The answer to these questions lies within the concept of **enthalpy of formation**. Below, we will consider these questions in greater detail. In addition, we will undertake a scientific survey of the field of chemical bonding energy. Please continue reading to find out more.

- First, we will look at the definition of enthalpy of formation.
- Then, we will look at a table containing the enthalpy of formation of a number of useful atoms and molecules.
- After, we will learn how to calculate enthalpy of formation.
- Subsequently, we will present an example utilizing the enthalpy of formation equation.
- Finally, we will talk about the enthalpy of formation of water

## Standard Enthalpy of Formation

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

**Enthalpy**, H - the potential energy contained within a chemical bond or chemical interaction when it is released as heat.

**Potential Energy** - the energy possessed by a molecule by virtue of the positions of its atoms at any given moment.

**Chemical bond** - the force of attraction that keeps the atoms within a molecule linked together in the appropriate orientation and valence.

**Bond Valence **- the number of electron pairs within a chemical bond.

The** standard enthalpy of formation**, ΔH_{f}^{°} - for a given molecule, or compound, is:

...the enthalpy change for the formation of one mole of the substance in its standard state from its elements in their reference form and in their standard states."

**Standard State** - standard thermodynamic conditions for compounds when listed in a table of thermodynamic data. Typical standard conditions are 1 atmosphere (atm) and 25 °C.

**Reference Form** - the stablest form of an element under standard conditions.

**Stoichiometric Coefficients** - the number in front of a chemical species in the balanced equation.

First, let's talk about where the **Standard Enthalpy of Formation, ****Δ***H _{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 Δ*H*° (Products)_{f}The

**standard enthalpy of formation of reactants**is, Δ*H*° (Reactants)_{f}

The summation symbol, Σ, instructs us to add: for example, \(\Sigma_{i=1}^4 a_i=a_1+a_2+a_3+a_4\), means the first term is, a, index, 1, added to term, a, index, 2, added to, a, index, 3, and finally we end with, a, index, 4.

## 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)

Formula | ΔH_{f }° (kJ mol^{-1}) | Formula | ΔH_{f }° (kJ mol^{-1}) |

e^{-}, Electron Gas (g) | 0 | C (g), graphite (elemental form) | 0 |

H^{+} (aq) | 0 | CO (g) | -110.5 |

H (g), Atomic Hydrogen gas | 218.0 | CO_{2} (g) | -393.5 |

H_{2} (g), Hydrogen gas | 0 (elemental form) | CH_{4} (g) | -74.9 |

Na (g) | 107.8 | C_{6}H_{6} (l) | 49.0 |

Na (s) | 0 (elemental form) | HCHO (g) | -116 |

NaCl (s) | -411.1 | HCN (g) | 135 |

NaHCO_{3} (s) | -947.7 | HCN (l) | 105 |

Na_{2}CO_{3} (s) | -1130.8 | CH_{3}CHO (g) | -166 |

Cl (g), Atomic Chlorine gas | 121.0 | CCl_{4} (l) | -139 |

Cl_{2} (g), Chlorine gas | 0 (elemental form) | HCl (g) | -92.3 |

Iodine Crystals, I_{2} (s) | 0 (elemental form) | Iodide gas, I^{- }(g) | -197.7 |

O_{2} (g), Oxygen gas | 0 (elemental form) | O (g) | 249.2 |

H_{2}O (g) | -241.8 | H_{2}O (l) | -285.8 |

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,

*ΔH*

*f*°(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, *CH*_{4} (g), and chlorine gas, *Cl*_{2} (g), to yield carbon tetrachloride liquid, *CCl*_{4} (l), and hydrochloric acid gas, *HCl* (g):

Here we have the following reaction between methane gas, *CH*_{4} (g), and chlorine gas, *Cl*_{2} (g), to yield carbon tetrachloride liquid, *CCl*_{4} (l), and hydrochloric acid gas, *HCl* (g) :

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

Then the enthalpy of formation equation will give the enthalpy of reaction:

\begin{align}\Delta{H^\circ}&=\Sigma_{i=1}^n,[q\Delta{H_f^\circ}(Products)]_i-\Sigma_{i=1}^m\,[r\Delta{H_f^\circ}(Reactants)]_i\\&=[\Delta{H_f^\circ}(CCl_4)+\Delta{H_f^\circ}(HCl)]-[\Delta{H_f^\circ}(CH_4)+\Delta{H_f^\circ}(Cl_2)]\\&=[1\cdot(-139\,kJ/mol)+4\cdot\,(-92.3\,kJ/mol)]-[1\cdot(-74.9\,kJ/mol)+4\cdot(0.0\,kJ/mol)]\\&=-433\,kJ/mol\end{align}

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, *H*_{2} (g), and chlorine gas, *Cl*_{2} (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, *H*_{2}(g), is equal to zero; *H*_{2}(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

*Cl*

_{2}(g):

*ΔH*° = 0.0 kJ/mol.

_{f}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 s**tandard 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:

Figure 1: Enthalpy diagram for the enthalpy of formation of hydrogen chloride, HCl.

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?"

**Hess's Law **- also known as Hess's Law of Constant Heat Summation, states that during all the steps of any chemical reaction the total enthalpy change, ΔH, does not depend on the order in which the steps, from reactants to intermediates to products, are taken but only depends on the sum of the enthalpies of all of the reactions in any order.

Let's consider again the following reaction between methane gas (*CH*_{4} (g)) and chlorine gas (*Cl*_{2} (g)) to yield carbon tetrachloride liquid (*CCl*_{4} (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 *CH*_{4} (g), *Cl*_{2} (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 CH_{4} (g) on the left-hand side while isolating carbon tetrachloride, *CCl*_{4}, and 4*HCl* (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, *H*_{2}*O* (l), from hydrogen gas, *H*_{2}, and oxygen gas, *O*_{2}. 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.

## Enthalpy of Formation - Key takeaways

**Enthalpy**is the potential energy contained within a chemical bond or chemical interaction when it is released as heat.- 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 and, q, and, r, are the stoichiometric coefficients of the balanced equation for products and reactants, respectively. The standard enthalpy of formation of products is, Δ

*Hf*° (Products), while the standard enthalpy of formation of reactants is, Δ*Hf*° (Reactants). - Standard enthalpy of formation is "...the enthalpy change for the formation of one mole of the substance in its standard state from its elements in their reference form and in their standard states."
^{1} - Hess's Law, also known as Hess's Law of Constant Heat Summation, states that during all the steps of any chemical reaction the total enthalpy change, ΔH, does not depend on the order in which the steps, from reactants to intermediates to products, are taken but only depends on the sum of the enthalpies of all of the reactions in any order.

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##### Frequently Asked Questions about Enthalpy of Formation

How to calculate enthalpy of formation?

To calculate the enthalpy of formation, we use the following formula:

Where:

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

What is enthalpy of formation?

Enthalpy of formation is the change in enthalpy during the formation of a species.

How to find standard enthalpy of formation

The standard enthalpy of formation can often be found in reference tables for common elements and compounds. Otherwise, it can be calculated using the formula:

Where:

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

How to calculate enthalpy of formation using Hess's law?

When calculating the enthalpy of formation for a series of reactions, you can use Hess's law. Hess's law states that the total enthalpy of formation is equal to the sum of the enthalpy of reaction for each step.

What best describes the enthalpy of formation of a substance?

Enthalpy of formation is the change in enthalpy during the formation of a species. It can tell us whether a species is thermodynamically favored or not. Species that have a negative enthalpy are thermodynamically favored.

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