find the enthalpy of formation of the reactants under standard conditions: ¹

Note that the standard enthalpy of formation of elemental hydrogen gas, H₂(g), is equal to zero; H₂(g): ΔH_f° = 0.0 kJ/mol. However, in the present case, the reaction involves breaking the molecular hydrogen bond, which yields atomic hydrogen gas, H (g).

The standard enthalpy of formation of atomic hydrogen gas is H (g): ΔH_f° = +218.0 kJ/mol. The same is true for chlorine - the enthalpy of formation of the elemental form of chlorine gas is Cl₂ (g): ΔH_f° = 0.0 kJ/mol.

Again, in the present case, the reaction involves the breaking of the bonds within the molecular gas, forming atomic chlorine gas; Cl (g): ΔH_f° = +121.0 kJ/mol. Then the actual reaction process is given by:

$$H_2\,(g)+Cl_2\,(s) \rightarrow 2H\,(g)+2Cl\,(g) \rightarrow 2HCl\,(g)$$

Notice that we must account for the stoichiometric coefficients of the balanced equation to get the standard enthalpy of formation of reactants, such that:

Standard Enthalpy of the Formation of Reactants:

\begin{align}\Delta{H_f^\circ}(Reactants)&=\Sigma_{i=1}^m\,[r\Delta{H_f^\circ}(Reactants)]_i\\&=2\cdot\Delta{H_f^\circ}(Atomic\,Hydrogen\,gas )+2\cdot\Delta{H_f^\circ}(Atomic\,Chlorine\,gas)]\\&=2\cdot(218.0\,kJ/mol)+2\cdot(121.0\,kJ/mol)\\&=678\,kJ/mol\end{align}

Thus, the standard enthalpy of formation for the production of 2 moles of hydrogen iodide, HCl, is given by:

$$2\cdot \Delta{H_f^\circ}(Hydrogen\,Chloride\,gas)=678\,kJ/mol$$

The enthalpy diagram for this reaction is:

Thus, the synthesis of hydrogen chloride from elemental hydrogen and chlorine is an energy-absorbing, or endothermic, reaction.

Now we may ask, "How to calculate the enthalpy of formation using Hess's law?"

Let's consider again the following reaction between methane gas (CH₄ (g)) and chlorine gas (Cl₂ (g)) to yield carbon tetrachloride liquid (CCl₄ (l)) and hydrochloric acid gas (HCl (g)):

$$1CH_4\,(g)+4Cl_2\,(g) \rightarrow 1CCl_4\,(g)+4HCl\,(g)$$

From the table for Standard Enthalpies of Formation ¹ (at 25°C), above, we pick out the enthalpies of formation for CH₄ (g), Cl₂ (g), and HCl (g). Then we can write the thermochemical equations in the following way:

\begin{align}C\,(Graphite)+2\,H_2\,(g) \rightarrow CH_4\,(g):\,\Delta{H_f^\circ}&=-74.9\,kJ/mol\,\,\,\,\,\,\,(1)\\C\,(Graphite)+Cl_2\,(g) \rightarrow CCl_4\,(l):\,\Delta{H_f^\circ}&=-139\,kJ/mol\,\,\,\,\,\,\,\,\,(2)\\\frac{1}{2}H_2+\frac{1}{2}Cl_2\,(g) \rightarrow HCl\,(g):\,\Delta{H_f^\circ}&=-92.3\,kJ/mol\,\,\,\,\,\,\,\,(3)\end{align}

Applying Hess's Law, our aim is to isolate CH₄ (g) on the left-hand side while isolating carbon tetrachloride, CCl₄, and 4HCl (g), on the right-hand side. Additionally, we want to eliminate all elements in their reference form, because these are equal to 0.0 kJ/mol and do not contribute to the heat summation. So what we do is reverse equation (1) (above), add equation (2) and multiply equation (3) by the number four. Thus:

\begin{align}CH_4\,(g) \rightarrow C\,(Graphite)+2\,H_2\,(g)&\,\,\,\,\,\,\,\,\,\,\,\,\,1\cdot (+74.9\,kJ/mol)\\C\,(Graphite)+Cl_2\,(g) \rightarrow CCl_4\,(l)&\,\,\,\,\,\,\,\,+1\cdot (-139\,kJ/mol)\\\underline{2H_2\,(g)+2Cl_2\,(g) \rightarrow 4HCl\,(g)}&\,\,\,\,\,\,\,\,\underline{+4\cdot (-92.3\,kJ/mol)}\\CH_4\,(g)+4Cl_2\,(g) \rightarrow CCl_4\,(l)+4HCl\,(g)&\,\,\,\,\,\,\,\Delta{H^\circ}=-433.3\,kJ/mol\end{align}

Enthalpy of Formation of Water

Let's consider the standard entahlpy of formation of water, H₂O (l), from hydrogen gas, H₂, and oxygen gas, O₂. The reaction is then:

$$2H_2\,(g)+O_2\,(g) \rightarrow 2H_2O\,(l):\,\Delta{H_f^\circ}=-571.6\,kJ/mol$$

Now, to write this for a reaction that produces 1 mole of water, we multiply this equation by a factor of 1/2:

$$\frac{1}{2}\cdot 2H_2\,(g)+\frac{1}{2}\cdot O_2\,(g) \rightarrow \frac{1}{2}\cdot2H_2O\,(l):\,\Delta{H_f^\circ}=\frac{1}{2}\cdot(-571.6\,kJ/mol)=-285.8kJ/mol$$

This is then the standard enthalpy of formation for 1 mole of liquid water.

Now we might ask, "What best describes the enthalpy of formation of a substance?"

• In all cases, the enthalpy of formation of a substance is associated with the potential energy that is released, as heat, from the breaking of a chemical bond within a compound.