# Standard Enthalpy of Formation

The standard enthalpy of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions.

### Introduction

The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements. The symbol of the standard enthalpy of formation is ΔHf.

• Δ = A change in enthalpy
• o = A degree signifies that it's a standard enthalpy change.
• f = The f indicates that the substance is formed from its elements

The equation for the standard enthalpy change of formation (originating from Enthalpy's being a State Function), shown below, is commonly used:

$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$

This equation essentially states that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.

Example $$\PageIndex{1}$$

Given a simple chemical equation with the variables A, B and C representing different compounds:

$$A + B \leftrightharpoons C$$

and the standard enthalpy of formation values:

• ΔHfo[A] = 433 KJ/mol
• ΔHfo[B] = -256 KJ/mol
• ΔHfo[C] = 523 KJ/mol

the equation for the standard enthalpy change of formation is as follows:

ΔHreactiono = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B])

ΔHreactiono = (1 mol)(523 kJ/mol) - ((1 mol)(433 kJ/mol) + (1 mol)(-256 kJ/mol)\)

Because there is one mole each of A, B and C,  the standard enthalpy of formation of each reactant and product is multiplied by 1 mole, which eliminates the mol denominator:

ΔHreactiono = 346 kJ

The result is 346 kJ, which is the standard enthalpy change of formation for the creation of variable "C".

The standard enthalpy of formation of a pure element is in its reference form its standard enthalpy formation is zero.

Carbon naturally exists as graphite and diamond. The enthalpy difference between graphite and diamond is too large for both to have a standard enthalpy of formation of zero. To determine which form is zero, the more stable form of carbon is chosen. This is also the form with the lowest enthalpy, so graphite has a standard enthalpy of formation equal to zero. Table 1 provides sample values of standard enthalpies of formation of various compounds.

Table 1: Sample Table of Standard Enthalpy of Formation Values. Table T1 is a more comprehensive table.
Compound  ΔHfo
O2(g) 0 kJ/mol
C(graphite)   0 kJ/mol
CO(g)  -110.5 kJ/mol
CO2(g)  -393.5 kJ/mol
H2(g)  0 kJ/mol
H2O(g)  -241.8 kJ/mol
HF(g)  -271.1 kJ/mol
NO(g)  90.25 kJ/mol
NO2(g) 33.18 kJ/mol
N2O4(g)  9.16 kJ/mol
SO2(g)  -296.8 kJ/mol
SO3(g)  -395.7 kJ/mol

All values have units of kJ/mol and physical conditions of 298.15 K and 1 atm, referred to as the "standard state." These are the conditions under which values of standard enthalpies of formation are typically given. Note that while the majority of the values of standard enthalpies of formation are exothermic, or negative, there are a few compounds such as NO(g) and N2O4(g) that actually require energy from its surroundings during its formation; these endothermic compounds are generally unstable.

Example $$\PageIndex{2}$$

Between Br2(l) and Br2(g) at 298.15 K, which substance has a nonzero standard enthalpy of formation?

SOLUTION

Br2(l) is the more stable form, which means it has the lower enthalpy; thus, Br2(l) has ΔHf = 0. Consequently, Br2(g) has a  nonzero standard enthalpy of formation.

Note: that the element phosphorus is a unique case. The reference form in phosphorus is not the most stable form, red phosphorus, but the less stable form, white phosphorus.

Recall that standard enthalpies of formation can be either positive or negative.

Example $$\PageIndex{3}$$

The enthalpy of formation of carbon dioxide at 298.15K is ΔHf = -393.5 kJ/mol CO2(g). Write the chemical equation for the formation of CO2.

SOLUTION

This equation must be written for one mole of CO2(g). In this case, the reference forms of the constituent elements are O2(g) and graphite for carbon.

$O_{2}(g) + C(graphite) \rightleftharpoons CO_{2}(g)$

The general equation for the standard enthalpy change of formation is given below:

$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$

Plugging in the equation for the formation of CO2 gives the following:

ΔHreactiono= ΔHfo[CO2(g)] - (ΔHfo[O2(g)] + ΔHfo[C(graphite)]

Because O2(g) and C(graphite) are in their most elementally stable forms, they each have a standard enthalpy of formation equal to 0:

ΔHreactiono= -393.5 kJ = ΔHfo[CO2(g)] - ((1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol))

ΔHfo[CO2(g)]= -393.5 kJ

Example $$\PageIndex{4}$$

Using the values in the above table of standard enthalpies of formation, calculate the ΔHreactiono for the formation of NO2(g).

SOLUTION

$$NO_{2(g)}$$ is formed from the combination of $$NO_{(g)}$$ and $$O_{2(g)}$$ in the following reaction:

$$2NO(g) + O_{2}(g) \leftrightharpoons 2NO_{2}(g)$$

To find the ΔHreactiono, use the formula for the standard enthalpy change of formation:

$\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}$

The relevant standard enthalpy of formation values from Table 1 are:

• O2(g): 0 kJ/mol
• NO(g): 90.25 kJ/mol
• NO2(g): 33.18 kJ/mol

Plugging these values into the formula above gives the following:

$ΔH_{reaction}^o= (2 \cancel{mol})(33.18\; kJ/\cancel{mol}) - \left[(2 \cancel{mol})(90.25\ kJ/\cancel{mol}) + (1 \cancel{mol})(0\; kJ/\cancel{mol})\right]$

$ΔH_{reaction}^o = -114.1\; kJ$

### Contributors

• Jonathan Nguyen (UCD), Garrett Larimer (UCD)