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Standard Enthalpy of Formation

The definition of the standard enthalpy of formation is the change in enthalpy when one mole of a substance, in the standard state of 1 atm of pressure and temperature of 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 = it is a reaction from a substance that's formed from its elements

For most chemistry problems involving standard enthalpies of formation, you will need the equation for the standard enthalpy change of formation:

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

Although this equation looks complicated, it essentially states that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products subtracted by the sum of the standard enthalpies of formation of the reactants.

Example 1

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

$$A + B \leftrightharpoons C$$

And we have the standard enthalpy of formation values as such:

• 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)\)

Since we have one mole of A, B and C, we multiply the standard enthalpy of formation of each reactant and product by 1 mole, which eliminates the mol denominator

ΔHreactiono = 346 kJ

We get the answer of 346 kJ, which is the standard enthalpy change of formation for the creation of variable "C".

An important point to be made about the standard enthalpy of formation is that when a pure element is in its reference form it is zero.

This can be seen with the element carbon. 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 we choose the reference form of the more stable form of carbon. Which is also the one with the lowest enthalpy, so this is why graphite has the standard enthalpy of formation equal to zero.

Sample Table of Standard Enthalpy of Formation Values

This table provides a few sample values of the standard enthalpies of formation of various compounds:

 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 are in the units of kJ/mol and the physical conditions of 298.15 K and under 1 atm of pressure, which is referred to as the "standard state", and are the conditions in which you will generally find values of standard enthalpies of formation.

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 2

Between Br2(l) and Br2(g) at 298.15 K which element would have a standard enthalpy of formation?

SOLUTION

We find that Br2(l) has the most stable form between Br2(l) and Br2(g), which means it has the lower enthalpy and thus Br2(l) has ΔHf = 0. The answer is that Br2(g) has a standard enthalpy of formation.

Note: that the element phosphorus is a unique case. The reference form in phosphorus isn't the most stable form. The solid white phosphorus is the reference form in this case. Even though red phosphorus is the most stable form, In this case white phosphorous ΔHf = 0 even though its not the most stable form.

Again, standard enthalpies of formation can be either positive or negative.

Example 2

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

SOLUTION

In writing this equation you have to make sure you set it up for one mole of CO2(g). You must find the stable forms of elements to. In this case they would be O2(g) and graphite for carbon.

$$O_{2}(g) + C(graphite) \rightleftharpoons CO_{2}(g)$$  ΔHf = -393.5 kJ

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

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

When we plug in our equation for the formation of CO2:

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

As O2(g) and C(graphite) are in the most elementally stable form, 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 3

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

SOLUTION

NO2(g) is formed from the combination of NO(g) and O2(g) in the reaction:

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

To find the ΔHreactiono, we'll need to 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 standard enthalpy of formation values we find from the data table are as follows:

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

When we plug these values into the formula for the standard enthalpy change of formation:

ΔHreactiono= (1 mol)(33.18 kJ/mol) - ((2 mol)(90.25 kJ/mol) + (1 mol)(0 kJ/mol))

ΔHreactiono = -147.32 kJ

References

1. Pisanty, Alejandro. "The electronic structure of graphite: A chemist's introduction to band theory." J. Chem. Educ. 1991, 68, 804.
2. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey 2007.

Contributors

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

10:42, 26 Dec 2013

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